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Simplify the following:
\begin{enumerate}
        \begin{multicols}{2}
    \item \((2 \sqrt{3}+\sqrt{2})^{2}\)
    \item \((2 \sqrt{5}+4 \sqrt{3})^{2}\)
    \item \((\sqrt{x y}+1)^{2}\)
    \item \((4 \sqrt{2}-3 \sqrt{7})^{2}\)
    \item \((\sqrt{x}-\sqrt{y})^{2}\)
    \item \((\sqrt{11}-2 \sqrt{22})^{2}\)
        \end{multicols}
        \begin{multicols}{2}
    \item If \(x=\sqrt{2}-1\) and \(y=\sqrt{2}+1\), find:
        \begin{enumerate}
            \item \(x y\)
            \item \(x^{2} y\)
            \item \(y^{2} x\)
            \item \(\frac{1}{x}+\frac{1}{y}\)
        \end{enumerate}
    \item Simplify:
        \begin{enumerate}
            \item \((\sqrt{5}+\sqrt{3})^{2}-(\sqrt{5}-\sqrt{3})^{2}\)
            \item \((\sqrt{3}+\sqrt{2})^{2}-(\sqrt{3}-\sqrt{2})^{2}\)
        \end{enumerate}
        \end{multicols}
    \item Simplify \((a \sqrt{b}+c \sqrt{d})(a \sqrt{b}-c \sqrt{d})\).
    \item For each of the figures shown, find:
        \begin{enumerate}
            \item the value of \(x\)
            \item the area of the triangle
            \item  the perimeter of the triangle
        \end{enumerate}
        \begin{multicols}{3}
            \begin{center}
            \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04(3)}
            \end{center}
            \begin{center}
            \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04(1)}
            \end{center}
            \begin{center}
            \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04(4)}
            \end{center}
        \end{multicols}
    \item For the 2 figures below, find:
        \begin{enumerate}
            \item the perimeter
            \item the area
        \end{enumerate}
        \begin{multicols}{2}
            \begin{center}
            \includegraphics[width=0.2\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04}
            \end{center}
            \begin{center}
            \includegraphics[width=0.2\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04(2)}
            \end{center}
        \end{multicols}
    \item Rationalise the following:
        \begin{enumerate}
                \begin{multicols}{2}
            \item \(\frac{14}{\sqrt{7}}\)
            \item \(\frac{\sqrt{14}}{\sqrt{7}}\)
            \item \(\frac{\sqrt{5}}{3 \sqrt{7}}\)
            \item \(\frac{\sqrt{3}}{4 \sqrt{6}}\)
            \item \(\frac{5}{\sqrt{3}}\)
            \item \(\frac{1}{\sqrt{18}}\)
            \item \(\frac{1}{\sqrt{2}}+\sqrt{2}\)
            \item \(\frac{\sqrt{72}}{\sqrt{3}}+\frac{3}{\sqrt{2}}-\frac{2}{2 \sqrt{2}}\)
            \item \(\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}+\frac{7}{2 \sqrt{3}}\)
                \end{multicols}
        \end{enumerate}
    \item If \(x=2 \sqrt{14}\) and \(y=4 \sqrt{2}\), find and rationalise the denominator.
        \begin{enumerate}
                \begin{multicols}{3}
                \item \(\frac{x}{y}\)
                \item \(\frac{y}{x}\)
                \item \(\frac{\sqrt{2} x}{\sqrt{3} y}\)
                \end{multicols}
        \end{enumerate}
    \item A bowl in the shape of a hemisphere of radius length \(5 \mathrm{~cm}\) is partially filled with water. The surface of the water is a circle of radius \(4 \mathrm{~cm}\) when the rim of the bowl is horizontal. Find the depth of the water.
    \item A bobbin for an industrial knitting machine is in the shape of a truncated cone. The diameter of the top is \(4 \mathrm{~cm}\), the diameter of the base is \(6 \mathrm{~cm}\) and the length of the slant is \(10 \mathrm{~cm}\). Find the height of the bobbin.
        \begin{center}
        \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-09}
        \end{center}
    \item Rationalise the following:
        \begin{multicols}{2}
        \begin{enumerate}
            \item \(\frac{1}{\sqrt{3}+2}\)
            \item \(\frac{1}{\sqrt{3}+\sqrt{2}}\)
        \end{enumerate}
            \end{multicols}
    \item Find the integers \(p\) and \(q\) such that \(\frac{\sqrt{5}}{\sqrt{5}-2}=p+q \sqrt{5}\).
    \item Simplify the following:
        \begin{multicols}{2}
        \begin{enumerate}
            \item \(\frac{3}{\sqrt{5}-2}+\frac{2}{\sqrt{5}+2}\)
            \item \(\frac{2}{6-3 \sqrt{3}}-\frac{1}{2 \sqrt{3}+3}\)
            \item \(\frac{5}{(\sqrt{7}-\sqrt{2})^{2}}\)
            \item \(0.0 \mathrm{i} \dot{6}\)
            \item \(0.3 \dot{2} \dot{4}\)
            \item \(0.51 \dot{2} \dot{6}\)
            \item \(0.001 \dot{1}\)
        \end{enumerate}
            \end{multicols}
    \item Challenge:
        \begin{enumerate}
            \item Show that \(\frac{138}{19}=7+\frac{1}{3+\frac{1}{1+\frac{1}{4}}}\)
The expression on the right is called a continued fraction.
            \item Express \(\frac{153}{11}\) as a continued fraction with all numerators 1 .
        \end{enumerate}
\end{enumerate}




|721806725|kitty|/Applications/kitty.app|0|
This week we shall continue with our study of the type of numbers which are created when we take some kind of root $ \sqrt{\,\,\,}$ of another \emph{positive} number. These quantities we now know to be called \line(1,0){100}, and last week we learned the basic arithmetic of these objects.\\

Before we begin our review / brain warm-up let us contextualise today's theory: We will begin by inserting \textbf{surds} into the algebraic structures of our first topic ($(a+b)^2 = a^2 + 2ab + b^2$, etc...) where $a$ and $b$ are now surd terms.\\

Then we will learn about something known as \emph{rationalising the denominator} which is an important simplification method in mathematics, and one of the earlier conventions adopted by the \textit{Babylonians} so that they would not be dividing by incomensurable quantities - i.e. what does it mean to divide $4$ apples into $\sqrt{2}$ quantities? That would be asking to divide the 4 apples into $1.41421356237\ldots$ parts, where the `dot, dot, dot' means the denominator never terminates!\\

After practising the above simplifications, we return to Pythagora's Theorem and Geometry. Specifically we attempt problems in 3 dimensions where the need for visualisation and mathematical stamina increase.\\

The final 2 sections then deal with rationalising \emph{binomial denominators} and converting irrational numbers to fractions respectively. The latter of these makes for a cool party trick: \textsc{What fraction is equal to the repeating decimal $0.81818181\ldots$?} Answer: bottom left
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\fancyfoot[L]{
    \begin{turn}{180}$\frac{9}{11}$\end{turn}
}
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% comment to self: always write your multicols before your enumerates!
\begin{exercisebox}
    \subsection*{Warm-up}
    \begin{doublespace}
    \begin{enumerate}
        \item Simplify:
            \begin{multicols}{2}
                \begin{enumerate}
                \item \(\sqrt{118}=\)\dotfill
                \item \(\sqrt{90}=\)\dotfill
                \item \(\sqrt{52}=\)\dotfill
                \item \(\sqrt{98}=\)\dotfill
                \end{enumerate}
            \end{multicols}
        \item Add or Subtract:
            \begin{multicols}{2}
                \begin{enumerate}
                    \item $3\sqrt{2}+2\sqrt{2}$=\dotfill
                    \item $\sqrt{32}-\sqrt{18}$=\dotfill
                    \item $\sqrt{45}+2\sqrt{5}-\sqrt{80}$=\dotfill
                    \item $\sqrt{28}+2\sqrt{63}-5\sqrt{7}$=\dotfill
                \end{enumerate}
            \end{multicols}
        \item Multiply or Divide:
            \begin{multicols}{2}
                \begin{enumerate}
                    \item $2\sqrt{3}\times 5\sqrt{6}$=\dotfill
                    \item $3\sqrt{5}\times 2\sqrt{10}$=\dotfill
                    \item $14\sqrt{40}\div 7\sqrt{5}$=\dotfill
                    \item $3\sqrt{2}\div \sqrt{2}$=\dotfill
                \end{enumerate}
            \end{multicols}
    \end{enumerate}
    \end{doublespace}
\end{exercisebox}

\newpage
\fancyfoot{}
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\documentclass[10pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage[version=4]{mhchem}
\usepackage{stmaryrd}
\usepackage{graphicx}
\usepackage[export]{adjustbox}
\graphicspath{ {./images/} }


examples

exercises


homework 

\section*{Solution}
The base is a square with side length \(4 \mathrm{~cm}\), so:

\[
\begin{aligned}
A C^{2} & =4^{2}+4^{2} \\
& =32 \\
A C & =\sqrt{32} \\
& =4 \sqrt{2}
\end{aligned}
\]

Hence \(E C=2 \sqrt{2} \quad\) (Leave in exact form to maintain accuracy.)

Triangle \(V E C\) is right-angled, so:

\[
\begin{aligned}
x^{2} & =V E^{2}+E C^{2} \\
& =52+\left(2 \sqrt{2)^{2}}\right. \\
& =25+8 \\
& =33 \\
x & =\sqrt{33} \\
& \approx 5.74 \quad \text { (correct to 2 decimal places) }
\end{aligned}
\]

The lenth of \(V C\) is \(5.74 \mathrm{~cm}\) correct to 2 decimal places.

\section*{Applications of Pythagoras' theorem in three dimensions}
\begin{itemize}
  \item Pythagoras' theorem can be used to find lengths in three-dimensional problems.

  \item Always draw a careful diagram identifying the appropriate right-angled triangle(s).

  \item To maintain accuracy, use exact values and only approximate using a calculator at the end of the problem if required.

\end{itemize}

\section*{Exercise 2G}

2 Find the length of the space diagonal of the rectangular prism whose length, width and height are:
a \(12 \mathrm{~cm}, 9 \mathrm{~cm}, 8 \mathrm{~cm}\)
b \(12 \mathrm{~cm}, 5 \mathrm{~cm}, 8 \mathrm{~cm}\)
c \(10 \mathrm{~cm}, 4 \mathrm{~cm}, 7 \mathrm{~cm}\)
d \(8 \mathrm{~cm}, 6 \mathrm{~cm}, 4 \mathrm{~cm}\)
e \(7 \mathrm{~cm}, 2 \mathrm{~cm}, 3 \mathrm{~cm}\)
f \(a \mathrm{~cm}, b \mathrm{~cm}, c \mathrm{~cm}\)



5 A builder needs to carry lengths of timber along a corridor in order to get them to where he is working. There is a right-angled bend in the corridor along the way. The corridor is \(3 \mathrm{~m}\) wide and the ceiling is \(2.6 \mathrm{~m}\) above the floor. What is the longest length of timber that the builder can take around the corner in the corridor? (Hint: Draw a diagram.)


7 For the rectangular prism shown opposite, \(E H=4 \mathrm{~cm}\) and \(H G=2 \mathrm{~cm}\).

a Find the exact length of \(E G\), giving your answer as a surd in simplest form.

b If \(A E=\frac{1}{2} E G\), find the exact value of \(A E\).

\includegraphics[max width=\textwidth, center]{2023_11_09_1c5231c22d14cc915b41g-09(2)}
c Find the length of:

i \(B E\)

ii \(B H\)

d What type of triangle is triangle \(B E H\) ?

e Show that if \(E H=2 a \mathrm{~cm}, H G=a \mathrm{~cm}\) and \(A E=\frac{1}{2} E G\), then the sides of the triangle \(B E H\) are in the ratio \(3: 4: 5\).

\section*{Binomial denominators}
Consider the expression \(\frac{1}{\sqrt{7}-\sqrt{5}}\).

How can we write this as a quotient with a rational denominator?

In the section on special products, we saw that:

\[
\begin{aligned}
(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5}) & =(\sqrt{7})^{2}-(\sqrt{5})^{2} \\
& =7-5 \\
& =2 \text {, which is rational, }
\end{aligned}
\]

so

\[
\begin{aligned}
\frac{1}{\sqrt{7}-\sqrt{5}} & =\frac{1}{\sqrt{7}-\sqrt{5}} \times \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}} \\
& =\frac{\sqrt{7}+\sqrt{5}}{7-5} \\
& =\frac{\sqrt{7}+\sqrt{5}}{2}
\end{aligned}
\]

Similarly,

\[
\begin{aligned}
(5 \sqrt{2}-4)(5 \sqrt{2}+4) & =(5 \sqrt{2})^{2}-4^{2} \\
& =34, \text { which again is rational, }
\end{aligned}
\]

so

\[
\begin{aligned}
\frac{3}{5 \sqrt{2}-4} & =\frac{3}{5 \sqrt{2}-4} \times \frac{5 \sqrt{2}+4}{5 \sqrt{2}+4} \\
& =\frac{15 \sqrt{2}+12}{34}
\end{aligned}
\]

Using the difference of two squares identity in this way is an important technique.

\section*{Example 28}
Simplify the following:

\section*{Solution}
a \(\frac{2 \sqrt{5}}{2 \sqrt{5}-2}=\frac{2 \sqrt{5}}{2 \sqrt{5}-2} \times \frac{2 \sqrt{5}+2}{2 \sqrt{5}+2}\)

b \(\frac{\sqrt{3}+\sqrt{2}}{3 \sqrt{2}+2 \sqrt{3}}=\frac{\sqrt{3}+\sqrt{2}}{3 \sqrt{2}+2 \sqrt{3}} \times \frac{3 \sqrt{2}-2 \sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}\)

\(=\frac{20+4 \sqrt{5}}{20-4}\)

\(=\frac{(\sqrt{3}+\sqrt{2})(3 \sqrt{2}-2 \sqrt{3})}{18-12}\)

\(=\frac{20+4 \sqrt{5}}{16}\)

\(=\frac{(3 \sqrt{6}-6+6-2 \sqrt{6})}{6}\)

\(=\frac{4(5+\sqrt{5})}{16}\)

\(=\frac{\sqrt{6}}{6}\)

\(=\frac{5+\sqrt{5}}{4}\)

\section*{Binomial denominators}
To rationalise a denominator which has two terms, we use the difference of two squares identity:

\begin{itemize}
  \item In an expression such as \(\frac{3}{5+\sqrt{3}}\), multiply top and bottom by \(5-\sqrt{3}\).

  \item In an expression such as \(\frac{\sqrt{2}}{7-3 \sqrt{2}}\), multiply top and bottom by \(7+3 \sqrt{2}\).

\end{itemize}

The surd \(7-\sqrt{3}\) is called the conjugate of \(7+\sqrt{3}\) and \(7+\sqrt{3}\) is the conjugate of \(7-\sqrt{3}\).

\section*{Exercise \(2 \mathrm{H}\)}
1 Simplify:
a \((\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})\)
b \((2 \sqrt{5}-\sqrt{3})(2 \sqrt{5}+\sqrt{3})\)
c \((7 \sqrt{2}+4 \sqrt{3})(7 \sqrt{2}-4 \sqrt{3})\)
d \((4-\sqrt{3})(4+\sqrt{3})\)

2 Rationalise the denominator in each expression.
a \(\frac{1}{\sqrt{6}+1}\)
d \(\frac{2}{3-\sqrt{5}}\)
g \(\frac{\sqrt{3}}{\sqrt{6}+\sqrt{5}}\)
h \(\frac{\sqrt{2}}{\sqrt{2}-\sqrt{3}}\)
i \(\frac{\sqrt{2}}{2 \sqrt{5}+\sqrt{2}}\)
j \(\frac{\sqrt{5}}{2 \sqrt{5}-\sqrt{3}}\)
k \(\frac{\sqrt{5}}{3 \sqrt{2}+4 \sqrt{3}}\)
l \(\frac{\sqrt{3}}{2 \sqrt{3}-3 \sqrt{6}}\)
m \(\frac{2 \sqrt{6}}{4 \sqrt{2}-3 \sqrt{7}}\)
n \(\frac{3 \sqrt{2}}{6 \sqrt{3}+11 \sqrt{5}}\)
o \(\frac{2 \sqrt{3}}{3 \sqrt{2}+10 \sqrt{3}}\)

3 Rationalise the denominators in these expressions and use the decimal approximations \(\sqrt{2} \approx 1.414\) and \(\sqrt{3} \approx 1.732\) to evaluate them correct to 2 decimal places.


5 Simplify:

6 Simplify:
a \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\)
b \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\)

\section*{Irrational numbers and surds}
We first recall some facts about fractions and decimals.

\section*{Fractions and decimals}
We know that some fractions can be written as decimals that terminate. For example,

\[
\frac{1}{4}=0.25, \frac{3}{16}=0.1875
\]

Some fractions have decimal representations that do not terminate. For example,

\(\frac{1}{3}=0.33333 \ldots\) which we write as \(0 . \dot{3}\).

The dot above the 3 indicates that the digit 3 is repeated forever.

Some fractions have decimal representations that eventually repeat. For example, \(\frac{1}{6}=0.1666 \ldots\) which is written as \(0.1 \dot{6}\).

Other fractions have decimal representations with more than one repeating digit. For example,

\(\frac{1}{11}=0.090909 \ldots\) which we write as \(0 . \dot{0} \dot{9}\)

with a dot above both of the repeating digits.

Another example is \(0.12 \dot{3} 45 \dot{6}=0.1234563456 \ldots\)

\section*{Converting decimals to fractions}
It is easy to write terminating decimals as fractions by using a denominator which is a power of 10 .

For example,

\[
0.14=\frac{14}{100}=\frac{7}{50}
\]

Decimals that have a repeated sequence of digits can also be written as fractions.

For example,

\[
0.12 \dot{3}=\frac{41}{333} \text { and } 0.67 \dot{1} \dot{2}=\frac{443}{660}
\]

A method for doing this is shown in the next two examples.

\section*{Example 29}

\section*{Example 30}

\section*{Rational numbers}
A rational number is a number that can be written as a fraction \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\).

All integers are rational numbers. For example,

\[
-3=\frac{-3}{1}
\]

\section*{Example 31}
Explain why each of these numbers is rational.
a \(5 \frac{3}{7}\)
b 2.1237
c \(0 . \dot{2} \dot{4}\)

\section*{Solution}
We will express each number in the form \(\frac{p}{q}\).

a \(5 \frac{3}{7}=\frac{38}{7}\)

b \(2.1237=2 \frac{1237}{10000}\)

\[
=\frac{21237}{10000}
\]

(continued over page)

\[
\begin{aligned}
S & =0 . \dot{2} \dot{4} \\
& =0.24242 \ldots \\
100 S & =24.24242 \ldots \quad \text { (Multipy by 100.) } \\
100 S & =24+S \\
99 S & =24 \\
\text { Hence } S & =\frac{24}{99} \\
& =\frac{8}{33} \\
0 . \dot{2} \dot{4} & =\frac{8}{33} \\
\text { Thus } \quad &
\end{aligned}
\]

Each number has been expressed as a fraction \(\frac{p}{q}\), where \(p\) and \(q\) are integers, so each number
is rational.

\section*{Irrational numbers}
Mathematicians up to about \(600 \mathrm{BCE}\) thought that all numbers were rational. However, when we apply Pythagoras' theorem, we encounter numbers such as \(\sqrt{2}\) that are not rational. The number \(\sqrt{2}\) is an example of an irrational number.

An irrational number is one that is not rational. Hence an irrational number cannot be written in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\). Nor can it be written as a terminating or repeating decimal.

In \(300 \mathrm{BCE}\), Euclid proved that making the assumption that \(\sqrt{2}\) is rational leads to a contradiction. Hence \(\sqrt{2}\) was shown to be irrational. The proof is outlined here.

Assume \(\sqrt{2}=\frac{p}{q}\) where \(p\) and \(q\) are integers with highest common factor 1 .

Square both sides of this equation to obtain \(2=\frac{p^{2}}{q^{2}}\). We can write \(p^{2}=2 q^{2}\). Hence \(p^{2}\) is even and thus \(p\) is even. We can now write, \(2=\frac{4 k^{2}}{q^{2}}\) For some whole number \(k\). From this, show \(q\) is also even which is a contradiction of our assumption that the highest common factor is 1 .

The decimal expansion of \(\sqrt{2}\) goes on forever but does not repeat. The value of \(\sqrt{2}\) can be approximated using a calculator. The same is true of other irrational numbers such as \(\sqrt{3}, \sqrt{14}\) and \(\sqrt{91}\). Try finding these numbers on your calculator and see what you get.

Other examples of irrational numbers include:

\[
\sqrt{3}, \sqrt[3]{2}, \pi, \pi^{3} \text { and } \sqrt{\pi}
\]

There are infinitely many rational numbers because every whole number is rational. We can also easily write down infinitely many other fractions. For example, \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots\)

There are infinitely many irrational numbers too. For example, \(\sqrt{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{3}, \frac{\sqrt{2}}{4}, \ldots\)

Every number, whether rational or irrational, is represented by a point on the number line.

Conversely, we can think of each point on the number line as a number.

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-16(1)}
\end{center}

\section*{Example 32}
Arrange these irrational numbers in order of size on the number line.
\(\sqrt{8}\)
\(\sqrt{2}\)
\(\sqrt[3]{60}\)
\(\sqrt[4]{30}\)

\section*{Solution}
Find on approximation of each number to 2 decimal places using a calculator.

\[
\begin{array}{ll}
\sqrt{8} \approx 2.83 & \sqrt{2} \approx 1.14 \\
\sqrt[3]{60} \approx 3.91 & \sqrt[4]{30} \approx 2.34
\end{array}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-16}
\end{center}

\section*{Surds}
In Year 8, we looked at squares, square roots, cubes and cube roots. For example:

\[
\begin{aligned}
& 5^{2}=25 \quad \text { and } \quad \sqrt{25}=5 \\
& 5^{3}=125 \quad \text { and } \quad \sqrt[3]{125}=5
\end{aligned}
\]

The use of this notation can be extended. For example:

\[
5^{4}=625 \quad \text { and } \quad \sqrt[4]{625}=5
\]

This is read as 'The fourth power of 5 is 625 and the fourth root of 625 is 5 '. We also have fifth roots, sixth roots and so on.

In general, we can take the \(n\)th root of any positive number \(a\). The \(n\)th root of \(a\) is the positive number whose \(n\)th power is \(a\).

The statement \(\sqrt[n]{a}=\mathrm{b}\) is equivalent to the statement \(b^{n}=a\).

Your calculator will give you approximations to the \(n\)th root of \(a\).

Note: For \(n\), a positive integer, \(0^{n}=0\) and \(\sqrt[n]{0}=0\).

An irrational number which can be expressed as \(\sqrt[n]{a}\), where \(a\) is a positive whole number, is called a surd.

For example, \(\sqrt{2}, \sqrt{7}\) and \(\sqrt[3]{5}\) are all examples of surds, while \(\sqrt{4}\) and \(\sqrt{9}\) are not surds since they are whole numbers. The number \(\pi\), although it is irrational, is not a surd. This is also difficult to prove.

\section*{Example 33}
Use Pythagoras' theorem to construct a line of length \(\sqrt{20}\).

\section*{Solution}
Find two perfect squares that sum to 20 .

\(4^{2}+2^{2}=20\)

Draw perpendicular line segments from a common point of lengths 2 units and 4 units. Connect their endpoints with a third line segment.

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-17}
\end{center}

\section*{Irrational numbers and surds}
\begin{itemize}
  \item Every fraction can be written as a terminating, or eventually repeating, decimal.

  \item Every terminating, or eventually repeating, decimal can be written as a fraction.

  \item A rational number is a number which can be expressed as \(\frac{p}{q}\) where \(p\) and \(q\) are integers and \(q \neq 0\). That is, a rational number is an integer or a fraction.

  \item There are numbers such as \(\pi\) and \(\sqrt{2}\) that are irrational (not rational).

  \item Each rational and each irrational number represents a point on the number line.

  \item Every point on the number line represents a rational or an irrational number.

  \item An irrational number which can be expressed as \(\sqrt[n]{a}\), where \(n\) and \(a\) are positive whole numbers, is called a surd. \(\pi\) is not a surd.

\end{itemize}

\section*{Exercise 2I}
1 Write each repeating decimal as a fraction.

2 Write each repeating decimal as a fraction.
e \(0 . \dot{6} 1 \dot{3}\)

3 Show that each number is rational by writing it as a fraction.
a \(3 \frac{2}{3}\)
b 5.15
c \(0 . \dot{4}\)
d \(0 . \dot{1} \dot{5}\)
e \(5 \frac{1}{7}\)
f 1.3


5 By approximating correct to 2 decimal places, place these real numbers on the same number line.
a \(\sqrt{7}\)
b 2.7
c \(\sqrt[3]{18}\)
d \(2 \pi\)
e \(\sqrt{2}\)

6 Which of these numbers are surds?
a 3
b \(\sqrt{5}\)
c 4
d \(\sqrt{9}\)
e \(\sqrt{7}\)
f \(\sqrt{16}\)
g \(\sqrt{10}\)
h \(\sqrt{1}\)
i \(\sqrt{3}\)
j \(\sqrt{15}\)
k 5
l \(\sqrt{25}\)

7 a Use the fact that \(1^{2}+2^{2}=5\) to construct a length \(\sqrt{5}\).

b Use the fact that \(4^{2}+5^{2}=41\) to construct a length \(\sqrt{41}\).

8 How would you construct an interval of length:
a \(\sqrt{73}\) ?
b \(\sqrt{12}\) ?
c \(\sqrt{21}\) ?


\section*{Review exercise}
1 For each right-angled triangle, find the value of the pronumeral.

a

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-18(3)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-18(1)}
\end{center}

2 For each right-angled triangle, find the value of the pronumeral.

a

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-18}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-18(2)}
\end{center}

3 A support bracket is to be placed under a shelf, as shown in the diagram. If \(A B=A C=20 \mathrm{~cm}\), find, correct to the nearest millimetre, the length of \(B C\).

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-19(1)}
\end{center}

4 A road runs in an east-west direction joining towns \(A\) and \(B\), which are \(40 \mathrm{~km}\) apart. \(A\) third town, \(C\), is situated \(20 \mathrm{~km}\) due north of \(B\). A straight road is built from \(C\), to the road between \(A\) and \(B\) and meets it at \(D\), which is equidistant from \(A\) and \(C\). Find the length of road \(C D\).

5 Circles of radius \(6 \mathrm{~cm}\) and \(3 \mathrm{~cm}\) are placed in a square, as shown in the diagram opposite. Find, correct to 2 decimal places:
a \(E B\)
c the length of the diagonal \(B D\)
d the side length of the square

b \(F D\)

6 Simplify:

\includegraphics[max width=\textwidth, center]{2023_11_09_1c5231c22d14cc915b41g-19}
a \(3 \sqrt{2}+2 \sqrt{2}\)
b \(\sqrt{32}-\sqrt{18}\)

7 Simplify:
a \(2 \sqrt{3} \times 5 \sqrt{6}\)
b \(3 \sqrt{5} \times 2 \sqrt{10}\)

8 Simplify:
a \(2 \sqrt{3}(3+\sqrt{3})\)
b \(5 \sqrt{2}(3 \sqrt{2}-2)\)

9 Expand and simplify:
a \((2 \sqrt{2}+1)(3 \sqrt{2}-2)\)
b \((5 \sqrt{3}-2)(2 \sqrt{3}-1)\)

10 Simplify:
a \(\sqrt{80}\)
b \(\sqrt{108}\)
c \(\sqrt{125}\)
d \(\sqrt{72}\)
e \(\sqrt{2048}\)
f \(\sqrt{448}\)
g \(\sqrt{800}\)
h \(\sqrt{112}\)

11 Simplify:
a \(\sqrt{45}+2 \sqrt{5}-\sqrt{80}\)
b \(\sqrt{28}+2 \sqrt{63}-5 \sqrt{7}\)
c \(\sqrt{44}+\sqrt{275}-4 \sqrt{11}\)
d \(\sqrt{162}-\sqrt{200}+\sqrt{288}\)

12 Express with a rational denominator.
a \(\frac{3}{\sqrt{11}}\)
b \(\frac{1}{5 \sqrt{15}}\)
c \(\frac{4}{7 \sqrt{7}}\)
d \(\frac{3}{\sqrt{17}}\)
e \(\frac{3}{\sqrt{3}}\)
f \(\frac{3}{5 \sqrt{15}}\)
g \(\frac{14}{\sqrt{7}}\)
h \(\frac{11}{\sqrt{3}}\)

13 Express with a rational denominator.
a \(\frac{3}{3-\sqrt{3}}\)
b \(\frac{22}{2+3 \sqrt{5}}\)
c \(\frac{24}{1-\sqrt{7}}\)
d \(\frac{24}{1+\sqrt{17}}\)
e \(\frac{3}{2-\sqrt{3}}\)
f \(\frac{30}{1-\sqrt{11}}\)
g \(\frac{15}{2-\sqrt{7}}\)
h \(\frac{10}{2-\sqrt{3}}\)

14 a Expand and simplify \((2 \sqrt{5}-\sqrt{3})^{2}\).

b Simplify \(\frac{2}{\sqrt{3}-2}+\frac{2}{\sqrt{3}}\), expressing your answer with a rational denominator.

15 The diagram opposite shows part of a skate board ramp. (It is a prism whose cross-section is a rightangled triangle.) Use the information in the diagram to find:
a \(B C\)
b \(A C\)

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-20}
\end{center}

16 Find the length of the space diagonal of a cube with side length \(5 \mathrm{~cm}\).

17 A motorist departs from town \(B\), which is \(8 \mathrm{~km}\) due south from another town, \(A\), and drives due east towards town \(C\), which is \(20 \mathrm{~km}\) from \(B\). After driving a distance of \(x \mathrm{~km}\), he notices that he is the same distance away from both towns \(A\) and \(C\).

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-20(2)}
\end{center}

a Express the motorist's distance from \(A\) in terms of \(x\) (that is, \(A D\) ).

b Express the motorist's distance from \(C\) in terms of \(x\).

c Find the distance he has driven from \(B\).

18 The diagram opposite shows the logo of a particular company. The large circle has centre \(O\) and radius \(12 \mathrm{~cm}\) and \(A B\) is a diameter. \(D\) is the centre of the middle-sized circle with diameter \(O B\). Finally, \(C\) is the centre of the smallest circle.

a What is the radius of the circle with centre \(D\) ?

b If the radius of the circle with centre \(C\) is \(r \mathrm{~cm}\), express these in terms of \(r\).
i \(O C\)
ii \(D C\)

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-20(1)}
\end{center}

c Find the value of \(r\).

19 In the diagram opposite, \(V A B C D\) is a square-based pyramid with \(A B=B C=C D=D A=10\) and \(V A=V B=V C=V D=10\).

a What type of triangle is triangle \(V B C\) ?

b If \(M\) is the midpoint of \(C B\), find the exact values of:

i \(V M\)

ii \(V E\), the height of the pyramid

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-21(1)}
\end{center}

20 Write each repeating decimal as a fraction.
a \(0 . \dot{8}\)
b \(0.1 \dot{5}\)
c \(0.7 \dot{2}\)
d \(0.0 \dot{3}\)
e \(0 . \dot{8} \dot{1}\)
f \(0 . \dot{9} \dot{6}\)
g \(0.10 \mathrm{i}\)
h 0.0015

21 Write \(\frac{67}{29}\) as a continued fraction. (See Exercise 2I, Question 9.)

22 Write \(0 . \dot{4} 79281\) as a fraction.

23 How would you construct intervals of the following lengths? Discuss with your teacher.
a \(\sqrt{3}\)
b \(\frac{1}{\sqrt{2}}\)
c \(\sqrt{8}\)
d \(\frac{1}{\sqrt{8}}\)

24 Find the length of the long diagonal of the rectangular prism whose length, width and height are:
a \(3+\sqrt{2}, 3-\sqrt{2}, 3 \sqrt{3}\)
b \(5+\sqrt{3}, 5-\sqrt{3}, 2 \sqrt{2}\)

25 For \(x=\sqrt{2}=1\) and \(y=\sqrt{5}-2\), find in simplest form.
a \(\frac{1}{x}\)
b \(\frac{1}{x}+\frac{1}{y}\)
c \(\frac{1}{x^{2}}+\frac{1}{y^{2}}\)
d \(\frac{1}{x}-\frac{1}{y}\)

\section*{Challenge exercise}
1 Prove that \(\sqrt[3]{2}\) is irrational.

2 In the triangle below, \(4^{2}+7^{2}=65>8^{2}\).

Is the angle \(\theta\) greater or less than \(90^{\circ}\) ? Explain your answer.

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-21}
\end{center}


\end{document}
|721806797|kitty|/Applications/kitty.app|0|
Special Products|721807506|kitty|/Applications/kitty.app|0|
Rationalising the Denominator|721807516|kitty|/Applications/kitty.app|0|
Pythagoras in 3-Dimensions!|721807522|kitty|/Applications/kitty.app|0|
Binomial Denominators|721807535|kitty|/Applications/kitty.app|0|
Irrational Numbers and Surds|721807549|kitty|/Applications/kitty.app|0|
\section{Compound Interest}
    \input{7-irrational}
|721807559|kitty|/Applications/kitty.app|0|
\section{Compound Interest}
|721807562|kitty|/Applications/kitty.app|0|
Compound Interest|721807567|kitty|/Applications/kitty.app|0|
7-irrational|721807573|kitty|/Applications/kitty.app|0|
    \input{7-irrational}
|721807578|kitty|/Applications/kitty.app|0|
 of practical financial topics. |721808268|kitty|/Applications/kitty.app|0|
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\end{questions}
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\documentclass{exam}
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\begin{document}
\begin{questions}
    \question What is the integral of \( f(x) = x^2 \)?

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        The integral of \( f(x) = x^2 \) is \( \frac{x^3}{3} + C \), where \( C \) is the constant of integration.
    \end{solutionorlines}
\end{questions}
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l.19     \begin{solutionorlines}[2cm]
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    \begin{solutionordottedlines}[2cm]
        The integral of \( f(x) = x^2 \) is \( \frac{x^3}{3} + C \), where \( C \) is the constant of integration.
    \end{solutionordottedlines}
|721911977|kitty|/Applications/kitty.app|0|
        The integral of \( f(x) = x^2 \) is \( \frac{x^3}{3} + C \), where \( C \) is the constant of integration.
|721912101|kitty|/Applications/kitty.app|0|
    \begin{solutionordottedlines}[1in]
        Consumer means somebody that takes something and uses it. By default humans are all consumers, and so being good at the arithmetic of consuming things is something that can be advantageous to all that are human.
    \end{solutionordottedlines}
|721912105|kitty|/Applications/kitty.app|0|
 By default humans are all consumers, and so being good at the arithmetic of consuming things is something that can be advantageous to all that are human.|721912134|kitty|/Applications/kitty.app|0|
        Consumer means somebody that takes something and uses it. By default humans are all consumers, and so being good at the arithmetic of consuming things is something that can be advantageous to all that are human.
|721912138|kitty|/Applications/kitty.app|0|
    \begin{solutionordottedlines}[1in] 
        By default humans are all consumers, and so being good at the arithmetic of consuming things is something that can be advantageous to all that are human.
    \end{solutionordottedlines}
|721912148|kitty|/Applications/kitty.app|0|
        By default humans are all consumers, and so being good at the arithmetic of consuming things is something that can be advantageous to all that are human.
|721912150|kitty|/Applications/kitty.app|0|
    In this lesson we will cover the \textbf{theory} of \textit{Consumer Arithmetic}.
|721912306|kitty|/Applications/kitty.app|0|
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        The integral of \( f(x) = x^2 \) is \( \frac{x^3}{3} + C \), where \( C \) is the constant of integration.
    \end{solutionordottedlines}
\end{questions}
\end{document}
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\documentclass{article}
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\begin{examplebox}
    \subsection*{Examples}
    \begin{questions}
        \question Convert the following from percentages to fractions
        \begin{part}
            \part[1] $42\%$
            \part[1] $12\frac{1}{4}\%$
        \end{part}
        \question Convert the following from fractions to percentages
        \begin{part}
            \part[1] $\frac{3}{5}$
            \part[1] $\frac{7}{20}$
        \end{part}
    \end{questions}


\end{examplebox}
|721914772|kitty|/Applications/kitty.app|0|
    \begin{questions}
        \question Convert the following from percentages to fractions
        \begin{part}
            \part[1] $42\%$
            \part[1] $12\frac{1}{4}\%$
        \end{part}
        \question Convert the following from fractions to percentages
        \begin{part}
            \part[1] $\frac{3}{5}$
            \part[1] $\frac{7}{20}$
        \end{part}
    \end{questions}
|721914870|kitty|/Applications/kitty.app|0|
        \begin{part}
            \part[1] $42\%$
            \part[1] $12\frac{1}{4}\%$
        \end{part}
        \question Convert the following from fractions to percentages
        \begin{part}
            \part[1] $\frac{3}{5}$
            \part[1] $\frac{7}{20}$
        \end{part}
|721914887|kitty|/Applications/kitty.app|0|
fillin[0.42][\hfill]|721915574|kitty|/Applications/kitty.app|0|
[\hfill]|721915628|kitty|/Applications/kitty.app|0|
{0.42}|721915712|kitty|/Applications/kitty.app|0|
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\documentclass[10pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage[version=4]{mhchem}
\usepackage{stmaryrd}
\usepackage{graphicx}
\usepackage[export]{adjustbox}
\graphicspath{ {./images/} }

\title{Consumer arithmetic }

\author{}
\date{}


\begin{document}
\maketitle
This chapter deals with some important practical financial topics such as investing and borrowing money, income tax and GST, inflation, depreciation, profits and losses, discounts and commissions. Mention is also made of topics such as population, rainfall and the composition of materials.

When buying a car, using a credit card, or deciding how good a particular 'bargain' is, we all need to be competent in our understanding of consumer arithmetic. Otherwise, we run the serious risk of wasting our money or, even worse, being at the mercy of unscrupulous individuals. This is particularly important when we come to the 'big ticket items' such as purchasing a car or a home, or planning superannuation. It is also important when completing our tax returns.

\section*{Using a calculator and approximations}
Everything in this chapter involves calculations with percentages. We are now assuming that you are using a calculator, so we have made little attempt to set questions where the numbers work out nicely. However, some of the problems can be done mentally. This chapter gives you the opportunity to learn how to efficiently use a calculator.

Nevertheless, you should always look over your work and check that the answers to your calculations are reasonable and sensible.

When the calculator displays numbers with many decimal places, you will need to round the answer in some way that is appropriate in the context of the question.

Whenever you round a number, you should be careful to use the symbol \(\approx\), which means approximately equals, rather than the symbol \(=\), which means exactly equals.

For example, if we are told that a grocer received \(\$ 1000\) in cash payments for goods that he had sold and that he banked one-third of this, we would write:

\[
\text { amount banked }=1000 \div 3
\]

\[
\approx \$ 333.33
\]

not

\[
\begin{aligned}
\text { amount banked } & =1000 \div 3 \\
& =\$ 333.3333 \ldots
\end{aligned}
\]

because the grocer could not have banked fractions of a cent.

\section*{Review of percentages}
A percentage such as \(35 \%\) is a rational number and can be rewritten as a decimal or as a fraction, as follows:

\[
\begin{aligned}
& \begin{aligned}
35 \% & =35 \div 100 & \text { or } & 35 \%=\frac{35}{100} \\
& =0.35 & &
\end{aligned} \\
& =\frac{7}{20}
\end{aligned}
\]

In this chapter, we encourage you to write percentages as decimals rather than fractions. Decimals are more commonly used than fractions when dealing with money.

\section*{Converting a percentage to a decimal or a fraction}
To convert a percentage to a decimal or a fraction, divide by 100 .

For example: \(42 \%=42 \div 100\)

\[
\begin{aligned}
12 \frac{1}{4} \% & =\frac{49}{4} \% \\
& =\frac{49}{4} \times \frac{1}{100} \\
& =\frac{49}{400}
\end{aligned}
\]

Conversely, to convert a decimal such as 1.24 , or a fraction such as \(\frac{3}{8}\), to a percentage, multiply
by \(100 \%\).

\[
\begin{aligned}
& 1.24=1.24 \times 100 \% \\
& \frac{3}{8}=\frac{3}{8} \times \frac{100}{1} \% \\
& =124 \% \\
& =37 \frac{1}{2} \%
\end{aligned}
\]

\section*{Converting a decimal or a fraction to a percentage}
To convert a decimal or a fraction to a percentage, multiply by \(100 \%\).

\section*{Percentages of a quantity}
To calculate a percentage of a quantity:

\begin{itemize}
  \item convert the percentage to a decimal or a fraction

  \item multiply the quantity by the fraction or decimal.

\end{itemize}

\section*{Example 1}
The Brilliant Light Bulb Company estimates that \(3.5 \%\) of its light bulbs are defective. If a shop owner buys 1250 light bulbs to light the shop, how many would he expect to be defective?

\section*{Solution}
Number of defective bulbs \(=1250 \times 3.5 \%\)

\[
\begin{aligned}
& =1250 \times 0.035 \\
& \approx 44 \quad(\text { Round } 43.75 \text { to } 44 .)
\end{aligned}
\]

\section*{Calculating a percentage}
We can express one quantity as a percentage of another. Remember that both quantities must be expressed in the same unit of measurement before calculating a percentage.

\section*{Example 2}
A typical computer weighs about \(25 \mathrm{~kg}\). When it is broken down as waste, it yields about \(3 \mathrm{~g}\) of arsenic. What percentage of the total is this?

\section*{Solution}
Using grams, the computer weighs \(25000 \mathrm{~g}\) and the arsenic weighs \(3 \mathrm{~g}\).

Hence percentage of arsenic \(=\frac{3}{25000} \times 100 \%\)

\[
=\frac{3}{250} \%
\]

\[
=0.012 \%
\]

\section*{Calculating a percentage}
To calculate the percentage that one quantity, \(a\), is of another quantity, \(b\) :

\begin{itemize}
  \item first convert both quantities to the same unit of measurement

  \item then form the fraction \(\frac{a}{b}\) and multiply it by \(100 \%\).

\end{itemize}

\section*{Exercise 3A}
1 Express each percentage as a decimal.
a \(72 \%\)
b \(7.6 \%\)
c \(98 \%\)
d \(16 \%\)
e \(8 \%\)
f \(6.25 \%\)
g \(175 \%\)
h \(0.6 \%\)
i \(77 \frac{3}{4} \%\)
j \(0.1 \%\)
k \(142.6 \%\)
l \(\frac{1}{4} \%\)

2 Express each percentage as a fraction in lowest terms.
a \(35 \%\)
b \(56 \%\)
c \(75 \%\)
d \(37 \frac{1}{2} \%\)
e \(33 \frac{1}{3} \%\)
f \(16 \frac{2}{3} \%\)
g \(7.25 \%\)
h \(6.4 \%\)
i \(210 \%\)
j \(125 \%\)
k \(112 \frac{1}{2} \%\)
l \(136 \%\)

3 Express each fraction or decimal as a percentage.
a \(\frac{3}{5}\)
b \(\frac{3}{8}\)
c \(\frac{9}{16}\)
d \(2 \frac{1}{4}\)
e \(\frac{7}{20}\)
f \(\frac{2}{3}\)
g \(\frac{4}{3}\)
h \(\frac{3}{400}\)
i 0.43
j 0.225
k 0.04
l 0.015
m 1.2
n 2.03
o 1.175
p 0.0075

4 Copy and complete this table.

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_16_3bbfc34a770529e9f285g-05}
\end{center}

5 Evaluate these amounts, correct to 2 decimal places where necessary.
a \(15 \%\) of 40
b \(57 \%\) of 1000
c \(26 \%\) of 264
d \(120 \%\) of 538
e \(15.8 \%\) of 972
f \(138.5 \%\) of 650
g \(2.8 \%\) of 318
h \(0.1 \%\) of 6000
i \(150 \%\) of 846

6 Evaluate these amounts, correct to the nearest cent where necessary.
a \(62 \%\) of \(\$ 10\)
b \(23.7 \%\) of \(\$ 960\)
c \(3.2 \%\) of \(\$ 1500\)
d \(110 \%\) of \(\$ 1280\)
e \(0.25 \%\) of \(\$ 800\)
f \(6 \frac{1}{2} \%\) of \(\$ 200\)
g \(7 \frac{3}{4} \%\) of \(\$ 1000\)
h \(\frac{1}{4} \%\) of \(\$ 840\)
i \(7.25 \%\) of \(\$ 1600\)

7 Find what percentage the first quantity is of the second quantity, correct to 1 decimal place where necessary.
a \(7 \mathrm{~km}, 50 \mathrm{~km}\)
b \(\$ 4, \$ 200\)
c \(14 \mathrm{~kg}, 400 \mathrm{~kg}\)
d \(70 \mathrm{~m}, 50 \mathrm{~m}\)
e 15 weeks, 60 weeks
f 60 weeks, 15 weeks

8 Find what percentage the first quantity is of the second quantity, correct to 2 decimal places where necessary. You will need to express both quantities in the same unit first.
a 68 cents, \(\$ 5.00\)
b \(3.4 \mathrm{~cm}, 2 \mathrm{~m}\)
c \(7 \mathrm{~g}, 3 \mathrm{~kg}\)
d 8 hours, 2 weeks
e 15 days, 3 years
f \(250 \mathrm{~m}, 4 \mathrm{~km}\)
g \(4 \mathrm{~km}, 250 \mathrm{~m}\)
h 1 day, 1 year
i 1 year, 1 day
j 33 weeks, 1 century
k \(56 \mathrm{~cm}, 2.4 \mathrm{~km}\)
l 5 apples, 16 dozen apples

9 Find what percentage the first quantity is of the second quantity, correct to 4 decimal places where necessary.
a \(48 \mathrm{~mm}, 1 \mathrm{~km}\)
b 1.5 hours, 3 years
c \(7.8 \mathrm{~g}, 60 \mathrm{~kg}\)
d 3.5 cents, \(\$ 1400\)

10 There are 740 students at a primary school, \(5 \%\) of whom have red hair. Calculate the number of students in the school who have red hair.

11 A sample of a certain alloy weighs \(1.6 \mathrm{~g}\).

a Aluminium makes up \(48 \%\) of the alloy. What is the weight of the aluminium in the sample?

b The percentage of lead in the alloy is \(0.23 \%\). What is the weight of the lead in the sample?

12 A soccer match lasted 92 minutes (including injury time). If team A was in possession for \(55 \%\) of the match, for how many minutes and seconds was team A in possession?

13 Carbon dioxide makes up \(0.059 \%\) of the mass of the Earth's atmosphere. The total mass of the atmosphere is about 5 million megatonnes. What is the total mass of the carbon dioxide in the atmosphere?

14 The label on a Sunnyvale tomato paste bottle says that in every \(25 \mathrm{~g}\) serving, there are \(3.6 \mathrm{~g}\) of carbohydrate, \(0.1 \mathrm{~g}\) of fat, and \(105 \mathrm{mg}\) of sodium.

a Express as a percentage of the \(25 \mathrm{~g}\) serving:

i the mass of carbohydrate

ii the mass of fat

iii the mass of sodium

b The Sunnyvale website claims that the percentage of protein is \(3.2 \%\). What mass of protein is that per \(25 \mathrm{~g}\) serving?

15 Mt Kosciusko has a height of \(2228 \mathrm{~m}\), while the height of Mt Everest is \(8848 \mathrm{~m}\). Calculate your answers to this question correct to 3 decimal places.

a What percentage is the height of Mt Everest of the height of Mt Kosciusko?

b The Earth's radius is about \(6400 \mathrm{~km}\). What percentage of the radius of the Earth is the height of Mt Everest?

16 In the Federal Parliament, there are 150 members in the House of Representatives, of whom 37 are from Victoria.

a Correct to 1 decimal place, what percentage of members are from Victoria?

b The population of Australia is about 22.6 million. What percentage of Australians are members of the House of Representatives?

17 The distance by air from Melbourne to Darwin is \(3346 \mathrm{~km}\), and from Melbourne to Singapore it is \(6021 \mathrm{~km}\). What percentage, correct to the nearest percent, is:

a the Melbourne-Darwin distance of the Melbourne-Singapore distance?

b the Melbourne-Singapore distance of the Melbourne-Darwin distance?

\section*{Using percentages}
Percentages are used extensively in finance, and are common in many other practical situations. The rest of this chapter will give examples of a few well-known situations, concentrating on financial applications. If you read a newspaper for a few days, you will find a great variety of further interesting uses of percentages.

First, however, we will introduce another important method that will be used with percentages throughout this chapter.

\section*{Reversing the process to find the original amount}
Suppose that \(15 \%\) of the total mass of a chicken roll is actually chicken. What mass of chicken rolls can be made with \(600 \mathrm{~g}\) of chicken?

\[
\text { Mass of chicken }=\text { mass of rolls } \times 15 \%
\]

Reversing this:

\[
\begin{aligned}
\text { mass of rolls } & =\text { mass of chicken } \div 15 \% \\
& =600 \div 0.15 \\
& =4000 \mathrm{~g}
\end{aligned}
\]

Hence \(4 \mathrm{~kg}\) of chicken rolls can be made with \(600 \mathrm{~g}\) of chicken.

\section*{Finding the original amount}
\begin{itemize}
  \item To find, for example, \(15 \%\) of a given amount, multiply by \(15 \%\).

  \item Conversely, to find the original amount given \(15 \%\) of it, divide by \(15 \%\).

\end{itemize}

\section*{Example 3}
Joshua saves \(12 \%\) of his after-tax salary every week. If he saves \(\$ 90\) a week, what is his after-tax salary?

\section*{Solution}
Savings \(=\) after-tax salary \(\times 12 \%\)

Reversing this:

\[
\begin{aligned}
\text { after-tax salary } & =\text { savings } \div 12 \% \\
& =90 \div 0.12 \\
& =\$ 750
\end{aligned}
\]

\section*{Example 4}
Sterling silver is an alloy that is made up of \(92.5 \%\) by mass silver and \(7.5 \%\) copper.

a How much sterling silver can be made with \(5 \mathrm{~kg}\) of silver and unlimited supplies of copper?

b How much sterling silver can be made with \(5 \mathrm{~kg}\) of copper and unlimited supplies of silver?

\section*{Solution}
a Mass of silver \(=\) mass of sterling silver \(\times 92.5 \%\)

Reversing this:

mass of sterling silver \(=\) mass of silver \(\div 92.5 \%\)

\[
\begin{aligned}
& =5 \div 0.925 \\
& \approx 5.405 \mathrm{~kg}
\end{aligned}
\]

b Mass of copper \(=\) mass of sterling silver \(\times 7.5 \%\)

Reversing this:

mass of sterling silver \(=\) mass of copper \(\div 7.5 \%\)

\[
\begin{aligned}
& =5 \div 0.075 \\
& \approx 66.667 \mathrm{~kg}
\end{aligned}
\]

Note: One reason for the choice of rounding to 3 decimal places is that there are \(1000 \mathrm{~g}\) in a kilogram. That is, we are calculating to the nearest gram.

\section*{Commission}
If a person takes a painting to a gallery to be sold, the gallery will usually charge the vendor or seller a percentage of the selling price as the fee for exhibiting, advertising and selling the painting. Such a fee is called a commission, and it applies whenever an agent sells goods or services such as a house or a car on behalf of someone else.

\section*{Example 5}
The Eureka Gallery charges a commission of \(9.2 \%\).

a The Australian painting Showing the Flag at Bakery Hill was sold recently for \(\$ 180000\). How much did the Gallery receive, and how much was left for the seller?

b The Gallery received a commission of \(\$ 7912\) for selling the painting Ned at the Glen. What was the selling price of the painting, and what did the seller actually receive?

\section*{Solution}
a Commission \(=180000 \times 9.2 \%\)

\[
\begin{aligned}
& =180000 \times 0.092 \\
& =\$ 16560
\end{aligned}
\]

Amount received by seller \(=180000-16560\)

\[
=\$ 163440
\]

b Commission \(=\) selling price \(\times 9.2 \%\)

Reversing this:

selling price \(=\) commission \(\div 9.2 \%\)

\[
\begin{aligned}
& =7912 \div 0.092 \\
& =\$ 86000
\end{aligned}
\]

Amount received by seller \(=86000-7912\)

\[
=\$ 78088
\]

\section*{Profit}
Businesses aim to make a profit on their investments. A profit equation can be formulated as: profit \(=\) total revenue \((\) sales \()-\) total costs

\section*{Profit and loss as percentages}
Is an annual profit of \(\$ 20000\) a great performance or a poor performance? For a business with annual turnover of \(\$ 100000\), such a profit would be considered very large. For a business with annual turnover of \(\$ 100000000\), however, it would be considered a very poor performance.

For this reason, it is often relevant to express profit or loss as a percentage of the total costs or the annual turnover.

\section*{Example 6}
The Budget Shoe Shop spent \(\$ 6600000\) last year buying shoes and paying salaries and other expenses. They made a \(2 \%\) profit on these costs.

a What was their profit last year?

b What was the total of their sales?

c In the previous year, their costs were \(\$ 5225000\) and their sales were only \(\$ 5145000\). What percentage loss did they make on their costs?

\section*{Solution}
a Profit \(=6600000 \times 2 \%\)

\[
\begin{aligned}
& =6600000 \times 0.02 \\
& =\$ 132000
\end{aligned}
\]

b Total sales \(=\) total costs + profit

\[
\begin{aligned}
& =6600000+132000 \\
& =\$ 6732000
\end{aligned}
\]

c Last year, loss \(=\) total costs - total sales

\[
\begin{aligned}
& =5225000-5145000 \\
& =\$ 80000
\end{aligned}
\]

Percentage loss \(=\left(\frac{80000}{5225000} \times \frac{100}{1}\right) \%\)

\[
\approx 1.53 \%
\]

\section*{Example 7}
Joe's tile shop made a profit of \(5.8 \%\) on total costs last year. If the actual profit was \(\$ 83000\), what were the total costs, and what were the total sales?

\section*{Solution}
\[
\text { Profit }=\text { costs } \times 5.8 \%
\]

Reversing this, costs \(=\) profit \(\div 5.8 \%\)

\[
=83000 \div 0.058
\]

\(\approx \$ 1431034\), correct to the nearest dollar.

Hence, total sales \(=\) profit + costs

\[
\approx 83000+1431034
\]

\[
=\$ 1514034
\]

\section*{Income tax}
Income tax rates are often progressive. This means that the more you earn, the higher the rate of tax on each extra dollar you earn.

\section*{Example 8}
Income tax in the nation of Immutatia is calculated as follows.

\begin{itemize}
  \item There is no tax on the first \(\$ 12000\) that a person earns in any one year.

  \item From \(\$ 12001\) to \(\$ 30000\), the tax rate is 15 c for each dollar over \(\$ 12000\).

  \item From \(\$ 30001\) to \(\$ 75000\), the tax rate is 25 c for each dollar over \(\$ 30000\).

  \item Over \(\$ 75000\), the tax rate is 35 c for each dollar over \(\$ 75000\).

\end{itemize}

Find the income tax payable by a person whose taxable income for the year is:
a \(\$ 10600\)
b \(\$ 25572\)
c \(\$ 62300\)
d \(\$ 455000\)

\section*{Solution}
a There is no tax on an income of \(\$ 10600\).

b Tax on first \(\$ 12000=\$ 0\)

Tax on remaining \(\$ 13572=13572 \times 0.15\)

\[
=\$ 2035.80
\]

This is the total tax payable.

c Tax on first \(\$ 12000=\$ 0\)

Tax on next \(\$ 18000=18000 \times 0.15\)

\[
=\$ 2700
\]

Tax on remaining \(\$ 32300=32300 \times 0.25\)

\[
=\$ 8075
\]

Total tax \(=2700+8075\)

\[
=\$ 10775
\]

d Tax on first \(\$ 12000=\$ 0\)

Tax on next \(\$ 18000=\$ 2700 \quad(\) see part c)

Tax on next \(\$ 45000=45000 \times 0.25\)

\[
=\$ 11250
\]

Tax on remaining \(\$ 380000=380000 \times 0.35\)

\[
=\$ 133000
\]

Total tax \(=2700+11250+133000\)

\[
=\$ 146950
\]

\section*{Exercise 3B}
Note that some of the questions can be done using the unitary method. For example, Question \(\mathbf{5}\) is one such question.

1 a Five per cent of a particular amount is \(\$ 12\). Find the amount by dividing

\(\$ 12\) by \(5 \%=0.05\).

b Check your answer to part a by taking \(5 \%\) of it.

2 a Twenty-two per cent of a sand pile is \(284 \mathrm{~kg}\). Find the mass of the sand pile.

b Check your answer to part a by taking \(22 \%\) of it.

3 Find the quantity, given that:
a \(2 \%\) of it is \(\$ 12\)
b \(6 \%\) of it is \(750 \mathrm{~g}\)
c \(30 \%\) of it is 36 minutes
d \(90 \%\) of it is \(54 \mathrm{~cm}\)

4 In each part, find the price if:
a a deposit of \(\$ 360\) is \(30 \%\) of the price
b a deposit of \(\$ 168\) is \(15 \%\) of the price

5 Find the original quantity, correct to a suitable number of decimal places, if:
a \(23 \%\) of it is \(100 \mathrm{~kg}\)
b \(0.2 \%\) of it is \(4 \mathrm{~mm}\)
c \(0.92 \%\) of it is 1.86 hectares
d \(97 \%\) of it is \(\$ 700\)

6 Cameron and Wendy together earn \(\$ 1156\) per week after tax. Of this, they pay \(\$ 460\) off their mortgage, \(\$ 185\) for groceries, and \(\$ 260\) for their car and transport, and they save \$124. a Express each amount as a percentage of their weekly income, correct to the nearest \(1 \%\).

b Find how much is unaccounted for in the list above, and what percentage it is of their weekly income.

7 What percentage of the total cost is a deposit of:
a \(\$ 33\) on a television valued at \(\$ 550\) ?
b \(\$ 124.10\) on a stove valued at \(\$ 1460\) ?

8 A book dealer sells rare books and charges a commission of \(8 \%\) on the selling price. Find, correct to the nearest cent, the commission charged on a book that sells for these prices, and the amount that the seller eventually receives.
a \(\$ 400\)
b \(\$ 1300\)
c \(\$ 575\)
d \(\$ 142.50\)

9 Shara the stockbroker charges \(0.15 \%\) commission on all shares that she sells for clients. In each case, find the price at which a parcel of shares was sold if her commission was:
a \(\$ 30.00\)
b \(\$ 67.35\)
c \(\$ 384.75\)
d \(\$ 36.51\)

10 Jeff works as a salesman selling second-hand tractors. He is paid a salary of \(\$ 35000\) a year, together with a \(6 \%\) commission on all the sales he makes. Find his total annual income if his sales for the year were:
a \(\$ 20000\)
b \(\$ 1000000\)
c \(\$ 126000\)
d \(\$ 3458000\)

11 A salesperson is paid a commission on her monthly sales. What is the percentage commission if she receives a payment of:
a \(\$ 168\) on sales of \(\$ 1200\)
b \(\$ 540\) on sales of \(\$ 6000\)
c \(\$ 1530\) on sales of \(\$ 18000\)
d \(\$ 1596\) on sales of \(\$ 42000\)

12 Find the selling price if the commission and the commission rate are as given.
a Commission \(\$ 35\), rate \(7 \%\)
b Commission \(\$ 646\), rate \(3.4 \%\)
c Commission \(\$ 16586.96\), rate \(5.2 \%\)
d Commission \(\$ 3518.61\), rate \(11.4 \%\)

13 Find, to 1 decimal place, the percentage profit or loss on costs in these situations.

a Costs \(\$ 16000\) and sales \(\$ 18000\)

b Costs \(\$ 162000\) and sales \(\$ 150000\)

c Costs \(\$ 2800000\) and sales \(\$ 3090000\)

d Costs \(\$ 289000000\) and sales \(\$ 268000000\)

Example 6 14 The Secure Locksmith Company had sales last year of \$568000 and costs of \$521000.

a What was their profit?

b What was the profit as a percentage of the cost price? (Calculate the percentage correct to 2 decimal places.)

c In the previous year, they made a loss of \(4 \%\) of their costs of \(\$ 250000\). Find their loss and their sales.

15 a A company made a profit of \(\$ 18000\), which was a \(2.4 \%\) profit on its costs. Find the costs and the total sales.

b A company made a loss of \(\$ 657000\), which was a \(4.5 \%\) loss on its costs. Find the costs and the total sales.

c A company made a loss of \(\$ 250800\), which was a \(3.8 \%\) loss on its costs. Find the costs and the total sales.

16 This question uses the income tax rates in the nation of Immutatia, which are as follows.

\begin{itemize}
  \item There is no tax on the first \(\$ 12000\) that a person earns in any one year.

  \item From \(\$ 12001\) to \(\$ 30000\), the tax rate is 15 c for each dollar over \(\$ 12000\).

  \item From \(\$ 30001\) to \(\$ 75000\), the tax rate is 25 c for each dollar over \(\$ 30000\).

  \item Over \(\$ 75000\), the tax rate is 35 c for each dollar over \(\$ 75000\).

\end{itemize}

a Find the income tax payable on:
i \(\$ 8000\)
ii \(\$ 14000\)
iii \(\$ 36000\)
iv \(\$ 200000\)

b What percentage, to 2 decimal places, of each person's income was paid in income tax in parts i-iv of part a?

c Find the income if the income tax on it was:
i \(\$ 1260\)
ii \(\$ 3420\)
iii \(\$ 13950\)
iv \(\$ 14650\)

\section*{Simple interest}
When money is lent by a bank, whoever borrows the money normally makes a payment, called interest, for the use of the money.

The amount of interest paid depends on:

\begin{itemize}
  \item the principal, which is the amount of money borrowed

  \item the rate at which interest is charged

  \item the time for which the money is borrowed.

\end{itemize}

Conversely, if a person invests money in a bank or elsewhere, the bank pays the person interest because the bank uses the money to finance its own investments.

This section will deal only with simple interest. In simple interest transactions, interest is paid on the original amount borrowed.

\section*{Formula for simple interest}
How much simple interest will I pay altogether if I borrow \(\$ 4000\) for 10 years at an interest rate of \(7 \%\) per annum? (The phrase per annum is Latin for 'for each year'; it is often abbreviated to p.a.)

Last year you probably learned to set out the working for simple interest in two successive steps, something like this:

\[
\begin{aligned}
\text { Interest paid at the end of each year } & =4000 \times 7 \% \\
& =4000 \times 0.07 \\
& =\$ 280
\end{aligned}
\]

Total interest paid over 10 years \(=280 \times 10\)

\[
=\$ 2800
\]

This working can all be done in one step if we can develop a suitable formula.

Suppose I borrow \(\$ P\) for \(T\) years at an interest rate \(R\). Using the same two-step approach as before:

Interest paid at the end of each year \(=P \times R\)

Total interest paid over T years \(=P \times R \times T\)

\[
=P R T
\]

This gives us the well-known simple interest formula:

\[
I=P R T \quad(\text { Interest }=\text { principal } \times \text { rate } \times \text { time })
\]

Using this formula, the calculation can now be set out in one step:

\[
\begin{aligned}
I & =P R T \\
& =4000 \times 7 \% \times 10 \quad(\text { Note }: \text { The interest rate } R \text { is } 7 \%, \text { not } 7 .) \\
& =4000 \times 0.07 \times 10 \\
& =\$ 2800
\end{aligned}
\]

Note: The interest rate is given per year, so the time must also be written in years. In some books \(R\) is written as \(r \%\).

\section*{Example 9}
Find the simple interest on \(\$ 8000\) for eight years at 9.5\% p.a.

\section*{Solution}
\[
\begin{aligned}
I & =P R T \\
& =8000 \times 9.5 \% \times 8 \\
& =8000 \times 0.095 \times 8 \\
& =\$ 6080
\end{aligned}
\]

\section*{Reverse use of the simple interest formula}
There are four pronumerals in the formula \(I=P R T\). If the values of any three are known, then substituting into the simple interest formula allows the fourth value to be found.

\section*{Example 10}
Jessie borrows \(\$ 3000\) from her parents to help buy a car. They agree that she should only pay simple interest. Five years later she pays them back \(\$ 3600\), which includes simple interest on the loan. What was the interest rate?

\section*{Solution}
The total interest paid was \(\$ 600\), the principal was \(\$ 3000\) and the time was 5 years.

\[
\begin{aligned}
I & =P R T \\
600 & =3000 \times R \times 5 \\
600 & =15000 \times R \\
R & =\frac{600}{15000} \times 100 \% \\
& =4 \%
\end{aligned}
\]

(Interest rates are normally written as percentages.)

\section*{Simple interest formula}
\begin{itemize}
  \item Suppose that a principal \(P\) is invested for \(T\) years at an interest rate \(R\) p.a. Then the total interest \(I\) is given by:
\end{itemize}

\[
I=P R T
\]

Remember that \(R\) is a percentage. If the interest rate is \(5 \%\), then \(R=0.05\).

\begin{itemize}
  \item If the interest rate \(R\) is given per year, the time \(T\) must be given in years.

  \item The formula has four pronumerals. If any three are known, the fourth can be found by substitution.

\end{itemize}

\section*{Exercise 3C}
\(1 \$ 12000\) is invested at \(7 \%\) p.a. simple interest for five years.

a How much interest will be earned each year?

b Find how much interest will be earned over the five-year period.

\(2 \$ 2000\) is invested at \(6.75 \%\) p.a. simple interest for three years.

a How much interest will be earned each year?

b Find how much interest will be earned over the three-year period.

3 Find the total simple interest earned in each of these investments.
a \(\$ 400\) for three years at \(6 \%\) p.a.
b \(\$ 850\) for six years at \(4.5 \%\) p.a.
c \(\$ 15000\) for 12 years at \(8.4 \%\) p.a.

4 Find the time \(T\) for \(\$ 2000\) of simple interest on a principal of \(\$ 8000\) at a rate of \(5 \%\) p.a.

5 Find the rate \(R\) p.a. for \(\$ 7200\) of simple interest on a principal of \(\$ 8000\) over 12 years.

6 Find the principal \(P\) for \(\$ 3500\) of simple interest at a rate of \(7 \%\) p.a. over 10 years.

7 Calculate the missing entries for these simple interest investments.

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
 & Principal & Rate p.a. & Time in years & Total interest \\
\hline
a & \(\$ 10000\) & \(8 \%\) &  & \(\$ 3200\) \\
\hline
b & \(\$ 4400000\) & \(7 \frac{1}{2} \%\) &  & \(\$ 3960000\) \\
\hline
c & \(\$ 5000\) &  & 6 & \(\$ 1350\) \\
\hline
d & \(\$ 260000\) &  & 8 & \(\$ 83200\) \\
\hline
e &  & \(6 \%\) & 5 & \(\$ 900\) \\
f &  & \(3.6 \%\) & 4 & \(\$ 115.20\) \\
\end{tabular}
\end{center}

8 Sartoro invested \(\$ 80000\) in a building society that pays \(6.5 \%\) p.a. simple interest. Over the years, the investment has paid him \(\$ 57200\) in interest. How many years has he had the investment?

9 Madeline has received \(\$ 168000\) in total simple interest payments on an investment of \(\$ 400000\) that she made six years ago. What rate of interest has the bank been paying?

10 An investor wishes to earn \(\$ 240000\) interest over a five-year period from an account that earns \(12.5 \%\) p.a. simple interest. How much does the investor have to deposit into the account?

11 Regan has arranged to borrow \(\$ 10000\) at \(9.5 \%\) p.a. for four years. She will pay simple interest to the bank every year for the loan, with the principal remaining unchanged. How much interest will Regan pay over the four years of the loan?

12 Tyler intends to live on the interest on an investment with the bank at \(8.6 \%\) p.a. simple interest. She will receive \(\$ 68000\) simple interest every year from the investment. How much money has she invested?

\section*{Percentage increase and decrease}
When a quantity is increased or decreased, the change is often expressed as a percentage of the original amount.

This section introduces a concise method of solving problems about percentage increase and decrease. This method will be applied in various ways throughout the remaining sections of the chapter.

\section*{Percentage increase}
This evening's news reported that shares in the Consolidated Nail Factory Pty Ltd were selling at \(\$ 12.00\) yesterday, but rose \(14.5 \%\) today.

Rather than calculating the price increase and adding it on, the calculation can be done in one step by using the fact that the new price is \(100 \%+14.5 \%=114.5 \%\) of the old price.

\[
\begin{aligned}
\text { New price } & =\text { old price } \times 114.5 \% \\
& =12 \times 114.5 \% \\
& =12 \times 1.145 \\
& =\$ 13.74
\end{aligned}
\]

\section*{Example 11}
The number of patients admitted to St Spyridon's Hospital this year suffering from pneumonia is \(56 \%\) greater than the number admitted for this condition last year. If 245 pneumonia patients were admitted last year, how many were admitted this year?

\section*{Solution}
This year's total is \(100 \%+56 \%=156 \%\) of last year's total.

This year's total \(=245 \times 156 \%\)

\(=245 \times 1.56\)

\(\approx 382\) (correct to the nearest whole number)

\section*{Inflation}
The prices of goods and services in Australia usually increase by a small amount every year.

This gradual rise in prices is called inflation, and is measured by taking the average percentage increase in the prices of a large range of goods and services.

Other things such as salaries and pensions are often adjusted automatically every year to take account of inflation.

\section*{Example 12}
The war-ravaged nation of Zerbai is experiencing inflation of \(35 \%\) p.a. as a result of overspending on its navy and air force. Inflation of \(35 \%\) means that, on average, prices are increasing by \(35 \%\) every year.

a If the price of water is adjusted in line with inflation, what will an annual bill of \(\$ 600\) become in the next year?

b What should an annual salary of \(\$ 169000\) in one year increase to in the following year if it is adjusted to keep pace with inflation?

\section*{Solution}
Next year's prices are \(100 \%+35 \%=135 \%\) of last year's prices.

a Next year's bill \(=600 \times 1.35\)

\[
=\$ 810
\]

b Next year's salary \(=169000 \times 1.35\)

\[
=\$ 228150
\]

\section*{Percentage decrease}
The same method can be used to calculate percentage decreases. For example, suppose that \(35 \%\) of a farmer's sheep station, which has an area of 7500 hectares, went under water during the recent floods.

We can calculate how much land remained above the water for his stock to graze:

\(100 \%-35 \%=65 \%\) of his land remained above water.

Area remaining above water \(=7500 \times 65 \%\)

\[
\begin{aligned}
& =7500 \times 0.65 \\
& =4875 \text { hectares }
\end{aligned}
\]

\section*{Example 13}
The company that Yuri Ivanov works for is going through hard times and has decreased all its salaries by \(12 \%\). Yuri is attempting to cut every one of his expenses by the same percentage.

a His family's weekly grocery bill averages 450 roubles. What should he try to reduce the weekly price of his groceries to?

b His monthly rental is 18000 roubles. If he moves apartments, what monthly rental should he try to find?

\section*{Solution}
Yuri's new salary is \(100 \%-12 \%=88 \%\) of his original salary.

a New weekly grocery bill \(=450 \times 0.88\)

b New monthly rental \(=18000 \times 0.88\)

\(=396\) roubles

\[
=15840 \text { roubles }
\]

Percentage increase and decrease

\begin{itemize}
  \item To increase an amount by, for example, \(15 \%\), multiply by \(1+0.15=1.15\).

  \item To decrease an amount by, for example, \(15 \%\), multiply by \(1-0.15=0.85\).

\end{itemize}

\section*{Finding the rate of increase or decrease}
\section*{Example 14}
Suppose that the rainfall has increased from \(480 \mathrm{~mm}\) p.a. to \(690 \mathrm{~mm}\) p.a. What rate of increase is this?

\section*{Solution}
\section*{Method 1}
Find the actual increase by subtraction, and then express the increase as a percentage of the original rainfall.

Increase \(=210 \mathrm{~mm}\)

\[
\begin{aligned}
\text { Percentage increase } & =\frac{\text { increase }}{\text { original rainfall }} \times 100 \% \\
& =\frac{210}{480} \times 100 \% \\
& =43.75 \%
\end{aligned}
\]

\section*{Method 2}
Express the new value as a percentage of the original value, and then subtract \(100 \%\).

\[
\begin{aligned}
\frac{\text { new rainfall }}{\text { old rainfall }} & =\frac{690}{480} \times 100 \% \\
& =143.75 \%
\end{aligned}
\]

So the rainfall has increased by \(43.75 \%\).

Note: Percentage decrease is sometimes represented as a negative percentage increase or, in other contexts, as a negative percentage change. This understanding is consistent with the formula, amount of change = new amount - old amount, where the new amount is less than the old amount in situations of decrease.

\section*{Reversing the process to find the original amount}
\section*{Example 15}
The Wind Energy Company recently announced that this year's profit of \(\$ 1400000\) constituted a \(35 \%\) increase on last year's profit. What was last year's profit?

\section*{Solution}
This year's profit is \(100 \%+35 \%=135 \%\) of last year's profit.

Hence this year's profit \(=\) last year's profit \(\times 1.35\)

Reversing this, last year's profit \(=\) this year's profit \(\div 1.35\)

\[
\begin{aligned}
& =1400000 \div 1.35 \\
& \approx \$ 1037037, \text { correct to the nearest dollar }
\end{aligned}
\]

Thus, to find the original amount, we divide by 1.35 , because dividing by 1.35 is the reverse of multiplying by 1.35 .

Exactly the same principle applies when an amount has been decreased by a percentage.

\section*{Example 16}
The price of shares in the Fountain Water Company has decreased by \(15 \%\) over the last month to \(\$ 52.70\). What was the price a month ago?

\section*{Solution}
The new share price is \(100 \%-15 \%=85 \%\) of the old share price.

Hence \(\quad\) new price \(=\) old price \(\times 0.85\)

Reversing this, old price \(=\) new price \(\div 0.85\)

\[
\begin{aligned}
& =52.70 \div 0.85 \\
& =\$ 62
\end{aligned}
\]

\section*{Finding the original amount}
\begin{itemize}
  \item To find the original amount after an increase of, for example, 15\%, divide by 1.15 .

  \item To find the original amount after a decrease of, for example, \(15 \%\), divide by 0.85 .

\end{itemize}

\section*{Discounts}
It is very common for a shop to discount the price of an item. This is done to sell stock of a slowmoving item more quickly, or simply to attract customers into the shop.

Discounts are normally expressed as a percentage of the original price.

\section*{Example 17}
The Elegant Shirt Shop is closing down and has discounted all its prices by \(35 \%\).

a What is the discounted price of a shirt whose original price is:
i \(\$ 120\) ?
ii \(\$ 75\) ?

b What was the original price of a shirt whose discounted price is \(\$ 92.30\) ?

\section*{Solution}
a The discounted price of each item is \(100 \%-35 \%=65 \%\) of the old price.

i Hence discounted price \(=\) old price \(\times 0.65\)

\[
\begin{aligned}
& =120 \times 0.65 \\
& =\$ 78
\end{aligned}
\]

ii discounted price \(=75 \times 0.65\)

\[
=\$ 48.75
\]

b From part a, discounted price \(=\) old price \(\times 0.65\)

Reversing this, \(\quad\) old price \(=\) discounted price \(\div 0.65\)

\[
\begin{aligned}
& =92.30 \div 0.65 \\
& =\$ 142
\end{aligned}
\]

\section*{The GST}
In 1999, the Australian Government introduced a Goods and Services Tax, or GST for short. This tax applies to nearly all goods and services in Australia.

The current rate is \(10 \%\) on the pre-tax price of the goods or service.

\begin{itemize}
  \item When GST applies, it is added to the pre-tax price. This is done by multiplying the pre-tax price by 1.10 .

  \item Conversely, if a quoted price already includes GST, the pre-tax price is obtained by dividing the quoted price by 1.10 .

\end{itemize}

\section*{Example 18}
The current GST rate is \(10 \%\) of the pre-tax price.

a If a domestic plumbing job costs \(\$ 630\) before GST, how much will it cost after adding GST, and how much tax is paid to the Government?

b I paid \(\$ 70\) for petrol recently. What was the price before adding GST, and what tax was paid to the Government?

\section*{Solution}
The after-tax price is \(110 \%\) of the pre-tax price.

a After-tax price \(=630 \times 1.10\)

\[
\begin{aligned}
& =\$ 693 \\
\text { Tax } & =693-630 \\
& =\$ 63
\end{aligned}
\]

Note: \(10 \%\) of \(\$ 630\) is \(\$ 63\).
b Pre-tax price \(=70 \div 1.10\) (Divide by 1.10 to reverse the process.)

\[
\begin{aligned}
& \approx \$ 63.64 \\
\mathrm{Tax} & \approx 70-63.64 \\
& =\$ 6.36
\end{aligned}
\]

\section*{Exercise 3D}
1 Traffic on all roads has increased by an average of \(8 \%\) during the past 12 months. By multiplying by \(108 \%=1.08\), estimate the number of vehicles now on a road given the number of vehicles the road carried a year ago was:
a 10000 per day
b 80000 per day
c 148000 per day

2 Prices have increased with inflation by an average of \(3.8 \%\) since the same time last year. Find today's price for an item that one year ago cost:
a \(\$ 200\)
b \(\$ 1.68\)
c \(\$ 345000\)
d \(\$ 9430\)

3 Rainfall across one state has decreased over the last five years by about \(24 \%\). By multiplying by \(76 \%=0.76\), estimate, correct to the nearest \(10 \mathrm{~mm}\), the annual rainfall this year at a place where the rainfall five years ago was:
a \(1000 \mathrm{~mm}\)
b \(250 \mathrm{~mm}\)
c \(680 \mathrm{~mm}\)
d \(146 \mathrm{~mm}\)

4 Admissions to different wards of St Luke's Hospital mostly rose from 2006 to 2007, but by quite different percentage amounts. Find the percentage increase or decrease in wards where the numbers during 2006 and 2007, respectively, were:
a 50 and 68
b 120 and 171
c 92 and 77
d 24 and 39

5 In another state, the percentage decrease in rainfall over the last five years has been quite variable, and in some cases, rainfall has actually increased. Find the rate of decrease or increase if the annual rainfall five years ago and now are, respectively:
a \(500 \mathrm{~mm}\) and \(410 \mathrm{~mm}\)
b \(920 \mathrm{~mm}\) and \(960 \mathrm{~mm}\)
c \(140 \mathrm{~mm}\) and \(155 \mathrm{~mm}\)
d \(420 \mathrm{~mm}\) and \(530 \mathrm{~mm}\)

6 Radix Holdings Pty Ltd recently issued bonus shares to its shareholders. Each shareholder received an extra \(12 \%\) of the number of shares currently held. Find the original holding of a shareholder who now holds:
a 672 shares
b 4816 shares
c 1000 shares
d 40200 shares

7 A clothing store is offering a 15\% discount on all its summer stock. What is the discounted price of an item with original price:
a \(\$ 80\) ?
b \(\$ 48\) ?
c \(\$ 680\) ?
d \(\$ 1.60\) ?

Example 17 8 A shoe store is offering a 35\% discount at its end-of-year sale. Find the original price of an item whose discounted price is:
a \(\$ 1820\)
b \(\$ 279.50\)
c \(\$ 1.56\)
d \(\$ 20.80\)

9 A research institute is trying to find out how much water Lake Grendel had 1000 years ago. The lake now contains 24000 megalitres, but there are various conflicting theories about the percentage change over the last 1000 years. Find how much water the lake had 1000 years ago, correct to the nearest 10 megalitres, if in the last 1000 years the volume has:
a risen by \(80 \%\)
b fallen by \(28 \%\)
c risen by \(140 \%\)
d fallen by \(4 \%\)

10 Mr Brown has a spreadsheet showing the value at which he bought his various parcels of shares, the value at 31 December last year, and the percentage increase or decrease in their value. (Decreases are shown with a negative sign.) Unfortunately, a virus has corrupted one entry in each row of his spreadsheet. Help him by calculating the missing values, correct to the nearest cent, and the missing percentages, correct to 2 decimal places.

\begin{center}
\begin{tabular}{r|c|c|c|}
\hline
 & Value at purchase & Value at 31 December & Percentage increase \\
\hline
\(\mathbf{a}\) & \(\$ 12000\) &  & \(30 \%\) \\
\hline
\(\mathbf{b}\) & \(\$ 28679.26\) &  & \(-62 \%\) \\
\hline
\(\mathbf{c}\) & \(\$ 5267.70\) &  & \(289.14 \%\) \\
\hline
\(\mathbf{d}\) &  & \(\$ 72000\) & \(20 \%\) \\
\hline
\(\mathbf{e}\) &  & \(\$ 26000\) & \(-22 \%\) \\
\hline
\(\mathbf{f}\) &  & \(\$ 112000\) & \(346.5 \%\) \\
\hline
\(\mathbf{g}\) &  & \(\$ 15934\) & \(-91.38 \%\) \\
\hline
\(\mathbf{h}\) & \(\$ 60000\) & \(\$ 81000\) &  \\
\hline
\(\mathbf{i}\) & \(\$ 98356.68\) & \(\$ 14321.57\) &  \\
\(\mathbf{j}\) & \(\$ 14294.12\) & \(\$ 2314.65\) &  \\
\hline
\end{tabular}
\end{center}

11 The GST is a tax on most goods and services at the rate of \(10 \%\) of the pre-tax price.

a Find the after-tax price on goods or services whose pre-tax price is:
i \(\$ 170\)
ii \(\$ 4624\)
iii \(\$ 68920\)
iv \(\$ 6.80\)

b Find the pre-tax price on goods or services whose after-tax price is:
i \(\$ 550\)
ii \(\$ 7821\)
iii \(\$ 192819\)
iv \(\$ 5.28\)

c Find the after-tax price on goods or services on which the GST is:
i \(\$ 60\)
ii \(\$ 678.20\)
iii \(\$ 54000\)
iv \(\$ 0.93\)

12 a A shirt originally priced at \(\$ 45\) was increased in price by \(100 \%\). What percentage discount will restore it to its original price?

b The daily passenger total of the Route 58 bus was 460 , and in one year, it increased by \(24 \%\). What percentage decrease next year would restore it to its original passenger total?

c Shafqat had savings of \(\$ 6000\), but he spent \(35 \%\) of this last year. By what percentage of the new amount must he increase his savings to restore them to their original value?
d The profit of the Arborville Gelatine Factory was \(\$ 86400\), but it then decreased by \(42 \%\). By what percentage must the profit increase to restore it to its original value?

13 a Find, correct to 2 decimal places, the percentage decrease necessary to restore a quantity to its original value if it has been increased by:
i \(10 \%\)
ii \(22 \%\)
iii \(240 \%\)
iv \(2.3 \%\)

b Find, correct to 2 decimal places, the percentage increase necessary to restore a quantity to its original value if it has been decreased by:
i \(10 \%\)
ii \(22 \%\)
iii \(75 \%\)
iv \(2.3 \%\)

\section*{Repeated increase and decrease}
The method introduced in the last section becomes very useful when two or more successive increases or decreases are applied, because the original amount can simply be multiplied successively by two or more factors. Here is a typical example.

\section*{Repeated increase}
\section*{Example 19}
The population of Abelsburg in the census three years ago was 46430 . In the three years after the census, however, its population has risen by \(6.2 \%, 8.5 \%\) and \(13.1 \%\), respectively.

a What was its population one year after the census?

b What was its population two years after the census?

c What is its population now, three years after the census?

d What was the percentage increase in population over the three years, correct to the nearest \(0.1 \%\) ?

\section*{Solution}
a One year after the census, the population was \(106.2 \%\) of its original value.

Hence population after one year \(=46430 \times 1.062\)

\[
\approx 49309 \text {, correct to the nearest whole number }
\]

b Two years afterwards, the population was \(108.5 \%\) of its value one year afterwards.

Hence population after two years \(=(46430 \times 1.062) \times 1.085\)

\[
\approx 53500 \text {, correct to the nearest whole number }
\]

Note: Do not use the approximation from part \(\mathbf{a}\) to calculate part \(\mathbf{b}\). Either start the calculation again, or use the unrounded value from part \(\mathbf{a}\).

(continued over page)
c Three years afterwards, the population was \(113.1 \%\) of its value two years afterwards.

Hence population after three years \(=(46430 \times 1.062 \times 1.085) \times 1.131\)

\[
\approx 60508 \text {, correct to the nearest whole number }
\]

d Population three years afterwards

\[
\begin{aligned}
& =\text { original population } \times 1.062 \times 1.085 \times 1.131 \\
& \approx \text { original population } \times 1.303 \\
& \approx \text { original population } \times 130.3 \%
\end{aligned}
\]

Hence the population has increased over the three years by about \(30.3 \%\).

Note: The percentage increase of \(30.3 \%\) is significantly larger than the sum of the three percentage increases,

\[
6.2 \%+8.5 \%+13.1 \%=27.8 \% .
\]

Note: The answer to part \(\mathbf{d}\) does not depend on what the original population was.

\section*{Repeated decrease}
The same method can be applied just as easily to percentage decreases, as demonstrated in the next example.

\section*{Example 20}
Teresa invested \(\$ 75000\) from her inheritance in a mining company that has not been very successful. In the first year, she lost \(55 \%\) of the money, and in the second year, she lost \(72 \%\) of what remained.

a How much does she have left after one year?

b How much does she have left after two years?

c What percentage of the original inheritance has she lost over the two years?

\section*{Solution}
a One year later, the percentage remaining was \(100 \%-55 \%=45 \%\).

Hence amount left after one year \(=75000 \times 0.45\)

\[
=\$ 33750
\]

b Two years later, she had \(100 \%-72 \%=28 \%\) of what she had after one year.

Hence amount left after two years \(=75000 \times 0.45 \times 0.28\)

\[
=\$ 9450
\]

c Amount left after two years \(=\) original investment \(\times 0.45 \times 0.28\)

\[
\begin{aligned}
& =\text { original investment } \times 0.126 \\
& =\text { original investment } \times 12.6 \%
\end{aligned}
\]

So she has lost \(100 \%-12.6 \%=87.4 \%\) of her investment over the two years.

\section*{Combinations of increases and decreases}
Some problems involve both increases and decreases. They can be solved in exactly the same way.

\section*{Example 21}
The volume of water in the Welcome Dam has varied considerably over the last three years. During the first year the volume rose by \(27 \%\), then it fell \(43 \%\) during the second year, and it rose \(16 \%\) in the third year.

a What is the percentage increase or decrease over the three years, correct to the nearest \(1 \%\) ?

b If there were 366500 megalitres of water in the dam three years ago, how much water is in the dam now, correct to the nearest 500 megalitres?

\section*{Solution}
a Final volume \(=\) original volume \(\times 1.27 \times 0.57 \times 1.16\)

\[
\approx \text { original volume } \times 0.84
\]

Since \(0.84<1\), the volume has decreased. The percentage decrease is about \(100 \%-84 \%=16 \%\) over the three years.

b Final volume \(=\) original volume \(\times 1.27 \times 0.57 \times 1.16\)

\[
\begin{aligned}
& =366500 \times 1.27 \times 0.57 \times 1.16 \\
& \approx 308000 \text { megalitres, correct to the nearest } 500 \text { megalitres. }
\end{aligned}
\]

This time the sum of the percentages is \(27 \%-43 \%+16 \%=0 \%\), but the volume of water has changed.

\section*{Repeated increases and decreases}
\begin{itemize}
  \item To apply successive increases of, for example, \(15 \%, 24 \%\) and \(38 \%\) to a quantity, multiply the quantity by \(1.15 \times 1.24 \times 1.38\).

  \item To apply successive decreases of, for example, 15\%, \(24 \%\) and \(38 \%\) to a quantity, multiply the quantity by \(0.85 \times 0.76 \times 0.62\).

\end{itemize}

\section*{Reversing the process to find the original amount}
As shown before, division reverses the process and allows us to find the original amount, as in the following example.

\section*{Example 22}
A clothing shop discounted a shirt by \(45 \%\) a month ago, and has now discounted the reduced price by \(20 \%\).

a What was the total discount on the shirt?

b If the shirt is now selling for \(\$ 61.60\), what was the original price of the shirt?

\section*{Solution}
a After the first discount, the price was \(100 \%-45 \%=55 \%\) of the original price.

After the second discount, the price was \(100 \%-20 \%=80 \%\) of the reduced price.

Thus final price \(=\) original price \(\times 0.55 \times 0.80\)

\[
=\text { original price } \times 0.44
\]

So the total discount is \(100 \%-44 \%=56 \%\).

b Reversing this, original price \(=\) final price \(\div 0.44\)

\[
=61.60 \div 0.44
\]

\[
=\$ 140
\]

\section*{Reversing repeated increases and decreases}
\begin{itemize}
  \item To find the original quantity after successive increases of, for example, \(15 \%, 24 \%\) and \(38 \%\), divide the final quantity by \((1.15 \times 1.24 \times 1.38)\).

  \item To find the original quantity after successive decreases of, for example, \(15 \%, 24 \%\) and \(38 \%\), divide the final quantity by \((0.85 \times 0.76 \times 0.62)\).

\end{itemize}

\section*{Successive divisions}
Calculations involving brackets can be tricky to handle when using the calculator.

For example, working with the figures in the summary box above, suppose that a population has increased by \(15 \%, 24 \%\) and \(38 \%\) in three successive years and is now 50000 . Then:

\[
\begin{aligned}
\text { original population } & =50000 \div(1.15 \times 1.24 \times 1.38) \\
& \approx 25408
\end{aligned}
\]

We suggest that it is easier to avoid brackets and divide 50000 successively by 1.15, 1.24 and 1.38. In effect, the working then goes like this:

\[
\begin{aligned}
\text { original population } & =50000 \div(1.15 \times 1.24 \times 1.38) \\
& =50000 \div 1.15 \div 1.24 \div 1.38 \\
& \approx 25408
\end{aligned}
\]

Try the calculation both ways and see which you find more natural.

\section*{Using the power button on the calculator}
When a quantity is repeatedly increased or decreased by the same percentage, the power button makes calculations quicker. Make sure that you can use it correctly by experimenting with simple calculations like \(3^{4}=81\) and \(2^{5}=32\).

\section*{Example 23}
The drought in Paradise Valley has been getting worse for years. Each year for the last five years, the rainfall has been \(8 \%\) less than the previous year's rainfall.

a What is the percentage decrease in rainfall over the five years?

b If the rainfall this year was \(458 \mathrm{~mm}\), what was the rainfall five years ago?

\section*{Solution}
Each year the rainfall is \(92 \%\) of the previous year's rainfall.

a Final rainfall \(=\) original rainfall \(\times 0.92 \times 0.92 \times 0.92 \times 0.92 \times 0.92\)

\[
\begin{aligned}
& =\text { original rainfall } \times(0.92)^{5} \\
& \approx \text { original rainfall } \times 0.659
\end{aligned}
\]

So rainfall has decreased by about \(100 \%-65.9 \%=34.1 \%\) over the five years.

b From part a, final rainfall \(=\) original rainfall \(\times(0.92)^{5}\)

Reversing this, original rainfall \(=\) final rainfall \(\div(0.92)^{5}\)

\[
\begin{aligned}
& =458 \div(0.92)^{5} \\
& \approx 695 \mathrm{~mm}
\end{aligned}
\]

\section*{Exercise 3E}
1 Oranges used to cost \(\$ 2.80\) per kg, but the price has increased by \(5 \%, 10 \%\) and \(12 \%\) in three successive years. Multiply by \(1.05 \times 1.1 \times 1.12\) to find their price now.

2 The dividend per share in the Electron Computer Software Company has risen over the last four years by 10\%, 15\%, 5\% and 12\%, respectively. Find the latest dividend received by a shareholder whose dividend four years ago was:
a \(\$ 1000\)
b \(\$ 1678\)
c \(\$ 28.46\)
d \(\$ 512.21\)

3 Rates in Bullimbamba Shire have risen 7\% every year for the last seven years. Find the rates now payable by a landowner whose rates seven years ago were:
a \(\$ 1000\)
b \(\$ 346\)
c \(\$ 2566.86\)
d \(\$ 788.27\)

4 A tree, whose original foliage was estimated to have a mass of \(1500 \mathrm{~kg}\), lost \(20 \%\) of its foliage in a storm, then lost \(15 \%\) of what was left in a storm the next day, then lost \(40 \%\) of what was left in a third storm. Estimate the remaining mass of foliage.

5 Shares in the Metropolitan Brickworks have been falling by \(23 \%\) per year for the last five years. Find the present worth of a parcel of shares whose original worth five years ago was:
a \(\$ 1000\)
b \(\$ 120000\)
c \(\$ 25660\)
d \(\$ 3860000\)

6 a A shirt is discounted by \(50 \%\) and the resulting price is then increased by \(50 \%\). By what percentage is the price increased or decreased from its original value?

b The price of a shirt is increased by \(50 \%\) and the resulting price is then decreased by \(50 \%\). By what percentage is the price increased or decreased from its original value?

c Can you explain the relationship between your answers to parts \(\mathbf{a}\) and \(\mathbf{b}\) ?

Example 22 7 A book shop has a \(50 \%\) sale on all stock, and has a container of books with the sale price reduced by a further factor of \(16 \%\).

a What was the total discount on each book in the container?

b If a book in the container is now selling for \(\$ 10.50\), what was its original price?

8 Calculate the total increase or decrease in a quantity when:

a it is increased by \(20 \%\) and then decreased by \(20 \%\)

b it is increased by \(80 \%\) and then decreased by \(80 \%\)

c it is increased by \(10 \%\) and then decreased by \(10 \%\)

d it is increased by \(30 \%\) and then decreased by \(30 \%\)

9 The price of gemfish has been rising. The price has risen by \(10 \%, 15 \%\) and \(35 \%\) in three successive years, and they now cost \(\$ 24\) per \(\mathrm{kg}\). Find:
a the price one year ago
b the price two years ago
c the original price three years ago

10 The crime rate in Gotham City has been rising each decade. In the last four decades the number of robberies has risen by \(64 \%, 223 \%, 75 \%\) and 12\%. If there are now 958 robberies per year, find how many robberies per year there were:
a one decade ago
b two decades ago
c three decades ago
d four decades ago

11 A particular strain of bacteria increases its population on a certain prepared Petri dish by \(34 \%\) every hour. Calculate the size of the original population four hours ago if there are now 56000 bacteria.

12 Flash Jim is desperate to attract customers to his used car yard. He has cut prices recently by \(5 \%\), then by \(10 \%\), then by \(24 \%\). Find, correct to the nearest \(\$ 100\), the original price of a used car now priced at:
a \(\$ 10000\)
b \(\$ 35000\)
c \(\$ 4600\)
d \(\$ 76800\)

13 The radioactivity of any sample of the element strontium- 90 decreases by \(90.75 \%\) every century. Find the percentage reduction in radioactivity over each of the periods given below. (Calculate percentages correct to 3 decimal places.)
a Two centuries
b Three centuries
c Five centuries

14 Here is a table of the annual inflation rate in Australia in the years ending 30 June 2001 to 30 June 2006 (from the Reserve Bank of Australia website).

\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|}
\hline
\begin{tabular}{l}
Year \\
\begin{tabular}{l}
Inflation \\
rate \\
\end{tabular} \\
\end{tabular} & 2001 & 2002 & 2003 & 2004 & 2005 & 2006 \\
\hline
\end{tabular}
\end{center}

Suppose the salary for certain jobs at Company \(\mathrm{X}\) rises on the 1 July every year, in line with Australia's inflation rate for the financial year just past (ending 30 June).
a A junior secretary earned \(\$ 40000\) from 1 July 2000 to 30 June 2001. Determine how much someone in that position would earn from:

i 1 July 2001 to 30 June 2002

ii 1 July 2006 to 30 June 2007

b A team manager was on an annual salary of \(\$ 100000\) from 1 July 2006 to 30 June 2007. Determine how much someone in that position would earn:

i in the previous financial year

ii from 1 July 2003 to 30 June 2004

15 At the start of the trading day, shares of a particular stock decrease in value by \(20 \%\).

However, by the end of the day the shares 'recover' and record a 15\% increase from its lowest value. Determine the percentage decrease in the value of the shares over the course of the day.

\section*{Compound interest}
In all the examples in this section, the interest is compounded annually. This means that at the end of each year, the interest earned is added to the principal invested or borrowed. That increased amount then becomes the amount on which interest is earned in the following year. This is called compound interest.

\section*{Example 24}
Gail has invested \(\$ 100000\) for six years with the Mountain Bank. The bank pays her interest at the rate of \(7.5 \%\) p.a., compounded annually.

a How much will the investment be worth at the end of one year?

b How much will the investment be worth at the end of two years?

c How much will the investment be worth at the end of six years?

d What is the percentage increase on her original investment at the end of six years?

e What is the total interest earned over the six years?

f What would the simple interest on the investment have been, assuming the same interest rate of \(7.5 \%\) p.a.?

\section*{Solution}
Each year the investment is worth \(107.5 \%\) of its value the previous year.

a Amount at the end of one year \(=100000 \times 1.075\)

\[
=\$ 107500
\]

b Amount at the end of two years \(=100000 \times 1.075 \times 1.075\)

\[
\begin{aligned}
& =100000 \times(1.075)^{2} \\
& =\$ 115562.50
\end{aligned}
\]

(continued over page)
c Amount at the end of six years \(=100000 \times(1.075)^{6}\)

\[
\approx \$ 154330.15
\]

d Final amount \(=\) original amount \(\times(1.075)^{6}\)

\[
\approx \text { original amount } \times 1.5433
\]

So the total increase over six years is about \(54.33 \%\).
e Total interest \(\approx 154330.15-100000\)

\[
=\$ 54330.15
\]

f Simple interest \(=P R T\)

\[
\begin{aligned}
& =100000 \times 0.075 \times 6 \\
& =\$ 45000
\end{aligned}
\]

Note: Compound interest for two or more years is always greater than simple interest for two or more years.

\section*{Compound interest on a loan}
Exactly the same principles apply when someone borrows money from a bank and the bank charges compound interest on the loan. If no repayments are made, the amount owing compounds in the same way, and can grow quite rapidly.

This is shown in the following example.

\section*{Example 25}
Hussain is setting up a plumbing business and needs to borrow \(\$ 150000\) from a bank to buy a truck and other equipment. The bank will charge him interest of \(11 \%\) p.a., compounded annually. Hussain will pay the whole loan off all at once four years later.

a How much will Hussain owe the bank at the end of four years?

b What is the percentage increase in the money owed at the end of four years?

c What is the total interest that Hussain will pay on the loan?

d What would the simple interest on the loan have been, assuming the same interest rate of \(11 \%\) p.a.?

\section*{Solution}
Each year Hussain owes \(111 \%\) of what he owed the previous year.

a Amount at the end of four years \(=150000 \times(1.11)^{4}\)

\[
\approx \$ 227710.56
\]

b Final amount \(=\) original amount \(\times(1.11)^{4}\)

\[
\approx \text { original amount } \times 1.5181
\]

So the total increase over four years is about \(51.81 \%\).

(continued over page)
c \(\quad\) Total interest \(\approx 227710.56-150000\)

\[
=\$ 77710.56
\]

d Simple interest \(=P R T\)

\[
\begin{aligned}
& =150000 \times 0.11 \times 4 \\
& =66000
\end{aligned}
\]

Note: Making no repayments on a loan that is accruing compound interest is a risky business practice because, as this example makes clear, the amount owing grows with increasing rapidity as time goes on. This is particularly relevant to credit card debt.

\section*{Compound interest}
Suppose that an amount \(P\) is invested at an interest rate, say \(7 \%\), compounded annually. The interest is calculated each year on the new balance. The new balance is obtained by adding on \(7 \%\) of the balance of the previous year. Thus:

\[
\begin{aligned}
\text { amount after four years } & =P \times 1.07 \times 1.07 \times 1.07 \times 1.07 \\
& =P \times(1.07)^{4}
\end{aligned}
\]

\section*{Reversing the process to find the original amount}
As always, division reverses the process to find the original amount, as in the following example.

\section*{Example 26}
Eleni wants to borrow money for three years to start a business, and then pay all the money back, with interest, at the end of that time. The bank will not allow her final debt, including interest, to exceed \(\$ 300000\). Interest is \(9 \%\) p.a., compounded annually. What is the maximum amount that Eleni can borrow?

\section*{Solution}
Each year Eleni will owe \(109 \%\) of what she owed the previous year.

Hence \(\quad\) final debt \(=\) original debt \(\times 1.09 \times 1.09 \times 1.09\)

\[
\text { final debt }=\text { original debt } \times(1.09)^{3}
\]

Reversing this, original debt \(=\) final debt \(\div(1.09)^{3}\)

\[
\begin{aligned}
& =300000 \div(1.09)^{3} \\
& \approx \$ 231655
\end{aligned}
\]

\section*{Exercise 3F}
1 a Christine invested \(\$ 100000\) for six years at 5\% p.a. interest, compounded annually.

i By multiplying by 1.05 , find the value of the investment after one year.

ii By multiplying by \((1.05)^{2}\), find the value of the investment after two years.

iii By multiplying by \((1.05)^{6}\), find the value of the investment after six years.

iv Find the percentage increase in the investment over the six years.

\(v\) Find the total interest earned over the six years.

b Find the simple interest on the principal of \(\$ 100000\) over the six years at the same rate of \(5 \%\) p.a.

2 a Gary has borrowed \(\$ 200000\) for six years at \(8 \%\) p.a. interest, compounded annually, in order to start his indoor decorating business. He intends to pay the whole amount back, plus interest, at the end of the six years.

i Find the amount owing after one year.

ii Find the amount owing after two years.

iii Find the amount owing after six years.

iv Find the percentage increase in the loan over the six years.

\(\mathbf{v}\) Find the total interest charged over the six years.

b Find the simple interest on the principal of \(\$ 200000\) over the six years at the same rate of \(8 \%\) p.a.

3 A couple take out a housing loan of \(\$ 320000\) over a period of 20 years. They make no repayments over the 20-year period of the loan. Compound interest is payable at \(6 \frac{1}{2} \%\) p.a., compounded annually. How much would they owe at the end of the 20-year period, and what is the total percentage increase?

4 The population of a city increases annually at a compound rate of \(3.2 \%\) for five years. If the population is 21000 initially, what is it at the end of the five-year period, and what is the total percentage increase?

5 a Find the compound interest on \(\$ 1000\) at 5\% p.a., compounded annually for 200 years.

b Find the simple interest on \(\$ 1000\) at \(5 \%\) p.a. for 200 years.

\(6 \$ 10000\) is borrowed for five years and compound interest at \(10 \%\) p.a. is charged by the lender.

a How much money is owed to the lender after the five-year period?

b How much of this amount is interest?

7 Money borrowed at \(8 \%\) p.a. interest, compounded annually, grew to \(\$ 100000\) in four years.

a Find the total percentage increase.

b Hence find the original amount invested.

8 Suzette wants to invest a sum of money now so that it will grow to \(\$ 180000\) in 10 years' time. How much should she invest now, given that the interest rate is \(6 \%\) compounded annually?

9 A bank offers \(8 \%\) p.a. compound interest. How much needs to be invested if the investment is to be worth \(\$ 100000\) in:
a 10 years?
b 20 years?
c 25 years?
d 100 years?

10 A man now owes the bank \(\$ 56000\), after having taken out a loan five years ago. Find the original amount that he borrowed if the rate of interest per annum, compounded annually, has been:
a \(3 \%\)
b \(5.6 \%\)
c \(9.25 \%\)
d \(15 \%\)

\(11 \mathrm{Mr}\) Brown has had further difficulties with the virus that attacks his spreadsheet entries. Here is the remains of a spreadsheet that he prepared in answer to questions from business friends. The spreadsheet calculated interest compounded annually, on various amounts, at various interest rates, for various periods of time. Help him reconstruct the missing entries.

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
 & Principal & Rate \(\mathbf{p} . \mathbf{a}\). & Time in years & Final amount & Total interest \\
\hline
\(\mathbf{a y y y y y}\) & \(\$ 4000\) & \(6 \%\) & 20 &  &  \\
\hline
b & \(\$ 10000\) & \(8.2 \%\) & 15 &  &  \\
\hline
c & \(\$ 2000000\) & \(4.8 \%\) & 10 &  &  \\
d &  & \(6 \%\) & 20 & \(\$ 4000\) &  \\
\hline
e &  & \(8.2 \%\) & 15 & \(\$ 10000\) &  \\
f &  & \(4.8 \%\) & 10 & \(\$ 2000000\) &  \\
\hline
\end{tabular}
\end{center}

12 Ms Smith invested \(\$ 50000\) at \(6 \%\) p.a. interest, compounded annually, for four years. The tax department wants to know exactly how much interest she earned each year. Calculate these figures for Ms Smith.

13 Mrs Robinson has taken out a loan of \(\$ 300000\) at \(8 \%\) p.a. interest, compounded annually, for four years. She wants to know exactly how much interest she will be charged each year so that she can include it as a tax deduction in her income tax return. Calculate these figures for Mrs Robinson.

14 a Find the total percentage growth in a compound interest investment:
i at \(15 \%\) for two years
ii at \(10 \%\) for three years
iii at \(6 \%\) for five years
iv at \(5 \%\) for six years
\(\mathbf{v}\) at \(3 \%\) for 10 years
vi at \(2 \%\) for 15 years

b What do you observe about these results?

15 A couple took out a six-year loan to start a business. For the first three years, compound interest of \(8 \%\) p.a. was charged. For the second three years, compound interest of \(12 \%\) p.a. was charged. Find the total percentage increase in the amount owing.

16 One six-year loan attracts compound interest calculated at 2\%, 4\%, 6\%, 8\%, 10\% and \(12 \%\) in successive years. Another six-year loan attracts compound interest calculated at \(12 \%, 10 \%, 8 \%, 6 \%, 4 \%\) and \(2 \%\) in successive years. Find the total percentage increase in money owing in both cases, compare the two results, and explain what has happened.

17 An investment at an interest rate of \(10 \%\) p.a., compounded annually, returned interest of \(\$ 40000\) after five years. Calculate the original amount invested.

\section*{Depreciation}
Depreciation occurs when the value of an asset reduces as time passes. For example, a company may buy a car for \(\$ 40000\), but after four years the car will be worth a lot less, because the motor will be worn, the car will be out of date, the body and interior may have a few scratches, and so forth.

Accountants usually make the assumption that an asset such as a car depreciates at the same rate every year. This rate is called the depreciation rate. It works like compound interest. The depreciation is applied each year to the current value. In the following example, the depreciation rate is taken to be \(20 \%\).

\section*{Example 27}
The Medicine Home Delivery Company bought a car four years ago for \(\$ 40000\), and assumed that the value of the car would depreciate at \(20 \%\) p.a.

a What value did the car have at the end of two years?

b What value does the car have now, after four years?

c What is the percentage decrease in value over the four years?

d What is the average depreciation on the car over the four years? (Express your answer in dollars per year)

\section*{Solution}
The value each year is taken to be \(100 \%-20 \%=80 \%\) of the value in the previous year.

a Value at the end of two years \(=40000 \times 0.80 \times 0.80\)

\[
\begin{aligned}
& =40000 \times(0.80)^{2} \\
& =\$ 25600
\end{aligned}
\]

b Value at the end of four years \(=40000 \times 0.80 \times 0.80 \times 0.80 \times 0.80\)

\[
\begin{aligned}
& =40000 \times(0.80)^{4} \\
& =\$ 16384
\end{aligned}
\]

(continued over page)
c \(\quad\) Final value \(=\) original value \(\times(0.80)^{4}\)

\[
=\text { original value } \times 0.4096
\]

Hence the percentage decrease over four years is \(100 \%-40.96 \%=59.04 \%\)

d Depreciation over four years \(=40000-16384\)

\[
=\$ 23616
\]

Average depreciation per year \(=23616 \div 4\)

\[
=\$ 5904 \text { per year }
\]

\section*{Depreciation}
Suppose that an asset with original value \(P\) depreciates at, for example, \(7 \%\) every year. To find the depreciated value, decrease the current value by \(7 \%\) each year. Thus:

\[
\begin{aligned}
\text { value after four years } & =P \times 0.93 \times 0.93 \times 0.93 \times 0.93 \\
& =P \times(0.93)^{4}
\end{aligned}
\]

\section*{Reversing the process to find the original amount}
If we are given a depreciated value and the rate of depreciation, we can find the original value by division.

\section*{Example 28}
A school buys new computers every four years. At the end of the four years, it offers them for sale to the students on the assumption that they have depreciated at \(35 \%\) p.a. (per annum). The school is presently advertising some computers at \(\$ 400\) each.

a What did each computer cost the school originally?

b What is the average depreciation on each computer, in dollars per year?

\section*{Solution}
a Each year a computer is worth \(100 \%-35 \%=65 \%\) of its value the previous year.

Hence final value \(=\) original value \(\times 0.65 \times 0.65 \times 0.65 \times 0.65\)

\[
\text { final value }=\text { original value } \times(0.65)^{4}
\]

Reversing this, original value \(=\) final value \(\div(0.65)^{4}\)

\[
\begin{aligned}
& =400 \div(0.65)^{4} \\
& \approx \$ 2241
\end{aligned}
\]

b Depreciation over four years \(\approx 2241-400\)

\[
=\$ 1841
\]

Average depreciation per year \(\approx 1841 \div 4\)

\[
\approx \$ 460
\]

\section*{Exercise 3G}
Note: The depreciation rates in this exercise are taken from the Australian Taxation Office's

Schedule of Depreciation. These are intended for income tax purposes. A company may have reasons to use different rates.

1 The landlord of a large block of home units purchased washing machines for its units four years ago for \(\$ 400000\), and is assuming a depreciation rate of \(30 \%\).

a By multiplying by 0.70 , find the value after one year.

b By multiplying by \((0.70)^{2}\), find the value after two years.

c By multiplying by \((0.70)^{3}\), find the value after three years.

d By multiplying by \((0.70)^{4}\), find the value after four years.

e What is the percentage decrease in value over the four years?

f What is the average depreciation on the washing machines, in dollars p.a. over the four years?

2 The Hungry Hour Cafe purchased an air-conditioning system six years ago for \(\$ 250000\), and is assuming a depreciation rate of \(20 \%\).

a Find the value after one year.

b Find the value after two years.

c Find the value after six years.

d What is the percentage decrease in value over the six years?

e What is the average depreciation, in dollars p.a., on the air-conditioning system over the six years?

3 A business spent \(\$ 560000\) installing alarms at its premises and then depreciated them at \(20 \%\) p.a. Find the value after five years, and the percentage depreciation of their value.

4 The population of a sea lion colony decreases at a compound rate of \(2 \%\) p.a. for 10 years. If the population is 8000 initially, what is it at the end of the 10-year period?

5 The Northern Start Bus Company bought a bus for \(\$ 480000\), depreciated it at \(30 \%\) p.a., and sold it again seven years later for \(\$ 60000\). Was the price that they obtained better or worse than the depreciated value, and by how much?

6 The Backyard Rubbish Experts bought a fleet of small trucks for \(\$ 1340000\) and depreciated them at \(22.5 \%\) p.a. Five years later they sold them for \(\$ 310000\). Was the price that they obtained better or worse than the depreciated value, and by how much?

7 A landlord spent \(\$ 3400\) on vacuum cleaners for his block of home units and depreciated them for taxation purposes at \(25 \%\) p.a. Find their value at the end of each of the first three years, and the amount of the depreciation that the landlord could claim against his taxable income for each of those three years.

8 Lara and Kate each received \(\$ 100000\) from their parents. Lara invested the money at \(6.2 \%\) p.a. compounded annually, whereas Kate bought a luxury car that depreciated at a rate of \(20 \%\) p.a. What were the values of their investments at the end of five years?

9 Taxis depreciate at \(50 \%\) p.a., and other cars depreciate at \(22.5 \%\) p.a.

a What is the total percentage depreciation on each type of vehicle after seven years?

b What is the difference in value, to the nearest dollar, after seven years of a fleet of taxis and a fleet of other cars, if both fleets cost \(\$ 1000000\) ?

Example 28 10 Mr Wong's 10-year-old used car is worth \(\$ 4000\), and has been depreciating at \(22.5 \%\) p.a. (Calculate amounts of money in whole dollars.)

a Use division by 0.775 to find how much it was worth a year ago.

b Find how much it was worth two years ago.

c Find how much it was worth 10 years ago.

d What is the total percentage depreciation on the car over the 10-year period?

e What was the average depreciation in dollars per year over the 10-year period?

11 St Scholasticus Grammar School bought photocopying machines six years ago, which it then depreciated at \(25 \%\) p.a. They are now worth \(\$ 72000\).

a How much were they worth one year ago?

b How much were they worth two years ago?

c How much were they worth six years ago?

d What is the total percentage depreciation on them over the six-year period?

e What was the average depreciation in dollars per year over the six-year period?

12 Ms Wu's seven-year-old car is worth \(\$ 5600\), and has been depreciating at \(22.5 \%\) p.a. Calculate your answers to the nearest dollar.

a How much was it worth four years ago?

b How much was it worth seven years ago?

c Ms Wu, however, only bought the car four years ago, at its depreciated value at that time. What has been Ms Wu's average depreciation in dollars over the four years she has owned the car?

d What was the average depreciation in dollars over the first three years of the car's life?

13 I take \(900 \mathrm{~mL}\) of a liquid and dilute it with \(100 \mathrm{~mL}\) of water. Then I take \(900 \mathrm{~mL}\) of the mixture and again dilute it with \(100 \mathrm{~mL}\) of water. I repeat this process 20 times.

a What proportion of the original liquid remains in the mixture at the end?

b How much mixture should I take if I want it to contain \(20 \mathrm{~mL}\) of the original liquid?

14 I take a sealed glass container and remove \(80 \%\) of the air. Then I remove \(80 \%\) of the remaining air. I do this process six times altogether. What percentage of the original air is left in the container?


\end{document}
|721917553|kitty|/Applications/kitty.app|0|
The Brilliant Light Bulb Company estimates that \(3.5 \%\) of its light bulbs are defective. If a shop owner buys 1250 light bulbs to light the shop, how many would he expect to be defective?

\section*{Solution}
Number of defective bulbs \(=1250 \times 3.5 \%\)

\[
\begin{aligned}
& =1250 \times 0.035 \\
& \approx 44 \quad(\text { Round } 43.75 \text { to } 44 .)
\end{aligned}
\]

|721917587|kitty|/Applications/kitty.app|0|
\section*{Solution}
Number of defective bulbs \(=1250 \times 3.5 \%\)

\[
\begin{aligned}
& =1250 \times 0.035 \\
& \approx 44 \quad(\text { Round } 43.75 \text { to } 44 .)
\end{aligned}
\]
|721917633|kitty|/Applications/kitty.app|0|
            \section*{Solution}
|721917641|kitty|/Applications/kitty.app|0|
A typical computer weighs about \(25 \mathrm{~kg}\). When it is broken down as waste, it yields about \(3 \mathrm{~g}\) of arsenic. What percentage of the total is this?

\section*{Solution}
Using grams, the computer weighs \(25000 \mathrm{~g}\) and the arsenic weighs \(3 \mathrm{~g}\).

Hence percentage of arsenic \(=\frac{3}{25000} \times 100 \%\)

\[
=\frac{3}{250} \%
\]

\[
=0.012 \%
\]
|721917965|kitty|/Applications/kitty.app|0|
Using grams, the computer weighs \(25000 \mathrm{~g}\) and the arsenic weighs \(3 \mathrm{~g}\).

Hence percentage of arsenic \(=\frac{3}{25000} \times 100 \%\)

\[
=\frac{3}{250} \%
\]

\[
=0.012 \%
\]
|721917989|kitty|/Applications/kitty.app|0|
        \begin{multicols}{4}
        \begin{parts}
            \part $72\%=$\fillin[]
            \part $7.6\%=$\fillin[]
            \part $77\frac{3}{4}\%=$\fillin[]
            \part $0.1\%=$\fillin[]
        \end{parts}
        \end{multicols}
|721918324|kitty|/Applications/kitty.app|0|
72\%=|721918328|kitty|/Applications/kitty.app|0|
            \part $125\%=$\fillin[]
|721918380|kitty|/Applications/kitty.app|0|
e \(\frac{7}{20}\)
|721918465|kitty|/Applications/kitty.app|0|
i 0.43
|721918468|kitty|/Applications/kitty.app|0|
i 0.43
m 1.2
|721918470|kitty|/Applications/kitty.app|0|
j 0.225
|721918476|kitty|/Applications/kitty.app|0|
j 0.225
n 2.03
|721918477|kitty|/Applications/kitty.app|0|
a \(\frac{3}{5}\)
e \(\frac{7}{20}\)
i 0.43
m 1.2
j 0.225
n 2.03
|721918480|kitty|/Applications/kitty.app|0|
a 
e 
i 
m 
j 
n 
|721918501|kitty|/Applications/kitty.app|0|
g \(175 \%\)
h \(0.6 \%\)
|721918588|kitty|/Applications/kitty.app|0|
g \(175 \%\)
h \(0.6 \%\)
k \(142.6 \%\)
l \(\frac{1}{4} \%\)
|721918592|kitty|/Applications/kitty.app|0|
1 Express each percentage as a decimal.
|721918607|kitty|/Applications/kitty.app|0|
1 Express each percentage as a decimal.
a \(72 \%\)
b \(7.6 \%\)
c \(98 \%\)
d \(16 \%\)
e \(8 \%\)
f \(6.25 \%\)
i \(77 \frac{3}{4} \%\)
j \(0.1 \%\)
|721918613|kitty|/Applications/kitty.app|0|
a \(35 \%\)
|721918618|kitty|/Applications/kitty.app|0|
e \(33 \frac{1}{3} \%\)
|721918620|kitty|/Applications/kitty.app|0|
g \(7.25 \%\)
|721918621|kitty|/Applications/kitty.app|0|
i \(210 \%\)
|721918625|kitty|/Applications/kitty.app|0|
j \(125 \%\)
|721918626|kitty|/Applications/kitty.app|0|
k \(112 \frac{1}{2} \%\)
|721918626|kitty|/Applications/kitty.app|0|
2 Express each percentage as a fraction in lowest terms.
b \(56 \%\)
c \(75 \%\)
d \(37 \frac{1}{2} \%\)
f \(16 \frac{2}{3} \%\)
h \(6.4 \%\)
l \(136 \%\)
|721918744|kitty|/Applications/kitty.app|0|
3 Express each fraction or decimal as a percentage.
b \(\frac{3}{8}\)
c \(\frac{9}{16}\)
d \(2 \frac{1}{4}\)
f \(\frac{2}{3}\)
g \(\frac{4}{3}\)
h \(\frac{3}{400}\)
k 0.04
l 0.015
o 1.175
p 0.0075
|721918762|kitty|/Applications/kitty.app|0|
\begin{table}[h!]
\centering
\begin{tabular}{|c|c|c|c|}
\hline
\rowcolor{headercolor}
& \textbf{Percentage} & \textbf{Fraction} & \textbf{Decimal} \\ \hline
\rowcolor{rowcolor}
a & 54\% & & \\ \hline
\rowcolor{rowcolor}
b & & $\frac{2}{5}$ & \\ \hline
\rowcolor{rowcolor}
c & 18.5\% & & 0.32 \\ \hline
\rowcolor{rowcolor}
d & & & \\ \hline
\rowcolor{rowcolor}
e & & $\frac{7}{8}$ & 0.06 \\ \hline
\rowcolor{rowcolor}
f & & & \\ \hline
\rowcolor{rowcolor}
g & 108.6\% & & 1.02 \\ \hline
\rowcolor{rowcolor}
h & & & \\ \hline
\rowcolor{rowcolor}
i & & $\frac{12}{5}$ & \\ \hline
\end{tabular}
\end{table}|721918927|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\definecolor{headercolor}{rgb}{0.6, 0.6, 1} % Light purple
\definecolor{rowcolor}{rgb}{0.9, 0.9, 1} % Very light purple|721919083|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
c \(26 \%\) of 264
|721919123|kitty|/Applications/kitty.app|0|
c \(26 \%\) of 264
f \(138.5 \%\) of 650
|721919126|kitty|/Applications/kitty.app|0|
5 Evaluate these amounts, correct to 2 decimal places where necessary.
a \(15 \%\) of 40
b \(57 \%\) of 1000
d \(120 \%\) of 538
e \(15.8 \%\) of 972
g \(2.8 \%\) of 318
h \(0.1 \%\) of 6000
|721919132|kitty|/Applications/kitty.app|0|
e 
g 
h 
|721919176|kitty|/Applications/kitty.app|0|
d \(70 \mathrm{~m}, 50 \mathrm{~m}\)
e 15 weeks, 60 weeks
f 60 weeks, 15 weeks
|721919281|kitty|/Applications/kitty.app|0|
7 Find what percentage the first quantity is of the second quantity, correct to 1 decimal place where necessary.
d \(70 \mathrm{~m}, 50 \mathrm{~m}\)
e 15 weeks, 60 weeks
f 60 weeks, 15 weeks
|721919284|kitty|/Applications/kitty.app|0|
7 Find what percentage the first quantity is of the second quantity, correct to 1 decimal place where necessary.
|721919290|kitty|/Applications/kitty.app|0|
 d
 e
 f
|721919306|kitty|/Applications/kitty.app|0|
b \(3.4 \mathrm{~cm}, 2 \mathrm{~m}\)
|721919355|kitty|/Applications/kitty.app|0|
b \(3.4 \mathrm{~cm}, 2 \mathrm{~m}\)
d 8 hours, 2 weeks
|721919359|kitty|/Applications/kitty.app|0|
b \(3.4 \mathrm{~cm}, 2 \mathrm{~m}\)
d 8 hours, 2 weeks
f \(250 \mathrm{~m}, 4 \mathrm{~km}\)
|721919366|kitty|/Applications/kitty.app|0|
b \(3.4 \mathrm{~cm}, 2 \mathrm{~m}\)
d 8 hours, 2 weeks
f \(250 \mathrm{~m}, 4 \mathrm{~km}\)
h 1 day, 1 year
|721919372|kitty|/Applications/kitty.app|0|
b \(3.4 \mathrm{~cm}, 2 \mathrm{~m}\)
d 8 hours, 2 weeks
f \(250 \mathrm{~m}, 4 \mathrm{~km}\)
h 1 day, 1 year
j 33 weeks, 1 century
|721919381|kitty|/Applications/kitty.app|0|
b \(3.4 \mathrm{~cm}, 2 \mathrm{~m}\)
d 8 hours, 2 weeks
f \(250 \mathrm{~m}, 4 \mathrm{~km}\)
h 1 day, 1 year
j 33 weeks, 1 century
l 5 apples, 16 dozen apples
|721919384|kitty|/Applications/kitty.app|0|
8 Find what percentage the first quantity is of the second quantity, correct to 2 decimal places where necessary. You will need to express both quantities in the same unit first.
a 68 cents, \(\$ 5.00\)
c \(7 \mathrm{~g}, 3 \mathrm{~kg}\)
e 15 days, 3 years
g \(4 \mathrm{~km}, 250 \mathrm{~m}\)
i 1 year, 1 day
k \(56 \mathrm{~cm}, 2.4 \mathrm{~km}\)
|721919390|kitty|/Applications/kitty.app|0|
a 68 cents, \(\$ 5.00\)
c \(7 \mathrm{~g}, 3 \mathrm{~kg}\)
e 15 days, 3 years
g \(4 \mathrm{~km}, 250 \mathrm{~m}\)
i 1 year, 1 day
k \(56 \mathrm{~cm}, 2.4 \mathrm{~km}\)
|721919399|kitty|/Applications/kitty.app|0|
8 Find what percentage the first quantity is of the second quantity, correct to 2 decimal places where necessary. You will need to express both quantities in the same unit first.

b \(3.4 \mathrm{~cm}, 2 \mathrm{~m}\)
d 8 hours, 2 weeks
f \(250 \mathrm{~m}, 4 \mathrm{~km}\)
h 1 day, 1 year
j 33 weeks, 1 century
l 5 apples, 16 dozen apples
|721919401|kitty|/Applications/kitty.app|0|
a \(62 \%\) of \(\$ 10\)
|721919416|kitty|/Applications/kitty.app|0|
d \(110 \%\) of \(\$ 1280\)
|721919417|kitty|/Applications/kitty.app|0|
g \(7 \frac{3}{4} \%\) of \(\$ 1000\)
|721919417|kitty|/Applications/kitty.app|0|
6 Evaluate these amounts, correct to the nearest cent where necessary.
b \(23.7 \%\) of \(\$ 960\)
c \(3.2 \%\) of \(\$ 1500\)
e \(0.25 \%\) of \(\$ 800\)
f \(6 \frac{1}{2} \%\) of \(\$ 200\)
h \(\frac{1}{4} \%\) of \(\$ 840\)
i \(7.25 \%\) of \(\$ 1600\)
|721919419|kitty|/Applications/kitty.app|0|
        \question Evaluate these amounts, correct to 2 decimal places where necessary.
|721919440|kitty|/Applications/kitty.app|0|
        \question Evaluate these amounts, correct to 2 decimal places where necessary.
c \(26 \%\) of 264
f \(138.5 \%\) of 650
i \(150 \%\) of 846
|721919458|kitty|/Applications/kitty.app|0|
        \question Find what percentage the first quantity is of the second quantity, correct to 1 decimal place.
|721919471|kitty|/Applications/kitty.app|0|
        \question Find what percentage the first quantity is of the second quantity, correct to 1 decimal place.

a \(7 \mathrm{~km}, 50 \mathrm{~km}\)
b \(\$ 4, \$ 200\)
c \(14 \mathrm{~kg}, 400 \mathrm{~kg}\)
|721919474|kitty|/Applications/kitty.app|0|
c 
e 
g 
i 
k 
|721919520|kitty|/Applications/kitty.app|0|
10 There are 740 students at a primary school, \(5 \%\) of whom have red hair. Calculate the number of students in the school who have red hair.
|721919552|kitty|/Applications/kitty.app|0|
10 There are 740 students at a primary school, \(5 \%\) of whom have red hair. Calculate the number of students in the school who have red hair.
12 A soccer match lasted 92 minutes (including injury time). If team A was in possession for \(55 \%\) of the match, for how many minutes and seconds was team A in possession?
|721919555|kitty|/Applications/kitty.app|0|
11 A sample of a certain alloy weighs \(1.6 \mathrm{~g}\).

a Aluminium makes up \(48 \%\) of the alloy. What is the weight of the aluminium in the sample?

b The percentage of lead in the alloy is \(0.23 \%\). What is the weight of the lead in the sample?


13 Carbon dioxide makes up \(0.059 \%\) of the mass of the Earth's atmosphere. The total mass of the atmosphere is about 5 million megatonnes. What is the total mass of the carbon dioxide in the atmosphere?
|721919610|kitty|/Applications/kitty.app|0|
question |721919636|kitty|/Applications/kitty.app|0|
\definecolor{headercolor}{rgb}{0.6, 0.6, 1} % Light purple
\definecolor{rowcolor}{rgb}{0.9, 0.9, 1} % Very light purple

\begin{table}[h!]
\centering
\begin{tabular}{|c|c|c|c|}
\hline
\rowcolor{headercolor}
& \textbf{Percentage} & \textbf{Fraction} & \textbf{Decimal} \\ \hline
\rowcolor{rowcolor}
a & 54\% & & \\ \hline
\rowcolor{rowcolor}
b & & $\frac{2}{5}$ & \\ \hline
\rowcolor{rowcolor}
c & 18.5\% & & 0.32 \\ \hline
\rowcolor{rowcolor}
d & & & \\ \hline
\rowcolor{rowcolor}
e & & $\frac{7}{8}$ & 0.06 \\ \hline
\rowcolor{rowcolor}
f & & & \\ \hline
\rowcolor{rowcolor}
g & 108.6\% & & 1.02 \\ \hline
\rowcolor{rowcolor}
h & & & \\ \hline
\rowcolor{rowcolor}
i & & $\frac{12}{5}$ & \\ \hline
\end{tabular}
\end{table}
|721919681|kitty|/Applications/kitty.app|0|
        \end{parts}
        \end{onehalfspacing}
|721919860|kitty|/Applications/kitty.app|0|
        \begin{multicols}{3}
        \begin{onehalfspacing}
|721919870|kitty|/Applications/kitty.app|0|
        \end{onehalfspacing}
        \end{multicols}
|721919872|kitty|/Applications/kitty.app|0|
    \subsection*{Exercises}
    \begin{questions}
        \question Express each percentage as a decimal:
        \begin{multicols}{2}
        \begin{parts}
            \part $72\%=$\fillin[]
            \part $7.6\%=$\fillin[]
            \part $77\frac{3}{4}\%=$\fillin[]
            \part $0.1\%=$\fillin[]
        \end{parts}
        \end{multicols}
        \question Express each percentage as a fraction:
        \begin{multicols}{3}
        \begin{onehalfspacing}
        \begin{parts}
            \part $35\%$\fillin[]
            \part $33\frac{1}{3}\%=$\fillin[]
            \part $210\%=$\fillin[]
            \part $125\%=$\fillin[]
            \part $7.25\%=$\fillin[]
            \part $112\frac{1}{2}\%=$\fillin[]
        \end{parts}
        \end{onehalfspacing}
        \end{multicols}
        \question Express each fraction or decimal as a percentage:
        \begin{multicols}{2}
        \begin{onehalfspacing}
        \begin{parts}
            \part \(\frac{3}{5}=\)\fillin[]
            \part \(\frac{7}{20}=\)\fillin[]
            \part $0.43=$\fillin[]
            \part $1.2=$\fillin[]
            \part $0.225=$\fillin[]
            \part $2.03=$\fillin[]
        \end{parts}
        \end{onehalfspacing}
        \end{multicols}
        \question Evaluate these amounts, correct to 2 decimal places where necessary.
        \begin{multicols}{2}
        \begin{onehalfspacing}
        \begin{parts}
            \part \(15 \%\) of 40 = \fillin[]
            \part \(57 \%\) of 1000 =\fillin[]
            \part \(120 \%\) of 538 =\fillin[]
            \part \(15.8 \%\) of 972 = \fillin[]
            \part \(2.8 \%\) of 318 =\fillin[]
            \part \(0.1 \%\) of 6000 = \fillin[]
        \end{parts}
        \end{onehalfspacing}
        \end{multicols}
        \question Find what percentage the first quantity is of the second quantity, correct to 1 decimal place.
        \begin{parts}
            \part[1] \(70 \mathrm{~m}, 50 \mathrm{~m}\)=\fillin[]
            \part[1] 15 weeks, 60 weeks=\fillin[]
            \part[1] 60 weeks, 15 weeks=\fillin[]
        \end{parts}
        \question Find what percentage the first quantity is of the second quantity, correct to 2 decimal places where necessary. You will need to express both quantities in the same unit first.
        \begin{multicols}{2}
        \begin{parts}
        \begin{onehalfspacing}
            \part 68 cents, \(\$ 5.00\)=\fillin[]
            \part \(7 \mathrm{~g}, 3 \mathrm{~kg}\)=\fillin[]
            \part 15 days, 3 years=\fillin[]
            \part \(4 \mathrm{~km}, 250 \mathrm{~m}\)=\fillin[]
            \part 1 year, 1 day=\fillin[]
            \part \(56 \mathrm{~cm}, 2.4 \mathrm{~km}\)=\fillin[]
        \end{onehalfspacing}
        \end{parts}
        \end{multicols}
        \question There are 740 students at a primary school, \(5 \%\) of whom have red hair. Calculate the number of students in the school who have red hair.
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \question 12 A soccer match lasted 92 minutes (including injury time). If team A was in possession for \(55 \%\) of the match, for how many minutes and seconds was team A in possession?
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
    \end{questions}
|722041394|kitty|/Applications/kitty.app|0|
    \subsection*{Exercises}
    \begin{questions}
        \question Express each percentage as a decimal:
        \begin{multicols}{2}
        \begin{parts}
            \part $72\%=$\fillin[]
            \part $7.6\%=$\fillin[]
            \part $77\frac{3}{4}\%=$\fillin[]
            \part $0.1\%=$\fillin[]
        \end{parts}
        \end{multicols}
        \question Express each percentage as a fraction:
        \begin{multicols}{3}
        \begin{onehalfspacing}
        \begin{parts}
            \part $35\%$\fillin[]
            \part $33\frac{1}{3}\%=$\fillin[]
            \part $210\%=$\fillin[]
            \part $125\%=$\fillin[]
            \part $7.25\%=$\fillin[]
            \part $112\frac{1}{2}\%=$\fillin[]
        \end{parts}
        \end{onehalfspacing}
        \end{multicols}
        \question Express each fraction or decimal as a percentage:
        \begin{multicols}{2}
        \begin{onehalfspacing}
        \begin{parts}
            \part \(\frac{3}{5}=\)\fillin[]
            \part \(\frac{7}{20}=\)\fillin[]
            \part $0.43=$\fillin[]
            \part $1.2=$\fillin[]
            \part $0.225=$\fillin[]
            \part $2.03=$\fillin[]
        \end{parts}
        \end{onehalfspacing}
        \end{multicols}
        \question Evaluate these amounts, correct to 2 decimal places where necessary.
        \begin{multicols}{2}
        \begin{onehalfspacing}
        \begin{parts}
            \part \(15 \%\) of 40 = \fillin[]
            \part \(57 \%\) of 1000 =\fillin[]
            \part \(120 \%\) of 538 =\fillin[]
            \part \(15.8 \%\) of 972 = \fillin[]
            \part \(2.8 \%\) of 318 =\fillin[]
            \part \(0.1 \%\) of 6000 = \fillin[]
        \end{parts}
        \end{onehalfspacing}
        \end{multicols}
        \question Find what percentage the first quantity is of the second quantity, correct to 1 decimal place.
        \begin{parts}
            \part[1] \(70 \mathrm{~m}, 50 \mathrm{~m}\)=\fillin[]
            \part[1] 15 weeks, 60 weeks=\fillin[]
            \part[1] 60 weeks, 15 weeks=\fillin[]
        \end{parts}
        \question Find what percentage the first quantity is of the second quantity, correct to 2 decimal places where necessary. You will need to express both quantities in the same unit first.
        \begin{multicols}{2}
        \begin{parts}
        \begin{onehalfspacing}
            \part 68 cents, \(\$ 5.00\)=\fillin[]
            \part \(7 \mathrm{~g}, 3 \mathrm{~kg}\)=\fillin[]
            \part 15 days, 3 years=\fillin[]
            \part \(4 \mathrm{~km}, 250 \mathrm{~m}\)=\fillin[]
            \part 1 year, 1 day=\fillin[]
            \part \(56 \mathrm{~cm}, 2.4 \mathrm{~km}\)=\fillin[]
        \end{onehalfspacing}
        \end{parts}
        \end{multicols}
        \question There are 740 students at a primary school, \(5 \%\) of whom have red hair. Calculate the number of students in the school who have red hair.
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \question 12 A soccer match lasted 92 minutes (including injury time). If team A was in possession for \(55 \%\) of the match, for how many minutes and seconds was team A in possession?
|722041405|kitty|/Applications/kitty.app|0|
    \begin{questions}
        \question Express each percentage as a decimal:
        \begin{multicols}{2}
        \begin{parts}
            \part $72\%=$\fillin[]
            \part $7.6\%=$\fillin[]
            \part $77\frac{3}{4}\%=$\fillin[]
            \part $0.1\%=$\fillin[]
        \end{parts}
        \end{multicols}
        \question Express each percentage as a fraction:
        \begin{multicols}{3}
        \begin{onehalfspacing}
        \begin{parts}
            \part $35\%$\fillin[]
            \part $33\frac{1}{3}\%=$\fillin[]
            \part $210\%=$\fillin[]
            \part $125\%=$\fillin[]
            \part $7.25\%=$\fillin[]
            \part $112\frac{1}{2}\%=$\fillin[]
        \end{parts}
        \end{onehalfspacing}
        \end{multicols}
        \question Express each fraction or decimal as a percentage:
        \begin{multicols}{2}
        \begin{onehalfspacing}
        \begin{parts}
            \part \(\frac{3}{5}=\)\fillin[]
            \part \(\frac{7}{20}=\)\fillin[]
            \part $0.43=$\fillin[]
            \part $1.2=$\fillin[]
            \part $0.225=$\fillin[]
            \part $2.03=$\fillin[]
        \end{parts}
        \end{onehalfspacing}
        \end{multicols}
        \question Evaluate these amounts, correct to 2 decimal places where necessary.
        \begin{multicols}{2}
        \begin{onehalfspacing}
        \begin{parts}
            \part \(15 \%\) of 40 = \fillin[]
            \part \(57 \%\) of 1000 =\fillin[]
            \part \(120 \%\) of 538 =\fillin[]
            \part \(15.8 \%\) of 972 = \fillin[]
            \part \(2.8 \%\) of 318 =\fillin[]
            \part \(0.1 \%\) of 6000 = \fillin[]
        \end{parts}
        \end{onehalfspacing}
        \end{multicols}
        \question Find what percentage the first quantity is of the second quantity, correct to 1 decimal place.
        \begin{parts}
            \part[1] \(70 \mathrm{~m}, 50 \mathrm{~m}\)=\fillin[]
            \part[1] 15 weeks, 60 weeks=\fillin[]
            \part[1] 60 weeks, 15 weeks=\fillin[]
        \end{parts}
        \question Find what percentage the first quantity is of the second quantity, correct to 2 decimal places where necessary. You will need to express both quantities in the same unit first.
        \begin{multicols}{2}
        \begin{parts}
        \begin{onehalfspacing}
            \part 68 cents, \(\$ 5.00\)=\fillin[]
            \part \(7 \mathrm{~g}, 3 \mathrm{~kg}\)=\fillin[]
            \part 15 days, 3 years=\fillin[]
            \part \(4 \mathrm{~km}, 250 \mathrm{~m}\)=\fillin[]
            \part 1 year, 1 day=\fillin[]
            \part \(56 \mathrm{~cm}, 2.4 \mathrm{~km}\)=\fillin[]
        \end{onehalfspacing}
        \end{parts}
        \end{multicols}
        \question There are 740 students at a primary school, \(5 \%\) of whom have red hair. Calculate the number of students in the school who have red hair.
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \question A soccer match lasted 92 minutes (including injury time). If team A was in possession for \(55 \%\) of the match, for how many minutes and seconds was team A in possession?
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
    \end{questions}
|722041415|kitty|/Applications/kitty.app|0|
\begin{questions}
    \question Express each percentage as a decimal:
    \begin{multicols}{2}
    \begin{parts}
        \part $72\%=$\fillin[0.72]
        \part $7.6\%=$\fillin[0.076]
        \part $77\frac{3}{4}\%=$\fillin[0.7775]
        \part $0.1\%=$\fillin[0.001]
    \end{parts}
    \end{multicols}
    \question Express each percentage as a fraction:
    \begin{multicols}{3}
    \begin{onehalfspacing}
    \begin{parts}
        \part $35\%$\fillin[\(\frac{7}{20}\)]
        \part $33\frac{1}{3}\%=$\fillin[\(\frac{1}{3}\)]
        \part $210\%=$\fillin[\(\frac{21}{10}\) or \(2\frac{1}{10}\)]
        \part $125\%=$\fillin[\(\frac{5}{4}\) or \(1\frac{1}{4}\)]
        \part $7.25\%=$\fillin[\(\frac{29}{400}\)]
        \part $112\frac{1}{2}\%=$\fillin[\(\frac{9}{8}\) or \(1\frac{1}{8}\)]
    \end{parts}
    \end{onehalfspacing}
    \end{multicols}
    \question Express each fraction or decimal as a percentage:
    \begin{multicols}{2}
    \begin{onehalfspacing}
    \begin{parts}
        \part \(\frac{3}{5}=\)\fillin[60\%]
        \part \(\frac{7}{20}=\)\fillin[35\%]
        \part $0.43=$\fillin[43\%]
        \part $1.2=$\fillin[120\%]
        \part $0.225=$\fillin[22.5\%]
        \part $2.03=$\fillin[203\%]
    \end{parts}
    \end{onehalfspacing}
    \end{multicols}
    \question Evaluate these amounts, correct to 2 decimal places where necessary.
    \begin{multicols}{2}
    \begin{onehalfspacing}
    \begin{parts}
        \part \(15 \%\) of 40 = \fillin[6]
        \part \(57 \%\) of 1000 =\fillin[570]
        \part \(120 \%\) of 538 =\fillin[645.60]
        \part \(15.8 \%\) of 972 = \fillin[153.57]
        \part \(2.8 \%\) of 318 =\fillin[8.90]
        \part \(0.1 \%\) of 6000 = \fillin[6]
    \end{parts}
    \end{onehalfspacing}
    \end{multicols}
    \question Find what percentage the first quantity is of the second quantity, correct to 1 decimal place.
    \begin{parts}
        \part[1] \(70 \mathrm{~m}, 50 \mathrm{~m}\)=\fillin[140.0\%]
        \part[1] 15 weeks, 60 weeks=\fillin[25.0\%]
        \part[1] 60 weeks, 15 weeks=\fillin[400.0\%]
    \end{parts}
    \question Find what percentage the first quantity is of the second quantity, correct to 2 decimal places where necessary. You will need to express both quantities in the same unit first.
    \begin{multicols}{2}
    \begin{parts}
    \begin{onehalfspacing}
        \part 68 cents, \(\$ 5.00\)=\fillin[13.60\%]
        \part \(7 \mathrm{~g}, 3 \mathrm{~kg}\)=\fillin[0.23\%]
        \part 15 days, 3 years=\fillin[\(\approx\) 1.37\%]
        \part \(4 \mathrm{~km}, 250 \mathrm{~m}\)=\fillin[1600.00\%]
        \part 1 year, 1 day=\fillin[\(\approx\) 0.27\%]
        \part \(56 \mathrm{~cm}, 2.4 \mathrm{~km}\)=\fillin[0.02\%]
    \end{onehalfspacing}
    \end{parts}
    \end{multicols}
    \question There are 740 students at a primary school, \(5 \%\) of whom have red hair. Calculate the number of students in the school who have red hair.
    \begin{solutionordottedlines}[1in]
    37 students
    \end{solutionordottedlines}
    \question A soccer match lasted 92 minutes (including injury time). If team A was in possession for \(55 \%\) of the match, for how many minutes and seconds was team A in possession?
    \begin{solutionordottedlines}[1in]
    50 minutes and 36 seconds
    \end{solutionordottedlines}
\end{questions}
|722041540|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
    \begin{questions}
        \question Express each percentage as a decimal:
        \begin{multicols}{2}
        \begin{parts}
            \part $72\%=$\fillin[]
            \part $7.6\%=$\fillin[]
            \part $77\frac{3}{4}\%=$\fillin[]
            \part $0.1\%=$\fillin[]
        \end{parts}
        \end{multicols}
        \question Express each percentage as a fraction:
        \begin{multicols}{3}
        \begin{onehalfspacing}
        \begin{parts}
            \part $35\%$\fillin[]
            \part $33\frac{1}{3}\%=$\fillin[]
            \part $210\%=$\fillin[]
            \part $125\%=$\fillin[]
            \part $7.25\%=$\fillin[]
            \part $112\frac{1}{2}\%=$\fillin[]
        \end{parts}
        \end{onehalfspacing}
        \end{multicols}
        \question Express each fraction or decimal as a percentage:
        \begin{multicols}{2}
        \begin{onehalfspacing}
        \begin{parts}
            \part \(\frac{3}{5}=\)\fillin[]
            \part \(\frac{7}{20}=\)\fillin[]
            \part $0.43=$\fillin[]
            \part $1.2=$\fillin[]
            \part $0.225=$\fillin[]
            \part $2.03=$\fillin[]
        \end{parts}
        \end{onehalfspacing}
        \end{multicols}
        \question Evaluate these amounts, correct to 2 decimal places where necessary.
        \begin{multicols}{2}
        \begin{onehalfspacing}
        \begin{parts}
            \part \(15 \%\) of 40 = \fillin[]
            \part \(57 \%\) of 1000 =\fillin[]
            \part \(120 \%\) of 538 =\fillin[]
            \part \(15.8 \%\) of 972 = \fillin[]
            \part \(2.8 \%\) of 318 =\fillin[]
            \part \(0.1 \%\) of 6000 = \fillin[]
        \end{parts}
        \end{onehalfspacing}
        \end{multicols}
        \question Find what percentage the first quantity is of the second quantity, correct to 1 decimal place.
        \begin{parts}
            \part[1] \(70 \mathrm{~m}, 50 \mathrm{~m}\)=\fillin[]
            \part[1] 15 weeks, 60 weeks=\fillin[]
            \part[1] 60 weeks, 15 weeks=\fillin[]
        \end{parts}
        \question Find what percentage the first quantity is of the second quantity, correct to 2 decimal places where necessary. You will need to express both quantities in the same unit first.
        \begin{multicols}{2}
        \begin{parts}
        \begin{onehalfspacing}
            \part 68 cents, \(\$ 5.00\)=\fillin[]
            \part \(7 \mathrm{~g}, 3 \mathrm{~kg}\)=\fillin[]
            \part 15 days, 3 years=\fillin[]
            \part \(4 \mathrm{~km}, 250 \mathrm{~m}\)=\fillin[]
            \part 1 year, 1 day=\fillin[]
            \part \(56 \mathrm{~cm}, 2.4 \mathrm{~km}\)=\fillin[]
        \end{onehalfspacing}
        \end{parts}
        \end{multicols}
        \question There are 740 students at a primary school, \(5 \%\) of whom have red hair. Calculate the number of students in the school who have red hair.
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \question A soccer match lasted 92 minutes (including injury time). If team A was in possession for \(55 \%\) of the match, for how many minutes and seconds was team A in possession?
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
|722041548|kitty|/Applications/kitty.app|0|
Why did we make you just do so many trivial percentage conversions?
|722041704|kitty|/Applications/kitty.app|0|
    \begin{example}
|722041858|kitty|/Applications/kitty.app|0|
Joshua saves \(12 \%\) of his after-tax salary every week. If he saves \(\$ 90\) a week, what is his after-tax salary?

\section*{Solution}
Savings \(=\) after-tax salary \(\times 12 \%\)

Reversing this:

\[
\begin{aligned}
\text { after-tax salary } & =\text { savings } \div 12 \% \\
& =90 \div 0.12 \\
& =\$ 750
\end{aligned}
\]
|722041974|kitty|/Applications/kitty.app|0|
Savings \(=\) after-tax salary \(\times 12 \%\)

Reversing this:

\[
\begin{aligned}
\text { after-tax salary } & =\text { savings } \div 12 \% \\
& =90 \div 0.12 \\
& =\$ 750
\end{aligned}
\]
|722041982|kitty|/Applications/kitty.app|0|
Sterling silver is an alloy that is made up of \(92.5 \%\) by mass silver and \(7.5 \%\) copper.

a How much sterling silver can be made with \(5 \mathrm{~kg}\) of silver and unlimited supplies of copper?

b How much sterling silver can be made with \(5 \mathrm{~kg}\) of copper and unlimited supplies of silver?

\section*{Solution}
a Mass of silver \(=\) mass of sterling silver \(\times 92.5 \%\)

Reversing this:

mass of sterling silver \(=\) mass of silver \(\div 92.5 \%\)

\[
\begin{aligned}
& =5 \div 0.925 \\
& \approx 5.405 \mathrm{~kg}
\end{aligned}
\]

b Mass of copper \(=\) mass of sterling silver \(\times 7.5 \%\)

Reversing this:

mass of sterling silver \(=\) mass of copper \(\div 7.5 \%\)

\[
\begin{aligned}
& =5 \div 0.075 \\
& \approx 66.667 \mathrm{~kg}
\end{aligned}
\]
|722042000|kitty|/Applications/kitty.app|0|
a How much sterling silver can be made with \(5 \mathrm{~kg}\) of silver and unlimited supplies of copper?
|722042036|kitty|/Applications/kitty.app|0|
b How much sterling silver can be made with \(5 \mathrm{~kg}\) of copper and unlimited supplies of silver?
|722042044|kitty|/Applications/kitty.app|0|
a Mass of silver \(=\) mass of sterling silver \(\times 92.5 \%\)

Reversing this:

mass of sterling silver \(=\) mass of silver \(\div 92.5 \%\)

\[
\begin{aligned}
& =5 \div 0.925 \\
& \approx 5.405 \mathrm{~kg}
\end{aligned}
\]

b Mass of copper \(=\) mass of sterling silver \(\times 7.5 \%\)

Reversing this:

mass of sterling silver \(=\) mass of copper \(\div 7.5 \%\)

\[
\begin{aligned}
& =5 \div 0.075 \\
& \approx 66.667 \mathrm{~kg}
\end{aligned}
\]
|722042062|kitty|/Applications/kitty.app|0|
The Eureka Gallery charges a commission of \(9.2 \%\).

a The Australian painting Showing the Flag at Bakery Hill was sold recently for \(\$ 180000\). How much did the Gallery receive, and how much was left for the seller?

b The Gallery received a commission of \(\$ 7912\) for selling the painting Ned at the Glen. What was the selling price of the painting, and what did the seller actually receive?

\section*{Solution}
a Commission \(=180000 \times 9.2 \%\)

\[
\begin{aligned}
& =180000 \times 0.092 \\
& =\$ 16560
\end{aligned}
\]

Amount received by seller \(=180000-16560\)

\[
=\$ 163440
\]

b Commission \(=\) selling price \(\times 9.2 \%\)

Reversing this:

selling price \(=\) commission \(\div 9.2 \%\)

\[
\begin{aligned}
& =7912 \div 0.092 \\
& =\$ 86000
\end{aligned}
\]

Amount received by seller \(=86000-7912\)

\[
=\$ 78088
\]
|722042092|kitty|/Applications/kitty.app|0|
a Commission \(=180000 \times 9.2 \%\)

\[
\begin{aligned}
& =180000 \times 0.092 \\
& =\$ 16560
\end{aligned}
\]

Amount received by seller \(=180000-16560\)

\[
=\$ 163440
\]

b Commission \(=\) selling price \(\times 9.2 \%\)

Reversing this:

selling price \(=\) commission \(\div 9.2 \%\)

\[
\begin{aligned}
& =7912 \div 0.092 \\
& =\$ 86000
\end{aligned}
\]

Amount received by seller \(=86000-7912\)

\[
=\$ 78088
\]
|722042122|kitty|/Applications/kitty.app|0|
        \begin{solutionordottedlines}[1in]
            a Commission \(=180000 \times 9.2 \%\)

            \[
            \begin{aligned}
            & =180000 \times 0.092 \\
            & =\$ 16560
            \end{aligned}
            \]

            Amount received by seller \(=180000-16560\)

            \[
            =\$ 163440
            \]

            b Commission \(=\) selling price \(\times 9.2 \%\)

            Reversing this:

            selling price \(=\) commission \(\div 9.2 \%\)

            \[
            \begin{aligned}
            & =7912 \div 0.092 \\
            & =\$ 86000
            \end{aligned}
            \]

            Amount received by seller \(=86000-7912\)

            \[
            =\$ 78088
            \]
        \end{solutionordottedlines}
|722042450|kitty|/Applications/kitty.app|0|
            a Commission \(=180000 \times 9.2 \%\)

            \[
            \begin{aligned}
            & =180000 \times 0.092 \\
            & =\$ 16560
            \end{aligned}
            \]

            Amount received by seller \(=180000-16560\)

            \[
            =\$ 163440
            \]

|722042456|kitty|/Applications/kitty.app|0|

            b Commission \(=\) selling price \(\times 9.2 \%\)

            Reversing this:

            selling price \(=\) commission \(\div 9.2 \%\)

            \[
            \begin{aligned}
            & =7912 \div 0.092 \\
            & =\$ 86000
            \end{aligned}
            \]

            Amount received by seller \(=86000-7912\)

            \[
            =\$ 78088
            \]
|722042464|kitty|/Applications/kitty.app|0|
        \begin{solutionordottedlines}[1in]
            a Mass of silver \(=\) mass of sterling silver \(\times 92.5 \%\)

            Reversing this:

            mass of sterling silver \(=\) mass of silver \(\div 92.5 \%\)

            \[
            \begin{aligned}
            & =5 \div 0.925 \\
            & \approx 5.405 \mathrm{~kg}
            \end{aligned}
            \]

            b Mass of copper \(=\) mass of sterling silver \(\times 7.5 \%\)

            Reversing this:

            mass of sterling silver \(=\) mass of copper \(\div 7.5 \%\)

            \[
            \begin{aligned}
            & =5 \div 0.075 \\
            & \approx 66.667 \mathrm{~kg}
            \end{aligned}
            \]
        \end{solutionordottedlines}
|722042540|kitty|/Applications/kitty.app|0|
        \begin{solutionordottedlines}[1in]
            Mass of silver \(=\) mass of sterling silver \(\times 92.5 \%\)

            Reversing this:

            mass of sterling silver \(=\) mass of silver \(\div 92.5 \%\)

            \[
            \begin{aligned}
            & =5 \div 0.925 \\
            & \approx 5.405 \mathrm{~kg}
            \end{aligned}
            \]

            b Mass of copper \(=\) mass of sterling silver \(\times 7.5 \%\)

            Reversing this:

            mass of sterling silver \(=\) mass of copper \(\div 7.5 \%\)

            \[
            \begin{aligned}
            & =5 \div 0.075 \\
            & \approx 66.667 \mathrm{~kg}
            \end{aligned}
            \]
        \end{solutionordottedlines}
|722042548|kitty|/Applications/kitty.app|0|
            Mass of silver \(=\) mass of sterling silver \(\times 92.5 \%\)

            Reversing this:

            mass of sterling silver \(=\) mass of silver \(\div 92.5 \%\)

            \[
            \begin{aligned}
            & =5 \div 0.925 \\
            & \approx 5.405 \mathrm{~kg}
            \end{aligned}
            \]

|722042553|kitty|/Applications/kitty.app|0|

            b Mass of copper \(=\) mass of sterling silver \(\times 7.5 \%\)

            Reversing this:

            mass of sterling silver \(=\) mass of copper \(\div 7.5 \%\)

            \[
            \begin{aligned}
            & =5 \div 0.075 \\
            & \approx 66.667 \mathrm{~kg}
            \end{aligned}
            \]
|722042560|kitty|/Applications/kitty.app|0|
\section*{Example 6}
The Budget Shoe Shop spent \(\$ 6600000\) last year buying shoes and paying salaries and other expenses. They made a \(2 \%\) profit on these costs.

a What was their profit last year?

b What was the total of their sales?

c In the previous year, their costs were \(\$ 5225000\) and their sales were only \(\$ 5145000\). What percentage loss did they make on their costs?

\section*{Solution}
a Profit \(=6600000 \times 2 \%\)

\[
\begin{aligned}
& =6600000 \times 0.02 \\
& =\$ 132000
\end{aligned}
\]

b Total sales \(=\) total costs + profit

\[
\begin{aligned}
& =6600000+132000 \\
& =\$ 6732000
\end{aligned}
\]

c Last year, loss \(=\) total costs - total sales

\[
\begin{aligned}
& =5225000-5145000 \\
& =\$ 80000
\end{aligned}
\]

Percentage loss \(=\left(\frac{80000}{5225000} \times \frac{100}{1}\right) \%\)

\[
\approx 1.53 \%
\]
|722042633|kitty|/Applications/kitty.app|0|
\section*{Example 6}
|722042672|kitty|/Applications/kitty.app|0|

a What was their profit last year?

b What was the total of their sales?

c In the previous year, their costs were \(\$ 5225000\) and their sales were only \(\$ 5145000\). What percentage loss did they make on their costs?

\section*{Solution}
a Profit \(=6600000 \times 2 \%\)

\[
\begin{aligned}
& =6600000 \times 0.02 \\
& =\$ 132000
\end{aligned}
\]

b Total sales \(=\) total costs + profit

\[
\begin{aligned}
& =6600000+132000 \\
& =\$ 6732000
\end{aligned}
\]

c Last year, loss \(=\) total costs - total sales

\[
\begin{aligned}
& =5225000-5145000 \\
& =\$ 80000
\end{aligned}
\]

Percentage loss \(=\left(\frac{80000}{5225000} \times \frac{100}{1}\right) \%\)

\[
\approx 1.53 \%
\]
|722042687|kitty|/Applications/kitty.app|0|
a Profit \(=6600000 \times 2 \%\)

\[
\begin{aligned}
& =6600000 \times 0.02 \\
& =\$ 132000
\end{aligned}
\]
|722042717|kitty|/Applications/kitty.app|0|
            \begin{solutionordottedlines}[1in]
                Profit \(=6600000 \times 2 \%\)

                \[
                \begin{aligned}
                & =6600000 \times 0.02 \\
                & =\$ 132000
                \end{aligned}
                \]
            \end{solutionordottedlines}
|722042730|kitty|/Applications/kitty.app|0|
                Profit \(=6600000 \times 2 \%\)

                \[
                \begin{aligned}
                & =6600000 \times 0.02 \\
                & =\$ 132000
                \end{aligned}
                \]
|722042738|kitty|/Applications/kitty.app|0|
Total sales \(=\) total costs + profit

\[
\begin{aligned}
& =6600000+132000 \\
& =\$ 6732000
\end{aligned}
\]
|722042744|kitty|/Applications/kitty.app|0|
Last year, loss \(=\) total costs - total sales

\[
\begin{aligned}
& =5225000-5145000 \\
& =\$ 80000
\end{aligned}
\]

Percentage loss \(=\left(\frac{80000}{5225000} \times \frac{100}{1}\right) \%\)

\[
\approx 1.53 \%
\]
|722042753|kitty|/Applications/kitty.app|0|

            \part 
            \part 
            \part 
|722042764|kitty|/Applications/kitty.app|0|
        \begin{parts}
            \part
|722042773|kitty|/Applications/kitty.app|0|
\section*{Example 7}
|722042785|kitty|/Applications/kitty.app|0|
Joe's tile shop made a profit of \(5.8 \%\) on total costs last year. If the actual profit was \(\$ 83000\), what were the total costs, and what were the total sales?
|722042787|kitty|/Applications/kitty.app|0|
\[
\text { Profit }=\text { costs } \times 5.8 \%
\]

Reversing this, costs \(=\) profit \(\div 5.8 \%\)

\[
=83000 \div 0.058
\]

\(\approx \$ 1431034\), correct to the nearest dollar.

Hence, total sales \(=\) profit + costs

\[
\approx 83000+1431034
\]

\[
=\$ 1514034
\]
|722042797|kitty|/Applications/kitty.app|0|
Income tax in the nation of Immutatia is calculated as follows.

\begin{itemize}
  \item There is no tax on the first \(\$ 12000\) that a person earns in any one year.

  \item From \(\$ 12001\) to \(\$ 30000\), the tax rate is 15 c for each dollar over \(\$ 12000\).

  \item From \(\$ 30001\) to \(\$ 75000\), the tax rate is 25 c for each dollar over \(\$ 30000\).

  \item Over \(\$ 75000\), the tax rate is 35 c for each dollar over \(\$ 75000\).

\end{itemize}

Find the income tax payable by a person whose taxable income for the year is:
a \(\$ 10600\)
b \(\$ 25572\)
c \(\$ 62300\)
d \(\$ 455000\)

\section*{Solution}
a There is no tax on an income of \(\$ 10600\).

b Tax on first \(\$ 12000=\$ 0\)

Tax on remaining \(\$ 13572=13572 \times 0.15\)

\[
=\$ 2035.80
\]

This is the total tax payable.

c Tax on first \(\$ 12000=\$ 0\)

Tax on next \(\$ 18000=18000 \times 0.15\)

\[
=\$ 2700
\]

Tax on remaining \(\$ 32300=32300 \times 0.25\)

\[
=\$ 8075
\]

Total tax \(=2700+8075\)

\[
=\$ 10775
\]

d Tax on first \(\$ 12000=\$ 0\)

Tax on next \(\$ 18000=\$ 2700 \quad(\) see part c)

Tax on next \(\$ 45000=45000 \times 0.25\)

\[
=\$ 11250
\]

Tax on remaining \(\$ 380000=380000 \times 0.35\)

\[
=\$ 133000
\]

Total tax \(=2700+11250+133000\)

\[
=\$ 146950
\]
|722043123|kitty|/Applications/kitty.app|0|
 Find the income tax payable by a person whose taxable income for the year is:|722043150|kitty|/Applications/kitty.app|0|
a There is no tax on an income of \(\$ 10600\).
|722043183|kitty|/Applications/kitty.app|0|
b Tax on first \(\$ 12000=\$ 0\)

Tax on remaining \(\$ 13572=13572 \times 0.15\)

\[
=\$ 2035.80
\]

This is the total tax payable.
|722043195|kitty|/Applications/kitty.app|0|
Tax on first \(\$ 12000=\$ 0\)

Tax on next \(\$ 18000=18000 \times 0.15\)

\[
=\$ 2700
\]

Tax on remaining \(\$ 32300=32300 \times 0.25\)

\[
=\$ 8075
\]

Total tax \(=2700+8075\)

\[
=\$ 10775
\]
|722043210|kitty|/Applications/kitty.app|0|
d Tax on first \(\$ 12000=\$ 0\)

Tax on next \(\$ 18000=\$ 2700 \quad(\) see part c)

Tax on next \(\$ 45000=45000 \times 0.25\)

\[
=\$ 11250
\]

Tax on remaining \(\$ 380000=380000 \times 0.35\)

\[
=\$ 133000
\]

Total tax \(=2700+11250+133000\)

\[
=\$ 146950
\]
|722043223|kitty|/Applications/kitty.app|0|
        \question[1]
        \begin{parts}
        \end{parts}
|722043601|kitty|/Applications/kitty.app|0|
3 Find the quantity, given that:
a \(2 \%\) of it is \(\$ 12\)
b \(6 \%\) of it is \(750 \mathrm{~g}\)
c \(30 \%\) of it is 36 minutes
d \(90 \%\) of it is \(54 \mathrm{~cm}\)

4 In each part, find the price if:
a a deposit of \(\$ 360\) is \(30 \%\) of the price
b a deposit of \(\$ 168\) is \(15 \%\) of the price

5 Find the original quantity, correct to a suitable number of decimal places, if:
a \(23 \%\) of it is \(100 \mathrm{~kg}\)
b \(0.2 \%\) of it is \(4 \mathrm{~mm}\)
c \(0.92 \%\) of it is 1.86 hectares
d \(97 \%\) of it is \(\$ 700\)

6 Cameron and Wendy together earn \(\$ 1156\) per week after tax. Of this, they pay \(\$ 460\) off their mortgage, \(\$ 185\) for groceries, and \(\$ 260\) for their car and transport, and they save \$124. a Express each amount as a percentage of their weekly income, correct to the nearest \(1 \%\).

b Find how much is unaccounted for in the list above, and what percentage it is of their weekly income.
|722043636|kitty|/Applications/kitty.app|0|
        \question[1]
        \begin{parts}
        \end{parts}
        \question[1]
        \begin{parts}
        \end{parts}
        \question[1]
        \begin{parts}
        \end{parts}
        \question[1]
        \begin{parts}
        \end{parts}
|722043872|kitty|/Applications/kitty.app|0|
\renewcommand{\questionlabel}{\thequestion. \ (\thepoints)}|722044477|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\renewcommand{\questionlabel}{\thequestion. \ (\thepoints)}
|722044568|kitty|/Applications/kitty.app|0|
[2]
[1]|722044648|kitty|/Applications/kitty.app|0|
[1]|722044774|kitty|/Applications/kitty.app|0|
    \question*[1]
|722045703|kitty|/Applications/kitty.app|0|
\question*[1]|722045734|kitty|/Applications/kitty.app|0|
\begin{questions}
    \question[2] Find the quantity, given that:
    \begin{parts}\begin{multicols}{2}
        \part \(2 \%\) of it is \(\$ 12\)
        \begin{solutionordottedlines}[0.5in]
            Quantity = \(\$ 12\) / \(0.02\) = \(\$ 600\)
        \end{solutionordottedlines}
        \part \(6 \%\) of it is \(750 \mathrm{~g}\)
        \begin{solutionordottedlines}[0.5in]
            Quantity = \(750 \mathrm{~g}\) / \(0.06\) = \(12500 \mathrm{~g}\) or \(12.5 \mathrm{~kg}\)
        \end{solutionordottedlines}
        \part \(30 \%\) of it is 36 minutes
        \begin{solutionordottedlines}[0.5in]
            Quantity = \(36 \text{ minutes}\) / \(0.3\) = \(120 \text{ minutes}\) or \(2 \text{ hours}\)
        \end{solutionordottedlines}
        \part \(90 \%\) of it is \(54 \mathrm{~cm}\)
        \begin{solutionordottedlines}[0.5in]
            Quantity = \(54 \mathrm{~cm}\) / \(0.9\) = \(60 \mathrm{~cm}\)
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[2] In each part, find the price if:
    \begin{parts}
        \part a deposit of \(\$ 360\) is \(30 \%\) of the price
        \begin{solutionordottedlines}[0.5in]
            Price = \(\$ 360\) / \(0.3\) = \(\$ 1200\)
        \end{solutionordottedlines}
        \part a deposit of \(\$ 168\) is \(15 \%\) of the price
        \begin{solutionordottedlines}[0.5in]
            Price = \(\$ 168\) / \(0.15\) = \(\$ 1120\)
        \end{solutionordottedlines}
    \end{parts}
    \question[2] Find the original quantity, correct to a suitable number of decimal places, if:
    \begin{parts}\begin{multicols}{2}
        \part \(23 \%\) of it is \(100 \mathrm{~kg}\)
        \begin{solutionordottedlines}[0.5in]
            Quantity = \(100 \mathrm{~kg}\) / \(0.23\) = \(434.78 \mathrm{~kg}\)
        \end{solutionordottedlines}
        \part \(0.2 \%\) of it is \(4 \mathrm{~mm}\)
        \begin{solutionordottedlines}[0.5in]
            Quantity = \(4 \mathrm{~mm}\) / \(0.002\) = \(2000 \mathrm{~mm}\) or \(2 \mathrm{~m}\)
        \end{solutionordottedlines}
        \part \(0.92 \%\) of it is 1.86 hectares
        \begin{solutionordottedlines}[0.5in]
            Quantity = \(1.86 \text{ hectares}\) / \(0.0092\) = \(20.22 \text{ hectares}\)
        \end{solutionordottedlines}
        \part \(97 \%\) of it is \(\$ 700\)
        \begin{solutionordottedlines}[0.5in]
            Quantity = \(\$ 700\) / \(0.97\) = \(\$ 721.65\)
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[2] Cameron and Wendy together earn \(\$ 1156\) per week after tax. Of this, they pay \(\$ 460\) off their mortgage, \(\$ 185\) for groceries, and \(\$ 260\) for their car and transport, and they save \$124. 
    \begin{parts}
        \part Express each amount as a percentage of their weekly income, correct to the nearest \(1 \%\).
        \begin{solutionordottedlines}[1in]
            Mortgage: \((\$ 460 / \$ 1156) \times 100 \approx 39.8\%\) (rounded to \(40\%\)) \\
            Groceries: \((\$ 185 / \$ 1156) \times 100 \approx 16.0\%\) (rounded to \(16\%\)) \\
            Car and transport: \((\$ 260 / \$ 1156) \times 100 \approx 22.5\%\) (rounded to \(23\%\)) \\
            Savings: \((\$ 124 / \$ 1156) \times 100 \approx 10.7\%\) (rounded to \(11\%\))
        \end{solutionordottedlines}
        \part Find how much is unaccounted for in the list above, and what percentage it is of their weekly income.
        \begin{solutionordottedlines}[1in]
            Unaccounted for: \( \$ 1156 - (\$ 460 + \$ 185 + \$ 260 + \$ 124) = \$ 1156 - \$ 1029 = \$ 127 \) \\
            Percentage: \((\$ 127 / \$ 1156) \times 100 \approx 11.0\%\) (rounded to \(11\%\))
        \end{solutionordottedlines}
    \end{parts}
\end{questions}
|722045891|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
    \begin{questions}
        \question[2] Find the quantity, given that:
        \begin{parts}\begin{multicols}{2}
            \part \(2 \%\) of it is \(\$ 12\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(6 \%\) of it is \(750 \mathrm{~g}\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(30 \%\) of it is 36 minutes
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(90 \%\) of it is \(54 \mathrm{~cm}\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[2] In each part, find the price if:
        \begin{parts}
            \part a deposit of \(\$ 360\) is \(30 \%\) of the price
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part a deposit of \(\$ 168\) is \(15 \%\) of the price
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \end{parts}
        \question[2] Find the original quantity, correct to a suitable number of decimal places, if:
        \begin{parts}\begin{multicols}{2}
            \part \(23 \%\) of it is \(100 \mathrm{~kg}\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(0.2 \%\) of it is \(4 \mathrm{~mm}\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(0.92 \%\) of it is 1.86 hectares
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(97 \%\) of it is \(\$ 700\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[2] Cameron and Wendy together earn \(\$ 1156\) per week after tax. Of this, they pay \(\$ 460\) off their mortgage, \(\$ 185\) for groceries, and \(\$ 260\) for their car and transport, and they save \$124. 
        \begin{parts}
            \part Express each amount as a percentage of their weekly income, correct to the nearest \(1 \%\).
            \begin{solutionordottedlines}[1in]
            \end{solutionordottedlines}
            \part Find how much is unaccounted for in the list above, and what percentage it is of their weekly income.
            \begin{solutionordottedlines}[1in]
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
|722045900|kitty|/Applications/kitty.app|0|
7 What percentage of the total cost is a deposit of:
a \(\$ 33\) on a television valued at \(\$ 550\) ?
b \(\$ 124.10\) on a stove valued at \(\$ 1460\) ?
|722046525|kitty|/Applications/kitty.app|0|
12 Find the selling price if the commission and the commission rate are as given.
a Commission \(\$ 35\), rate \(7 \%\)
b Commission \(\$ 646\), rate \(3.4 \%\)
c Commission \(\$ 16586.96\), rate \(5.2 \%\)
d Commission \(\$ 3518.61\), rate \(11.4 \%\)

13 Find, to 1 decimal place, the percentage profit or loss on costs in these situations.

a Costs \(\$ 16000\) and sales \(\$ 18000\)

b Costs \(\$ 162000\) and sales \(\$ 150000\)

c Costs \(\$ 2800000\) and sales \(\$ 3090000\)

d Costs \(\$ 289000000\) and sales \(\$ 268000000\)
|722046552|kitty|/Applications/kitty.app|0|
\end{onehalfspacing}

\|722047221|kitty|/Applications/kitty.app|0|

\fbox{
\quote Compound interest is the eighth wonder of the world. S/he who understands it, earns it, s/he who doesn't, pays it. - Albert Einstein.
}
|722047253|kitty|/Applications/kitty.app|0|
Compound interest is the eighth wonder of the world. S/he who understands it, earns it, s/he who doesn't, pays it. - Albert Einstein.|722047336|kitty|/Applications/kitty.app|0|
\begin{flushright}
\begin{minipage}{8cm}
    \begin{flushleft}
        \emph{It's better to feed one cat than many mice.}
    \end{flushleft}
    \begin{flushright}--- Norwegian proverb\end{flushright}
\end{minipage}
\end{flushright}
|722047328|kitty|/Applications/kitty.app|0|
It's better to feed one cat than many mice.|722047333|kitty|/Applications/kitty.app|0|
 - Albert Einstein.|722047339|kitty|/Applications/kitty.app|0|
. Norwegian proverb|722047346|kitty|/Applications/kitty.app|0|
\quote \fbox{}
|722047349|kitty|/Applications/kitty.app|0|
\begin{flushleft}
|722047374|kitty|/Applications/kitty.app|0|
\end{flushright}
|722047376|kitty|/Applications/kitty.app|0|
    }
|722047415|kitty|/Applications/kitty.app|0|
    \fbox{
|722047417|kitty|/Applications/kitty.app|0|
Find the simple interest on \(\$ 8000\) for eight years at 9.5\% p.a.

\section*{Solution}
\[
\begin{aligned}
I & =P R T \\
& =8000 \times 9.5 \% \times 8 \\
& =8000 \times 0.095 \times 8 \\
& =\$ 6080
\end{aligned}
\]
|722047517|kitty|/Applications/kitty.app|0|
        \

\section*{Solution}
|722047537|kitty|/Applications/kitty.app|0|
Jessie borrows \(\$ 3000\) from her parents to help buy a car. They agree that she should only pay simple interest. Five years later she pays them back \(\$ 3600\), which includes simple interest on the loan. What was the interest rate?

\section*{Solution}
The total interest paid was \(\$ 600\), the principal was \(\$ 3000\) and the time was 5 years.

\[
\begin{aligned}
I & =P R T \\
600 & =3000 \times R \times 5 \\
600 & =15000 \times R \\
R & =\frac{600}{15000} \times 100 \% \\
& =4 \%
\end{aligned}
\]
|722047578|kitty|/Applications/kitty.app|0|
\section*{Simple interest formula}
\begin{itemize}
  \item Suppose that a principal \(P\) is invested for \(T\) years at an interest rate \(R\) p.a. Then the total interest \(I\) is given by:
\end{itemize}

\[
I=P R T
\]

Remember that \(R\) is a percentage. If the interest rate is \(5 \%\), then \(R=0.05\).

\begin{itemize}
  \item If the interest rate \(R\) is given per year, the time \(T\) must be given in years.

  \item The formula has four pronumerals. If any three are known, the fourth can be found by substitution.

\end{itemize}
|722047629|kitty|/Applications/kitty.app|0|
\(1 \$ 12000\) is invested at \(7 \%\) p.a. simple interest for five years.

a How much interest will be earned each year?

b Find how much interest will be earned over the five-year period.

\(2 \$ 2000\) is invested at \(6.75 \%\) p.a. simple interest for three years.

a How much interest will be earned each year?

b Find how much interest will be earned over the three-year period.

3 Find the total simple interest earned in each of these investments.
a \(\$ 400\) for three years at \(6 \%\) p.a.
b \(\$ 850\) for six years at \(4.5 \%\) p.a.
c \(\$ 15000\) for 12 years at \(8.4 \%\) p.a.

4 Find the time \(T\) for \(\$ 2000\) of simple interest on a principal of \(\$ 8000\) at a rate of \(5 \%\) p.a.

5 Find the rate \(R\) p.a. for \(\$ 7200\) of simple interest on a principal of \(\$ 8000\) over 12 years.

6 Find the principal \(P\) for \(\$ 3500\) of simple interest at a rate of \(7 \%\) p.a. over 10 years.
|722047650|kitty|/Applications/kitty.app|0|
    k
|722047745|kitty|/Applications/kitty.app|0|
    \begin{questions}
    \question[2] \(\$ 12000\) is invested at \(7 \%\) p.a. simple interest for five years.
    \begin{parts}
        \part How much interest will be earned each year?
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \part Find how much interest will be earned over the five-year period.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
    \end{parts}
    \question \(\$ 2000\) is invested at \(6.75 \%\) p.a. simple interest for three years.
    \begin{parts}
        \part How much interest will be earned each year?
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \part Find how much interest will be earned over the three-year period.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
    \end{parts}
    \question[3] Find the total simple interest earned in each of these investments.
    \begin{parts}
        \part \(\$ 400\) for three years at \(6 \%\) p.a.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \part \(\$ 850\) for six years at \(4.5 \%\) p.a.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \part \(\$ 15000\) for 12 years at \(8.4 \%\) p.a.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
    \end{parts}
    \question[1] Find the time \(T\) for \(\$ 2000\) of simple interest on a principal of \(\$ 8000\) at a rate of \(5 \%\) p.a.
    \begin{solutionordottedlines}[1in]
    \end{solutionordottedlines}
    \question[1] Find the rate \(R\) p.a. for \(\$ 7200\) of simple interest on a principal of \(\$ 8000\) over 12 years.
    \begin{solutionordottedlines}[1in]
    \end{solutionordottedlines}
    \question Find the principal \(P\) for \(\$ 3500\) of simple interest at a rate of \(7 \%\) p.a. over 10 years.
    \begin{solutionordottedlines}[1in]
    \end{solutionordottedlines}
    \end{questions}
|722048189|kitty|/Applications/kitty.app|0|
7 Calculate the missing entries for these simple interest investments.

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
 & Principal & Rate p.a. & Time in years & Total interest \\
\hline
a & \(\$ 10000\) & \(8 \%\) &  & \(\$ 3200\) \\
\hline
b & \(\$ 4400000\) & \(7 \frac{1}{2} \%\) &  & \(\$ 3960000\) \\
\hline
c & \(\$ 5000\) &  & 6 & \(\$ 1350\) \\
\hline
d & \(\$ 260000\) &  & 8 & \(\$ 83200\) \\
\hline
e &  & \(6 \%\) & 5 & \(\$ 900\) \\
f &  & \(3.6 \%\) & 4 & \(\$ 115.20\) \\
\end{tabular}
\end{center}

8 Sartoro invested \(\$ 80000\) in a building society that pays \(6.5 \%\) p.a. simple interest. Over the years, the investment has paid him \(\$ 57200\) in interest. How many years has he had the investment?

9 Madeline has received \(\$ 168000\) in total simple interest payments on an investment of \(\$ 400000\) that she made six years ago. What rate of interest has the bank been paying?

10 An investor wishes to earn \(\$ 240000\) interest over a five-year period from an account that earns \(12.5 \%\) p.a. simple interest. How much does the investor have to deposit into the account?

|722048236|kitty|/Applications/kitty.app|0|
11 Regan has arranged to borrow \(\$ 10000\) at \(9.5 \%\) p.a. for four years. She will pay simple interest to the bank every year for the loan, with the principal remaining unchanged. How much interest will Regan pay over the four years of the loan?

12 Tyler intends to live on the interest on an investment with the bank at \(8.6 \%\) p.a. simple interest. She will receive \(\$ 68000\) simple interest every year from the investment. How much money has she invested?
|722048265|kitty|/Applications/kitty.app|0|
    \question[2] Regan has arranged to borrow \(\$ 10000\) at \(9.5 \%\) p.a. for four years. She will pay simple interest to the bank every year for the loan, with the principal remaining unchanged. How much interest will Regan pay over the four years of the loan?
    \begin{solutionordottedlines}[1in]
    \end{solutionordottedlines}
    \question[2] Tyler intends to live on the interest on an investment with the bank at \(8.6 \%\) p.a. simple interest. She will receive \(\$ 68000\) simple interest every year from the investment. How much money has she invested?
    \begin{solutionordottedlines}[1in]
    \end{solutionordottedlines}
|722048306|kitty|/Applications/kitty.app|0|
\begin{questions}
    \question[2] \(\$ 12000\) is invested at \(7 \%\) p.a. simple interest for five years.
    \begin{parts}
        \part How much interest will be earned each year?
            \begin{solutionordottedlines}[0.5in]
                Annual interest = \(\$ 12000 \times 7\% = \$ 840\)
            \end{solutionordottedlines}
        \part Find how much interest will be earned over the five-year period.
            \begin{solutionordottedlines}[0.5in]
                Total interest = \(\$ 840 \times 5 = \$ 4200\)
            \end{solutionordottedlines}
    \end{parts}
    \question \(\$ 2000\) is invested at \(6.75 \%\) p.a. simple interest for three years.
    \begin{parts}
        \part How much interest will be earned each year?
            \begin{solutionordottedlines}[0.5in]
                Annual interest = \(\$ 2000 \times 6.75\% = \$ 135\)
            \end{solutionordottedlines}
        \part Find how much interest will be earned over the three-year period.
            \begin{solutionordottedlines}[0.5in]
                Total interest = \(\$ 135 \times 3 = \$ 405\)
            \end{solutionordottedlines}
    \end{parts}
    \question[3] Find the total simple interest earned in each of these investments.
    \begin{parts}
        \part \(\$ 400\) for three years at \(6 \%\) p.a.
            \begin{solutionordottedlines}[0.5in]
                Total interest = \(\$ 400 \times 6\% \times 3 = \$ 72\)
            \end{solutionordottedlines}
        \part \(\$ 850\) for six years at \(4.5 \%\) p.a.
            \begin{solutionordottedlines}[0.5in]
                Total interest = \(\$ 850 \times 4.5\% \times 6 = \$ 229.50\)
            \end{solutionordottedlines}
        \part \(\$ 15000\) for 12 years at \(8.4 \%\) p.a.
            \begin{solutionordottedlines}[0.5in]
                Total interest = \(\$ 15000 \times 8.4\% \times 12 = \$ 15120\)
            \end{solutionordottedlines}
    \end{parts}
    \question[1] Find the time \(T\) for \(\$ 2000\) of simple interest on a principal of \(\$ 8000\) at a rate of \(5 \%\) p.a.
    \begin{solutionordottedlines}[1in]
        \(2000 = 8000 \times 5\% \times T\)
        \(T = \frac{2000}{8000 \times 5\%} = 5\) years
    \end{solutionordottedlines}
    \question[1] Find the rate \(R\) p.a. for \(\$ 7200\) of simple interest on a principal of \(\$ 8000\) over 12 years.
    \begin{solutionordottedlines}[1in]
        \(7200 = 8000 \times R \times 12\)
        \(R = \frac{7200}{8000 \times 12} = 0.075\) or \(7.5\% \) p.a.
    \end{solutionordottedlines}
    \question Find the principal \(P\) for \(\$ 3500\) of simple interest at a rate of \(7 \%\) p.a. over 10 years.
    \begin{solutionordottedlines}[1in]
        \(3500 = P \times 7\% \times 10\)
        \(P = \frac{3500}{7\% \times 10} = \$ 5000\)
    \end{solutionordottedlines}
\end{questions}
|722048322|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
    \begin{questions}
    \question[2] \(\$ 12000\) is invested at \(7 \%\) p.a. simple interest for five years.
    \begin{parts}
        \part How much interest will be earned each year?
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \part Find how much interest will be earned over the five-year period.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
    \end{parts}
    \question \(\$ 2000\) is invested at \(6.75 \%\) p.a. simple interest for three years.
    \begin{parts}
        \part How much interest will be earned each year?
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \part Find how much interest will be earned over the three-year period.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
    \end{parts}
    \question[3] Find the total simple interest earned in each of these investments.
    \begin{parts}
        \part \(\$ 400\) for three years at \(6 \%\) p.a.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \part \(\$ 850\) for six years at \(4.5 \%\) p.a.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \part \(\$ 15000\) for 12 years at \(8.4 \%\) p.a.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
    \end{parts}
    \question[1] Find the time \(T\) for \(\$ 2000\) of simple interest on a principal of \(\$ 8000\) at a rate of \(5 \%\) p.a.
    \begin{solutionordottedlines}[1in]
    \end{solutionordottedlines}
    \question[1] Find the rate \(R\) p.a. for \(\$ 7200\) of simple interest on a principal of \(\$ 8000\) over 12 years.
    \begin{solutionordottedlines}[1in]
    \end{solutionordottedlines}
    \question Find the principal \(P\) for \(\$ 3500\) of simple interest at a rate of \(7 \%\) p.a. over 10 years.
    \begin{solutionordottedlines}[1in]
    \end{solutionordottedlines}
|722048330|kitty|/Applications/kitty.app|0|
    \end{questions}
|722048340|kitty|/Applications/kitty.app|0|
        \question[2] Regan has arranged to borrow \(\$ 10000\) at \(9.5 \%\) p.a. for four years. She will pay simple interest to the bank every year for the loan, with the principal remaining unchanged. How much interest will Regan pay over the four years of the loan?
        \begin{solutionordottedlines}[1in]

        \end{solutionordottedlines}
        \question[2] Tyler intends to live on the interest on an investment with the bank at \(8.6 \%\) p.a. simple interest. She will receive \(\$ 68000\) simple interest every year from the investment. How much money has she invested?
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
|722048357|kitty|/Applications/kitty.app|0|
\question[2] Regan has arranged to borrow \(\$ 10000\) at \(9.5 \%\) p.a. for four years. She will pay simple interest to the bank every year for the loan, with the principal remaining unchanged. How much interest will Regan pay over the four years of the loan?
    \begin{solutionordottedlines}[1in]
        Annual interest = \(\$ 10000 \times 9.5\% = \$ 950\) \\
        Total interest over 4 years = \(\$ 950 \times 4 = \$ 3800\)
    \end{solutionordottedlines}
    \question[2] Tyler intends to live on the interest on an investment with the bank at \(8.6 \%\) p.a. simple interest. She will receive \(\$ 68000\) simple interest every year from the investment. How much money has she invested?
    \begin{solutionordottedlines}[1in]
        Principal = \(\$ 68000\) / \(8.6\%\) \\
        Principal = \(\$ 68000\) / \(0.086 \approx \$ 790697.67\)
    \end{solutionordottedlines}|722048366|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
    \subsection{Examples:}
|722048545|kitty|/Applications/kitty.app|0|
Examples|722048548|kitty|/Applications/kitty.app|0|
The number of patients admitted to St Spyridon's Hospital this year suffering from pneumonia is \(56 \%\) greater than the number admitted for this condition last year. If 245 pneumonia patients were admitted last year, how many were admitted this year?

\section*{Solution}
This year's total is \(100 \%+56 \%=156 \%\) of last year's total.

This year's total \(=245 \times 156 \%\)

\(=245 \times 1.56\)

\(\approx 382\) (correct to the nearest whole number)
|722048625|kitty|/Applications/kitty.app|0|
The war-ravaged nation of Zerbai is experiencing inflation of \(35 \%\) p.a. as a result of overspending on its navy and air force. Inflation of \(35 \%\) means that, on average, prices are increasing by \(35 \%\) every year.

a If the price of water is adjusted in line with inflation, what will an annual bill of \(\$ 600\) become in the next year?

b What should an annual salary of \(\$ 169000\) in one year increase to in the following year if it is adjusted to keep pace with inflation?

\section*{Solution}
Next year's prices are \(100 \%+35 \%=135 \%\) of last year's prices.

a Next year's bill \(=600 \times 1.35\)

\[
=\$ 810
\]

b Next year's salary \(=169000 \times 1.35\)

\[
=\$ 228150
\]
|722048767|kitty|/Applications/kitty.app|0|
Next year's prices are \(100 \%+35 \%=135 \%\) of last year's prices.

a Next year's bill \(=600 \times 1.35\)

\[
=\$ 810
\]
|722048807|kitty|/Applications/kitty.app|0|
b Next year's salary \(=169000 \times 1.35\)

\[
=\$ 228150
\]
|722048825|kitty|/Applications/kitty.app|0|


\section*{Solution}
|722048833|kitty|/Applications/kitty.app|0|
The company that Yuri Ivanov works for is going through hard times and has decreased all its salaries by \(12 \%\). Yuri is attempting to cut every one of his expenses by the same percentage.

a His family's weekly grocery bill averages 450 roubles. What should he try to reduce the weekly price of his groceries to?

b His monthly rental is 18000 roubles. If he moves apartments, what monthly rental should he try to find?

\section*{Solution}
Yuri's new salary is \(100 \%-12 \%=88 \%\) of his original salary.

a New weekly grocery bill \(=450 \times 0.88\)

b New monthly rental \(=18000 \times 0.88\)

\(=396\) roubles

\[
=15840 \text { roubles }
\]
|722048888|kitty|/Applications/kitty.app|0|
Yuri's new salary is \(100 \%-12 \%=88 \%\) of his original salary.

a New weekly grocery bill \(=450 \times 0.88\)
|722049030|kitty|/Applications/kitty.app|0|
b New monthly rental \(=18000 \times 0.88\)
|722049039|kitty|/Applications/kitty.app|0|

\section*{Solution}


|722049048|kitty|/Applications/kitty.app|0|
\(=396\) roubles

\[
=15840 \text { roubles }
\]
|722049052|kitty|/Applications/kitty.app|0|
word8|722049172|kitty|/Applications/kitty.app|0|
{word9}|722049175|kitty|/Applications/kitty.app|0|
    \randomword{#9}%
|722049192|kitty|/Applications/kitty.app|0|
\newcommand{\randomwords}[8]{%
|722049236|kitty|/Applications/kitty.app|0|
  \fbox{%
  \makebox[\textwidth]{
    \randomword{#1}%
    \randomword{#2}%
    \randomword{#3}%
    \randomword{#4}%
    \randomword{#5}%
    \randomword{#6}%
    \randomword{#7}%
    \randomword{#8}%
    }
  }
|722049252|kitty|/Applications/kitty.app|0|
#2|722049266|kitty|/Applications/kitty.app|0|
#3|722049269|kitty|/Applications/kitty.app|0|
#4|722049276|kitty|/Applications/kitty.app|0|
#5|722049280|kitty|/Applications/kitty.app|0|
#6|722049285|kitty|/Applications/kitty.app|0|
#7|722049289|kitty|/Applications/kitty.app|0|
#8|722049292|kitty|/Applications/kitty.app|0|
\randomwords{appreciation}{GST}{simple interest}{compound interest}{commissions}{inflation}{depreciation}{discounts}
|722049301|kitty|/Applications/kitty.app|0|
The Wind Energy Company recently announced that this year's profit of \(\$ 1400000\) constituted a \(35 \%\) increase on last year's profit. What was last year's profit?

\section*{Solution}
This year's profit is \(100 \%+35 \%=135 \%\) of last year's profit.

Hence this year's profit \(=\) last year's profit \(\times 1.35\)

Reversing this, last year's profit \(=\) this year's profit \(\div 1.35\)

\[
\begin{aligned}
& =1400000 \div 1.35 \\
& \approx \$ 1037037, \text { correct to the nearest dollar }
\end{aligned}
\]

Thus, to find the original amount, we divide by 1.35 , because dividing by 1.35 is the reverse of multiplying by 1.35 .

Exactly the same principle applies when an amount has been decreased by a percentage.
|722049392|kitty|/Applications/kitty.app|0|
This year's profit is \(100 \%+35 \%=135 \%\) of last year's profit.

Hence this year's profit \(=\) last year's profit \(\times 1.35\)

Reversing this, last year's profit \(=\) this year's profit \(\div 1.35\)

\[
\begin{aligned}
& =1400000 \div 1.35 \\
& \approx \$ 1037037, \text { correct to the nearest dollar }
\end{aligned}
\]

Thus, to find the original amount, we divide by 1.35 , because dividing by 1.35 is the reverse of multiplying by 1.35 .

Exactly the same principle applies when an amount has been decreased by a percentage.
|722049426|kitty|/Applications/kitty.app|0|
                Thus, to find the original amount, we divide by 1.35 , because dividing by 1.35 is the reverse of multiplying by 1.35 .

                Exactly the same principle applies when an amount has been decreased by a percentage.
|722049441|kitty|/Applications/kitty.app|0|
The price of shares in the Fountain Water Company has decreased by \(15 \%\) over the last month to \(\$ 52.70\). What was the price a month ago?
|722049456|kitty|/Applications/kitty.app|0|
            \question[1]
|722049468|kitty|/Applications/kitty.app|0|
The new share price is \(100 \%-15 \%=85 \%\) of the old share price.

Hence \(\quad\) new price \(=\) old price \(\times 0.85\)

Reversing this, old price \(=\) new price \(\div 0.85\)

\[
\begin{aligned}
& =52.70 \div 0.85 \\
& =\$ 62
\end{aligned}
\]
|722049481|kitty|/Applications/kitty.app|0|
The current GST rate is \(10 \%\) of the pre-tax price.

a If a domestic plumbing job costs \(\$ 630\) before GST, how much will it cost after adding GST, and how much tax is paid to the Government?

b I paid \(\$ 70\) for petrol recently. What was the price before adding GST, and what tax was paid to the Government?

|722049574|kitty|/Applications/kitty.app|0|
a After-tax price \(=630 \times 1.10\)

\[
\begin{aligned}
& =\$ 693 \\
\text { Tax } & =693-630 \\
& =\$ 63
\end{aligned}
\]

Note: \(10 \%\) of \(\$ 630\) is \(\$ 63\).
|722049601|kitty|/Applications/kitty.app|0|
b Pre-tax price \(=70 \div 1.10\) (Divide by 1.10 to reverse the process.)

\[
\begin{aligned}
& \approx \$ 63.64 \\
\mathrm{Tax} & \approx 70-63.64 \\
& =\$ 6.36
\end{aligned}
\]
|722049611|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
    \end{questions}

\end{exercisebox}
|722050059|kitty|/Applications/kitty.app|0|
1 Traffic on all roads has increased by an average of \(8 \%\) during the past 12 months. By multiplying by \(108 \%=1.08\), estimate the number of vehicles now on a road given the number of vehicles the road carried a year ago was:
a 10000 per day
b 80000 per day
c 148000 per day

2 Prices have increased with inflation by an average of \(3.8 \%\) since the same time last year. Find today's price for an item that one year ago cost:
a \(\$ 200\)
b \(\$ 1.68\)
c \(\$ 345000\)
d \(\$ 9430\)
|722050370|kitty|/Applications/kitty.app|0|
a 10000 per day
b 80000 per day
c 148000 per day
|722050396|kitty|/Applications/kitty.app|0|
7 A clothing store is offering a 15\% discount on all its summer stock. What is the discounted price of an item with original price:
a \(\$ 80\) ?
b \(\$ 48\) ?
c \(\$ 680\) ?
d \(\$ 1.60\) ?
|722050476|kitty|/Applications/kitty.app|0|
Example 17 8 A shoe store is offering a 35\% discount at its end-of-year sale. Find the original price of an item whose discounted price is:
a \(\$ 1820\)
b \(\$ 279.50\)
c \(\$ 1.56\)
d \(\$ 20.80\)
|722050551|kitty|/Applications/kitty.app|0|
\fbox{
|722050645|kitty|/Applications/kitty.app|0|
\makebox[\textwidth]{\fbox{An aside on \textbf{GST}. Let's clear the air and make sure everyone understands what this is: it is a \textbf{G}oods and \textbf{S}ervice \textbf{T}ax. This means that whenever you purchase something in Australia, the government receives 10\% of the cost of your item. This makes sense since the government makes living for you possible in the first place...Who pays for the police and firemen? This tax is always included in the cost of your item. i.e. if the banana says it costs \$1.10, then it must have really been \$1.00 plus 10c in tax. In other countries you should be careful though because they might not have included the tax in the displayed price!}}
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3 Rainfall across one state has decreased over the last five years by about \(24 \%\). By multiplying by \(76 \%=0.76\), estimate, correct to the nearest \(10 \mathrm{~mm}\), the annual rainfall this year at a place where the rainfall five years ago was:
a \(1000 \mathrm{~mm}\)
b \(250 \mathrm{~mm}\)
c \(680 \mathrm{~mm}\)
d \(146 \mathrm{~mm}\)

4 Admissions to different wards of St Luke's Hospital mostly rose from 2006 to 2007, but by quite different percentage amounts. Find the percentage increase or decrease in wards where the numbers during 2006 and 2007, respectively, were:
a 50 and 68
b 120 and 171
c 92 and 77
d 24 and 39
|722051036|kitty|/Applications/kitty.app|0|
3 Rainfall across one state has decreased over the last five years by about \(24 \%\). By multiplying by \(76 \%=0.76\), estimate, correct to the nearest \(10 \mathrm{~mm}\), the annual rainfall this year at a place where the rainfall five years ago was:
a \(1000 \mathrm{~mm}\)
b \(250 \mathrm{~mm}\)
c \(680 \mathrm{~mm}\)
d \(146 \mathrm{~mm}\)

4 Admissions to different wards of St Luke's Hospital mostly rose from 2006 to 2007, but by quite different percentage amounts. Find the percentage increase or decrease in wards where the numbers during 2006 and 2007, respectively, were:
a 50 and 68
b 120 and 171
c 92 and 77
d 24 and 39

6 Radix Holdings Pty Ltd recently issued bonus shares to its shareholders. Each shareholder received an extra \(12 \%\) of the number of shares currently held. Find the original holding of a shareholder who now holds:
a 672 shares
b 4816 shares
c 1000 shares
d 40200 shares
|722051041|kitty|/Applications/kitty.app|0|
9 A research institute is trying to find out how much water Lake Grendel had 1000 years ago. The lake now contains 24000 megalitres, but there are various conflicting theories about the percentage change over the last 1000 years. Find how much water the lake had 1000 years ago, correct to the nearest 10 megalitres, if in the last 1000 years the volume has:
a risen by \(80 \%\)
b fallen by \(28 \%\)
c risen by \(140 \%\)
d fallen by \(4 \%\)
|722051054|kitty|/Applications/kitty.app|0|
\begin{questions}
    \question[3] Traffic on all roads has increased by an average of \(8 \%\) during the past 12 months. By multiplying by \(108 \%=1.08\), estimate the number of vehicles now on a road given the number of vehicles the road carried a year ago was:
    \begin{parts}
        \part 10000 per day
        \begin{solutionordottedlines}[0.5in]
            \(10000 \times 1.08 = 10800\) vehicles per day
        \end{solutionordottedlines}
        \part 80000 per day
        \begin{solutionordottedlines}[0.5in]
            \(80000 \times 1.08 = 86400\) vehicles per day
        \end{solutionordottedlines}
        \part 148000 per day
        \begin{solutionordottedlines}[0.5in]
            \(148000 \times 1.08 = 159840\) vehicles per day
        \end{solutionordottedlines}
    \end{parts}
    \question[4] Prices have increased with inflation by an average of \(3.8 \%\) since the same time last year. Find today's price for an item that one year ago cost:
    \begin{parts}\begin{multicols}{2}
        \part \(\$ 200\)
        \begin{solutionordottedlines}[0.5in]
            \(\$ 200 \times 1.038 = \$ 207.60\)
        \end{solutionordottedlines}
        \part \(\$ 1.68\)
        \begin{solutionordottedlines}[0.5in]
            \(\$ 1.68 \times 1.038 = \$ 1.74\)
        \end{solutionordottedlines}
        \part \(\$ 345000\)
        \begin{solutionordottedlines}[0.5in]
            \(\$ 345000 \times 1.038 = \$ 358110\)
        \end{solutionordottedlines}
        \part \(\$ 9430\)
        \begin{solutionordottedlines}[0.5in]
            \(\$ 9430 \times 1.038 = \$ 9790.14\)
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[4] A clothing store is offering a 15\% discount on all its summer stock. What is the discounted price of an item with original price:
    \begin{parts}\begin{multicols}{2}
        \part \(\$ 80\)
        \begin{solutionordottedlines}[0.5in]
            \(\$ 80 \times 0.85 = \$ 68\)
        \end{solutionordottedlines}
        \part \(\$ 48\)
        \begin{solutionordottedlines}[0.5in]
            \(\$ 48 \times 0.85 = \$ 40.80\)
        \end{solutionordottedlines}
        \part \(\$ 680\)
        \begin{solutionordottedlines}[0.5in]
            \(\$ 680 \times 0.85 = \$ 578\)
        \end{solutionordottedlines}
        \part \(\$ 1.60\)
        \begin{solutionordottedlines}[0.5in]
            \(\$ 1.60 \times 0.85 = \$ 1.36\)
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[4] A shoe store is offering a 35\% discount at its end-of-year sale. Find the original price of an item whose discounted price is:
    \begin{parts}\begin{multicols}{2}
        \part \(\$ 1820\)
        \begin{solutionordottedlines}[0.5in]
            Original Price = \(\$ 1820 / 0.65 = \$ 2800\)
        \end{solutionordottedlines}
        \part \(\$ 279.50\)
        \begin{solutionordottedlines}[0.5in]
            Original Price = \(\$ 279.50 / 0.65 = \$ 430\)
        \end{solutionordottedlines}
        \part \(\$ 1.56\)
        \begin{solutionordottedlines}[0.5in]
            Original Price = \(\$ 1.56 / 0.65 = \$ 2.40\)
        \end{solutionordottedlines}
        \part \(\$ 20.80\)
        \begin{
|722051083|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\begin{questions}
    \question[3] Traffic on all roads has increased by an average of \(8 \%\) during the past 12 months. By multiplying by \(108 \%=1.08\), estimate the number of vehicles now on a road given the number of vehicles the road carried a year ago was:
    \begin{parts}
        \part 10000 per day
        \begin{solutionordottedlines}[0.5in]
            \(10000 \times 1.08 = 10800\) vehicles per day
        \end{solutionordottedlines}
        \part 80000 per day
        \begin{solutionordottedlines}[0.5in]
            \(80000 \times 1.08 = 86400\) vehicles per day
        \end{solutionordottedlines}
        \part 148000 per day
        \begin{solutionordottedlines}[0.5in]
            \(148000 \times 1.08 = 159840\) vehicles per day
        \end{solutionordottedlines}
    \end{parts}
    \question[4] Prices have increased with inflation by an average of \(3.8 \%\) since the same time last year. Find today's price for an item that one year ago cost:
    \begin{parts}\begin{multicols}{2}
        \part \(\$ 200\)
        \begin{solutionordottedlines}[0.5in]
            \(\$ 200 \times 1.038 = \$ 207.60\)
        \end{solutionordottedlines}
        \part \(\$ 1.68\)
        \begin{solutionordottedlines}[0.5in]
            \(\$ 1.68 \times 1.038 = \$ 1.74\)
        \end{solutionordottedlines}
        \part \(\$ 345000\)
        \begin{solutionordottedlines}[0.5in]
            \(\$ 345000 \times 1.038 = \$ 358110\)
        \end{solutionordottedlines}
        \part \(\$ 9430\)
        \begin{solutionordottedlines}[0.5in]
            \(\$ 9430 \times 1.038 = \$ 9790.14\)
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[4] A clothing store is offering a 15\% discount on all its summer stock. What is the discounted price of an item with original price:
    \begin{parts}\begin{multicols}{2}
        \part \(\$ 80\)
        \begin{solutionordottedlines}[0.5in]
            \(\$ 80 \times 0.85 = \$ 68\)
        \end{solutionordottedlines}
        \part \(\$ 48\)
        \begin{solutionordottedlines}[0.5in]
            \(\$ 48 \times 0.85 = \$ 40.80\)
        \end{solutionordottedlines}
        \part \(\$ 680\)
        \begin{solutionordottedlines}[0.5in]
            \(\$ 680 \times 0.85 = \$ 578\)
        \end{solutionordottedlines}
        \part \(\$ 1.60\)
        \begin{solutionordottedlines}[0.5in]
            \(\$ 1.60 \times 0.85 = \$ 1.36\)
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[4] A shoe store is offering a 35\% discount at its end-of-year sale. Find the original price of an item whose discounted price is:
    \begin{parts}\begin{multicols}{2}
        \part \(\$ 1820\)
        \begin{solutionordottedlines}[0.5in]
            Original Price = \(\$ 1820 / 0.65 = \$ 2800\)
        \end{solutionordottedlines}
        \part \(\$ 279.50\)
        \begin{solutionordottedlines}[0.5in]
            Original Price = \(\$ 279.50 / 0.65 = \$ 430\)
        \end{solutionordottedlines}
        \part \(\$ 1.56\)
        \begin{solutionordottedlines}[0.5in]
            Original Price = \(\$ 1.56 / 0.65 = \$ 2.40\)
        \end{solutionordottedlines}
        \part \(\$ 20.80\)
        \begin{solutionordottedlines}[0.5in]
            Original Price = \(\$ 20.80 / 0.65 = \$ 32\)
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
\end{questions}
|722051124|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
    \begin{questions}
        \question[3] Traffic on all roads has increased by an average of \(8 \%\) during the past 12 months. By multiplying by \(108 \%=1.08\), estimate the number of vehicles now on a road given the number of vehicles the road carried a year ago was:
        \begin{parts}
            \part 10000 per day
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part 80000 per day
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part 148000 per day
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \end{parts}
        \question[4] Prices have increased with inflation by an average of \(3.8 \%\) since the same time last year. Find today's price for an item that one year ago cost:
        \begin{parts}\begin{multicols}{2}
            \part \(\$ 200\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(\$ 1.68\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(\$ 345000\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(\$ 9430\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[4] A clothing store is offering a 15\% discount on all its summer stock. What is the discounted price of an item with original price:
        \begin{parts}\begin{multicols}{2}
            \part \(\$ 80\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(\$ 48\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(\$ 680\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(\$ 1.60\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[4] A shoe store is offering a 35\% discount at its end-of-year sale. Find the original price of an item whose discounted price is:
        \begin{parts}\begin{multicols}{2}
            \part \(\$ 1820\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(\$ 279.50\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(\$ 1.56\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(\$ 20.80\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \end{questions}
|722051126|kitty|/Applications/kitty.app|0|
The population of Abelsburg in the census three years ago was 46430 . In the three years after the census, however, its population has risen by \(6.2 \%, 8.5 \%\) and \(13.1 \%\), respectively.

a What was its population one year after the census?

b What was its population two years after the census?

c What is its population now, three years after the census?

d What was the percentage increase in population over the three years, correct to the nearest \(0.1 \%\) ?
|722052325|kitty|/Applications/kitty.app|0|
a One year after the census, the population was \(106.2 \%\) of its original value.

Hence population after one year \(=46430 \times 1.062\)

\[
\approx 49309 \text {, correct to the nearest whole number }
\]
|722052404|kitty|/Applications/kitty.app|0|
b Two years afterwards, the population was \(108.5 \%\) of its value one year afterwards.

Hence population after two years \(=(46430 \times 1.062) \times 1.085\)

\[
\approx 53500 \text {, correct to the nearest whole number }
\]

Note: Do not use the approximation from part \(\mathbf{a}\) to calculate part \(\mathbf{b}\). Either start the calculation again, or use the unrounded value from part \(\mathbf{a}\).
|722052421|kitty|/Applications/kitty.app|0|
c Three years afterwards, the population was \(113.1 \%\) of its value two years afterwards.

Hence population after three years \(=(46430 \times 1.062 \times 1.085) \times 1.131\)

\[
\approx 60508 \text {, correct to the nearest whole number }
\]
|722052435|kitty|/Applications/kitty.app|0|

\[
\begin{aligned}
& =\text { original population } \times 1.062 \times 1.085 \times 1.131 \\
& \approx \text { original population } \times 1.303 \\
& \approx \text { original population } \times 130.3 \%
\end{aligned}
\]

Hence the population has increased over the three years by about \(30.3 \%\).

Note: The percentage increase of \(30.3 \%\) is significantly larger than the sum of the three percentage increases,

\[
6.2 \%+8.5 \%+13.1 \%=27.8 \% .
\]

Note: The answer to part \(\mathbf{d}\) does not depend on what the original population was.
|722052447|kitty|/Applications/kitty.app|0|
Teresa invested \(\$ 75000\) from her inheritance in a mining company that has not been very successful. In the first year, she lost \(55 \%\) of the money, and in the second year, she lost \(72 \%\) of what remained.

a How much does she have left after one year?

b How much does she have left after two years?

c What percentage of the original inheritance has she lost over the two years?
|722052471|kitty|/Applications/kitty.app|0|
a One year later, the percentage remaining was \(100 \%-55 \%=45 \%\).

Hence amount left after one year \(=75000 \times 0.45\)

\[
=\$ 33750
\]

b Two years later, she had \(100 \%-72 \%=28 \%\) of what she had after one year.

Hence amount left after two years \(=75000 \times 0.45 \times 0.28\)

\[
=\$ 9450
\]

c Amount left after two years \(=\) original investment \(\times 0.45 \times 0.28\)

\[
\begin{aligned}
& =\text { original investment } \times 0.126 \\
& =\text { original investment } \times 12.6 \%
\end{aligned}
\]

So she has lost \(100 \%-12.6 \%=87.4 \%\) of her investment over the two years.

\section*{Combinations of increases and decreases}
Some problems involve both increases and decreases. They can be solved in exactly the same way.
|722052517|kitty|/Applications/kitty.app|0|
b Two years later, she had \(100 \%-72 \%=28 \%\) of what she had after one year.

Hence amount left after two years \(=75000 \times 0.45 \times 0.28\)

\[
=\$ 9450
\]

c Amount left after two years \(=\) original investment \(\times 0.45 \times 0.28\)

\[
\begin{aligned}
& =\text { original investment } \times 0.126 \\
& =\text { original investment } \times 12.6 \%
\end{aligned}
\]
|722052528|kitty|/Applications/kitty.app|0|
c Amount left after two years \(=\) original investment \(\times 0.45 \times 0.28\)

\[
\begin{aligned}
& =\text { original investment } \times 0.126 \\
& =\text { original investment } \times 12.6 \%
\end{aligned}
\]
|722052535|kitty|/Applications/kitty.app|0|
So she has lost \(100 \%-12.6 \%=87.4 \%\) of her investment over the two years.
|722052585|kitty|/Applications/kitty.app|0|
\section*{Combinations of increases and decreases}
|722052604|kitty|/Applications/kitty.app|0|



Some problems involve both increases and decreases. They can be solved in exactly the same way.
|722052612|kitty|/Applications/kitty.app|0|
The volume of water in the Welcome Dam has varied considerably over the last three years. During the first year the volume rose by \(27 \%\), then it fell \(43 \%\) during the second year, and it rose \(16 \%\) in the third year.

a What is the percentage increase or decrease over the three years, correct to the nearest \(1 \%\) ?

b If there were 366500 megalitres of water in the dam three years ago, how much water is in the dam now, correct to the nearest 500 megalitres?

\section*{Solution}
a Final volume \(=\) original volume \(\times 1.27 \times 0.57 \times 1.16\)

\[
\approx \text { original volume } \times 0.84
\]

Since \(0.84<1\), the volume has decreased. The percentage decrease is about \(100 \%-84 \%=16 \%\) over the three years.

b Final volume \(=\) original volume \(\times 1.27 \times 0.57 \times 1.16\)

\[
\begin{aligned}
& =366500 \times 1.27 \times 0.57 \times 1.16 \\
& \approx 308000 \text { megalitres, correct to the nearest } 500 \text { megalitres. }
\end{aligned}
\]

This time the sum of the percentages is \(27 \%-43 \%+16 \%=0 \%\), but the volume of water has changed.
|722052660|kitty|/Applications/kitty.app|0|
a Final volume \(=\) original volume \(\times 1.27 \times 0.57 \times 1.16\)

\[
\approx \text { original volume } \times 0.84
\]

Since \(0.84<1\), the volume has decreased. The percentage decrease is about \(100 \%-84 \%=16 \%\) over the three years.
|722052683|kitty|/Applications/kitty.app|0|
b Final volume \(=\) original volume \(\times 1.27 \times 0.57 \times 1.16\)

\[
\begin{aligned}
& =366500 \times 1.27 \times 0.57 \times 1.16 \\
& \approx 308000 \text { megalitres, correct to the nearest } 500 \text { megalitres. }
\end{aligned}
\]

This time the sum of the percentages is \(27 \%-43 \%+16 \%=0 \%\), but the volume of water has changed.
|722052703|kitty|/Applications/kitty.app|0|
1 Oranges used to cost \(\$ 2.80\) per kg, but the price has increased by \(5 \%, 10 \%\) and \(12 \%\) in three successive years. Multiply by \(1.05 \times 1.1 \times 1.12\) to find their price now.

2 The dividend per share in the Electron Computer Software Company has risen over the last four years by 10\%, 15\%, 5\% and 12\%, respectively. Find the latest dividend received by a shareholder whose dividend four years ago was:
a \(\$ 1000\)
b \(\$ 1678\)
c \(\$ 28.46\)
d \(\$ 512.21\)

3 Rates in Bullimbamba Shire have risen 7\% every year for the last seven years. Find the rates now payable by a landowner whose rates seven years ago were:
a \(\$ 1000\)
b \(\$ 346\)
c \(\$ 2566.86\)
d \(\$ 788.27\)

4 A tree, whose original foliage was estimated to have a mass of \(1500 \mathrm{~kg}\), lost \(20 \%\) of its foliage in a storm, then lost \(15 \%\) of what was left in a storm the next day, then lost \(40 \%\) of what was left in a third storm. Estimate the remaining mass of foliage.
|722052732|kitty|/Applications/kitty.app|0|
5 Shares in the Metropolitan Brickworks have been falling by \(23 \%\) per year for the last five years. Find the present worth of a parcel of shares whose original worth five years ago was:
a \(\$ 1000\)
b \(\$ 120000\)
c \(\$ 25660\)
d \(\$ 3860000\)

6 a A shirt is discounted by \(50 \%\) and the resulting price is then increased by \(50 \%\). By what percentage is the price increased or decreased from its original value?

b The price of a shirt is increased by \(50 \%\) and the resulting price is then decreased by \(50 \%\). By what percentage is the price increased or decreased from its original value?

c Can you explain the relationship between your answers to parts \(\mathbf{a}\) and \(\mathbf{b}\) ?

Example 22 7 A book shop has a \(50 \%\) sale on all stock, and has a container of books with the sale price reduced by a further factor of \(16 \%\).

a What was the total discount on each book in the container?

b If a book in the container is now selling for \(\$ 10.50\), what was its original price?
|722052875|kitty|/Applications/kitty.app|0|
11 A particular strain of bacteria increases its population on a certain prepared Petri dish by \(34 \%\) every hour. Calculate the size of the original population four hours ago if there are now 56000 bacteria.
|722052883|kitty|/Applications/kitty.app|0|
\begin{questions}
    \question[1] Oranges used to cost \(\$ 2.80\) per kg, but the price has increased by \(5 \%, 10 \%\) and \(12 \%\) in three successive years. Multiply by \(1.05 \times 1.1 \times 1.12\) to find their price now.
    \begin{solutionordottedlines}[1in]
        New price = \(\$ 2.80 \times 1.05 \times 1.1 \times 1.12 \approx \$ 3.50\) per kg
    \end{solutionordottedlines}
    \question[4] The dividend per share in the Electron Computer Software Company has risen over the last four years by 10\%, 15\%, 5\% and 12\%, respectively. Find the latest dividend received by a shareholder whose dividend four years ago was:
    \begin{parts}\begin{multicols}{2}
        \part \(\$ 1000\)
        \begin{solutionordottedlines}[0.5in]
            Latest dividend = \(\$ 1000 \times 1.1 \times 1.15 \times 1.05 \times 1.12 \approx \$ 1408.62\)
        \end{solutionordottedlines}
        \part \(\$ 1678\)
        \begin{solutionordottedlines}[0.5in]
            Latest dividend = \(\$ 1678 \times 1.1 \times 1.15 \times 1.05 \times 1.12 \approx \$ 2363.62\)
        \end{solutionordottedlines}
        \part \(\$ 28.46\)
        \begin{solutionordottedlines}[0.5in]
            Latest dividend = \(\$ 28.46 \times 1.1 \times 1.15 \times 1.05 \times 1.12 \approx \$ 40.15\)
        \end{solutionordottedlines}
        \part \(\$ 512.21\)
        \begin{solutionordottedlines}[0.5in]
            Latest dividend = \(\$ 512.21 \times 1.1 \times 1.15 \times 1.05 \times 1.12 \approx \$ 722.15\)
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[4] Rates in Bullimbamba Shire have risen 7\% every year for the last seven years. Find the rates now payable by a landowner whose rates seven years ago were:
    \begin{parts}\begin{multicols}{2}
        \part \(\$ 1000\)
        \begin{solutionordottedlines}[0.5in]
            Current rates = \(\$ 1000 \times (1.07)^7 \approx \$ 1605.78\)
        \end{solutionordottedlines}
        \part \(\$ 346\)
        \begin{solutionordottedlines}[0.5in]
            Current rates = \(\$ 346 \times (1.07)^7 \approx \$ 556.14\)
        \end{solutionordottedlines}
        \part \(\$ 2566.86\)
        \begin{solutionordottedlines}[0.5in]
            Current rates = \(\$ 2566.86 \times (1.07)^7 \approx \$ 4121.98\)
        \end{solutionordottedlines}
        \part \(\$ 788.27\)
        \begin{solutionordottedlines}[0.5in]
            Current rates = \(\$ 788.27 \times (1.07)^7 \approx \$ 1265.76\)
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[2] A tree, whose original foliage was estimated to have a mass of \(1500 \mathrm{~kg}\), lost \(20 \%\) of its foliage in a storm, then lost \(15 \%\) of what was left in a storm the next day, then lost \(40 \%\) of what was left in a third storm. Estimate the remaining mass of foliage.
    \begin{solutionordottedlines}[1in]
        Remaining mass = \(1500 \times 0.8 \times 0.85 \times 0.6 \approx 612 \mathrm{~kg}\)
    \end{solutionordottedlines}
\end{questions}
|722053169|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
    \begin{questions}
        \question[1] Oranges used to cost \(\$ 2.80\) per kg, but the price has increased by \(5 \%, 10 \%\) and \(12 \%\) in three successive years. Multiply by \(1.05 \times 1.1 \times 1.12\) to find their price now.
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \question[4] The dividend per share in the Electron Computer Software Company has risen over the last four years by 10\%, 15\%, 5\% and 12\%, respectively. Find the latest dividend received by a shareholder whose dividend four years ago was:
        \begin{parts}\begin{multicols}{2}
            \part \(\$ 1000\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(\$ 1678\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(\$ 28.46\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(\$ 512.21\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[4] Rates in Bullimbamba Shire have risen 7\% every year for the last seven years. Find the rates now payable by a landowner whose rates seven years ago were:
        \begin{parts}\begin{multicols}{2}
            \part \(\$ 1000\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(\$ 346\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(\$ 2566.86\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part \(\$ 788.27\)
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[2] A tree, whose original foliage was estimated to have a mass of \(1500 \mathrm{~kg}\), lost \(20 \%\) of its foliage in a storm, then lost \(15 \%\) of what was left in a storm the next day, then lost \(40 \%\) of what was left in a third storm. Estimate the remaining mass of foliage.
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
    \end{questions}
|722053175|kitty|/Applications/kitty.app|0|
    \subsection{Exercises}
|722053405|kitty|/Applications/kitty.app|0|
\begin{examplebox}
    \subsection{Examples:}
    \begin{questions}

    \end{questions}


\end{examplebox}

\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}

    \end{questions}

\end{exercisebox}
|722053408|kitty|/Applications/kitty.app|0|
Gail has invested \(\$ 100000\) for six years with the Mountain Bank. The bank pays her interest at the rate of \(7.5 \%\) p.a., compounded annually.

a How much will the investment be worth at the end of one year?

b How much will the investment be worth at the end of two years?

c How much will the investment be worth at the end of six years?

d What is the percentage increase on her original investment at the end of six years?

e What is the total interest earned over the six years?

f What would the simple interest on the investment have been, assuming the same interest rate of \(7.5 \%\) p.a.?

\section*{Solution}
Each year the investment is worth \(107.5 \%\) of its value the previous year.

a Amount at the end of one year \(=100000 \times 1.075\)

\[
=\$ 107500
\]

b Amount at the end of two years \(=100000 \times 1.075 \times 1.075\)

\[
\begin{aligned}
& =100000 \times(1.075)^{2} \\
& =\$ 115562.50
\end{aligned}
\]

(continued over page)
c Amount at the end of six years \(=100000 \times(1.075)^{6}\)

\[
\approx \$ 154330.15
\]

d Final amount \(=\) original amount \(\times(1.075)^{6}\)

\[
\approx \text { original amount } \times 1.5433
\]

So the total increase over six years is about \(54.33 \%\).
e Total interest \(\approx 154330.15-100000\)

\[
=\$ 54330.15
\]

f Simple interest \(=P R T\)

\[
\begin{aligned}
& =100000 \times 0.075 \times 6 \\
& =\$ 45000
\end{aligned}
\]

Note: Compound interest for two or more years is always greater than simple interest for two or more years.
|722053444|kitty|/Applications/kitty.app|0|
Each year the investment is worth \(107.5 \%\) of its value the previous year.

a Amount at the end of one year \(=100000 \times 1.075\)

\[
=\$ 107500
\]
|722053493|kitty|/Applications/kitty.app|0|
b Amount at the end of two years \(=100000 \times 1.075 \times 1.075\)

\[
\begin{aligned}
& =100000 \times(1.075)^{2} \\
& =\$ 115562.50
\end{aligned}
\]

|722053498|kitty|/Applications/kitty.app|0|
c Amount at the end of six years \(=100000 \times(1.075)^{6}\)

\[
\approx \$ 154330.15
\]
|722053504|kitty|/Applications/kitty.app|0|
d Final amount \(=\) original amount \(\times(1.075)^{6}\)

\[
\approx \text { original amount } \times 1.5433
\]

So the total increase over six years is about \(54.33 \%\).
|722053513|kitty|/Applications/kitty.app|0|
e Total interest \(\approx 154330.15-100000\)

\[
=\$ 54330.15
\]
|722053518|kitty|/Applications/kitty.app|0|
f Simple interest \(=P R T\)

\[
\begin{aligned}
& =100000 \times 0.075 \times 6 \\
& =\$ 45000
\end{aligned}
\]

Note: Compound interest for two or more years is always greater than simple interest for two or more years.
|722053524|kitty|/Applications/kitty.app|0|

\section*{Solution}

(continued over page)



|722053580|kitty|/Applications/kitty.app|0|
    \
|722053583|kitty|/Applications/kitty.app|0|
Hussain is setting up a plumbing business and needs to borrow \(\$ 150000\) from a bank to buy a truck and other equipment. The bank will charge him interest of \(11 \%\) p.a., compounded annually. Hussain will pay the whole loan off all at once four years later.

a How much will Hussain owe the bank at the end of four years?

b What is the percentage increase in the money owed at the end of four years?

c What is the total interest that Hussain will pay on the loan?

d What would the simple interest on the loan have been, assuming the same interest rate of \(11 \%\) p.a.?

\section*{Solution}
Each year Hussain owes \(111 \%\) of what he owed the previous year.

a Amount at the end of four years \(=150000 \times(1.11)^{4}\)

\[
\approx \$ 227710.56
\]

b Final amount \(=\) original amount \(\times(1.11)^{4}\)

\[
\approx \text { original amount } \times 1.5181
\]

So the total increase over four years is about \(51.81 \%\).

(continued over page)
c \(\quad\) Total interest \(\approx 227710.56-150000\)

\[
=\$ 77710.56
\]

d Simple interest \(=P R T\)

\[
\begin{aligned}
& =150000 \times 0.11 \times 4 \\
& =66000
\end{aligned}
\]

Note: Making no repayments on a loan that is accruing compound interest is a risky business practice because, as this example makes clear, the amount owing grows with increasing rapidity as time goes on. This is particularly relevant to credit card debt.
|722053617|kitty|/Applications/kitty.app|0|
Each year Hussain owes \(111 \%\) of what he owed the previous year.

a Amount at the end of four years \(=150000 \times(1.11)^{4}\)

\[
\approx \$ 227710.56
\]

|722053648|kitty|/Applications/kitty.app|0|
b Final amount \(=\) original amount \(\times(1.11)^{4}\)

\[
\approx \text { original amount } \times 1.5181
\]

So the total increase over four years is about \(51.81 \%\).

|722053653|kitty|/Applications/kitty.app|0|
c \(\quad\) Total interest \(\approx 227710.56-150000\)

\[
=\$ 77710.56
\]
|722053657|kitty|/Applications/kitty.app|0|
d Simple interest \(=P R T\)

\[
\begin{aligned}
& =150000 \times 0.11 \times 4 \\
& =66000
\end{aligned}
\]

Note: Making no repayments on a loan that is accruing compound interest is a risky business practice because, as this example makes clear, the amount owing grows with increasing rapidity as time goes on. This is particularly relevant to credit card debt.
|722053663|kitty|/Applications/kitty.app|0|
(continued over page)
|722053688|kitty|/Applications/kitty.app|0|
Eleni wants to borrow money for three years to start a business, and then pay all the money back, with interest, at the end of that time. The bank will not allow her final debt, including interest, to exceed \(\$ 300000\). Interest is \(9 \%\) p.a., compounded annually. What is the maximum amount that Eleni can borrow?

\section*{Solution}
Each year Eleni will owe \(109 \%\) of what she owed the previous year.

Hence \(\quad\) final debt \(=\) original debt \(\times 1.09 \times 1.09 \times 1.09\)

\[
\text { final debt }=\text { original debt } \times(1.09)^{3}
\]

Reversing this, original debt \(=\) final debt \(\div(1.09)^{3}\)

\[
\begin{aligned}
& =300000 \div(1.09)^{3} \\
& \approx \$ 231655
\end{aligned}
\]
|722053702|kitty|/Applications/kitty.app|0|
Each year Eleni will owe \(109 \%\) of what she owed the previous year.

Hence \(\quad\) final debt \(=\) original debt \(\times 1.09 \times 1.09 \times 1.09\)

\[
\text { final debt }=\text { original debt } \times(1.09)^{3}
\]

Reversing this, original debt \(=\) final debt \(\div(1.09)^{3}\)

\[
\begin{aligned}
& =300000 \div(1.09)^{3} \\
& \approx \$ 231655
\end{aligned}
\]

|722053714|kitty|/Applications/kitty.app|0|
1 a Christine invested \(\$ 100000\) for six years at 5\% p.a. interest, compounded annually.

i By multiplying by 1.05 , find the value of the investment after one year.

ii By multiplying by \((1.05)^{2}\), find the value of the investment after two years.

iii By multiplying by \((1.05)^{6}\), find the value of the investment after six years.

iv Find the percentage increase in the investment over the six years.

\(v\) Find the total interest earned over the six years.

b Find the simple interest on the principal of \(\$ 100000\) over the six years at the same rate of \(5 \%\) p.a.
|722133615|kitty|/Applications/kitty.app|0|
        \question 
|722133666|kitty|/Applications/kitty.app|0|
\(v\) Find the total interest earned over the six years.
|722133670|kitty|/Applications/kitty.app|0|
2 a Gary has borrowed \(\$ 200000\) for six years at \(8 \%\) p.a. interest, compounded annually, in order to start his indoor decorating business. He intends to pay the whole amount back, plus interest, at the end of the six years.

i Find the amount owing after one year.

ii Find the amount owing after two years.

iii Find the amount owing after six years.

iv Find the percentage increase in the loan over the six years.

\(\mathbf{v}\) Find the total interest charged over the six years.
|722133760|kitty|/Applications/kitty.app|0|
b Find the simple interest on the principal of \(\$ 200000\) over the six years at the same rate of \(8 \%\) p.a.
|722133895|kitty|/Applications/kitty.app|0|

3 A couple take out a housing loan of \(\$ 320000\) over a period of 20 years. They make no repayments over the 20-year period of the loan. Compound interest is payable at \(6 \frac{1}{2} \%\) p.a., compounded annually. How much would they owe at the end of the 20-year period, and what is the total percentage increase?

4 The population of a city increases annually at a compound rate of \(3.2 \%\) for five years. If the population is 21000 initially, what is it at the end of the five-year period, and what is the total percentage increase?
|722133914|kitty|/Applications/kitty.app|0|
5 a Find the compound interest on \(\$ 1000\) at 5\% p.a., compounded annually for 200 years.

b Find the simple interest on \(\$ 1000\) at \(5 \%\) p.a. for 200 years.

\(6 \$ 10000\) is borrowed for five years and compound interest at \(10 \%\) p.a. is charged by the lender.

a How much money is owed to the lender after the five-year period?

b How much of this amount is interest?

7 Money borrowed at \(8 \%\) p.a. interest, compounded annually, grew to \(\$ 100000\) in four years.

a Find the total percentage increase.

b Hence find the original amount invested.
|722134041|kitty|/Applications/kitty.app|0|

8 Suzette wants to invest a sum of money now so that it will grow to \(\$ 180000\) in 10 years' time. How much should she invest now, given that the interest rate is \(6 \%\) compounded annually?

9 A bank offers \(8 \%\) p.a. compound interest. How much needs to be invested if the investment is to be worth \(\$ 100000\) in:
a 10 years?
b 20 years?
c 25 years?
d 100 years?
|722134052|kitty|/Applications/kitty.app|0|
\begin{questions}
    \question[5] Christine invested \(\$ 100000\) for six years at 5\% p.a. interest, compounded annually.
    \begin{parts}
        \part By multiplying by 1.05, find the value of the investment after one year.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 100000 \times 1.05 = \$ 105000\)
        \end{solutionordottedlines}
        \part By multiplying by \((1.05)^{2}\), find the value of the investment after two years.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 100000 \times (1.05)^{2} = \$ 110250\)
        \end{solutionordottedlines}
        \part By multiplying by \((1.05)^{6}\), find the value of the investment after six years.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 100000 \times (1.05)^{6} \approx \$ 134009.56\)
        \end{solutionordottedlines}
        \part Find the percentage increase in the investment over the six years.
        \begin{solutionordottedlines}[0.5in]
            Percentage increase = \(\frac{\$ 134009.56 - \$ 100000}{\$ 100000} \times 100 \approx 34.01\%\)
        \end{solutionordottedlines}
        \part Find the total interest earned over the six years.
        \begin{solutionordottedlines}[0.5in]
            Total interest = \(\$ 134009.56 - \$ 100000 = \$ 34009.56\)
        \end{solutionordottedlines}
    \end{parts}
    \question[1] Find the simple interest on the principal of \(\$ 100000\) over the six years at the same rate of \(5 \%\) p.a.
    \begin{solutionordottedlines}[1in]
        Simple interest = \(\$ 100000 \times 5\% \times 6 = \$ 30000\)
    \end{solutionordottedlines}
    \question[5] Gary has borrowed \(\$ 200000\) for six years at \(8 \%\) p.a. interest, compounded annually, in order to start his indoor decorating business. He intends to pay the whole amount back, plus interest, at the end of the six years.
    \begin{parts}
        \part Find the amount owing after one year.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 200000 \times 1.08 = \$ 216000\)
        \end{solutionordottedlines}
        \part Find the amount owing after two years.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 200000 \times (1.08)^{2} \approx \$ 233280\)
        \end{solutionordottedlines}
        \part Find the amount owing after six years.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 200000 \times (1.08)^{6} \approx \$ 318111.47\)
        \end{solutionordottedlines}
        \part Find the percentage increase in the loan over the six years.
        \begin{solutionordottedlines}[0.5in]
            Percentage increase = \(\frac{\$ 318111.47 - \$ 200000}{\$ 200000} \times 100 \approx 59.06\%\)
        \end{solutionordottedlines}
        \part Find the total interest charged over the six years.
        \begin{solutionordottedlines}[0.5in]
            Total interest = \(\$ 318111.47 - \$ 200000 = \$ 118111.47\)
        \end{solutionordottedlines}
    \end{parts}
    \question[1] Find the simple interest on the principal of \(\$ 200000\) over the six years at the same rate of \(8 \%\) p.a.
    \begin{solutionordottedlines}[1in]
        Simple interest = \(\$ 200000 \times 8\% \times 6 = \$ 96000\)
    \end{solutionordottedlines}
    \question[1] A couple take out a housing loan of \(\$ 320000\) over a period of 20 years. They make no repayments over the 20-year period of the loan. Compound interest is payable at \(6 \frac{1}{2} \%\) p.a., compounded annually. How much would they owe at the end of the 20-year period, and what is the total percentage increase?
    \begin{solutionordottedlines}[0.5in]
        Owing amount = \(\$ 320000 \times (1 + 6.5\%)^{20} \approx \$ 1126970.22\) \\
        Percentage increase = \(\frac{\$ 1126970.22 - \$ 320000}{\$ 320000} \times 100 \approx 252.18\%\)
    \end{solutionordottedlines}
    \question[1] The population of a city increases annually at a compound rate of \(3.2 \%\) for five years. If the population is 21000 initially, what is it at the end of the five-year period, and what is the total percentage increase?
    \begin{solutionordottedlines}[0.5in]
        Final population = \(21000 \times (1 + 3.2\%)^{5} \approx 24423.60\) \\
        Percentage increase = \(\frac{24423.60 - 21000}{21000} \times 100 \approx 16.3\%\)
    \end{solutionordottedlines}
    \question[1] Suzette wants to invest a sum of money now so that it will grow to \(\$ 180000\) in 10 years' time. How much should she invest now, given that the interest rate is \(6 \%\) compounded annually?
    \begin{solutionordottedlines}[0.5in]
        Present value = \(\$ 180000 / (1.06)^{10} \approx \$ 100927.08\)
    \end{solutionordottedlines}
    \question[4] A bank offers \(8 \%\) p.a. compound interest. How much needs to be invested if the investment is to be worth \(\$ 100000\) in:
    \begin{parts}\begin{multicols}{2}
        \part 10 years?
        \begin{solutionordottedlines}[0.5in]
            Present value = \(\$ 100000 / (1.08)^{10} \approx \$ 46319.38\)
        \end{solutionordottedlines}
        \part 20 years?
        \begin{solutionordottedlines}[0.5in]
            Present value = \(\$ 100000 / (1.08)^{20} \approx \$ 21454.70\)
        \end{solutionordottedlines}
        \part 25 years?
        \begin{solutionordottedlines}[0.5in]
            Present value = \(\$ 100000 / (1.08)^{25} \approx \$ 14693.57\)
        \end{solutionordottedlines}
        \part 100 years?
        \begin{solutionordottedlines}[0.5in]
            Present value = \(\$ 100000 / (1.08)^{100} \approx \$ 214.55\)
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
\end{questions}
|722134355|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
    \begin{questions}
        \question[5] Christine invested \(\$ 100000\) for six years at 5\% p.a. interest, compounded annually.
        \begin{parts}
            \part By multiplying by 1.05 , find the value of the investment after one year.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part By multiplying by \((1.05)^{2}\), find the value of the investment after two years.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part By multiplying by \((1.05)^{6}\), find the value of the investment after six years.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part Find the percentage increase in the investment over the six years.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part Find the total interest earned over the six years.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \end{parts}
        \question[1] Find the simple interest on the principal of \(\$ 100000\) over the six years at the same rate of \(5 \%\) p.a.
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \question[5] Gary has borrowed \(\$ 200000\) for six years at \(8 \%\) p.a. interest, compounded annually, in order to start his indoor decorating business. He intends to pay the whole amount back, plus interest, at the end of the six years.
        \begin{parts}
            \part Find the amount owing after one year.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part Find the amount owing after two years.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part Find the amount owing after six years.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part Find the percentage increase in the loan over the six years.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part Find the total interest charged over the six years.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \end{parts}
        \question[1] Find the simple interest on the principal of \(\$ 200000\) over the six years at the same rate of \(8 \%\) p.a.
        \question[1] A couple take out a housing loan of \(\$ 320000\) over a period of 20 years. They make no repayments over the 20-year period of the loan. Compound interest is payable at \(6 \frac{1}{2} \%\) p.a., compounded annually. How much would they owe at the end of the 20-year period, and what is the total percentage increase?
        \begin{solutionordottedlines}[0.5in]
        \end{solutionordottedlines}
        \question[1] The population of a city increases annually at a compound rate of \(3.2 \%\) for five years. If the population is 21000 initially, what is it at the end of the five-year period, and what is the total percentage increase?
        \begin{solutionordottedlines}[0.5in]
        \end{solutionordottedlines}
        \question[1] Suzette wants to invest a sum of money now so that it will grow to \(\$ 180000\) in 10 years' time. How much should she invest now, given that the interest rate is \(6 \%\) compounded annually?
        \begin{solutionordottedlines}[0.5in]
        \end{solutionordottedlines}
        \question[4] A bank offers \(8 \%\) p.a. compound interest. How much needs to be invested if the investment is to be worth \(\$ 100000\) in:
        \begin{parts}\begin{multicols}{2}
            \part 10 years?
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part 20 years?
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part 25 years?
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part 100 years?
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \end{questions}
|722134364|kitty|/Applications/kitty.app|0|
10 A man now owes the bank \(\$ 56000\), after having taken out a loan five years ago. Find the original amount that he borrowed if the rate of interest per annum, compounded annually, has been:
a \(3 \%\)
b \(5.6 \%\)
c \(9.25 \%\)
d \(15 \%\)
|722134572|kitty|/Applications/kitty.app|0|
12 Ms Smith invested \(\$ 50000\) at \(6 \%\) p.a. interest, compounded annually, for four years. The tax department wants to know exactly how much interest she earned each year. Calculate these figures for Ms Smith.

13 Mrs Robinson has taken out a loan of \(\$ 300000\) at \(8 \%\) p.a. interest, compounded annually, for four years. She wants to know exactly how much interest she will be charged each year so that she can include it as a tax deduction in her income tax return. Calculate these figures for Mrs Robinson.

|722134585|kitty|/Applications/kitty.app|0|
The Medicine Home Delivery Company bought a car four years ago for \(\$ 40000\), and assumed that the value of the car would depreciate at \(20 \%\) p.a.

a What value did the car have at the end of two years?

b What value does the car have now, after four years?

c What is the percentage decrease in value over the four years?

d What is the average depreciation on the car over the four years? (Express your answer in dollars per year)
|722134617|kitty|/Applications/kitty.app|0|
The value each year is taken to be \(100 \%-20 \%=80 \%\) of the value in the previous year.

a Value at the end of two years \(=40000 \times 0.80 \times 0.80\)

\[
\begin{aligned}
& =40000 \times(0.80)^{2} \\
& =\$ 25600
\end{aligned}
\]

b Value at the end of four years \(=40000 \times 0.80 \times 0.80 \times 0.80 \times 0.80\)

\[
\begin{aligned}
& =40000 \times(0.80)^{4} \\
& =\$ 16384
\end{aligned}
\]

(continued over page)
c \(\quad\) Final value \(=\) original value \(\times(0.80)^{4}\)

\[
=\text { original value } \times 0.4096
\]

Hence the percentage decrease over four years is \(100 \%-40.96 \%=59.04 \%\)

d Depreciation over four years \(=40000-16384\)

\[
=\$ 23616
\]

Average depreciation per year \(=23616 \div 4\)

\[
=\$ 5904 \text { per year }
\]
|722134688|kitty|/Applications/kitty.app|0|
b Value at the end of four years \(=40000 \times 0.80 \times 0.80 \times 0.80 \times 0.80\)

\[
\begin{aligned}
& =40000 \times(0.80)^{4} \\
& =\$ 16384
\end{aligned}
\]

(continued over page)
c \(\quad\) Final value \(=\) original value \(\times(0.80)^{4}\)

\[
=\text { original value } \times 0.4096
\]

Hence the percentage decrease over four years is \(100 \%-40.96 \%=59.04 \%\)

d Depreciation over four years \(=40000-16384\)

\[
=\$ 23616
\]

Average depreciation per year \(=23616 \div 4\)

\[
=\$ 5904 \text { per year }
\]

|722134694|kitty|/Applications/kitty.app|0|
c \(\quad\) Final value \(=\) original value \(\times(0.80)^{4}\)

\[
=\text { original value } \times 0.4096
\]

Hence the percentage decrease over four years is \(100 \%-40.96 \%=59.04 \%\)

d Depreciation over four years \(=40000-16384\)

\[
=\$ 23616
\]

Average depreciation per year \(=23616 \div 4\)

\[
=\$ 5904 \text { per year }
\]
|722134701|kitty|/Applications/kitty.app|0|
d Depreciation over four years \(=40000-16384\)

\[
=\$ 23616
\]

Average depreciation per year \(=23616 \div 4\)

\[
=\$ 5904 \text { per year }
\]
|722134708|kitty|/Applications/kitty.app|0|
A school buys new computers every four years. At the end of the four years, it offers them for sale to the students on the assumption that they have depreciated at \(35 \%\) p.a. (per annum). The school is presently advertising some computers at \(\$ 400\) each.

a What did each computer cost the school originally?

b What is the average depreciation on each computer, in dollars per year?
|722134929|kitty|/Applications/kitty.app|0|
a Each year a computer is worth \(100 \%-35 \%=65 \%\) of its value the previous year.

Hence final value \(=\) original value \(\times 0.65 \times 0.65 \times 0.65 \times 0.65\)

\[
\text { final value }=\text { original value } \times(0.65)^{4}
\]

Reversing this, original value \(=\) final value \(\div(0.65)^{4}\)

\[
\begin{aligned}
& =400 \div(0.65)^{4} \\
& \approx \$ 2241
\end{aligned}
\]

b Depreciation over four years \(\approx 2241-400\)

\[
=\$ 1841
\]

Average depreciation per year \(\approx 1841 \div 4\)

\[
\approx \$ 460
\]

|722134966|kitty|/Applications/kitty.app|0|
b Depreciation over four years \(\approx 2241-400\)

\[
=\$ 1841
\]

Average depreciation per year \(\approx 1841 \div 4\)

\[
\approx \$ 460
\]
|722134980|kitty|/Applications/kitty.app|0|
1 The landlord of a large block of home units purchased washing machines for its units four years ago for \(\$ 400000\), and is assuming a depreciation rate of \(30 \%\).

a By multiplying by 0.70 , find the value after one year.

b By multiplying by \((0.70)^{2}\), find the value after two years.

c By multiplying by \((0.70)^{3}\), find the value after three years.

d By multiplying by \((0.70)^{4}\), find the value after four years.

e What is the percentage decrease in value over the four years?

f What is the average depreciation on the washing machines, in dollars p.a. over the four years?
|722135025|kitty|/Applications/kitty.app|0|
2 The Hungry Hour Cafe purchased an air-conditioning system six years ago for \(\$ 250000\), and is assuming a depreciation rate of \(20 \%\).

a Find the value after one year.

b Find the value after two years.

c Find the value after six years.

d What is the percentage decrease in value over the six years?

e What is the average depreciation, in dollars p.a., on the air-conditioning system over the six years?
|722135064|kitty|/Applications/kitty.app|0|

3 A business spent \(\$ 560000\) installing alarms at its premises and then depreciated them at \(20 \%\) p.a. Find the value after five years, and the percentage depreciation of their value.

4 The population of a sea lion colony decreases at a compound rate of \(2 \%\) p.a. for 10 years. If the population is 8000 initially, what is it at the end of the 10-year period?

5 The Northern Start Bus Company bought a bus for \(\$ 480000\), depreciated it at \(30 \%\) p.a., and sold it again seven years later for \(\$ 60000\). Was the price that they obtained better or worse than the depreciated value, and by how much?
|722135077|kitty|/Applications/kitty.app|0|
6 The Backyard Rubbish Experts bought a fleet of small trucks for \(\$ 1340000\) and depreciated them at \(22.5 \%\) p.a. Five years later they sold them for \(\$ 310000\). Was the price that they obtained better or worse than the depreciated value, and by how much?

7 A landlord spent \(\$ 3400\) on vacuum cleaners for his block of home units and depreciated them for taxation purposes at \(25 \%\) p.a. Find their value at the end of each of the first three years, and the amount of the depreciation that the landlord could claim against his taxable income for each of those three years.

8 Lara and Kate each received \(\$ 100000\) from their parents. Lara invested the money at \(6.2 \%\) p.a. compounded annually, whereas Kate bought a luxury car that depreciated at a rate of \(20 \%\) p.a. What were the values of their investments at the end of five years?

9 Taxis depreciate at \(50 \%\) p.a., and other cars depreciate at \(22.5 \%\) p.a.

a What is the total percentage depreciation on each type of vehicle after seven years?

b What is the difference in value, to the nearest dollar, after seven years of a fleet of taxis and a fleet of other cars, if both fleets cost \(\$ 1000000\) ?
|722135553|kitty|/Applications/kitty.app|0|
10 Mr Wong's 10-year-old used car is worth \(\$ 4000\), and has been depreciating at \(22.5 \%\) p.a. (Calculate amounts of money in whole dollars.)

a Use division by 0.775 to find how much it was worth a year ago.

b Find how much it was worth two years ago.

c Find how much it was worth 10 years ago.

d What is the total percentage depreciation on the car over the 10-year period?

e What was the average depreciation in dollars per year over the 10-year period?
|722135569|kitty|/Applications/kitty.app|0|
10 Mr Wong's 10-year-old used car is worth \(\$ 4000\), and has been depreciating at \(22.5 \%\) p.a. (Calculate amounts of money in whole dollars.)|722135581|kitty|/Applications/kitty.app|0|
                (continued over page)
|722136264|kitty|/Applications/kitty.app|0|
\begin{questions}
    \question[6] The landlord of a large block of home units purchased washing machines for its units four years ago for \(\$ 400000\), and is assuming a depreciation rate of \(30 \%\).
    \begin{parts}
        \part By multiplying by 0.70, find the value after one year.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 400000 \times 0.70 = \$ 280000\)
        \end{solutionordottedlines}
        \part By multiplying by \((0.70)^{2}\), find the value after two years.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 400000 \times (0.70)^{2} = \$ 196000\)
        \end{solutionordottedlines}
        \part By multiplying by \((0.70)^{3}\), find the value after three years.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 400000 \times (0.70)^{3} = \$ 137200\)
        \end{solutionordottedlines}
        \part By multiplying by \((0.70)^{4}\), find the value after four years.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 400000 \times (0.70)^{4} = \$ 96040\)
        \end{solutionordottedlines}
        \part What is the percentage decrease in value over the four years?
        \begin{solutionordottedlines}[0.5in]
            Percentage decrease = \((1 - \frac{\$ 96040}{\$ 400000}) \times 100 \approx 76\%\)
        \end{solutionordottedlines}
        \part What is the average depreciation on the washing machines, in dollars p.a. over the four years?
        \begin{solutionordottedlines}[0.5in]
            Average depreciation = \(\frac{\$ 400000 - \$ 96040}{4} \approx \$ 76049\) per year
        \end{solutionordottedlines}
    \end{parts}
    \question[1] A business spent \(\$ 560000\) installing alarms at its premises and then depreciated them at \(20 \%\) p.a. Find the value after five years, and the percentage depreciation of their value.
        \begin{solutionordottedlines}[0.5in]
            Value after five years = \(\$ 560000 \times (0.80)^{5} \approx \$ 180224\) \\
            Percentage depreciation = \((1 - \frac{\$ 180224}{\$ 560000}) \times 100 \approx 67.82\%\)
        \end{solutionordottedlines}
    \question[1] The population of a sea lion colony decreases at a compound rate of \(2 \%\) p.a. for 10 years. If the population is 8000 initially, what is it at the end of the 10-year period?
        \begin{solutionordottedlines}[0.5in]
            Final population = \(8000 \times (0.98)^{10} \approx 6612\) sea lions
        \end{solutionordottedlines}
    \question[1] The Northern Start Bus Company bought a bus for \(\$ 480000\), depreciated it at \(30 \%\) p.a., and sold it again seven years later for \(\$ 60000\). Was the price that they obtained better or worse than the depreciated value, and by how much?
        \begin{solutionordottedlines}[0.5in]
            Depreciated value = \(\$ 480000 \times (0.70)^{7} \approx \$ 33684\) \\
            The sale price was better by \(\$ 60000 - \$ 33684 = \$ 26316\)
        \end{solutionordottedlines}
    \question[5] Mr Wong's 10-year-old used car is worth \(\$ 4000\), and has been depreciating at \(22.5 \%\) p.a. (Calculate amounts of money in whole dollars.)
    \begin{parts}
        \part Use division by 0.775 to find how much it was worth a year ago.
        \begin{solutionordottedlines}[0.5in]
            Value a year ago = \(\$ 4000 / 0.775 \approx \$ 5161\)
        \end{solutionordottedlines}
        \part Find how much it was worth two years ago.
        \begin{solutionordottedlines}[0.5in]
            Value two years ago = \(\$ 5161 / 0.775 \approx \$ 6660\)
        \end{solutionordottedlines}
        \part Find how much it was worth 10 years ago.
        \begin{solutionordottedlines}[0.5in]
            Value 10 years ago = \(\$ 4000 / (0.775)^{10} \approx \$ 20860\)
        \end{solutionordottedlines}
        \part What is the total percentage depreciation on the car over the 10-year period?
        \begin{solutionordottedlines}[0.5in]
            Percentage depreciation = \((1 - \frac{\$ 4000}{\$ 20860}) \times 100 \approx 80.8\%\)
        \end{solutionordottedlines}
        \part What was the average depreciation in dollars per year over the 10-year period?
        \begin{solutionordottedlines}[0.5in]
            Average depreciation = \(\frac{\$ 20860 - \$ 4000}{10} \approx \$ 1686\) per year
        \end{solutionordottedlines}
    \end{parts}
\end{questions}
|722136495|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
    \begin{questions}
        \question[6] The landlord of a large block of home units purchased washing machines for its units four years ago for \(\$ 400000\), and is assuming a depreciation rate of \(30 \%\).
        \begin{parts}
            \part By multiplying by 0.70 , find the value after one year.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part By multiplying by \((0.70)^{2}\), find the value after two years.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part By multiplying by \((0.70)^{3}\), find the value after three years.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part By multiplying by \((0.70)^{4}\), find the value after four years.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part What is the percentage decrease in value over the four years?
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part What is the average depreciation on the washing machines, in dollars p.a. over the four years?
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \end{parts}
        \question[1] A business spent \(\$ 560000\) installing alarms at its premises and then depreciated them at \(20 \%\) p.a. Find the value after five years, and the percentage depreciation of their value.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \question[1] The population of a sea lion colony decreases at a compound rate of \(2 \%\) p.a. for 10 years. If the population is 8000 initially, what is it at the end of the 10-year period?
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \question[1] The Northern Start Bus Company bought a bus for \(\$ 480000\), depreciated it at \(30 \%\) p.a., and sold it again seven years later for \(\$ 60000\). Was the price that they obtained better or worse than the depreciated value, and by how much?
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \question[5] Mr Wong's 10-year-old used car is worth \(\$ 4000\), and has been depreciating at \(22.5 \%\) p.a. (Calculate amounts of money in whole dollars.)
        \begin{parts}
            \part Use division by 0.775 to find how much it was worth a year ago.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part Find how much it was worth two years ago.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part Find how much it was worth 10 years ago.
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part What is the total percentage depreciation on the car over the 10-year period?
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
            \part What was the average depreciation in dollars per year over the 10-year period?
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
|722136502|kitty|/Applications/kitty.app|0|
\section*{Section 2}
|722136562|kitty|/Applications/kitty.app|0|
b \(56 \%\)
c \(75 \%\)
d \(37 \frac{1}{2} \%\)
f \(16 \frac{2}{3} \%\)
h \(6.4 \%\)
l \(136 \%\)
|722136645|kitty|/Applications/kitty.app|0|
\begin{multicols}{2}
|722136742|kitty|/Applications/kitty.app|0|
\section*{Section 3}
|722137129|kitty|/Applications/kitty.app|0|
\section*{Section 4}
|722137138|kitty|/Applications/kitty.app|0|
\section*{Section 5}
|722137145|kitty|/Applications/kitty.app|0|
\section*{Section 6}
|722137154|kitty|/Applications/kitty.app|0|
S|722137186|kitty|/Applications/kitty.app|0|
Section 7|722137202|kitty|/Applications/kitty.app|0|
Section 5|722137218|kitty|/Applications/kitty.app|0|
Section 6|722137225|kitty|/Applications/kitty.app|0|
Section 4|722137258|kitty|/Applications/kitty.app|0|
Section 3|722137270|kitty|/Applications/kitty.app|0|
Section 2|722137292|kitty|/Applications/kitty.app|0|
            \item \(136 \%\)
|722137797|kitty|/Applications/kitty.app|0|
            \item 0.0075
|722137800|kitty|/Applications/kitty.app|0|
\end{enumera}
|722137943|kitty|/Applications/kitty.app|0|
        \begin{center}
        \begin{tabular}{|c|c|c|c|c|}
        \hline
        & Principal & Rate p.a. & Time in years & Total interest \\
        \hline
        a & \(\$ 10000\) & \(8 \%\) &  & \(\$ 3200\) \\
        \hline
        b & \(\$ 4400000\) & \(7 \frac{1}{2} \%\) &  & \(\$ 3960000\) \\
        \hline
        c & \(\$ 5000\) &  & 6 & \(\$ 1350\) \\
        \hline
        d & \(\$ 260000\) &  & 8 & \(\$ 83200\) \\
        \hline
        e &  & \(6 \%\) & 5 & \(\$ 900\) \\
        f &  & \(3.6 \%\) & 4 & \(\$ 115.20\) \\
        \end{tabular}
        \end{center}
|722138221|kitty|/Applications/kitty.app|0|
\begin{enumerate}
|722138355|kitty|/Applications/kitty.app|0|
*
%
%|722138899|kitty|/Applications/kitty.app|0|
Shirt|722139305|kitty|/Applications/kitty.app|0|
\documentclass{letter}
\usepackage[utf8]{inputenc}
\usepackage[colorlinks]{hyperref}
\usepackage[left=1in,top=1in,right=1in,bottom=1in]{geometry} % Document margins
\usepackage{graphicx}
\usepackage{tabularx}
\usepackage{multirow}
\usepackage{ragged2e}
\usepackage{hhline}
\usepackage{array}

\hypersetup{
    urlcolor=blue
}

\newcolumntype{R}[1]{!{\raggedleft\let\newline\\\\arraybackslash\hspace{0pt}}m{#1}}

\begin{document}
    
\thispagestyle{empty}

% Header, for company, invoice info
\begin{tabularx}{\textwidth}{l X l}
    \multirow{5}{*} & \textbf{Private Tutoring} & \hskip12pt\multirow{5}{*}{\begin{tabular}{r}\footnotesize\bf INVOICE \\[-0.8ex] \footnotesize \MakeUppercase{period} \\[-0.4ex] \footnotesize\bf ISSUE DATE \\[-0.8ex] \footnotesize \MakeUppercase{\today} \\[-0.4ex] \footnotesize\bf DUE DATE\\[-0.8ex] \footnotesize \MakeUppercase{due date} \end{tabular}}\hspace{-6pt} \\
   & Aayush Bajaj & \\
   & \hrulefill \\
   & Email: j@abaj.io & \\
   & Ph:  0481 910 408 & \\
   & ABN: 64257676141
\end{tabularx} 

\vspace{1 cm}

BILL TO

% Recipient name
\Large\textbf{recipient}\normalsize

% Table of fees
\begin{tabularx}{\linewidth}{l X X X c}
    \hline
    & & & &\\[0.25ex]
    \centering{\bf{Service}} & \centering{\bf{Rate}} & \centering{\bf{Quantity}} & \centering{\bf{Discount}} & \bf Payment due\\[2.5ex]\hline
    & & & &\\
    service & \centering\$hourly rate/hr & \centering quantity & \centering -\$discount & \$due\\[2.5ex]\hline
    & & & &\\
    & & & \bf Total & \$total\\[2.5ex]\hhline{~~~--}
    & & & & \\
    & & & \bf Received & \$received\\[2.5ex]\hhline{~~~--}
    & & & & \\
    & & & \bf Balance due & \$total due\\[2.5ex]\hhline{~~~==}
\end{tabularx}

\vspace{1 cm}

\Large\textbf{Payment instructions}\normalsize

\vspace{0.1 cm}

\textbf{Account Transfer}\\
BSB: 633-123\\
Account Number: 174201004

\textbf{Pay-ID}\\
0481 910 408

\end{document}
|722139418|kitty|/Applications/kitty.app|0|
\documentclass{letter}
\usepackage[utf8]{inputenc}
\usepackage[colorlinks]{hyperref}
\usepackage[left=1in,top=1in,right=1in,bottom=1in]{geometry} % Document margins
\usepackage{graphicx}
\usepackage{tabularx}
\usepackage{multirow}
\usepackage{ragged2e}
\usepackage{hhline}
\usepackage{array}

\hypersetup{
    urlcolor=blue
}

\newcolumntype{R}[1]{!{\raggedleft\let\newline\\\\arraybackslash\hspace{0pt}}m{#1}}

\begin{document}
    
\thispagestyle{empty}

% Header, for company, invoice info
\begin{tabularx}{\textwidth}{l X l}
    \multirow{5}{*} & \textbf{Resource Creation} & \hskip12pt\multirow{5}{*}{\begin{tabular}{r}\footnotesize\bf INVOICE \\[-0.8ex] \footnotesize \MakeUppercase{Deposit} \\[-0.4ex] \footnotesize\bf ISSUE DATE \\[-0.8ex] \footnotesize \MakeUppercase{\today} \\[-0.4ex] \footnotesize\bf DUE DATE\\[-0.8ex] \footnotesize \MakeUppercase{November 8, 2023} \end{tabular}}\hspace{-6pt} \\
   & Aayush Bajaj & \\
   & \hrulefill \\
   & Email: j@abaj.io & \\
   & Ph:  0481 910 408 & \\
   & ABN: 64257676141
\end{tabularx} 

\vspace{1 cm}

BILL TO

% Recipient name
\Large\textbf{PEN Education}\normalsize

% Table of fees
\begin{tabularx}{\linewidth}{l X X X c}
    \hline
    & & & &\\[0.25ex]
    \centering{\bf{Service}} & \centering{\bf{Rate}} & \centering{\bf{Quantity}} & \centering{\bf{Discount}} & \bf Payment due\\[2.5ex]\hline
    & & & &\\
    Year 9 Mathematics Booklets Deposit & \centering\$300 & \centering 1 & \centering -\$0 & \$300\\[2.5ex]\hline
    & & & &\\
    & & & \bf Total & \$300\\[2.5ex]\hhline{~~~--}
    & & & & \\
    & & & \bf Received & \$0\\[2.5ex]\hhline{~~~--}
    & & & & \\
    & & & \bf Due & \$300\\[2.5ex]\hhline{~~~==}
\end{tabularx}

\vspace{1 cm}

\Large\textbf{Payment instructions}\normalsize

\vspace{0.1 cm}

\textbf{Account Transfer}\\
BSB: 633-123\\
Account Number: 174201004

\textbf{Pay-ID}\\
0481 910 408

\end{document}
|722139424|kitty|/Applications/kitty.app|0|
    Year 9 Mathematics Pythagoras Booklet & \centering\$100 & \centering 1 & \centering -\$0 & \$100\\[2.5ex]\hline
    & & & &\\
|722139462|kitty|/Applications/kitty.app|0|
Deposit|722139499|kitty|/Applications/kitty.app|0|
A*hTY@sTfv$9#h9g@n3iKJjW|722153972|kitty|/Applications/kitty.app|0|
DSC00058.JPG
DSC00057.JPG
DSC00056.JPG
DSC00055.JPG
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DSC00053.JPG
DSC00052.JPG
DSC00051.JPG
DSC00050.JPG
DSC00049.JPG
DSC00048.JPG
DSC00060.JPG
DSC00059.JPG|722170598|Finder|/System/Library/CoreServices/Finder.app|0|
Hickeys and Rhythm|722170665|kitty|/Applications/kitty.app|0|
{{< image-gallery gallery_dir="album/23-11-11-cricket-pitch" >}}
|722170729|kitty|/Applications/kitty.app|0|
sudo htpasswd -c /etc/nginx/.htpasswd username|722171198|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
The|722171610|kitty|/Applications/kitty.app|0|
mirac|722171632|kitty|/Applications/kitty.app|0|
\begin{questions}
    \question[1] Convert the following from percentages to decimals:
    \begin{multicols}{2}
    \begin{parts}
        \part $58\%=$\fillin[$0.58$][0.3\textwidth]
        \part $17\frac{3}{4}\%=$\fillin[$\frac{71}{400}$][0.3\textwidth]
    \end{parts}
    \end{multicols}

    \question[1] Convert the following from fractions to percentages:
    \begin{multicols}{2}
    \begin{parts}
        \part $\frac{4}{7}=$\fillin[$\approx 57.14\%$][0.3\textwidth]
        \part $\frac{11}{25}=$\fillin[$44\%$][0.3\textwidth]
    \end{parts}
    \end{multicols}

    \question[1] The Efficient Appliance Company finds that \(4.2 \%\) of its refrigerators need adjustments after delivery. If 950 refrigerators are sold, how many might require adjustments?
    \begin{solutionordottedlines}[1in]
        Number of refrigerators needing adjustments \(= 950 \times 4.2 \%\)
        \[
        \begin{aligned}
        & = 950 \times 0.042 \\
        & = 39.9 \approx 40 \quad(\text{Round to nearest whole number})
        \end{aligned}
        \]
    \end{solutionordottedlines}

    \question[1] An average smartphone has a mass of \(200 \mathrm{~g}\). If it contains \(0.25 \mathrm{~g}\) of rare earth metals, what percentage of the total mass does this represent?
    \begin{solutionordottedlines}[1in]
        Total mass of the smartphone in grams = \(200 \mathrm{~g}\), rare earth metals = \(0.25 \mathrm{~g}\).

        Hence, percentage of rare earth metals \(= \frac{0.25}{200} \times 100 \%\)
        \[
        = \frac{1}{800} \%
        \]
        \[
        \approx 0.125 \%
        \]
    \end{solutionordottedlines}
\end{questions}
|722223150|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
    \begin{questions}
        \question[1] Convert the following from percentages to fractions
        \begin{multicols}{2}
        \begin{parts}
            \part $42\%=$\fillin[$0.42$][0.3\textwidth]
            \part $12\frac{1}{4}\%=$\fillin[$\frac{49}{400}$][0.3\textwidth]
        \end{parts}
        \end{multicols}
        \question[1] Convert the following from fractions to percentages
        \begin{multicols}{2}
        \begin{parts}
            \part $\frac{3}{5}=$\fillin[$60\%$][0.3\textwidth]
            \part $\frac{7}{20}=$\fillin[$35\%$][0.3\textwidth]
        \end{parts}
        \end{multicols}
        \question[1] The Brilliant Light Bulb Company estimates that \(3.5 \%\) of its light bulbs are defective. If a shop owner buys 1250 light bulbs to light the shop, how many would he expect to be defective?
        \begin{solutionordottedlines}[1in]
            Number of defective bulbs \(=1250 \times 3.5 \%\)
            \[
            \begin{aligned}
            & =1250 \times 0.035 \\
            & \approx 44 \quad(\text { Round } 43.75 \text { to } 44 .)
            \end{aligned}
            \]
        \end{solutionordottedlines}
        \question[1] A typical computer weighs about \(25 \mathrm{~kg}\). When it is broken down as waste, it yields about \(3 \mathrm{~g}\) of arsenic. What percentage of the total is this?
        \begin{solutionordottedlines}[1in]
            Using grams, the computer weighs \(25000 \mathrm{~g}\) and the arsenic weighs \(3 \mathrm{~g}\).

            Hence percentage of arsenic \(=\frac{3}{25000} \times 100 \%\)

            \[
            =\frac{3}{250} \%
            \]

            \[
            =0.012 \%
            \]
        \end{solutionordottedlines}
    \end{questions}
|722223157|kitty|/Applications/kitty.app|0|
\begin{questions}
    \question[2] Express each percentage as a decimal:
    \begin{multicols}{2}
    \begin{parts}
        \part $53\%=$\fillin[0.53]
        \part $9.5\%=$\fillin[0.095]
        \part $85\frac{1}{2}\%=$\fillin[0.855]
        \part $0.25\%=$\fillin[0.0025]
    \end{parts}
    \end{multicols}

    \question[3] Express each percentage as a fraction:
    \begin{multicols}{3}
    \begin{onehalfspacing}
    \begin{parts}
        \part $50\%=$\fillin[\(\frac{1}{2}\)]
        \part $40\%=$\fillin[\(\frac{2}{5}\)]
        \part $150\%=$\fillin[\(\frac{3}{2}\) or \(1\frac{1}{2}\)]
        \part $20\%=$\fillin[\(\frac{1}{5}\)]
        \part $9.75\%=$\fillin[\(\frac{39}{400}\)]
        \part $175\%=$\fillin[\(\frac{7}{4}\) or \(1\frac{3}{4}\)]
    \end{parts}
    \end{onehalfspacing}
    \end{multicols}

    \question[3] Express each fraction or decimal as a percentage:
    \begin{multicols}{2}
    \begin{onehalfspacing}
    \begin{parts}
        \part \(\frac{4}{7}=\)\fillin[\(\approx 57.14\%\)]
        \part \(\frac{1}{8}=\)\fillin[12.5\%]
        \part $0.75=$\fillin[75\%]
        \part $2.5=$\fillin[250\%]
        \part $0.125=$\fillin[12.5\%]
        \part $1.75=$\fillin[175\%]
    \end{parts}
    \end{onehalfspacing}
    \end{multicols}

    \question[3] Evaluate these amounts, correct to 2 decimal places where necessary.
    \begin{multicols}{2}
    \begin{onehalfspacing}
    \begin{parts}
        \part \(20 \%\) of 50 = \fillin[10]
        \part \(45 \%\) of 800 =\fillin[360]
        \part \(110 \%\) of 250 =\fillin[275]
        \part \(12.5 \%\) of 480 = \fillin[60]
        \part \(3.2 \%\) of 250 =\fillin[8]
        \part \(0.5 \%\) of 5000 = \fillin[25]
    \end{parts}
    \end{onehalfspacing}
    \end{multicols}

    \question[3] Find what percentage the first quantity is of the second quantity, correct to 1 decimal place.
    \begin{parts}
        \part \(80 \mathrm{~m}, 40 \mathrm{~m}\)=\fillin[200.0\%]
        \part 10 weeks, 40 weeks=\fillin[25.0\%]
        \part 30 weeks, 10 weeks=\fillin[300.0\%]
    \end{parts}

    \question[3] Find what percentage the first quantity is of the second quantity, correct to 2 decimal places where necessary. You will need to express both quantities in the same unit first.
    \begin{multicols}{2}
    \begin{parts}
    \begin{onehalfspacing}
        \part 50 cents, \(\$ 4.00\)=\fillin[12.50\%]
        \part \(5 \mathrm{~g}, 2 \mathrm{~kg}\)=\fillin[0.25\%]
        \part 10 days, 2 years=\fillin[\(\approx\) 1.37\%]
        \part \(2 \mathrm{~km}, 100 \mathrm{~m}\)=\fillin[2000.00\%]
        \part 2 years, 1 day=\fillin[\(\approx\) 0.55\%]
        \part \(20 \mathrm{~cm}, 1 \mathrm{~km}\)=\fillin[0.02\%]
    \end{parts}
    \end{onehalfspacing}
    \end{multicols}

    \question[2] There are 820 students at a secondary school, \(10 \%\) of whom are left-handed. Calculate the number of left-handed students in the school.
    \begin{solutionordottedlines}[1in]
    82 students
    \end{solutionordottedlines}

    \question[2] A football match lasted 120 minutes (including extra time). If team B was in possession for \(65 \%\) of the match, for how many minutes and seconds was team B in possession?
    \begin{solutionordottedlines}[1in]
    78 minutes and 0 seconds
    \end{solutionordottedlines}
\end{questions}
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    \begin{questions}
        \question[2] Express each percentage as a decimal:
        \begin{multicols}{2}
        \begin{parts}
            \part $72\%=$\fillin[0.72]
            \part $7.6\%=$\fillin[0.076]
            \part $77\frac{3}{4}\%=$\fillin[0.7775]
            \part $0.1\%=$\fillin[0.001]
        \end{parts}
        \end{multicols}
        \question[3] Express each percentage as a fraction:
        \begin{multicols}{3}
        \begin{onehalfspacing}
        \begin{parts}
            \part $35\%$\fillin[\(\frac{7}{20}\)]
            \part $33\frac{1}{3}\%=$\fillin[\(\frac{1}{3}\)]
            \part $210\%=$\fillin[\(\frac{21}{10}\) or \(2\frac{1}{10}\)]
            \part $125\%=$\fillin[\(\frac{5}{4}\) or \(1\frac{1}{4}\)]
            \part $7.25\%=$\fillin[\(\frac{29}{400}\)]
            \part $112\frac{1}{2}\%=$\fillin[\(\frac{9}{8}\) or \(1\frac{1}{8}\)]
        \end{parts}
        \end{onehalfspacing}
        \end{multicols}
        \question[3] Express each fraction or decimal as a percentage:
        \begin{multicols}{2}
        \begin{onehalfspacing}
        \begin{parts}
            \part \(\frac{3}{5}=\)\fillin[60\%]
            \part \(\frac{7}{20}=\)\fillin[35\%]
            \part $0.43=$\fillin[43\%]
            \part $1.2=$\fillin[120\%]
            \part $0.225=$\fillin[22.5\%]
            \part $2.03=$\fillin[203\%]
        \end{parts}
        \end{onehalfspacing}
        \end{multicols}
        \question[3] Evaluate these amounts, correct to 2 decimal places where necessary.
        \begin{multicols}{2}
        \begin{onehalfspacing}
        \begin{parts}
            \part \(15 \%\) of 40 = \fillin[6]
            \part \(57 \%\) of 1000 =\fillin[570]
            \part \(120 \%\) of 538 =\fillin[645.60]
            \part \(15.8 \%\) of 972 = \fillin[153.57]
            \part \(2.8 \%\) of 318 =\fillin[8.90]
            \part \(0.1 \%\) of 6000 = \fillin[6]
        \end{parts}
        \end{onehalfspacing}
        \end{multicols}
        \question[3] Find what percentage the first quantity is of the second quantity, correct to 1 decimal place.
        \begin{parts}
            \part \(70 \mathrm{~m}, 50 \mathrm{~m}\)=\fillin[140.0\%]
            \part 15 weeks, 60 weeks=\fillin[25.0\%]
            \part 60 weeks, 15 weeks=\fillin[400.0\%]
        \end{parts}
        \question[3] Find what percentage the first quantity is of the second quantity, correct to 2 decimal places where necessary. You will need to express both quantities in the same unit first.
        \begin{multicols}{2}
        \begin{parts}
        \begin{onehalfspacing}
            \part 68 cents, \(\$ 5.00\)=\fillin[13.60\%]
            \part \(7 \mathrm{~g}, 3 \mathrm{~kg}\)=\fillin[0.23\%]
            \part 15 days, 3 years=\fillin[\(\approx\) 1.37\%]
            \part \(4 \mathrm{~km}, 250 \mathrm{~m}\)=\fillin[1600.00\%]
            \part 1 year, 1 day=\fillin[\(\approx\) 0.27\%]
            \part \(56 \mathrm{~cm}, 2.4 \mathrm{~km}\)=\fillin[0.02\%]
        \end{onehalfspacing}
        \end{parts}
        \end{multicols}
        \question[2] There are 740 students at a primary school, \(5 \%\) of whom have red hair. Calculate the number of students in the school who have red hair.
        \begin{solutionordottedlines}[1in]
        37 students
        \end{solutionordottedlines}
        \question[2] A soccer match lasted 92 minutes (including injury time). If team A was in possession for \(55 \%\) of the match, for how many minutes and seconds was team A in possession?
        \begin{solutionordottedlines}[1in]
        50 minutes and 36 seconds
        \end{solutionordottedlines}
    \end{questions}
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    \end{onehalfspacing}
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\begin{examplebox}
    \subsection{Examples}
    \begin{questions}
        \question[1] Joshua saves \(12 \%\) of his after-tax salary every week. If he saves \(\$ 90\) a week, what is his after-tax salary?
        \begin{solutionordottedlines}[1in]
            Savings \(=\) after-tax salary \(\times 12 \%\)

            Reversing this:

            \[
            \begin{aligned}
            \text { after-tax salary } & =\text { savings } \div 12 \% \\
            & =90 \div 0.12 \\
            & =\$ 750
            \end{aligned}
            \]
        \end{solutionordottedlines}

        \question[2] Sterling silver is an alloy that is made up of \(92.5 \%\) by mass silver and \(7.5 \%\) copper.
        \begin{parts}
            \part How much sterling silver can be made with \(5 \mathrm{~kg}\) of silver and unlimited supplies of copper?
        \begin{solutionordottedlines}[1in]
            Mass of silver \(=\) mass of sterling silver \(\times 92.5 \%\)

            Reversing this:

            mass of sterling silver \(=\) mass of silver \(\div 92.5 \%\)

            \[
            \begin{aligned}
            & =5 \div 0.925 \\
            & \approx 5.405 \mathrm{~kg}
            \end{aligned}
            \]
        \end{solutionordottedlines}
            \part How much sterling silver can be made with \(5 \mathrm{~kg}\) of copper and unlimited supplies of silver?
        \begin{solutionordottedlines}[1in]
            Mass of copper \(=\) mass of sterling silver \(\times 7.5 \%\)

            Reversing this:

            mass of sterling silver \(=\) mass of copper \(\div 7.5 \%\)

            \[
            \begin{aligned}
            & =5 \div 0.075 \\
            & \approx 66.667 \mathrm{~kg}
            \end{aligned}
            \]
        \end{solutionordottedlines}
        \end{parts}
        \titledquestion{Commission}[2] The Eureka Gallery charges a commission of \(9.2 \%\).
        \begin{parts}
            \part The Australian painting Showing the Flag at Bakery Hill was sold recently for \(\$ 180000\). How much did the Gallery receive, and how much was left for the seller?
        \begin{solutionordottedlines}[1in]
            Commission \(=180000 \times 9.2 \%\)

            \[
            \begin{aligned}
            & =180000 \times 0.092 \\
            & =\$ 16560
            \end{aligned}
            \]

            Amount received by seller \(=180000-16560\)

            \[
            =\$ 163440
            \]
        \end{solutionordottedlines}
            \part The Gallery received a commission of \(\$ 7912\) for selling the painting Ned at the Glen. What was the selling price of the painting, and what did the seller actually receive?
        \begin{solutionordottedlines}[1in]
            Commission \(=\) selling price \(\times 9.2 \%\)

            Reversing this:

            selling price \(=\) commission \(\div 9.2 \%\)

            \[
            \begin{aligned}
            & =7912 \div 0.092 \\
            & =\$ 86000
            \end{aligned}
            \]

            Amount received by seller \(=86000-7912\)

            \[
            =\$ 78088
            \]
        \end{solutionordottedlines}
        \end{parts}
        \titledquestion{Profit}[3] The Budget Shoe Shop spent \(\$ 6600000\) last year buying shoes and paying salaries and other expenses. They made a \(2 \%\) profit on these costs.
        \begin{parts}
            \part What was their profit last year?
            \begin{solutionordottedlines}[1in]
                Profit \(=6600000 \times 2 \%\)

                \[
                \begin{aligned}
                & =6600000 \times 0.02 \\
                & =\$ 132000
                \end{aligned}
                \]
            \end{solutionordottedlines}
            \part What was the total of their sales?
            \begin{solutionordottedlines}[1in]
                Total sales \(=\) total costs + profit

                \[
                \begin{aligned}
                & =6600000+132000 \\
                & =\$ 6732000
                \end{aligned}
                \]
            \end{solutionordottedlines}
            \part In the previous year, their costs were \(\$ 5225000\) and their sales were only \(\$ 5145000\). What percentage loss did they make on their costs?
            \begin{solutionordottedlines}[1in]
                Last year, loss \(=\) total costs - total sales

                \[
                \begin{aligned}
                & =5225000-5145000 \\
                & =\$ 80000
                \end{aligned}
                \]

                Percentage loss \(=\left(\frac{80000}{5225000} \times \frac{100}{1}\right) \%\)

                \[
                \approx 1.53 \%
                \]
            \end{solutionordottedlines}
        \end{parts}
        \question[1] Joe's tile shop made a profit of \(5.8 \%\) on total costs last year. If the actual profit was \(\$ 83000\), what were the total costs, and what were the total sales?
        \begin{solutionordottedlines}[1in]
            \[
            \text { Profit }=\text { costs } \times 5.8 \%
            \]

            Reversing this, costs \(=\) profit \(\div 5.8 \%\)

            \[
            =83000 \div 0.058
            \]

            \(\approx \$ 1431034\), correct to the nearest dollar.

            Hence, total sales \(=\) profit + costs

            \[
            \approx 83000+1431034
            \]

            \[
            =\$ 1514034
            \]
        \end{solutionordottedlines}
        \titledquestion{Income Tax}[4]
        Income tax in the nation of Immutatia is calculated as follows.

        \begin{itemize}
          \item There is no tax on the first \(\$ 12000\) that a person earns in any one year.

          \item From \(\$ 12001\) to \(\$ 30000\), the tax rate is 15 c for each dollar over \(\$ 12000\).

          \item From \(\$ 30001\) to \(\$ 75000\), the tax rate is 25 c for each dollar over \(\$ 30000\).

          \item Over \(\$ 75000\), the tax rate is 35 c for each dollar over \(\$ 75000\).

        \end{itemize}
         Find the income tax payable by a person whose taxable income for the year is:
        \begin{parts}
            \part \(\$ 10600\)
            \begin{solutionordottedlines}[1in]
                There is no tax on an income of \(\$ 10600\).
            \end{solutionordottedlines}
            \part \(\$ 25572\)
            \begin{solutionordottedlines}[1in]
                Tax on first \(\$ 12000=\$ 0\)

                Tax on remaining \(\$ 13572=13572 \times 0.15\)

                \[
                =\$ 2035.80
                \]

                This is the total tax payable.
            \end{solutionordottedlines}
            \part \(\$ 62300\)
            \begin{solutionordottedlines}[1in]
                Tax on first \(\$ 12000=\$ 0\)

                Tax on next \(\$ 18000=18000 \times 0.15\)

                \[
                =\$ 2700
                \]

                Tax on remaining \(\$ 32300=32300 \times 0.25\)

                \[
                =\$ 8075
                \]

                Total tax \(=2700+8075\)

                \[
                =\$ 10775
                \]
            \end{solutionordottedlines}
            \part \(\$ 455000\)
            \begin{solutionordottedlines}[1in]
                Tax on first \(\$ 12000=\$ 0\)

                Tax on next \(\$ 18000=\$ 2700 \quad(\) see part c)

                Tax on next \(\$ 45000=45000 \times 0.25\)

                \[
                =\$ 11250
                \]

                Tax on remaining \(\$ 380000=380000 \times 0.35\)

                \[
                =\$ 133000
                \]

                Total tax \(=2700+11250+133000\)

                \[
                =\$ 146950
                \]
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{examplebox}

\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \question[2] Find the quantity, given that:
        \begin{parts}\begin{multicols}{2}
            \part \(2 \%\) of it is \(\$ 12\)
            \begin{solutionordottedlines}[0.5in]
                Quantity = \(\$ 12\) / \(0.02\) = \(\$ 600\)
            \end{solutionordottedlines}
            \part \(6 \%\) of it is \(750 \mathrm{~g}\)
            \begin{solutionordottedlines}[0.5in]
                Quantity = \(750 \mathrm{~g}\) / \(0.06\) = \(12500 \mathrm{~g}\) or \(12.5 \mathrm{~kg}\)
            \end{solutionordottedlines}
            \part \(30 \%\) of it is 36 minutes
            \begin{solutionordottedlines}[0.5in]
                Quantity = \(36 \text{ minutes}\) / \(0.3\) = \(120 \text{ minutes}\) or \(2 \text{ hours}\)
            \end{solutionordottedlines}
            \part \(90 \%\) of it is \(54 \mathrm{~cm}\)
            \begin{solutionordottedlines}[0.5in]
                Quantity = \(54 \mathrm{~cm}\) / \(0.9\) = \(60 \mathrm{~cm}\)
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[2] In each part, find the price if:
        \begin{parts}
            \part a deposit of \(\$ 360\) is \(30 \%\) of the price
            \begin{solutionordottedlines}[0.5in]
                Price = \(\$ 360\) / \(0.3\) = \(\$ 1200\)
            \end{solutionordottedlines}
            \part a deposit of \(\$ 168\) is \(15 \%\) of the price
            \begin{solutionordottedlines}[0.5in]
                Price = \(\$ 168\) / \(0.15\) = \(\$ 1120\)
            \end{solutionordottedlines}
        \end{parts}
        \question[2] Find the original quantity, correct to a suitable number of decimal places, if:
        \begin{parts}\begin{multicols}{2}
            \part \(23 \%\) of it is \(100 \mathrm{~kg}\)
            \begin{solutionordottedlines}[0.5in]
                Quantity = \(100 \mathrm{~kg}\) / \(0.23\) = \(434.78 \mathrm{~kg}\)
            \end{solutionordottedlines}
            \part \(0.2 \%\) of it is \(4 \mathrm{~mm}\)
            \begin{solutionordottedlines}[0.5in]
                Quantity = \(4 \mathrm{~mm}\) / \(0.002\) = \(2000 \mathrm{~mm}\) or \(2 \mathrm{~m}\)
            \end{solutionordottedlines}
            \part \(0.92 \%\) of it is 1.86 hectares
            \begin{solutionordottedlines}[0.5in]
                Quantity = \(1.86 \text{ hectares}\) / \(0.0092\) = \(20.22 \text{ hectares}\)
            \end{solutionordottedlines}
            \part \(97 \%\) of it is \(\$ 700\)
            \begin{solutionordottedlines}[0.5in]
                Quantity = \(\$ 700\) / \(0.97\) = \(\$ 721.65\)
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[2] Cameron and Wendy together earn \(\$ 1156\) per week after tax. Of this, they pay \(\$ 460\) off their mortgage, \(\$ 185\) for groceries, and \(\$ 260\) for their car and transport, and they save \$124. 
        \begin{parts}
            \part Express each amount as a percentage of their weekly income, correct to the nearest \(1 \%\).
            \begin{solutionordottedlines}[1in]
                Mortgage: \((\$ 460 / \$ 1156) \times 100 \approx 39.8\%\) (rounded to \(40\%\)) \\
                Groceries: \((\$ 185 / \$ 1156) \times 100 \approx 16.0\%\) (rounded to \(16\%\)) \\
                Car and transport: \((\$ 260 / \$ 1156) \times 100 \approx 22.5\%\) (rounded to \(23\%\)) \\
                Savings: \((\$ 124 / \$ 1156) \times 100 \approx 10.7\%\) (rounded to \(11\%\))
            \end{solutionordottedlines}
            \part Find how much is unaccounted for in the list above, and what percentage it is of their weekly income.
            \begin{solutionordottedlines}[1in]
                Unaccounted for: \( \$ 1156 - (\$ 460 + \$ 185 + \$ 260 + \$ 124) = \$ 1156 - \$ 1029 = \$ 127 \) \\
                Percentage: \((\$ 127 / \$ 1156) \times 100 \approx 11.0\%\) (rounded to \(11\%\))
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{exercisebox}

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\begin{examplebox}
    \subsection{Examples}
    \begin{questions}
        \question[1] Emily sets aside \(15 \%\) of her monthly income for leisure activities. If she spends \(\$ 180\) on leisure each month, what is her monthly income?
        \begin{solutionordottedlines}[1in]
            Leisure spending \(=\) monthly income \(\times 15 \%\)

            Reversing this:

            \[
            \begin{aligned}
            \text { monthly income } & =\text { leisure spending } \div 15 \% \\
            & =180 \div 0.15 \\
            & =\$ 1200
            \end{aligned}
            \]
        \end{solutionordottedlines}

        \titledquestion{Metal Alloy Composition}[2] A metal alloy is composed of \(60 \%\) iron and \(40 \%\) chromium.
        \begin{parts}
            \part How much alloy can be made with \(4 \mathrm{~kg}\) of iron and unlimited chromium?
            \begin{solutionordottedlines}[1in]
                Mass of iron \(=\) mass of alloy \(\times 60 \%\)

                Reversing this:

                mass of alloy \(=\) mass of iron \(\div 60 \%\)

                \[
                \begin{aligned}
                & =4 \div 0.60 \\
                & \approx 6.667 \mathrm{~kg}
                \end{aligned}
                \]
            \end{solutionordottedlines}
            \part How much alloy can be made with \(4 \mathrm{~kg}\) of chromium and unlimited iron?
            \begin{solutionordottedlines}[1in]
                Mass of chromium \(=\) mass of alloy \(\times 40 \%\)

                Reversing this:

                mass of alloy \(=\) mass of chromium \(\div 40 \%\)

                \[
                \begin{aligned}
                & =4 \div 0.40 \\
                & =10 \mathrm{~kg}
                \end{aligned}
                \]
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{examplebox}

\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \question[2] Find the original amount, given that:
        \begin{parts}\begin{multicols}{2}
            \part \(4 \%\) of it is \(\$ 24\)
            \begin{solutionordottedlines}[0.5in]
                Original amount = \(\$ 24\) / \(0.04\) = \(\$ 600\)
            \end{solutionordottedlines}
            \part \(5 \%\) of it is \(500 \mathrm{~g}\)
            \begin{solutionordottedlines}[0.5in]
                Original amount = \(500 \mathrm{~g}\) / \(0.05\) = \(10000 \mathrm{~g}\) or \(10 \mathrm{~kg}\)
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[2] Calculate the price, considering:
        \begin{parts}
            \part a down payment of \(\$ 450\) is \(25 \%\) of the total price
            \begin{solutionordottedlines}[0.5in]
                Total price = \(\$ 450\) / \(0.25\) = \(\$ 1800\)
            \end{solutionordottedlines}
            \part a down payment of \(\$ 200\) is \(20 \%\) of the total price
            \begin{solutionordottedlines}[0.5in]
                Total price = \(\$ 200\) / \(0.20\) = \(\$ 1000\)
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{exercisebox}
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\begin{examplebox}
    \subsection{Examples}
    \begin{questions}
        \question[1] Joshua saves \(12 \%\) of his after-tax salary every week. If he saves \(\$ 90\) a week, what is his after-tax salary?
        \begin{solutionordottedlines}[1in]
            Savings \(=\) after-tax salary \(\times 12 \%\)

            Reversing this:

            \[
            \begin{aligned}
            \text { after-tax salary } & =\text { savings } \div 12 \% \\
            & =90 \div 0.12 \\
            & =\$ 750
            \end{aligned}
            \]
        \end{solutionordottedlines}

        \question[2] Sterling silver is an alloy that is made up of \(92.5 \%\) by mass silver and \(7.5 \%\) copper.
        \begin{parts}
            \part How much sterling silver can be made with \(5 \mathrm{~kg}\) of silver and unlimited supplies of copper?
        \begin{solutionordottedlines}[1in]
            Mass of silver \(=\) mass of sterling silver \(\times 92.5 \%\)

            Reversing this:

            mass of sterling silver \(=\) mass of silver \(\div 92.5 \%\)

            \[
            \begin{aligned}
            & =5 \div 0.925 \\
            & \approx 5.405 \mathrm{~kg}
            \end{aligned}
            \]
        \end{solutionordottedlines}
            \part How much sterling silver can be made with \(5 \mathrm{~kg}\) of copper and unlimited supplies of silver?
        \begin{solutionordottedlines}[1in]
            Mass of copper \(=\) mass of sterling silver \(\times 7.5 \%\)

            Reversing this:

            mass of sterling silver \(=\) mass of copper \(\div 7.5 \%\)

            \[
            \begin{aligned}
            & =5 \div 0.075 \\
            & \approx 66.667 \mathrm{~kg}
            \end{aligned}
            \]
        \end{solutionordottedlines}
        \end{parts}
        \question[2] \textbf{Commission:}\\The Eureka Gallery charges a commission of \(9.2 \%\).
        \begin{parts}
            \part The Australian painting Showing the Flag at Bakery Hill was sold recently for \(\$ 180000\). How much did the Gallery receive, and how much was left for the seller?
        \begin{solutionordottedlines}[1in]
            Commission \(=180000 \times 9.2 \%\)

            \[
            \begin{aligned}
            & =180000 \times 0.092 \\
            & =\$ 16560
            \end{aligned}
            \]

            Amount received by seller \(=180000-16560\)

            \[
            =\$ 163440
            \]
        \end{solutionordottedlines}
            \part The Gallery received a commission of \(\$ 7912\) for selling the painting Ned at the Glen. What was the selling price of the painting, and what did the seller actually receive?
        \begin{solutionordottedlines}[1in]
            Commission \(=\) selling price \(\times 9.2 \%\)

            Reversing this:

            selling price \(=\) commission \(\div 9.2 \%\)

            \[
            \begin{aligned}
            & =7912 \div 0.092 \\
            & =\$ 86000
            \end{aligned}
            \]

            Amount received by seller \(=86000-7912\)

            \[
            =\$ 78088
            \]
        \end{solutionordottedlines}
        \end{parts}
        \titledquestion{Profit}[3] The Budget Shoe Shop spent \(\$ 6600000\) last year buying shoes and paying salaries and other expenses. They made a \(2 \%\) profit on these costs.
        \begin{parts}
            \part What was their profit last year?
            \begin{solutionordottedlines}[1in]
                Profit \(=6600000 \times 2 \%\)

                \[
                \begin{aligned}
                & =6600000 \times 0.02 \\
                & =\$ 132000
                \end{aligned}
                \]
            \end{solutionordottedlines}
            \part What was the total of their sales?
            \begin{solutionordottedlines}[1in]
                Total sales \(=\) total costs + profit

                \[
                \begin{aligned}
                & =6600000+132000 \\
                & =\$ 6732000
                \end{aligned}
                \]
            \end{solutionordottedlines}
            \part In the previous year, their costs were \(\$ 5225000\) and their sales were only \(\$ 5145000\). What percentage loss did they make on their costs?
            \begin{solutionordottedlines}[1in]
                Last year, loss \(=\) total costs - total sales

                \[
                \begin{aligned}
                & =5225000-5145000 \\
                & =\$ 80000
                \end{aligned}
                \]

                Percentage loss \(=\left(\frac{80000}{5225000} \times \frac{100}{1}\right) \%\)

                \[
                \approx 1.53 \%
                \]
            \end{solutionordottedlines}
        \end{parts}
        \question[1] Joe's tile shop made a profit of \(5.8 \%\) on total costs last year. If the actual profit was \(\$ 83000\), what were the total costs, and what were the total sales?
        \begin{solutionordottedlines}[1in]
            \[
            \text { Profit }=\text { costs } \times 5.8 \%
            \]

            Reversing this, costs \(=\) profit \(\div 5.8 \%\)

            \[
            =83000 \div 0.058
            \]

            \(\approx \$ 1431034\), correct to the nearest dollar.

            Hence, total sales \(=\) profit + costs

            \[
            \approx 83000+1431034
            \]

            \[
            =\$ 1514034
            \]
        \end{solutionordottedlines}
        \titledquestion{Income Tax}[4]
        Income tax in the nation of Immutatia is calculated as follows.

        \begin{itemize}
          \item There is no tax on the first \(\$ 12000\) that a person earns in any one year.

          \item From \(\$ 12001\) to \(\$ 30000\), the tax rate is 15 c for each dollar over \(\$ 12000\).

          \item From \(\$ 30001\) to \(\$ 75000\), the tax rate is 25 c for each dollar over \(\$ 30000\).

          \item Over \(\$ 75000\), the tax rate is 35 c for each dollar over \(\$ 75000\).

        \end{itemize}
         Find the income tax payable by a person whose taxable income for the year is:
        \begin{parts}
            \part \(\$ 10600\)
            \begin{solutionordottedlines}[1in]
                There is no tax on an income of \(\$ 10600\).
            \end{solutionordottedlines}
            \part \(\$ 25572\)
            \begin{solutionordottedlines}[1in]
                Tax on first \(\$ 12000=\$ 0\)

                Tax on remaining \(\$ 13572=13572 \times 0.15\)

                \[
                =\$ 2035.80
                \]

                This is the total tax payable.
            \end{solutionordottedlines}
            \part \(\$ 62300\)
            \begin{solutionordottedlines}[1in]
                Tax on first \(\$ 12000=\$ 0\)

                Tax on next \(\$ 18000=18000 \times 0.15\)

                \[
                =\$ 2700
                \]

                Tax on remaining \(\$ 32300=32300 \times 0.25\)

                \[
                =\$ 8075
                \]

                Total tax \(=2700+8075\)

                \[
                =\$ 10775
                \]
            \end{solutionordottedlines}
            \part \(\$ 455000\)
            \begin{solutionordottedlines}[1in]
                Tax on first \(\$ 12000=\$ 0\)

                Tax on next \(\$ 18000=\$ 2700 \quad(\) see part c)

                Tax on next \(\$ 45000=45000 \times 0.25\)

                \[
                =\$ 11250
                \]

                Tax on remaining \(\$ 380000=380000 \times 0.35\)

                \[
                =\$ 133000
                \]

                Total tax \(=2700+11250+133000\)

                \[
                =\$ 146950
                \]
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{examplebox}

\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \question[2] Find the quantity, given that:
        \begin{parts}\begin{multicols}{2}
            \part \(2 \%\) of it is \(\$ 12\)
            \begin{solutionordottedlines}[0.5in]
                Quantity = \(\$ 12\) / \(0.02\) = \(\$ 600\)
            \end{solutionordottedlines}
            \part \(6 \%\) of it is \(750 \mathrm{~g}\)
            \begin{solutionordottedlines}[0.5in]
                Quantity = \(750 \mathrm{~g}\) / \(0.06\) = \(12500 \mathrm{~g}\) or \(12.5 \mathrm{~kg}\)
            \end{solutionordottedlines}
            \part \(30 \%\) of it is 36 minutes
            \begin{solutionordottedlines}[0.5in]
                Quantity = \(36 \text{ minutes}\) / \(0.3\) = \(120 \text{ minutes}\) or \(2 \text{ hours}\)
            \end{solutionordottedlines}
            \part \(90 \%\) of it is \(54 \mathrm{~cm}\)
            \begin{solutionordottedlines}[0.5in]
                Quantity = \(54 \mathrm{~cm}\) / \(0.9\) = \(60 \mathrm{~cm}\)
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[2] In each part, find the price if:
        \begin{parts}
            \part a deposit of \(\$ 360\) is \(30 \%\) of the price
            \begin{solutionordottedlines}[0.5in]
                Price = \(\$ 360\) / \(0.3\) = \(\$ 1200\)
            \end{solutionordottedlines}
            \part a deposit of \(\$ 168\) is \(15 \%\) of the price
            \begin{solutionordottedlines}[0.5in]
                Price = \(\$ 168\) / \(0.15\) = \(\$ 1120\)
            \end{solutionordottedlines}
        \end{parts}
        \question[2] Find the original quantity, correct to a suitable number of decimal places, if:
        \begin{parts}\begin{multicols}{2}
            \part \(23 \%\) of it is \(100 \mathrm{~kg}\)
            \begin{solutionordottedlines}[0.5in]
                Quantity = \(100 \mathrm{~kg}\) / \(0.23\) = \(434.78 \mathrm{~kg}\)
            \end{solutionordottedlines}
            \part \(0.2 \%\) of it is \(4 \mathrm{~mm}\)
            \begin{solutionordottedlines}[0.5in]
                Quantity = \(4 \mathrm{~mm}\) / \(0.002\) = \(2000 \mathrm{~mm}\) or \(2 \mathrm{~m}\)
            \end{solutionordottedlines}
            \part \(0.92 \%\) of it is 1.86 hectares
            \begin{solutionordottedlines}[0.5in]
                Quantity = \(1.86 \text{ hectares}\) / \(0.0092\) = \(20.22 \text{ hectares}\)
            \end{solutionordottedlines}
            \part \(97 \%\) of it is \(\$ 700\)
            \begin{solutionordottedlines}[0.5in]
                Quantity = \(\$ 700\) / \(0.97\) = \(\$ 721.65\)
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[2] Cameron and Wendy together earn \(\$ 1156\) per week after tax. Of this, they pay \(\$ 460\) off their mortgage, \(\$ 185\) for groceries, and \(\$ 260\) for their car and transport, and they save \$124. 
        \begin{parts}
            \part Express each amount as a percentage of their weekly income, correct to the nearest \(1 \%\).
            \begin{solutionordottedlines}[1in]
                Mortgage: \((\$ 460 / \$ 1156) \times 100 \approx 39.8\%\) (rounded to \(40\%\)) \\
                Groceries: \((\$ 185 / \$ 1156) \times 100 \approx 16.0\%\) (rounded to \(16\%\)) \\
                Car and transport: \((\$ 260 / \$ 1156) \times 100 \approx 22.5\%\) (rounded to \(23\%\)) \\
                Savings: \((\$ 124 / \$ 1156) \times 100 \approx 10.7\%\) (rounded to \(11\%\))
            \end{solutionordottedlines}
            \part Find how much is unaccounted for in the list above, and what percentage it is of their weekly income.
            \begin{solutionordottedlines}[1in]
                Unaccounted for: \( \$ 1156 - (\$ 460 + \$ 185 + \$ 260 + \$ 124) = \$ 1156 - \$ 1029 = \$ 127 \) \\
                Percentage: \((\$ 127 / \$ 1156) \times 100 \approx 11.0\%\) (rounded to \(11\%\))
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{exercisebox}


|722224427|kitty|/Applications/kitty.app|0|
\begin{examplebox}
    \subsection{Examples}
    \begin{questions}
        \question[1] Alice allocates \(10 \%\) of her monthly net income for vacation savings. If she saves \(\$ 120\) monthly, what is her net monthly income?
        \begin{solutionordottedlines}[1in]
            Vacation savings \(=\) net monthly income \(\times 10 \%\)

            Calculating her income:

            \[
            \begin{aligned}
            \text { Net monthly income } & = \text { savings } \div 10 \% \\
            & = \$ 120 \div 0.10 \\
            & = \$ 1200
            \end{aligned}
            \]
        \end{solutionordottedlines}

        \question[2] Bronze is made up of \(88 \%\) copper and \(12 \%\) tin.
        \begin{parts}
            \part How much bronze can be made with \(8 \mathrm{~kg}\) of copper, assuming unlimited tin?
            \begin{solutionordottedlines}[1in]
                Mass of copper \(=\) mass of bronze \(\times 88 \%\)

                Calculating the mass of bronze:

                \[
                \begin{aligned}
                \text{Mass of bronze} & = \text{mass of copper} \div 88 \% \\
                & = 8 \div 0.88 \\
                & \approx 9.09 \mathrm{~kg}
                \end{aligned}
                \]
            \end{solutionordottedlines}
            \part How much bronze can be produced with \(8 \mathrm{~kg}\) of tin, given enough copper?
            \begin{solutionordottedlines}[1in]
                Mass of tin \(=\) mass of bronze \(\times 12 \%\)

                Calculating the mass of bronze:

                \[
                \begin{aligned}
                \text{Mass of bronze} & = \text{mass of tin} \div 12 \% \\
                & = 8 \div 0.12 \\
                & \approx 66.67 \mathrm{~kg}
                \end{aligned}
                \]
            \end{solutionordottedlines}
        \end{parts}
        \question[2] \textbf{Art Gallery Commission:}\\The Artistic Vision Gallery charges a commission of \(10 \%\).
        \begin{parts}
            \part An artwork titled "Dawn of Nature" was sold for \(\$ 200000\). How much was the gallery's commission, and what was the artist's share?
            \begin{solutionordottedlines}[1in]
                Commission \(= \$ 200000 \times 10 \%\)

                \[
                \begin{aligned}
                & = \$ 200000 \times 0.10 \\
                & = \$ 20000
                \end{aligned}
                \]

                Artist's share \(= \$ 200000 - \$ 20000\)

                \[
                = \$ 180000
                \]
            \end{solutionordottedlines}
            \part For a sale of \(\$ 9500\) by the gallery, how much did the artist receive?
            \begin{solutionordottedlines}[1in]
                Commission \(=\) sale amount \(\times 10 \%\)

                \[
                \begin{aligned}
                & = \$ 9500 \times 0.10 \\
                & = \$ 950
                \end{aligned}
                \]

                Artist's share \(= \$ 9500 - \$ 950\)

                \[
                = \$ 8550
                \]
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{examplebox}

\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \question[2] Determine the original amount based on the given percentage:
        \begin{parts}\begin{multicols}{2}
            \part \(3 \%\) of it is \(\$ 18\)
            \begin{solutionordottedlines}[0.5in]
                Original amount = \(\$ 18\) / \(0.03\) = \(\$ 600\)
            \end{solutionordottedlines}
            \part \(5 \%\) of it is \(500 \mathrm{~ml}\)
            \begin{solutionordottedlines}[0.5in]
                Original amount = \(500 \mathrm{~ml}\) / \(0.05\) = \(10000 \mathrm{~ml}\) or \(10 \mathrm{~L}\)
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[2] Calculate the total cost given a partial payment:
        \begin{parts}
            \part A deposit of \(\$ 400\) is \(20 \%\) of the total cost
            \begin{solutionordottedlines}[0.5in]
                Total cost = \(\$ 400\) / \(0.20\) = \(\$ 2000\)
            \end{solutionordottedlines}
            \part A deposit of \(\$ 250\) is \(25 \%\) of the total cost
            \begin{solutionordottedlines}[0.5in]
                Total cost = \(\$ 250\) / \(0.25\) = \(\$ 1000\)
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{exercisebox}
|722224579|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\gradetable[v][pages]
|722230952|kitty|/Applications/kitty.app|0|
https://www.youtube.com/watch?v=5LOrxdwld7A|722259513|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
https://youtube.com/shorts/qKW3d2FK2oY?si=jIEWGqDqlrQDT3RT|722260085|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
https://youtube.com/shorts/OXcaE-n9QCA?si=V9mlUUVNyLfVnvpK|722260588|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
2023

2024
K chillin
AJ trynna pass
    - fin degree

2025
K Doctorate????????????????????
AJ ballin


|722309494|kitty|/Applications/kitty.app|0|
\documentclass[12pt,twoside,addpoints]{exam}|722311426|kitty|/Applications/kitty.app|0|
Image: 225x72 (66.6 KB)|722582846|Preview|/System/Applications/Preview.app|1|56433b93aa4c25d6ee56bf499bf6cae6c746afb2.tiff
file:///Users/aayushbajaj/door.png|722583041|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Image: 564x912 (2.0 MB)|722583647|Preview|/System/Applications/Preview.app|1|d94918ff3c3946fcc20081244b0f7a7afa0d73fa.tiff
Image: 576x151 (343.1 KB)|722583695|Preview|/System/Applications/Preview.app|1|7e13fbfcda255938720efd14fe56a3cff6d4f211.tiff
Cisco Certified Network Associate (CCNA)|722588446|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Image: 12x77 (18.7 KB)|722589079|Brave Browser Beta|/Applications/Brave Browser Beta.app|1|73e2f8cfb4a870bfcc5c5386a9f68fd573397fc5.tiff
A host with a private IP address 192.168.0.2 opens a TCP socket
on its local port 4567 and connects to a web server at 34.5.6.7.
The NAT’s public IP address is 22.33.44.55. Suppose the NAT
created the mapping [22.33.44.55, 3967]à[192.168.0.2, 4567]
as a result. What are the source and destination port numbers
in the SYN-ACK response from the server?|722668718|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Lzzy joins a BitTorrent torrent to download a file and connects to 6* peers – Alice, Bob, Joan, Lemmy, Axl and Ozzy. The file Lzzy wishes to download is divided into 6 chunks. Each peer tracks its availability of chunks using a vector of 6 bits. When the bit value is 1, the peer holds the chunk on the disk. When the bit value is 0, the chunk is not on the peer’s disk. For example, (1, 0, 1, 0, 1, 0) indicates Chunks 1, 3 and 5 are available at this particular peer. 

 

The vectors of Alice, Bob, Joan, Lemmy, Axl, and Ozzy are shown below: 

Alice:          (0, 1, 1, 1, 1, 0)

Bob:           (0, 0, 1, 1, 1, 0)

Joan:          (0, 1, 1, 1, 1, 1)
Lemmy:     (1, 1, 1, 0, 1, 0)
Axl:             (0, 0, 1, 0, 0, 0) 
Ozzy:          (1, 0, 1, 1, 1, 0)

 

Assume that Alice, Bob, Joan, Lemmy, Axl, and Ozzy are not interested in the file anymore and stop downloading the remaining chunks but continue to participate in the torrent and service requests for chunks. 

 

What is the order in which Lzzy requests the chunks for downloading the file. Select the appropriate options in the right section. 

 

* Assume that this version's BitTorrent client allows a user to select the number of peers that he/she wishes to connect to.

 |722731597|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
MPORTANT: This question focuses on generic pipelined reliable delivery protocols (GBN and SR) and not specifically on TCP. Keep this in mind when providing your answers.

 

Assume that we are using 5 bit sequence numbers and the window size is 16. Assume that data only flows in one direction from a sender to a receiver. Supposed that the receiver has received all packets up to and including sequence number 29 and next receives packets with sequence numbers 28, 30 and 31, in that order.

 

(a) Assume that Go-Back-N is used (assume that the Go-Back-N implementation buffers out-of-order packet in the receiver). What are the sequence numbers in the ACK(s) sent out by the receiver in response to the above noted packets. Simply enter the answers in the space provided below. No explanation is necessary. 0.5 mark for each answer.

 

ACK sent in response to packet with sequence number 28 = 

ACK sent in response to packet with sequence number 30 = 

ACK sent in response to packet with sequence number 31 = 

 

(b) Now assume that instead of Go-Back-N, Selective Repeat is used. What are the sequence numbers in the ACK(s) sent out by the receiver in response to the above noted packets. Simply enter the answers in the space provided below. No explanation is necessary. 0.5 mark for each answer. 

ACK sent in response to packet with sequence number 28 =

ACK sent in response to packet with sequence number 30 = 

ACK sent in response to packet with sequence number 31 =|722731673|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Image: 1562x1054 (4.7 MB)|722731762|Brave Browser Beta|/Applications/Brave Browser Beta.app|1|dc1729ad54257d1ce08ee88c47ba3a2232d0abbe.tiff
Consider the sequence of segments exchanged over a TCP Reno connection between Host A and Host B as depicted in the picture below. Assume that all segments sent prior to the sequence of segments shown below have been correctly received at both hosts. Neglect connection setup and teardown. A total of 8 segments are shown of which one segment sent from Host A  (with sequence number c) and  one segment sent from Host B  (with sequence number g) are lost. No other segments are lost. The relevant timeout periods are indicated for you to determine if any one of the segments may be a retransmission. In the event of a timeout, the oldest unacknowledged segment is immediately retransmitted. Disregard TCP congestion control.|722731858|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Assume that the TCP connection has an MSS of 1000 bytes and all packets are 1 MSS. Assume that the the round-trip-time between sender and receiver is 100 milliseconds and fixed. Assume at time 0 the sender attempts to open the connection and it takes 1 round trip time to setup the connection. Assume that the transmission time of packets is negligible. Assume that there is no other traffic on the network. Note that, 1KBytes = 1000 bytes. (1 mark)
How much time has progressed between Point A and B? Explain briefly (2 - 3 sentences).|722733569|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Assume that the TCP connection has an MSS of 1000 bytes and all packets are 1 MSS. Assume that the the round-trip-time between sender and receiver is 100 milliseconds and fixed. Assume at time 0 the sender attempts to open the connection and it takes 1 round trip time to setup the connection. Assume that the transmission time of packets is negligible. Assume that there is no other traffic on the network. Note that, 1KBytes = 1000 bytes. (1 mark)
How much time has progressed between Points C and D? Explain briefly (2 - 3 sentences).|722733859|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Image: 1640x740 (3.5 MB)|722733875|Brave Browser Beta|/Applications/Brave Browser Beta.app|1|7be5dd1dec72db56da0f429c0f8c1ecf3244c14c.tiff
Assume that the TCP connection has an MSS of 1000 bytes and all packets are 1 MSS. Assume that the the round-trip-time between sender and receiver is 100 milliseconds and fixed. Assume at time 0 the sender attempts to open the connection and it takes 1 round trip time to setup the connection. Assume that the transmission time of packets is negligible. Assume that there is no other traffic on the network. Note that, 1KBytes = 1000 bytes. (1 mark)
How much time has progressed between Points D and E? Explain briefly (2 - 3 sentences).|722734479|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
(1) Assume that an ISP has 4 subscribers which have been allocated the following IP address blocks: 

 

192.122.252.0/24

192.122.253.0/24

192.122.254.0/24

192.122.255.0/24

 

The ISP would like to aggregate the above blocks into a single address block and advertise this block for the purpose of routing. The advertised IP address block should not contain IP addresses that do not belong to the above 4 blocks of addresses.

 

Note down the advertised IP address block in the space provided below in the a.b.c.d/x format. No explanation is required. |722734681|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
(2) An organisation has a class C network address block 220.94.95.0 and wishes to form 2 equally sized subnets. 

 

Note down the two advertised IP address blocks in the spaces provided below in the a.b.c.d/x format. No explanation is required.

 

Block 1: (Please note that the order of the blocks won't matter.)


 

Block 2: (Please note that the order of the blocks won't matter.) 

|722734874|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
The figure below shows a local (private) network with a NAT router and its translation table. In Step 1, host  192.168.101.11 sends a datagram from its port 2222 to a destination host 129.99.10.101’s port 80 (not shown in the figure). The NAT router will modify certain header fields after examining the NAT translation table.

Background image

 

In Step 2, the destination IP address and port number of the outgoing datagram are provided. What are the source IP address and port number within this datagram?
Source IP address (in a.b.c.d format): 
Source port number: 
In Step 3, the destination host 129.99.10.101 will return a datagram from its port 80 back to   192.168.101.11.  The source IP address and port number of the returning datagram are provided. What are the destination IP address and port number within this datagram before it arrives the NAT router?
Destination IP address (in a.b.c.d format): 
  
Destination port number: 
In Step 4, The NAT router will notify certain header field of the returning datagram from  129.99.10.101 before forwarding it to the local network. What are the destination IP address and port number within this datagram?
Destination IP address (in a.b.c.d format): 
Destination port number: |722735507|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
An IP datagram which is itself a fragment (of a larger datagram) is of size 1,080 bytes (inclusive of the IP header) and has an offset field of 90. It arrives at a router which has to forward it on an outgoing link with MTU of 780 bytes, and thus needs to create two fragments. What are the offset fields of these two fragments (fill in the values in the provided spaces)? Assume that IP headers are always 20 bytes (i.e. no options are used).

 

Offset for first fragment: 

Offset for second fragment: |722735553|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Assume that Host A connected to an Ethernet network has an INCORRECT implementation of the exponential back-off algorithm such that after each collision, the value of K will be chosen from the set {0, 1, ..., n - 1} as opposed to the set {0, 1, … 2n-1} used in Ethernet (n corresponds to the number of consecutive collisions). Now, another Host B, which is from another manufacturer and obeys the Ethernet back-off algorithm, is trying to send a frame at the same time as host A. Assume that A and B have already collided three times consecutively. What is the probability that host B will succeed in transmitting a frame before host A in its next attempted transmission? Explain your answer.|722735682|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Consider the network shown in the figure below which comprises 6 hosts (H1-H6), 4 Ethernet switches (S1-S4) and a router R. Suppose that the switch tables in all switches are empty and that the router forwarding tables are correctly configured. Once an entry is added to a switch table, you may assume that it persists forever.

 Background image

Answer the following questions.

(1) Suppose H4 sends a frame to H5. Which switches learn where H4 is? The Ethernet adapters of which hosts other than H5 may receive this frame? Explain your answer. (1.5 marks)

 

(2)Next, suppose H2 wants to send an IP datagram to H4 and knows H4’s IP address. Must H2 also know H4’s MAC address to send the datagram to H4? If so, how does H2 get this information? If not, explain why not. (1.5 marks)

 

(3) When Router R sends the frame encapsulating the datagram from question 2 to H4, the Ethernet adapters of which hosts other than H4 may receive this frame? Which switches (in the entire network) learn where H2 is? (2 marks)

|722736216|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
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What are the potential hidden terminals and exposed terminals?|722736818|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Image: 1516x838 (3.6 MB)|722737444|Brave Browser Beta|/Applications/Brave Browser Beta.app|1|e551cbf0c39d55f327d94274cb35a78435650c60.tiff
Image: 644x197 (1.9 MB)|722738297|Brave Browser Beta|/Applications/Brave Browser Beta.app|1|8782e8fb9e3a287512502e75e0500af165c437d4.tiff
Image: 1052x698 (2.1 MB)|722738922|Brave Browser Beta|/Applications/Brave Browser Beta.app|1|f0c0810d61d971e8117b9bca49436592c1af0754.tiff
   A   B   C   D   E   F   G   H
A  0   4   0   0   0   0   2   0
B  4   0  10   0   0   0   0   4
C  0  10   0   1   0   0   0   0
D  0   0   1   0   2   0   0   0
E  0   0   0   2   0   2   0   2
F  0   0   0   0   2   0   6   4
G  2   0   0   0   0   6   0   1
H  0   4   0   0   2   4   1   0
|722739325|||0|
Based on the execution of the Dijkstra's algorithm in the above question, draw the forwarding

table for node F, which contains the outgoing link for reaching every other node in the network. A link between two nodes x and y should be denoted as (x, y).|722739446|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
ere is the updated link-state routing table from node F to every other node, formatted as specified:

Step	N	D(A), p(A)	D(B), p(B)	D(C), p(C)	D(D), p(D)	D(E), p(E)	D(F), p(F)	D(G), p(G)	D(H), p(H)
0	FE	-	-	-	-	2, F	0, F	-	-
1	FEH	-	-	-	-	2, F	0, F	-	4, E
2	FEHD	-	-	-	4, E	2, F	0, F	-	4, E
3	FEHDC	-	-	5, D	4, E	2, F	0, F	-	4, E
4	FEHDCG	-	-	5, D	4, E	2, F	0, F	5, H	4, E
5	FEHDCGB	-	8, H	5, D	4, E	2, F	0, F	5, H	4, E
6	FEHDCGBA	7, G	8, H	5, D	4, E	2, F	0, F	5, H	4, E|722739592|||0|
ere is the updated link-state routing table from node F to every other node, formatted as specified:
|722739653|kitty|/Applications/kitty.app|0|
Step	N	D(A), p(A)	D(B), p(B)	D(C), p(C)	D(D), p(D)	D(E), p(E)	D(F), p(F)	D(G), p(G)	D(H), p(H)
0	FE	-	-	-	-	2, F	0, F	-	-
1	FEH	-	-	-	-	2, F	0, F	-	4, E
2	FEHD	-	-	-	4, E	2, F	0, F	-	4, E
3	FEHDC	-	-	5, D	4, E	2, F	0, F	-	4, E
4	FEHDCG	-	-	5, D	4, E	2, F	0, F	5, H	4, E
5	FEHDCGB	-	8, H	5, D	4, E	2, F	0, F	5, H	4, E
6	FEHDCGBA	7, G	8, H	5, D	4, E	2, F	0, F	5, H	4, E|722739669|||0|
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DSC00114.JPG|722985976|Finder|/System/Library/CoreServices/Finder.app|0|
{{< image-gallery gallery_dir="album/23-11-19-malatang" >}}
|722986054|kitty|/Applications/kitty.app|0|
- 12pm 
|722988773|kitty|/Applications/kitty.app|0|
- 12pm aj guitar
|722988774|kitty|/Applications/kitty.app|0|
This one was a mess.
|722989282|kitty|/Applications/kitty.app|0|
- 
|722993505|kitty|/Applications/kitty.app|0|
To books. 

.|722995519|kitty|/Applications/kitty.app|0|
Hey baby, I'm sorry I'm dead.

It's not too bad over here though, it turns out heaven is IN a cloud! Not a terrible deal after all.

Poseidon grants and denies visiting requests down to Earth. I've put my name into the queue, the others say it takes 3 human months before he gets to them - heaven is pretty busy these days.

Oh, you could be of help actually; just keep saving everyone down there to stop overcrowding up here! DR. SANOOOOOOOOO.

I do miss you though. Nobody knows what Othello is up here, they think I'm talking about the play 🙄️.

Once they make the mistake of letting me down I'll run away mi amor. Come and find you, be your guardian angel once again. I'd finally be able to do that density thing. Slip through the floor into the living room, steal jewellery for you, walk through walls, float into a cloud. But fuck, you wouldn't be able to see me. It's okay, you've seen enough anyways, you always kept your eyes open when we kissed. Now you can close them and you know exactly what I look like 🤠️.

I love you Kiyo. I'll wait forever for you.

Yours and no-one elses,
AJ
|722995661|kitty|/Applications/kitty.app|0|
Hey kiddo, this is my private letter to you once I die. A modicum of closure.
|722995691|kitty|/Applications/kitty.app|0|
Te esperare por siempre|722996202|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
https://abaj.io/kyo/|722997067|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Hey baby, I'm sorry I'm dead.

It's not too bad over here though, it turns out heaven is IN a cloud! Not a terrible deal after all.

Poseidon grants and denies visiting requests down to Earth. I've put my name into the queue, the others say it takes 3 human months before he gets to them - heaven is pretty busy these days.

Oh, you could be of help actually; just keep saving everyone down there to stop overcrowding up here! DR. SANOOOOOOOOO.

I do miss _you_ though. Nobody knows what Othello is up here, they think I'm talking about the play 🙄️.

Once they make the mistake of letting me down I'll run away mi amor. Come and find you, be your guardian angel once again. I'd finally be able to do that density thing. Slip through the floor into the living room, steal jewellery for you, walk through walls, float into a cloud. But fuck, you wouldn't be able to see me. It's okay, you've seen enough anyways, you always kept your eyes open when we kissed. Now you can close them and you'll know exactly what I look like 🤠️.

I love you Kiyo. Te esperare por siempre.

Yours and no-one elses,
AJ

|722997876|kitty|/Applications/kitty.app|0|
\begin{tabular}{ |p{1.5cm}|p{3cm}|p{2cm}|p{6cm}|p{3cm}|  }
    \hline
    \textbf{Status} & \textbf{Booklet Number} & \textbf{ICE-EM 9} & \textbf{Topic} & \textbf{Delivery} \\
    \hline
    \large{\Square} & 1 & CH:1A-1H & Algebra & October '23\\
    \large{\Square} & 2 & CH:2A-D & Pythagoras \& Surds & October '23\\
    \large{\Square} & 3 & CH:2E-I & Pythagoras \& Surds & November '23 \\
    \large{\Square} & 4 & CH:3A-G & Consumer Arithmetic & November '23 \\
    \large{\Square} & 5 & CH:3 & Consumer arithmetic \textbf{Workshop} & November '23 \\
    \large{\Square} & 6 & CH:4A-E & Factorisation & November '23 \\
    \hline
    \large{\Square} & 7 & N/A & \textbf{Class Quiz 1} & November '23 \\
    \hline
    \large{\Square} & 8 & CH:5A-D & Linear Equations \& Inequalities & December '23\\
    \large{\Square} & 9 & CH:5E-H & Linear Equations \& Inequalities & December '23\\
    \large{\Square} & 10 & CH:6A-C & Formulas & December '23\\
    \hline
    \large{\Square} & 11 & N/A & \textbf{Class Quiz 2} & December '23\\
    \hline
    \large{\Square} & 12 & CH:7A-E & Congruence and Special Quadrilaterals & December '23\\
    \large{\Square} & 13 & CH:7 & Congruence Problems \textbf{Workshop} & December '23\\
    \large{\Square} & 14 & CH:8A-E & Index Laws & December '23\\
    \large{\Square} & 15 & CH:9A-E & Enlargements and Similarity & December '23\\
    \large{\Square} & 16 & CH:9 & Similarity \textbf{Workshop} & December '23\\
    \hline
    \large{\Square} & 17 & N/A & \textbf{Class Quiz 3} & December '23\\
    \hline
    \large{\Square} & 18 & CH:11A-D & Coordinate Geometry & January '24\\
    \large{\Square} & 19 & CH:11E-H & Coordinate Geometry & January '24\\
    \large{\Square} & 20 & CH:12A-E & Probability & January '24\\
    \large{\Square} & 21 & CH:12 & Probability \textbf{Workshop} & January '24\\
    \hline
    \large{\Square} & 22 & N/A & \textbf{Class Quiz 4} & January '24\\
    \hline
    \large{\Square} & 23 & CH:13A-E & Trigonometry & January '24\\
    \large{\Square} & 24 & CH:13F & Trigonometry + \textbf{Workshop} & January '24\\
    \large{\Square} & 25 & CH:14A-D & Simultaneous Linear Equations & January '24\\
    \large{\Square} & 26 & CH:14E-F & Simultaneous Linear Equations & January '24\\
    \large{\Square} & 27 & CH:15A-E & Further Factorisation & January '24\\
    \hline
    \large{\Square} & 28 & N/A & \textbf{Class Quiz 5} & January '24\\
    \hline
    \large{\Square} & 29 & CH:16A-C & Measurement - \small{Areas, Volumes and Time} & January '24 \\
    \large{\Square} & 30 & CH:16D-F &  Measurement - \small{Areas, Volumes and Time} &  January '24\\
    \large{\Square} & 31 & CH:16 & Measurement \textbf{Workshop} & January '24\\
    \large{\Square} & 32 & CH:17A-D & Quadratic Equations & January '24\\
    \large{\Square} & 33 & CH:17E-G & Quadratic Equations & January '24\\
    \hline
    \large{\Square} & 34 & N/A & \textbf{Class Quiz 6} & January '24\\
    \hline
    \large{\Square} & 35 & CH:18A-B & Rates and Direct Proportion & January '24\\
    \large{\Square} & 36 & CH:19A-C & Statistics & January '24\\
    \large{\Square} & 37 & CH:19D-E & Statistics & January '24\\
    \hline
    \large{\Square} & 38 & All & Yearly Open Review & January '24\\
    \large{\Square} & 39 & N/A & \textbf{Final Exam} & January '24\\
    \hline
    \large{\Square} & 40 & N/A & Spare Booklet 1 & February '24\\
    \large{\Square} & 41 & N/A & Spare Booklet 2 & February '24\\
    \large{\Square} & 42 & N/A & Spare Booklet 3 & February '24\\
    \large{\Square} & 43 & N/A & Spare Booklet 4 & February '24\\
    \large{\Square} & 44 & N/A & Spare Booklet 5 & February '24\\
    \large{\Square} & 45 & N/A & Spare Booklet 6 & February '24\\
    \large{\Square} & 46 & N/A & Spare Booklet 7 & February '24\\
    \large{\Square} & 47 & N/A & Spare Booklet 8 & February '24\\
    \large{\Square} & 48 & N/A & Spare Booklet 9 & February '24\\
    \large{\Square} & 49 & N/A & Spare Booklet 10 & February '24\\
    \hline
\end{tabular}
|723009540|kitty|/Applications/kitty.app|0|
  **Status**   **Booklet Number**   **ICE-EM 9**   **Topic**                               **Delivery**
  ------------ -------------------- -------------- --------------------------------------- --------------
               1                    CH:1A-1H       Algebra                                 October '23
               2                    CH:2A-D        Pythagoras & Surds                      October '23
               3                    CH:2E-I        Pythagoras & Surds                      November '23
               4                    CH:3A-G        Consumer Arithmetic                     November '23
               5                    CH:3           Consumer arithmetic **Workshop**        November '23
               6                    CH:4A-E        Factorisation                           November '23
               7                    N/A            **Class Quiz 1**                        November '23
               8                    CH:5A-D        Linear Equations & Inequalities         December '23
               9                    CH:5E-H        Linear Equations & Inequalities         December '23
               10                   CH:6A-C        Formulas                                December '23
               11                   N/A            **Class Quiz 2**                        December '23
               12                   CH:7A-E        Congruence and Special Quadrilaterals   December '23
               13                   CH:7           Congruence Problems **Workshop**        December '23
               14                   CH:8A-E        Index Laws                              December '23
               15                   CH:9A-E        Enlargements and Similarity             December '23
               16                   CH:9           Similarity **Workshop**                 December '23
               17                   N/A            **Class Quiz 3**                        December '23
               18                   CH:11A-D       Coordinate Geometry                     January '24
               19                   CH:11E-H       Coordinate Geometry                     January '24
               20                   CH:12A-E       Probability                             January '24
               21                   CH:12          Probability **Workshop**                January '24
               22                   N/A            **Class Quiz 4**                        January '24
               23                   CH:13A-E       Trigonometry                            January '24
               24                   CH:13F         Trigonometry + **Workshop**             January '24
               25                   CH:14A-D       Simultaneous Linear Equations           January '24
               26                   CH:14E-F       Simultaneous Linear Equations           January '24
               27                   CH:15A-E       Further Factorisation                   January '24
               28                   N/A            **Class Quiz 5**                        January '24
               29                   CH:16A-C       Measurement - Areas, Volumes and Time   January '24
               30                   CH:16D-F       Measurement - Areas, Volumes and Time   January '24
               31                   CH:16          Measurement **Workshop**                January '24
               32                   CH:17A-D       Quadratic Equations                     January '24
               33                   CH:17E-G       Quadratic Equations                     January '24
               34                   N/A            **Class Quiz 6**                        January '24
               35                   CH:18A-B       Rates and Direct Proportion             January '24
               36                   CH:19A-C       Statistics                              January '24
               37                   CH:19D-E       Statistics                              January '24
               38                   All            Yearly Open Review                      January '24
               39                   N/A            **Final Exam**                          January '24
               40                   N/A            Spare Booklet 1                         February '24
               41                   N/A            Spare Booklet 2                         February '24
               42                   N/A            Spare Booklet 3                         February '24
               43                   N/A            Spare Booklet 4                         February '24
               44                   N/A            Spare Booklet 5                         February '24
               45                   N/A            Spare Booklet 6                         February '24
               46                   N/A            Spare Booklet 7                         February '24
               47                   N/A            Spare Booklet 8                         February '24
               48                   N/A            Spare Booklet 9                         February '24
               49                   N/A            Spare Booklet 10                        February '24
|723009268|kitty|/Applications/kitty.app|0|
| Status | Booklet Number | ICE-EM 9 | Topic | Delivery |
|--------|----------------|----------|-------|----------|
| ☐      | 1              | CH:1A-1H | Algebra | October '23 |
| ☐      | 2              | CH:2A-D  | Pythagoras & Surds | October '23 |
| ☐      | 3              | CH:2E-I  | Pythagoras & Surds | November '23 |
| ☐      | 4              | CH:3A-G  | Consumer Arithmetic | November '23 |
| ☐      | 5              | CH:3     | Consumer arithmetic **Workshop** | November '23 |
| ☐      | 6              | CH:4A-E  | Factorisation | November '23 |
| ☐      | 7              | N/A      | **Class Quiz 1** | November '23 |
| ☐      | 8              | CH:5A-D  | Linear Equations & Inequalities | December '23 |
| ☐      | 9              | CH:5E-H  | Linear Equations & Inequalities | December '23 |
| ☐      | 10             | CH:6A-C  | Formulas | December '23 |
| ☐      | 11             | N/A      | **Class Quiz 2** | December '23 |
| ☐      | 12             | CH:7A-E  | Congruence and Special Quadrilaterals | December '23 |
| ☐      | 13             | CH:7     | Congruence Problems **Workshop** | December '23 |
| ☐      | 14             | CH:8A-E  | Index Laws | December '23 |
| ☐      | 15             | CH:9A-E  | Enlargements and Similarity | December '23 |
| ☐      | 16             | CH:9     | Similarity **Workshop** | December '23 |
| ☐      | 17             | N/A      | **Class Quiz 3** | December '23 |
| ☐      | 18             | CH:11A-D | Coordinate Geometry | January '24 |
| ☐      | 19             | CH:11E-H | Coordinate Geometry | January '24 |
| ☐      | 20             | CH:12A-E | Probability | January '24 |
| ☐      | 21             | CH:12    | Probability **Workshop** | January '24 |
| ☐      | 22             | N/A      | **Class Quiz 4** | January '24 |
| ☐      | 23             | CH:13A-E | Trigonometry | January '24 |
| ☐      | 24             | CH:13F   | Trigonometry + **Workshop** | January '24 |
| ☐      | 25             | CH:14A-D | Simultaneous Linear Equations | January '24 |
| ☐      | 26             | CH:14E-F | Simultaneous Linear Equations | January '24 |
| ☐      | 27             | CH:15A-E | Further Factorisation | January '24 |
| ☐      | 28             | N/A      | **Class Quiz 5** | January '24 |
| ☐      | 29             | CH:16A-C | Measurement - Areas, Volumes and Time | January '24 |
| ☐      | 30             | CH:16D-F | Measurement - Areas, Volumes and Time | January '24 |
| ☐      | 31             | CH:16    | Measurement **Workshop** | January '24 |
| ☐      | 32             | CH:17A-D | Quadratic Equations | January '24 |
| ☐      | 33             | CH:17E-G | Quadratic Equations | January '24 |
| ☐      | 34             | N/A      | **Class Quiz 6** | January '24 |
| ☐      | 35             | CH:18A-B | Rates and Direct Proportion | January '24 |
| ☐      | 36             | CH:19A-C | Statistics | January '24 |
| ☐      | 37             | CH:19D-E | Statistics | January '24 |
| ☐      | 38             | All      | Yearly Open Review | January '24 |
| ☐      | 39             | N/A      | **Final Exam** | January '24 |
| ☐      | 40             | N/A      | Spare Booklet 1 | February '24 |
| ☐      | 41             | N/A      | Spare Booklet 2 | February '24 |
| ☐      | 42             | N/A      | Spare Booklet 3 | February '24 |
| ☐      | 43             | N/A      | Spare Booklet 4 | February '24 |
| ☐      | 44             | N/A      | Spare Booklet 5 | February '24 |
| ☐      | 45             | N/A      | Spare Booklet 6 | February '24 |
| ☐      | 46             | N/A      | Spare Booklet 7 | February '24 |
| ☐      | 47             | N/A      | Spare Booklet 8 | February '24 |
| ☐      | 48             | N/A      | Spare Booklet 9 | February '24 |
| ☐      | 49             | N/A      | Spare Booklet 10 | February '24 |
|723009695|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
  **Status**   **Booklet Number**   **ICE-EM 9**   **Topic**                               **Delivery**
  ------------ -------------------- -------------- --------------------------------------- --------------
               1                    CH:1A-1H       Algebra                                 October '23
               2                    CH:2A-D        Pythagoras & Surds                      October '23
               3                    CH:2E-I        Pythagoras & Surds                      November '23
               4                    CH:3A-G        Consumer Arithmetic                     November '23
               5                    CH:3           Consumer arithmetic **Workshop**        November '23
               6                    CH:4A-E        Factorisation                           November '23
               7                    N/A            **Class Quiz 1**                        November '23
               8                    CH:5A-D        Linear Equations & Inequalities         December '23
               9                    CH:5E-H        Linear Equations & Inequalities         December '23
               10                   CH:6A-C        Formulas                                December '23
               11                   N/A            **Class Quiz 2**                        December '23
               12                   CH:7A-E        Congruence and Special Quadrilaterals   December '23
               13                   CH:7           Congruence Problems **Workshop**        December '23
               14                   CH:8A-E        Index Laws                              December '23
               15                   CH:9A-E        Enlargements and Similarity             December '23
               16                   CH:9           Similarity **Workshop**                 December '23
               17                   N/A            **Class Quiz 3**                        December '23
               18                   CH:11A-D       Coordinate Geometry                     January '24
               19                   CH:11E-H       Coordinate Geometry                     January '24
               20                   CH:12A-E       Probability                             January '24
               21                   CH:12          Probability **Workshop**                January '24
               22                   N/A            **Class Quiz 4**                        January '24
               23                   CH:13A-E       Trigonometry                            January '24
               24                   CH:13F         Trigonometry + **Workshop**             January '24
               25                   CH:14A-D       Simultaneous Linear Equations           January '24
               26                   CH:14E-F       Simultaneous Linear Equations           January '24
               27                   CH:15A-E       Further Factorisation                   January '24
               28                   N/A            **Class Quiz 5**                        January '24
               29                   CH:16A-C       Measurement - Areas, Volumes and Time   January '24
               30                   CH:16D-F       Measurement - Areas, Volumes and Time   January '24
               31                   CH:16          Measurement **Workshop**                January '24
               32                   CH:17A-D       Quadratic Equations                     January '24
               33                   CH:17E-G       Quadratic Equations                     January '24
               34                   N/A            **Class Quiz 6**                        January '24
               35                   CH:18A-B       Rates and Direct Proportion             January '24
               36                   CH:19A-C       Statistics                              January '24
               37                   CH:19D-E       Statistics                              January '24
               38                   All            Yearly Open Review                      January '24
               39                   N/A            **Final Exam**                          January '24
               40                   N/A            Spare Booklet 1                         February '24
               41                   N/A            Spare Booklet 2                         February '24
               42                   N/A            Spare Booklet 3                         February '24
               43                   N/A            Spare Booklet 4                         February '24
               44                   N/A            Spare Booklet 5                         February '24
               45                   N/A            Spare Booklet 6                         February '24
               46                   N/A            Spare Booklet 7                         February '24
               47                   N/A            Spare Booklet 8                         February '24
               48                   N/A            Spare Booklet 9                         February '24
               49                   N/A            Spare Booklet 10                        February '24



|723009698|kitty|/Applications/kitty.app|0|
| ☐      | 1              | CH:1A-1H | Algebra | October '23 |
|723009714|kitty|/Applications/kitty.app|0|
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<|723012268|kitty|/Applications/kitty.app|0|
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-|723012269|kitty|/Applications/kitty.app|0|
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|723012272|kitty|/Applications/kitty.app|0|
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body {
	font-family: sans-serif ;
	background: #110000 ;
	color: #ccc ;
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main {
	max-width: 800px ;
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header h1 {
	text-align: center ;
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footer {
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	clear: both ;
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/* For TAGLIST.HTML */
.taglist {
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/* For NEXTPREV.HTML */
#nextprev {
	/* The container for both the previous and next articles. */
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#prevart {
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#nextart,#prevart {
	max-width: 33% ;
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body {
	font-family: sans-serif ;
	background: #110000 ;
	color: #ccc ;
}

main {
	max-width: 800px ;
	margin: auto ;
}

img {
	max-width: 100% ;
}

header h1 {
	text-align: center ;
}

footer {
	text-align: center ;
	clear: both ;
}

/* For TAGLIST.HTML */
.taglist {
	text-align: center ;
	clear: both ;
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/* For NEXTPREV.HTML */
#nextprev {
	/* The container for both the previous and next articles. */
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#prevart {
	float: left ;
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#nextart {
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#nextart,#prevart {
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main {
	max-width: 800px ;
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:root {
    --main-width: 1200px;
}
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/
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    <link rel='stylesheet' type='text/css' href='/style.css'>
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style|723015895|kitty|/Applications/kitty.app|0|
    <link rel='stylesheet' type='text/css' href="{{ "/custom.css" | resources.Get }}">
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main {
	max-width: 1000px ;
	margin: auto ;
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Goodluck|723016838|kitty|/Applications/kitty.app|0|
. Factorisation|723017034|kitty|/Applications/kitty.app|0|
Another booklet
|723017372|kitty|/Applications/kitty.app|0|

| Status | Booklet Number | ICE-EM 9 | Topic | Delivery |
|--------|----------------|----------|-------|----------|
| [X]      | 1              | CH:1A-1H | Algebra | October '23 |
| [X]      | 2              | CH:2A-D  | Pythagoras & Surds | October '23 |
| [X]      | 3              | CH:2E-I  | Pythagoras & Surds | November '23 |
| [X]      | 4              | CH:3A-G  | Consumer Arithmetic | November '23 |
| [ ]      | 5              | CH:3     | Consumer arithmetic **Workshop** | November '23 |
|723017379|kitty|/Applications/kitty.app|0|
# 6. Factorisation
|723017429|kitty|/Applications/kitty.app|0|
# 6. Class Quiz 1
|723017434|kitty|/Applications/kitty.app|0|
| [ ]      | 6              | CH:4A-E  | Factorisation | November '23 |
| [ ]      | 7              | N/A      | **Class Quiz 1** | November '23 |
| [ ]      | 8              | CH:5A-D  | Linear Equations & Inequalities | December '23 |
| [ ]      | 9              | CH:5E-H  | Linear Equations & Inequalities | December '23 |
|723017468|kitty|/Applications/kitty.app|0|
| [ ]      | 10             | CH:6A-C  | Formulas | December '23 |
|723017518|kitty|/Applications/kitty.app|0|
| [ ]      | 11             | N/A      | **Class Quiz 2** | December '23 |
| [ ]      | 12             | CH:7A-E  | Congruence and Special Quadrilaterals | December '23 |
| [ ]      | 13             | CH:7     | Congruence Problems **Workshop** | December '23 |
| [ ]      | 14             | CH:8A-E  | Index Laws | December '23 |
| [ ]      | 15             | CH:9A-E  | Enlargements and Similarity | December '23 |
| [ ]      | 16             | CH:9     | Similarity **Workshop** | December '23 |
|723017619|kitty|/Applications/kitty.app|0|
# 17. Coordinate Geometry
|723017636|kitty|/Applications/kitty.app|0|

| [ ]      | 17             | N/A      | **Class Quiz 3** | December '23 |
| [ ]      | 18             | CH:11A-D | Coordinate Geometry | January '24 |
| [ ]      | 19             | CH:11E-H | Coordinate Geometry | January '24 |
| [ ]      | 20             | CH:12A-E | Probability | January '24 |
| [ ]      | 21             | CH:12    | Probability **Workshop** | January '24 |
|723017730|kitty|/Applications/kitty.app|0|
| [ ]      | 22             | N/A      | **Class Quiz 4** | January '24 |
| [ ]      | 23             | CH:13A-E | Trigonometry | January '24 |
| [ ]      | 24             | CH:13F   | Trigonometry + **Workshop** | January '24 |
|723017778|kitty|/Applications/kitty.app|0|
一二三|723017956|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
二三|723017960|kitty|/Applications/kitty.app|0|
三|723017972|kitty|/Applications/kitty.app|0|
| [ ]      | 25             | CH:14A-D | Simultaneous Linear Equations | January '24 |
| [ ]      | 26             | CH:14E-F | Simultaneous Linear Equations | January '24 |
| [ ]      | 27             | CH:15A-E | Further Factorisation | January '24 |
| [ ]      | 28             | N/A      | **Class Quiz 5** | January '24 |
| [ ]      | 29             | CH:16A-C | Measurement - Areas, Volumes and Time | January '24 |
| [ ]      | 30             | CH:16D-F | Measurement - Areas, Volumes and Time | January '24 |
| [ ]      | 31             | CH:16    | Measurement **Workshop** | January '24 |
| [ ]      | 32             | CH:17A-D | Quadratic Equations | January '24 |
| [ ]      | 33             | CH:17E-G | Quadratic Equations | January '24 |
| [ ]      | 34             | N/A      | **Class Quiz 6** | January '24 |
|723018110|kitty|/Applications/kitty.app|0|
| [ ]      | 35             | CH:18A-B | Rates and Direct Proportion | January '24 |
| [ ]      | 36             | CH:19A-C | Statistics | January '24 |
| [ ]      | 37             | CH:19D-E | Statistics | January '24 |
|723018114|kitty|/Applications/kitty.app|0|
| [ ]      | 38             | All      | Yearly Open Review | January '24 |
| [ ]      | 39             | N/A      | **Final Exam** | January '24 |
| [ ]      | 40             | N/A      | Spare Booklet 1 | February '24 |
| [ ]      | 41             | N/A      | Spare Booklet 2 | February '24 |
| [ ]      | 42             | N/A      | Spare Booklet 3 | February '24 |
| [ ]      | 43             | N/A      | Spare Booklet 4 | February '24 |
| [ ]      | 44             | N/A      | Spare Booklet 5 | February '24 |
| [ ]      | 45             | N/A      | Spare Booklet 6 | February '24 |
| [ ]      | 46             | N/A      | Spare Booklet 7 | February '24 |
| [ ]      | 47             | N/A      | Spare Booklet 8 | February '24 |
| [ ]      | 48             | N/A      | Spare Booklet 9 | February '24 |
| [ ]      | 49             | N/A      | Spare Booklet 10 | February '24 |
|723018141|kitty|/Applications/kitty.app|0|
[student folder](/penbooklets/student/)
|723018406|kitty|/Applications/kitty.app|0|
student|723018407|kitty|/Applications/kitty.app|0|
    location /yourfolder {
        root /path/to/your/directory;|723018663|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Theory|723021032|kitty|/Applications/kitty.app|0|
Review of Percentages|723027742|kitty|/Applications/kitty.app|0|
Using Percentages|723027755|kitty|/Applications/kitty.app|0|
Simple Interest|723027764|kitty|/Applications/kitty.app|0|
Percentage Increase \& Decrease|723027799|kitty|/Applications/kitty.app|0|
5-percentage-increase|723027808|kitty|/Applications/kitty.app|0|
Repeated Increase \& Decrease|723027819|kitty|/Applications/kitty.app|0|
\section{Compound Interest}
    \input{7-compound-interest}
\section{Depreciation}
    \input{8-depreciation}
|723027936|kitty|/Applications/kitty.app|0|
Here is a box with a number of \textbf{buzz words} from today's class. As you come to understand these words, cross them off here.

\fbox{%
\makebox[\textwidth]{
\randomword{appreciation}%
\randomword{GST}%
\randomword{simple interest}%
\randomword{compound interest}%
\randomword{commissions}%
\randomword{inflation}%
\randomword{depreciation}%
\randomword{discounts}%
}
}
|723032910|kitty|/Applications/kitty.app|0|
appreciation|723032920|kitty|/Applications/kitty.app|0|
GST|723032926|kitty|/Applications/kitty.app|0|
simple interest|723032939|kitty|/Applications/kitty.app|0|
compound interest|723032947|kitty|/Applications/kitty.app|0|
commissions|723032953|kitty|/Applications/kitty.app|0|
inflation|723032963|kitty|/Applications/kitty.app|0|
depreciation|723032978|kitty|/Applications/kitty.app|0|
discounts|723032982|kitty|/Applications/kitty.app|0|
\randomword{}%
|723032996|kitty|/Applications/kitty.app|0|
\randomword{common factor}%
|723033095|kitty|/Applications/kitty.app|0|
 factor|723033098|kitty|/Applications/kitty.app|0|
common |723033100|kitty|/Applications/kitty.app|0|
\randomword{difference of two squares}%
|723033109|kitty|/Applications/kitty.app|0|
difference |723033111|kitty|/Applications/kitty.app|0|
of |723033114|kitty|/Applications/kitty.app|0|
two |723033114|kitty|/Applications/kitty.app|0|
 squares|723033115|kitty|/Applications/kitty.app|0|
 quadratic|723033123|kitty|/Applications/kitty.app|0|
\randomword{non-monic}%
|723033123|kitty|/Applications/kitty.app|0|
 square|723033130|kitty|/Applications/kitty.app|0|
\randomword{perfect}%
|723033131|kitty|/Applications/kitty.app|0|
perfect|723033132|kitty|/Applications/kitty.app|0|
\section*{Example 2}
Factorise: a \(12 x^{2}+3 x\)

b \(36 a b-27 a\)

\section*{Solution}
a \(12 x^{2}+3 x=3\left(4 x^{2}+x\right)\)
b \(36 a b-27 a=9(4 a b-3 a)\)
\(=3 x(4 x+1)\)
\(=9 a(4 b-3)\)

Note: It is more efficient to take out all the common factors in one step, as shown in the following examples.

\section*{Example 3}
Factorise:
a \(3 x+9\)
b \(2 x^{2}+4 x\)
c \(-7 a^{2}-49\)
d \(7 a^{2}+63 a b\)

Solution

a \(3 x+9=3(x+3)\)

b \(2 x^{2}+4 x=2 x(x+2)\)

c \(-7 a^{2}-49=-7 \times a^{2}+(-7) \times 7\)

d \(7 a^{2}+63 a b=7 a(a+9 b)\)

\(=-7\left(a^{2}+7\right)\)

\(7\left(-a^{2}-7\right)\) is also correct.

\section*{Example 4}
Factorise:
a \(5 p q^{2}+10 p^{2} q+25 p^{2} q^{2}\)
b \(16 a b+10 b^{2}-2 a^{2} b\)

\section*{Solution}
a \(5 p q^{2}+10 p^{2} q+25 p^{2} q^{2}=5 p q(q+2 p+5 p q)\)

b \(16 a b+10 b^{2}-2 a^{2} b=2 b\left(8 a+5 b-a^{2}\right)\)
|723033324|kitty|/Applications/kitty.app|0|
c \(36 a b=9 a \times \square\)
|723033351|kitty|/Applications/kitty.app|0|
c \(36 a b=9 a \times \square\)
f \(6 y^{2}=3 y \times \square\)
|723033353|kitty|/Applications/kitty.app|0|
c \(36 a b=9 a \times \square\)
f \(6 y^{2}=3 y \times \square\)
i \(8 a^{2} b=2 a b \times \square\)
|723033357|kitty|/Applications/kitty.app|0|
1 Complete each factorisation.
a \(12 x=12 \times \square\)
b \(24 a=12 \times \square\)
d \(15 a c=5 c \times \square\)
e \(y^{2}=y \times \square\)
g \(24 a^{2}=6 a \times \square\)
h \(-6 b^{2}=2 b \times \square\)
j \(4 x^{2} y=2 x \times \square\)
k \(12 m^{2} n=3 m n \times \square\)
|723033364|kitty|/Applications/kitty.app|0|
\section*{Exercise 4A}
c \(36 a b=9 a \times \square\)
f \(6 y^{2}=3 y \times \square\)
i \(8 a^{2} b=2 a b \times \square\)
l \(25 a^{2} b^{2}=5 a b \times \square\)
|723033381|kitty|/Applications/kitty.app|0|
b \(15 p-10=5 \times\)
|723033389|kitty|/Applications/kitty.app|0|
b \(15 p-10=5 \times\)
d \(20 m n-15 n=5 n \times \square\)
|723033394|kitty|/Applications/kitty.app|0|
b \(15 p-10=5 \times\)
d \(20 m n-15 n=5 n \times \square\)
f \(b^{2}-10 b=b \times\)
|723033400|kitty|/Applications/kitty.app|0|
b \(15 p-10=5 \times\)
d \(20 m n-15 n=5 n \times \square\)
f \(b^{2}-10 b=b \times\)
h \(6 y z^{2}-18 y z=y z \times \square\)
|723033404|kitty|/Applications/kitty.app|0|
b \(15 p-10=5 \times\)
d \(20 m n-15 n=5 n \times \square\)
f \(b^{2}-10 b=b \times\)
h \(6 y z^{2}-18 y z=y z \times \square\)
j \(6 y z^{2}-18 y z=6 y z \times \square\)
|723033406|kitty|/Applications/kitty.app|0|
2 Fill in the blanks by finding the missing factors.
a \(12 a+18=6 \times\)
c \(20 m n-15 n=5 \times \square\)
e \(a^{2}+4 a=(a+4) \times \square\)
g \(6 y z^{2}-18 y z=3 z \times \square\)
i \(6 y z^{2}-18 y z=-3 y z \times \square\)
|723033414|kitty|/Applications/kitty.app|0|
e \(y^{2}+x y\)
|723033422|kitty|/Applications/kitty.app|0|
f \(4 x+24\)
|723033444|kitty|/Applications/kitty.app|0|
i \(y^{2}-3 y\)
|723033451|kitty|/Applications/kitty.app|0|
c \(a c+5 c\)
d \(a^{2}+a\)
g \(7 a-63\)
h \(9 a+36\)
k \(-6 y-9\)
l \(-4-12 b\)
|723033500|kitty|/Applications/kitty.app|0|
c \(18 m^{2} n+9 m n^{2}\)
|723033512|kitty|/Applications/kitty.app|0|
c \(18 m^{2} n+9 m n^{2}\)
f \(8 a^{2}+12 a b\)
|723033515|kitty|/Applications/kitty.app|0|
3 Factorise:
a \(6 x+24\)
b \(5 a+15\)
e \(y^{2}+x y\)
f \(4 x+24\)
i \(y^{2}-3 y\)
j \(-14a - 21\)
|723033752|kitty|/Applications/kitty.app|0|
4 Factorise:
a \(4 a b+16 a\)
b \(12 a^{2}+8 a\)
d \(15 a^{2} b^{2}+10 a b^{2}\)
e \(4 a^{2}+6 a\)
|723033758|kitty|/Applications/kitty.app|0|
c \(9 m n-12 m^{2} n\)
|723033898|kitty|/Applications/kitty.app|0|
c \(9 m n-12 m^{2} n\)
f \(6 x y-4 x^{2}\)
i \(10 a b^{2}-25 a^{2} b\)
|723033908|kitty|/Applications/kitty.app|0|
5 Factorise:
a \(3 b-6 b^{2}\)
b \(4 x^{2}-6 x y\)
d \(18 y-9 y^{2}\)
e \(4 a-6 a b^{2}\)
g \(14 m n^{2}-21 m^{2} n\)
h \(6 p q^{2}-21 q p^{2}\)
|723033918|kitty|/Applications/kitty.app|0|
c \(-x^{2} y-3 x y\)
|723033930|kitty|/Applications/kitty.app|0|
c \(-x^{2} y-3 x y\)
f \(18 p^{2}-4 p q\)
|723033937|kitty|/Applications/kitty.app|0|
c \(-x^{2} y-3 x y\)
f \(18 p^{2}-4 p q\)
i \(-25 m^{2} n^{2}-10 m n^{2}\)
|723033939|kitty|/Applications/kitty.app|0|
6 Factorise:
|723033953|kitty|/Applications/kitty.app|0|
        \question[1]
|723034060|kitty|/Applications/kitty.app|0|
        \question[1]
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
|723034061|kitty|/Applications/kitty.app|0|
\
\begin{multicols}{3}
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            \part \begin{tikzpicture} \end{tikzpicture}
|723034246|kitty|/Applications/kitty.app|0|
            \part 
            \part 
|723034248|kitty|/Applications/kitty.app|0|
    \begin{questions}
|723034257|kitty|/Applications/kitty.app|0|
\begin{multicols}{3}
|723034258|kitty|/Applications/kitty.app|0|
        \question[1]
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \question[1]
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
    \end{questions}
|723034292|kitty|/Applications/kitty.app|0|
\end{multicols}
|723034292|kitty|/Applications/kitty.app|0|
\question[3] In each part, an expression for the area of the rectangle has been given. Find an expression for the missing side length.
    \begin{subparts}\begin{multicols}{3}
        \part 
            \begin{tikzpicture} 
                \draw (0,0) -- (3,0) -- (3,2) -- (0,2) -- cycle;
                \node at (1.5, 2) [above] {$2a + 3$};
                \node at (1.5, 1) {Area = $8a + 12$};
            \end{tikzpicture}
            \begin{solutionordottedlines}[1in]
                $4$
            \end{solutionordottedlines}
        \part \begin{tikzpicture} \end{tikzpicture}
            \begin{solutionordottedlines}[1in]
                $2b+3$
            \end{solutionordottedlines}
        \part \begin{tikzpicture} \end{tikzpicture}
            \begin{solutionordottedlines}[1in]
                $2a + b$
            \end{solutionordottedlines}
    \end{multicols}\end{subparts}
|723034614|kitty|/Applications/kitty.app|0|

7 In each part, an expression for the area of the rectangle has been given. Find an expression for the missing side length.

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_30_d36dc84f51b69993a9a7g-04(1)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_30_d36dc84f51b69993a9a7g-04}
\end{center}

d
\includegraphics[max width=\textwidth, center]{2023_11_30_d36dc84f51b69993a9a7g-04(2)}

f

\(3 q^{2}-2\)

Area \(=24 p^{2} q^{2}-16 p^{2}\)
|723034676|kitty|/Applications/kitty.app|0|
Example 48 Factorise:
|723034678|kitty|/Applications/kitty.app|0|
b \(4 m^{2} n-4 m n+16 n^{2}\)
|723034682|kitty|/Applications/kitty.app|0|
b \(4 m^{2} n-4 m n+16 n^{2}\)
d \(2 m^{2}+4 m n+6 n\)
|723034686|kitty|/Applications/kitty.app|0|
b \(4 m^{2} n-4 m n+16 n^{2}\)
d \(2 m^{2}+4 m n+6 n\)
f \(6 a+8 a b+10 a b^{2}\)
|723034690|kitty|/Applications/kitty.app|0|

a \(4 a^{2} b-2 a b+8 a b^{2}\)


c \(7 a b+14 a^{2}+21 b\)


e \(5 a^{2} b+3 a b+4 a b^{2}\)


g \(5 p^{2} q^{2}+10 p q^{2}+15 p^{2} q\)

|723034700|kitty|/Applications/kitty.app|0|
\section*{\(\triangle \mathrm{B}\) Factorisation using the difference of two squares}
You will recall from Section \(1 \mathrm{G}\) the important identity

\[
(a+b)(a-b)=a^{2}-b^{2},
\]

which is called the difference of two squares. We can now use this result the other way around to factorise an expression that is the difference of two squares.

That is:

\[
a^{2}-b^{2}=(a+b)(a-b)
\]

That is, the factors of \(a^{2}-b^{2}\) are \(a+b\) and \(a-b\).
|723034789|kitty|/Applications/kitty.app|0|
Factorise:
a \(x^{2}-9\)
b \(25-y^{2}\)
c \(4 x^{2}-9\)

\section*{Solution}
a \(x^{2}-9=x^{2}-3^{2}\)

\[
=(x+3)(x-3)
\]

Check the answer by expanding \((x+3)(x-3)\) to see that \(x^{2}-9\) is obtained.

b \(25-y^{2}=5^{2}-y^{2}\)

\[
=(5+y)(5-y)
\]

c \(4 x^{2}-9=(2 x)^{2}-3^{2}\)

\[
=(2 x+3)(2 x-3)
\]

\section*{Example 6}
Factorise:

a \(3 a^{2}-27\)

b \(-16+9 x^{2}\)

\section*{Solution}
a \(3 a^{2}-27=3\left(a^{2}-9\right)\)
b \(-16+9 x^{2}=9 x^{2}-16\)
\(=3\left(a^{2}-3^{2}\right)\)
\(=(3 x)^{2}-4^{2}\)
\(=3(a+3)(a-3)\)
\(=(3 x-4)(3 x+4)\)

Factorisation using the difference of two squares identity

\[
a^{2}-b^{2}=(a+b)(a-b)
\]
|723035040|kitty|/Applications/kitty.app|0|
a \(x^{2}-16\)
|723035065|kitty|/Applications/kitty.app|0|
a \(x^{2}-16\)
e \((2 x)^{2}-25\)
|723035068|kitty|/Applications/kitty.app|0|
a \(x^{2}-16\)
e \((2 x)^{2}-25\)
i \(9 x^{2}-4\)
|723035070|kitty|/Applications/kitty.app|0|
1 Factorise:
b \(x^{2}-49\)
f \((3 x)^{2}-16\)
j \(16 y^{2}-49\)
n \(9-16 y^{2}\)

c \(a^{2}-121\)

g \((4 x)^{2}-1\)

k \(100 a^{2}-49 b^{2}\)

o \(25 a^{2}-100 b^{2}\) d \(d^{2}-400\)
|723035200|kitty|/Applications/kitty.app|0|
h \((5 m)^{2}-9\)

l \(64 m^{2}-81 p^{2}\)

p \(-9+x^{2}\)
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\section*{Exercise 4B}

h \((5 m)^{2}-9\)

l \(64 m^{2}-81 p^{2}\)

p \(-9+x^{2}\)
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a \(x^{2}-16\)
e \((2 x)^{2}-25\)
i \(9 x^{2}-4\)
m \(1-4 a^{2}\)
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\title{Factorisation }

\author{}
\date{}


\begin{document}
\maketitle
Number and Algebra

The distributive law can be used to rewrite a product involving brackets as an expression without brackets. For instance, the product \(3(a+2)\) can be rewritten as \(3 a+6\); this is called the expanded form of the expression, and the process is called expansion.

The process of writing an algebraic expression as a product of two or more algebraic factors is called factorisation. Factorisation is the reverse process to expansion. For example, we can write \(3 a+6\) as \(3(a+2)\). This is called the factorised form.

In this chapter we will look at how to factorise expressions such as \(x^{2}+7 x+12\).

We recall that such an expression is called a quadratic.

\section*{Factorisation using common factors}
If each term in the algebraic expression to be factorised contains a common factor, then this common factor is a factor of the entire expression. To find the other factor, divide each term by the common factor. The common factor is placed outside brackets. For this reason the procedure is sometimes called 'taking the common factor outside the brackets'.


Notice that the answer can be checked by expanding \(4(a+3)\).

In general, take out as many common factors as possible. The common factors may involve both numbers and pronumerals. It may be easier if you first take out the common factor of the numbers and then the common factor of the pronumerals.



\section*{Example 5}


2 Factorise:

\begin{center}
\begin{tabular}{ll}
a \(3 x^{2}-48\) & b \(4 x^{2}-100\) \\
e \(10 x^{2}-1000\) & f \(7 x^{2}-63\) \\
i \(3-12 b^{2}\) & j \(20-5 y^{2}\) \\
m \(45 m^{2}-125 n^{2}\) & n \(27 a^{2}-192 l^{2}\) \\
\end{tabular}
\end{center}

c \(5 x^{2}-45\)
g \(8 x^{2}-50\)
k \(27 a^{2}-12 b^{2}\)
o \(-8 x^{2}+32 y^{2}\)

d \(6 x^{2}-24\)

h \(12 m^{2}-75\)

l \(16 x^{2}-100 y^{2}\)

\(\mathbf{p}-200 p^{2}+32 q^{2}\)

3 Factorise:
a \(-25+x^{2}\)
d \(-81 x^{2}+16\)
g \(-12 x^{2}+27\)

b \(-4+9 x^{2}\)

c \(-9 x^{2}+4\)

e \(-100 x^{2}+9\)

f \(-18+50 x^{2}\)

h \(-36 x^{2}+400\)

i \(-175+28 x^{2}\)

4 Use the factorisation of the difference of two squares to evaluate the following. One has been done for you.

\[
\begin{aligned}
17^{2}-3^{2} & =(17+3)(17-3) \\
& =20 \times 14 \\
& =280
\end{aligned}
\]

a \(23^{2}-7^{2}\)
b \(23^{2}-3^{2}\)
c \(36^{2}-6^{2}\)
d \(94^{2}-6^{2}\)
e \(1.8^{2}-0.2^{2}\)
f \(28^{2}-2.2^{2}\)
g \(11.3^{2}-8.7^{2}\)
h \(92.6^{2}-7.4^{2}\)
i \(3.214^{2}-2.214^{2}\)

5 a Evaluate:
i \(4^{2}-3^{2}\)
ii \(5^{2}-4^{2}\)
iii \(6^{2}-5^{2}\)
iv \(7^{2}-6^{2}\)
v \(8^{2}-7^{2}\)
vi \(9^{2}-8^{2}\)
vii \(10^{2}-9^{2}\)
viii \(101^{2}-100^{2}\)
b What do you notice?

c Use the factorisation of \((n+1)^{2}-n^{2}\) to prove the result of part \(\mathbf{b}\).

6 The following leads you through the geometrical proof that \(a^{2}-b^{2}=(a-b)(a+b)\).

In the diagram opposite, square \(E C F G\), of side length \(b\), is cut out of square \(A B C D\), which has side length \(a\).

a What is the area of hexagon \(A B E G F D\) ?

b What is the length of \(B E\) ?

c What is the length of \(D F\) ?

d The rectangle \(H G F D\) is moved so that \(D F\) is placed on top of \(B E\) (see diagram).

i What is the area of the large shaded rectangle?

ii What have we proved?
\includegraphics[max width=\textwidth, center]{2023_11_30_d36dc84f51b69993a9a7g-06}

\section*{Factorisation of simple quadratics}
A simple quadratic expression is an expression of the form \(x^{2}+b x+c\), where \(b\) and \(c\) are given numbers. When we expand \((x+3)(x+4)\), we obtain a simple quadratic.

\[
\begin{gathered}
x(x+4)+3(x+4)=x^{2}+4 x+3 x+12 \\
=x^{2}+7 x+12
\end{gathered}
\]

We want to develop a method for reversing this process.

In the expansion \((x+3)(x+4)=x^{2}+7 x+12\), notice that the coefficient of \(x\) is \(3+4=7\). The term that is independent of \(x\), the constant term, is \(3 \times 4=12\). This suggests a method of factorising.

In general, when we expand \((x+p)(x+q)\), we obtain

\[
x^{2}+p x+q x+p q=x^{2}+(p+q) x+p q
\]

The coefficient of \(x\) is the sum of \(p\) and \(q\), and the constant term is the product of \(p\) and \(q\).

\section*{Factorisation of simple quadratics}
To factorise a simple quadratic, look for two numbers that add to give the coefficient of \(x\), and that multiply together to give the constant term.

For example, to factorise \(x^{2}+8 x+15\), we look for two numbers that multiply to give 15 and add to give 8 . Of the pairs that multiply to give \(15(15 \times 1,5 \times 3,-5 \times(-3)\) and \(-15 \times(-1))\), only 5 and 3 add to give 8 .

Therefore, \(x^{2}+8 x+15=(x+3)(x+5)\)

The result can be checked by expanding \((x+3)(x+5)\).

\section*{Example 7}
Factorise \(x^{2}-3 x-18\).

\section*{Solution}
We are looking for two numbers that multiply to give -18 and add to give -3 . The numbers -6 and 3 satisfy both conditions.

\(x^{2}-3 x-18=(x-6)(x+3)\)

Consider factorising \(x^{2}-36\).

Since the constant term is -36 and the coefficient of \(x\) is 0 , we are looking for two numbers that multiply to give -36 and add to give 0 . The numbers -6 and 6 satisfy both conditions.

Thus \(x^{2}-36=(x-6)(x+6)\).

Note: It is not really sensible to do the example in this way - it is best to recognise \(x^{2}-36\) as a difference of squares. However, it does show that the method of factorisation in this section is consistent with the earlier technique.

\section*{Example 8}
Factorise \(x^{2}+8 x+16\)

\section*{Solution}
We are looking for two numbers that multiply to give 16 and add to give 8 . The numbers 4 and 4 satisfy both conditions.

Thus \(x^{2}+8 x+16=(x+4)(x+4)\)

\[
=(x+4)^{2}
\]

\section*{Exercise 4C}
Example 7 Factorise these quadratric expressions.
a \(x^{2}+5 x+6\)
b \(x^{2}+11 x+18\)
c \(x^{2}+7 x+10\)
d \(x^{2}+11 x+30\)
e \(x^{2}+9 x+14\)
f \(x^{2}+19 x+90\)
g \(x^{2}+9 x+20\)
h \(x^{2}+7 x+12\)
i \(x^{2}+12 x+32\)
j \(x^{2}+13 x+40\)
k \(x^{2}+20 x+75\)
l \(x^{2}+28 x+27\)
m \(x^{2}+15 x+56\)
n \(x^{2}+18 x+56\)

2 Factorise these quadratric expressions.
a \(x^{2}-5 x+6\)
b \(x^{2}-14 x+33\)
c \(x^{2}-17 x+30\)
d \(x^{2}-13 x+42\)
e \(x^{2}-9 x+14\)
f \(x^{2}-47 x+90\)
g \(x^{2}-15 x+44\)
h \(x^{2}-25 x+100\)
i \(x^{2}-18 x+80\)
j \(x^{2}-21 x+80\)
k \(x^{2}-14 x+40\)
l \(x^{2}-11 x+24\)
m \(x^{2}-30 x+56\)
n \(x^{2}-14 x+24\)

3 Factorise these quadratric expressions.
a \(x^{2}+x-6\)
b \(x^{2}-8 x-33\)
c \(x^{2}+x-30\)
d \(x^{2}-19 x-42\)
e \(x^{2}-5 x-14\)
f \(x^{2}-9 x-90\)
g \(x^{2}-7 x-44\)

i \(x^{2}+2 x-80\)

k \(x^{2}+3 x-40\)

\(\mathbf{m} x^{2}+4 x-21\)

o \(x^{2}-x+56\)

4 Factorise these quadratic expressions.

a \(x^{2}-3 x+2\)

c \(x^{2}-3 x-10\)

e \(x^{2}-5 x-14\)

g \(x^{2}-5 x+4\)

i \(x^{2}-x-12\)

k \(x^{2}+3 x-10\) h \(x^{2}+15 x-100\)

j \(x^{2}-7 x-60\)

l \(x^{2}-10 x-24\)

n \(x^{2}+2 x-15\)

p \(x^{2}+5 x-24\)

b \(x^{2}+8 x+12\)

d \(x^{2}+11 x+30\)

f \(x^{2}-9 x-90\)

h \(x^{2}-7 x-18\)

j \(x^{2}-11 x+28\)

l \(x^{2}+x-90\)

b \(x^{2}+14 x+49\)

d \(x^{2}-18 x+81\)

f \(x^{2}+12 x+36\)

h \(x^{2}-16 x+64\)

j \(x^{2}-8 x+16\)

\section*{1. Factorisation using perfect squares}
In Chapter 4, Example 8, we used the methods discussed previously to show that \(x^{2}+8 x+16=(x+4)^{2}\). This form of quadratic factorisation is called a perfect square.

We recall from Chapter 1 that a perfect square is an expression such as \((x+3)^{2},(x-5)^{2}\) or \((2 x+7)^{2}\).

The expansion of a perfect square has a special form. For example:

\[
\begin{aligned}
(x+3)^{2} & =(x+3)(x+3) \\
& =x^{2}+6 x+9 \\
& =x^{2}+2 \times(3 x)+3^{2}
\end{aligned}
\]

Note that the constant term 9 is the square of half the coefficient of \(x\).

The quadratic \(x^{2}+10 x+25\) is a perfect square since the "constant term is equal to the square of half of the coefficient of \(x\) '. So:

\[
x^{2}+10 x+25=(x+5)^{2}
\]

We recognise a perfect square such as this in the following way: the constant term is the square of half of the coefficient of \(x\). For example:

\[
\begin{array}{ll}
x^{2}+12 x+36=(x+6)^{2} & x^{2}+7 x+\frac{49}{4}=\left(x+\frac{7}{2}\right)^{2} \\
x^{2}-14 x+49=(x-7)^{2} & x^{2}-9 x+\frac{81}{4}=\left(x-\frac{9}{2}\right)^{2}
\end{array}
\]

\section*{Factorising using the perfect square identities}
\begin{itemize}
  \item In general, \(a^{2}+2 a b+b^{2}=(a+b)^{2}\)
  \item Similarly, \(a^{2}-2 a b+b^{2}=(a-b)^{2}\)
\end{itemize}

When a quadratic expression has the form of a perfect square, factorisation can occur immediately by application of the relevant identity.

\section*{Example 9}
Factorise:
a \(x^{2}+8 x+16\)
b \(x^{2}-10 x+25\)
c \(x^{2}+11 x+\frac{121}{4}\)

\section*{Solution}
a \(x^{2}+8 x+16=x^{2}+2 \times 4 x+4^{2}\) \(=(x+4)^{2}\)

c \(x^{2}+11 x+\frac{121}{4}=x^{2}+2 \times \frac{11}{2} x+\left(\frac{11}{2}\right)^{2}\)

\[
=\left(x+\frac{11}{2}\right)^{2}
\]

b \(x^{2}-10 x+25=x^{2}-2 \times 5 x+5^{2}\)

\[
=(x-5)^{2}
\]

\section*{Exercise 4D}
Example 9 1 Factorise using the appropriate perfect square identity.
a \(x^{2}+12 x+36\)
b \(x^{2}-8 x+16\)
d \(a^{2}-4 a+4\)
e \(m^{2}-26 m+169\)
c \(x^{2}+10 x+25\)
g \(x^{2}-9 x+\frac{81}{4}\)
h \(x^{2}+13 x+\frac{169}{4}\)
f \(a^{2}+28 a+196\)
i \(x^{2}-11 x+\frac{121}{4}\)

2 Copy and complete:
a \(x^{2}+8 x+16=(x+\)
b \(x^{2}-10 x+\)
\(=(x-5)^{2}\)
c \(x^{2}-\ldots+\ldots=(x-9)^{2}\)
e \(x^{2}-9 x+\)
\(=(x-\)
\(\begin{aligned} & \text { d } x^{2}+ \\ & \text { f } x^{2}-\frac{5 x}{2}+\square\end{aligned}+=(x-\)
\(+\ldots=\left(x+\frac{7}{2}\right)^{2}\)
)\(^{2}\)

3 Identify the simple quadratic expression that cannot be factorised as a perfect square.
a i \(x^{2}+4 x+4\)
ii \(x^{2}-6 x+12\)
iii \(x^{2}-12 x+36\)
iv \(x^{2}-10 x+25\)
b i \(x^{2}+6 x+9\)
ii \(x^{2}+5 x+\frac{25}{4}\)
iii \(x^{2}-8 x-16\)
iv \(x^{2}-14 x+49\)
c i \(x^{2}+\frac{2 x}{3}+\frac{1}{9}\)
ii \(x^{2}-3 x+\frac{9}{4}\)
iii \(x^{2}-\frac{5 x}{3}+\frac{25}{36}\)
iv \(x^{2}+\frac{7 x}{4}+\frac{49}{16}\)
d i \(x^{2}-\frac{11 x}{2}+\frac{121}{16}\)
ii \(x^{2}-\frac{4 x}{5}+\frac{4}{25}\)
iii \(x^{2}-\frac{3 x}{2}-\frac{9}{16}\)
iv \(x^{2}-\frac{9 x}{2}+\frac{81}{16}\)

4 A brick company provides rectangular and square pavers.

a Draw a diagram to show how two different square pavers of side lengths \(a\) and \(b\) respectively, and two identical rectangular pavers with dimensions \(a \times b\), can be arranged into a square.

b How many of each type of paver enables you to pave a square area of side length \(a+3 b\) ? Draw a diagram to illustrate how this can be done.

\section*{\(4 \leftleftarrows\) Quadratics with common factors}
Sometimes a common factor can be taken out of a quadratic expression so that the expression inside the brackets becomes a simple quadratic that can be factorised.

\section*{Example 10}
Factorise:

a \(3 x^{2}+9 x+6\)

b \(6 x^{2}-54\)

c \(-x^{2}-x+2\)

\section*{Solution}
a \(3 x^{2}+9 x+6=3\left(x^{2}+3 x+2\right) \quad\) (Take out the common factor.)

\[
=3(3 x+2)(3 x+1)
\]

b \(6 x^{2}-54=6\left(x^{2}-9\right)\)

\[
=6(x+3)(x-3)
\]

c \(-x^{2}-x+2=-\left(x^{2}+x-2\right) \quad\) (Factor -1 from each term.)

\[
=-(x+2)(x-1)
\]

\section*{Exercise 4E}
1 Factorise:
a \(2 x^{2}+14 x+24\)
b \(3 x^{2}+24 x+36\)
c \(3 x^{2}-27 x+24\)
d \(4 x^{2}-24 x+36\)
e \(7 x^{2}+14 x+7\)
f \(5 x^{2}-5 x-30\)
g \(4 x^{2}-4 x+48\)
h \(2 x^{2}-18 x+36\)
i \(5 x^{2}+40 x+35\)
j \(3 x^{2}+9 x-120\)
k \(3 x^{2}-3 x-90\)
l \(5 x^{2}+60 x+180\)
m \(2 x^{2}-4 x-96\)
n \(5 x^{2}+65 x+180\)
o \(3 x^{2}+30 x-72\)
p \(3 x^{2}-18 x+27\)
q \(5 x^{2}-20 x+20\)
r \(3 x^{2}-24 x+36\)

2 Factorise:
a \(4 x^{2}-16\)
d \(3 a^{2}-27\)
g \(27 x^{2}-3 y^{2}\)
j \(128-2 x^{2}\)
\(\mathbf{m} \frac{1}{4} a^{2}-9\)

b \(2 x^{2}-18\)

c \(3 x^{2}-48\)

e \(6 x^{2}-600\)

f \(3 a^{2}-27 b^{2}\)

h \(45-5 b^{2}\)

k \(\frac{1}{2} a^{2}-2 b^{2}\)

i \(12-3 m^{2}\)

n \(\frac{1}{5} x^{2}-20\)

l \(27 x^{2}-\frac{1}{3} y^{2}\)

o \(\frac{1}{4} x^{2}-y^{2}\)

3 Factorise:
a \(-x^{2}-8 x-12\)
d \(9+8 x-x^{2}\)
g \(-x^{2}+3 x+40\)
j \(11 x-x^{2}-24\)
\(\mathbf{m}-16 x-63-x^{2}\)

\[
\begin{aligned}
& \text { b } 12-11 x-x^{2} \\
& \text { e }-x^{2}-4 x-4 \\
& \text { h } 42+x-x^{2} \\
& \mathbf{k}-3 x^{2}-30 x+72 \\
& \mathbf{n}-x^{2}-35+12 x
\end{aligned}
\]

c \(7-6 x-x^{2}\)

f \(-x^{2}-14 x-45\)

i \(22 x-x^{2}-40\)

l \(-56-x^{2}-15 x\)

o \(7 x+18-x^{2}\)

\section*{Review exercise}
1 Factorise:
a \(4 x+16\)
b \(7 x-21\)
c \(6 a-9\)
d \(4 a b+7 a\)
e \(6 p q-11 p\)
f \(5 m n-10 n\)
g \(4 u v-8 v\)
h \(a^{2}+9 a\)
i \(4 m^{2} n-12 m n\)
j \(a^{2} b-4 a b^{2}\)
k \(3 p q-6 p^{2}\)
l \(6 p^{3} q-18 p q\)

2 Factorise:
a \(x^{2}-9\)
b \(x^{2}-16\)
c \(9 a^{2}-25\)
d \(16 m^{2}-1\)
e \(9-4 b^{2}\)
f \(100-81 b^{2}\)
g \(16 x^{2}-y^{2}\)
h \(2 m^{2}-50\)
i \(3 a^{2}-27\)
j \(1-36 b^{2}\)
k \(4 y^{2}-\frac{1}{4}\)
l \(p^{2} q^{2}-1\)

3 Factorise:

a \(x^{2}+8 x+12\)

b \(x^{2}+9 x+18\)

c \(x^{2}+11 x+30\)

d \(x^{2}-11 x+24\)

e \(x^{2}-10 x+24\)

f \(x^{2}-14 x+24\)

g \(x^{2}-25 x+24\)

h \(x^{2}+x-20\)

i \(x^{2}-2 x-48\)

j \(x^{2}-4 x-12\)

k \(x^{2}+3 x-40\)

l \(x^{2}-7 x-8\)

m \(x^{2}-x-132\)

n \(a^{2}+19 a+60\)

o \(x^{2}-50 x+96\)

4 Factorise:
a \(a^{2}-22 a+121\)
b \(m^{2}-14 m+49\)
c \(s^{2}+8 s+16\)
d \(a^{2}+24 a+144\)
e \(a^{2}-12 a+36\)
f \(z^{2}-40 z+400\)
g \(x^{2}+5 x+\frac{25}{4}\)
h \(y^{2}-\frac{2 y}{3}+\frac{1}{9}\)
i \(a^{2}+\frac{3 a}{2}+\frac{9}{16}\)

5 Factorise:
a \(2 x^{2}+18 x+40\)
b \(3 x^{2}-30 x+63\)
c \(5 x^{2}-50 x+120\)
d \(2 x^{2}+8 x-90\)
e \(3 x^{2}-6 x-105\)
f \(2 x^{2}-6 x-260\)

6 Factorise:

a \(25 a^{2}-16 b^{2}\)

b \(a^{2}+14 a+49\)

c \(a^{2}-a-20\)

d \(1-36 m^{2}\)

e \(4-9 x^{2} y^{2}\)

f \(\frac{1}{9}-\frac{a^{2}}{25}\)

g \(m^{2}-\frac{1}{4}\)

h \(x^{3}-49 x y^{2}\)

i \(3 a^{2}-75\)

j \(b^{2}-20 b+96\)

k \(n^{2}-31 n+150\)

l \(m^{2}+20 m+91\)

\(\mathbf{m} a^{2}-7 a b-98 b^{2}\)

n \(m^{2}-4 m-165\)

o \(x^{2}+3 x y-4 y^{2}\)

p \(20 m^{2} n-5 n^{3}\)

q \(a^{2}-2 a-63\)

r \(5 q^{2}-5 p q-30 p^{2}\)

s \(x^{2}-3 x-130\)

t \(42-x-x^{2}\)

u \(x^{2}+7 x-18\)

\section*{Challenge exercise}
1 Factorise:
a \(x^{4}-1\)
b \(x^{4}-16\)
c \(x^{2}-3\)
d \(x^{2}-5\)
e \(x^{2}-2 \sqrt{2} x+2\)
f \(x^{2}+2 \sqrt{2} x+2\)
g \(x^{2}+x+\frac{1}{4}\)
h \(x^{2}+3 x+\frac{9}{4}\)
i \(x^{2}-x+\frac{1}{4}\)
j \(x^{2}-3 x+\frac{9}{4}\)
k \(x^{2}-x y-2 y^{2}\)
l \(x^{2}+x y-2 y^{2}\)

2 Simplify:

a \(\frac{x^{2}+x-2}{x^{2}-x-20} \times \frac{x^{2}+5 x+4}{x^{2}-x} \div\left(\frac{x^{2}+3 x+2}{x^{2}-2 x-15} \times \frac{x+3}{x^{2}}\right)\)

b \(\frac{x^{2}-64}{x^{2}+24 x+128} \times \frac{x^{2}+12 x-64}{x^{2}-16} \div \frac{x^{2}-16 x+64}{x^{2}-10 x+16}\)

c \(\frac{x^{2}-18 x+80}{x^{2}-5 x-50} \times \frac{x^{2}-6 x-7}{x^{2}-15 x+56} \div \frac{x-1}{x+5}\)

3 Factorise:
a \((x+2)^{2}-8 x\)
\(\mathbf{b}(x-3)^{2}+12 x\)
c \((x+a)^{2}-4 a x\)
d \((x-a)^{2}+4 a x\)

4 Simplify \(\left(\frac{a}{b}+\frac{c}{d}\right) \div\left(\frac{a}{b}-\frac{c}{d}\right)\).

5 Factorise:
a \(\left(x^{2}+1\right)^{2}-4\)
b \(\left(x^{2}-2\right)^{2}-4\)
c \(x^{2}+4 x+4-y^{2}\)
d \(x^{2}+8 x+16-a^{2}\)
e \(m^{2}-2 m+1-n^{2}\)
f \(p^{2}-5 p+\frac{25}{4}-q^{2}\)

6 a By adding and subtracting \(4 x^{2}\), factorise \(x^{4}+4\).

b Factorise \(x^{4}+4 a^{4}\).

7 A swimming pool is designed in an L-shape with dimensions in metres as shown.

The pool is enlarged or reduced depending on the value of \(x\).

a Find, in terms of \(x\), the length of:
i \(A F\)
ii \(C D\)

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_30_d36dc84f51b69993a9a7g-14}
\end{center}

b Show that the perimeter is equal to \((10 x+50) \mathrm{m}\).

c What is the perimeter if \(x=3\) ?

d Find the area of the swimming pool in terms of \(x\). Expand and simplify your answer.

e It is decided that a square swimming pool would be a better use of space.

i By factorising your answer to part \(\mathbf{d}\), find the dimensions, in terms of \(x\), of a square swimming pool with the same area as the L-shaped swimming pool.

ii What is the perimeter, in terms of \(x\), of this square swimming pool?

8 For positive whole numbers \(a\) and \(b\), prove that:
a if \(\frac{a}{b}<1\) then \(\frac{a+1}{b+1}>\frac{a}{b}\)
b if \(\frac{a}{b}>1\) then \(\frac{a+1}{b+1}<\frac{a}{b}\)

9 A right-angled triangle has a hypotenuse of length \(b \mathrm{~cm}\) and one other side of length \(a \mathrm{~cm}\). If \(b-a=1\), find the length of the third side in terms of \(a\) and \(b\).

10 Factorise \(\left(1+\frac{y^{2}+z^{2}-x^{2}}{2 y z}\right) \div\left(1-\frac{x^{2}+y^{2}-z^{2}}{2 x y}\right)\).


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\title{Factorisation }

\author{}
\date{}


\begin{document}
\maketitle
Number and Algebra

The distributive law can be used to rewrite a product involving brackets as an expression without brackets. For instance, the product \(3(a+2)\) can be rewritten as \(3 a+6\); this is called the expanded form of the expression, and the process is called expansion.

The process of writing an algebraic expression as a product of two or more algebraic factors is called factorisation. Factorisation is the reverse process to expansion. For example, we can write \(3 a+6\) as \(3(a+2)\). This is called the factorised form.

In this chapter we will look at how to factorise expressions such as \(x^{2}+7 x+12\).

We recall that such an expression is called a quadratic.

\section*{Factorisation using common factors}
If each term in the algebraic expression to be factorised contains a common factor, then this common factor is a factor of the entire expression. To find the other factor, divide each term by the common factor. The common factor is placed outside brackets. For this reason the procedure is sometimes called 'taking the common factor outside the brackets'.

\section*{Example 1}
Factorise \(4 a+12\)

\section*{Solution}
4 is a common factor of \(4 a\) and 12.

Thus \(4 a+12=4(a+3)\)

Notice that the answer can be checked by expanding \(4(a+3)\).

In general, take out as many common factors as possible. The common factors may involve both numbers and pronumerals. It may be easier if you first take out the common factor of the numbers and then the common factor of the pronumerals.

\section*{Example 2}
\(\begin{array}{ll}\text { Factorise: } \mathbf{a} 12 x^{2}+3 x & \text { b } 36 a b-27 a\end{array}\)

Solution
a \(12 x^{2}+3 x=3\left(4 x^{2}+x\right)\)
b \(36 a b-27 a=9(4 a b-3 a)\)
\(=3 x(4 x+1)\)
\(=9 a(4 b-3)\)

Note: It is more efficient to take out all the common factors in one step, as shown in the following examples.

\section*{Example 3}
Factorise:
a \(3 x+9\)
b \(2 x^{2}+4 x\)
c \(-7 a^{2}-49\)
d \(7 a^{2}+63 a b\)

Solution

a \(3 x+9=3(x+3)\)

b \(2 x^{2}+4 x=2 x(x+2)\)

c \(-7 a^{2}-49=-7 \times a^{2}+(-7) \times 7\)

d \(7 a^{2}+63 a b=7 a(a+9 b)\)

\(=-7\left(a^{2}+7\right)\)

\(7\left(-a^{2}-7\right)\) is also correct.

\section*{Example 4}
Factorise:
a \(5 p q^{2}+10 p^{2} q+25 p^{2} q^{2}\)
b \(16 a b+10 b^{2}-2 a^{2} b\)

Solution

a \(5 p q^{2}+10 p^{2} q+25 p^{2} q^{2}=5 p q(q+2 p+5 p q)\)

b \(16 a b+10 b^{2}-2 a^{2} b=2 b\left(8 a+5 b-a^{2}\right)\)

\section*{Exercise 4A}
1 Complete each factorisation.
a \(12 x=12 \times \square\)
b \(24 a=12 \times \square\)
c \(36 a b=9 a \times\)
d \(15 a c=5 c \times \square\)
e \(y^{2}=y \times \square\)
f \(6 y^{2}=3 y \times \square\)
g \(24 a^{2}=6 a \times \square\)
h \(-6 b^{2}=2 b \times \square\)
i \(8 a^{2} b=2 a b \times\)
j \(4 x^{2} y=2 x \times \square\)
k \(12 m^{2} n=3 m n \times \square\)
l \(25 a^{2} b^{2}=5 a b \times \square\)

2 Fill in the blanks by finding the missing factors.
a \(12 a+18=6 \times\)
b \(15 p-10=5 \times\)
c \(20 m n-15 n=5 \times \square\)
d \(20 m n-15 n=5 n \times \square\)
e \(a^{2}+4 a=(a+4) \times \square\)
f \(b^{2}-10 b=b \times \square\)
g \(6 y z^{2}-18 y z=3 z \times \square\)
h \(6 y z^{2}-18 y z=y z \times \square\)
i \(6 y z^{2}-18 y z=-3 y z \times \square\)
j \(6 y z^{2}-18 y z=6 y z \times \square\)

3 Factorise:
a \(6 x+24\)
b \(5 a+15\)
c \(a c+5 c\)
d \(a^{2}+a\)
e \(y^{2}+x y\)
f \(4 x+24\)
g \(7 a-63\)
h \(9 a+36\)
i \(y^{2}-3 y\)
\(\mathbf{j}-14 a-21\)
\(\mathbf{k}-6 y-9\)
l \(-4-12 b\)

4 Factorise:
a \(4 a b+16 a\)
b \(12 a^{2}+8 a\)
c \(18 m^{2} n+9 m n^{2}\)
d \(15 a^{2} b^{2}+10 a b^{2}\)
e \(4 a^{2}+6 a\)
f \(8 a^{2}+12 a b\)

5 Factorise:
a \(3 b-6 b^{2}\)
b \(4 x^{2}-6 x y\)
c \(9 m n-12 m^{2} n\)
d \(18 y-9 y^{2}\)
e \(4 a-6 a b^{2}\)
f \(6 x y-4 x^{2}\)
g \(14 m n^{2}-21 m^{2} n\)
h \(6 p q^{2}-21 q p^{2}\)
i \(10 a b^{2}-25 a^{2} b\)

6 Factorise:
\(\mathbf{a}-10 b^{2}+5 b\)
\(\mathbf{b}-16 a^{2} b-8 a b\)
c \(-x^{2} y-3 x y\)
d \(-4 p q+16 p^{2}\)
e \(-5 x^{2} y+30 x\)
f \(18 p^{2}-4 p q\)
\(\mathbf{g}-8 a^{2} b^{2}-2 a b\)
h \(12 x y^{2}-3 x^{2} y\)
i \(-25 m^{2} n^{2}-10 m n^{2}\)

7 In each part, an expression for the area of the rectangle has been given. Find an expression for the missing side length.

a

\(2 a+3\)

Area \(=8 a+12\)

c

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_30_c996b59107099c56822bg-04(1)}
\end{center}

e \(6 a\)

\[
\text { Area }=12 a^{2}+6 a b
\]

b

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_30_c996b59107099c56822bg-04}
\end{center}

d

\(a\)

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_30_c996b59107099c56822bg-04(2)}
\end{center}

\(\mathbf{f}\) \(3 q^{2}-2\)

Area \(=24 p^{2} q^{2}-16 p^{2}\)

Example 4 8 Factorise:

a \(4 a^{2} b-2 a b+8 a b^{2} \quad\) b \(4 m^{2} n-4 m n+16 n^{2}\)

c \(7 a b+14 a^{2}+21 b\)

d \(2 m^{2}+4 m n+6 n\)

e \(5 a^{2} b+3 a b+4 a b^{2}\)

f \(6 a+8 a b+10 a b^{2}\)

g \(5 p^{2} q^{2}+10 p q^{2}+15 p^{2} q\)

h \(5 \ell^{2}-15 \ell m-20 m^{2}\)

\section*{1 B Factorisation using the difference of two squares}
You will recall from Section \(1 \mathrm{G}\) the important identity

\[
(a+b)(a-b)=a^{2}-b^{2},
\]

which is called the difference of two squares. We can now use this result the other way around to factorise an expression that is the difference of two squares.

That is:

\[
a^{2}-b^{2}=(a+b)(a-b)
\]

That is, the factors of \(a^{2}-b^{2}\) are \(a+b\) and \(a-b\).

\section*{Example 5}
Factorise:
a \(x^{2}-9\)
b \(25-y^{2}\)
c \(4 x^{2}-9\)

\section*{Solution}
a \(x^{2}-9=x^{2}-3^{2}\)

\[
=(x+3)(x-3)
\]

Check the answer by expanding \((x+3)(x-3)\) to see that \(x^{2}-9\) is obtained.

b \(25-y^{2}=5^{2}-y^{2}\)

\[
=(5+y)(5-y)
\]

c \(4 x^{2}-9=(2 x)^{2}-3^{2}\)

\[
=(2 x+3)(2 x-3)
\]

\section*{Example 6}
Factorise:
a \(3 a^{2}-27\)
b \(-16+9 x^{2}\)

\section*{Solution}
a \(3 a^{2}-27=3\left(a^{2}-9\right)\)
b \(-16+9 x^{2}=9 x^{2}-16\)
\(=3\left(a^{2}-3^{2}\right)\)
\(=(3 x)^{2}-4^{2}\)
\(=3(a+3)(a-3)\)
\(=(3 x-4)(3 x+4)\)

Factorisation using the difference of two squares identity

\(a^{2}-b^{2}=(a+b)(a-b)\)

1 Factorise:
a \(x^{2}-16\)
b \(x^{2}-49\)
c \(a^{2}-121\)
d \(d^{2}-400\)
e \((2 x)^{2}-25\)
f \((3 x)^{2}-16\)
g \((4 x)^{2}-1\)
h \((5 m)^{2}-9\)
i \(9 x^{2}-4\)
j \(16 y^{2}-49\)
k \(100 a^{2}-49 b^{2}\)
l \(64 m^{2}-81 p^{2}\)
m \(1-4 a^{2}\)
n \(9-16 y^{2}\)
o \(25 a^{2}-100 b^{2}\)
p \(-9+x^{2}\)

Example 6a 2 Factorise:
a \(3 x^{2}-48\)
b \(4 x^{2}-100\)
c \(5 x^{2}-45\)
d \(6 x^{2}-24\)
e \(10 x^{2}-1000\)
f \(7 x^{2}-63\)
g \(8 x^{2}-50\)
h \(12 m^{2}-75\)
i \(3-12 b^{2}\)
j \(20-5 y^{2}\)
k \(27 a^{2}-12 b^{2}\)
l \(16 x^{2}-100 y^{2}\)
m \(45 m^{2}-125 n^{2}\)
n \(27 a^{2}-192 l^{2}\)
o \(-8 x^{2}+32 y^{2}\)
\(\mathbf{p}-200 p^{2}+32 q^{2}\)

Example 6b 3 Factorise:
a \(-25+x^{2}\)
b \(-4+9 x^{2}\)
c \(-9 x^{2}+4\)
d \(-81 x^{2}+16\)
e \(-100 x^{2}+9\)
f \(-18+50 x^{2}\)
g \(-12 x^{2}+27\)
h \(-36 x^{2}+400\)
i \(-175+28 x^{2}\)

4 Use the factorisation of the difference of two squares to evaluate the following. One has been done for you.

\[
\begin{aligned}
17^{2}-3^{2} & =(17+3)(17-3) \\
& =20 \times 14 \\
& =280
\end{aligned}
\]

a \(23^{2}-7^{2}\)
b \(23^{2}-3^{2}\)
c \(36^{2}-6^{2}\)
d \(94^{2}-6^{2}\)
e \(1.8^{2}-0.2^{2}\)
f \(28^{2}-2.2^{2}\)
g \(11.3^{2}-8.7^{2}\)
h \(92.6^{2}-7.4^{2}\)
i \(3.214^{2}-2.214^{2}\)

5 a Evaluate:
i \(4^{2}-3^{2}\)
ii \(5^{2}-4^{2}\)
iii \(6^{2}-5^{2}\)
iv \(7^{2}-6^{2}\)
v \(8^{2}-7^{2}\)
vi \(9^{2}-8^{2}\)
vii \(10^{2}-9^{2}\)
viii \(101^{2}-100^{2}\)

b What do you notice?

c Use the factorisation of \((n+1)^{2}-n^{2}\) to prove the result of part \(\mathbf{b}\).

6 The following leads you through the geometrical proof that \(a^{2}-b^{2}=(a-b)(a+b)\).

In the diagram opposite, square \(E C F G\), of side length \(b\), is cut out of square \(A B C D\), which has side length \(a\).

a What is the area of hexagon \(A B E G F D\) ?

b What is the length of \(B E\) ?

c What is the length of \(D F\) ?

d The rectangle \(H G F D\) is moved so that \(D F\) is placed on top of \(B E\) (see diagram).

i What is the area of the large shaded rectangle?

ii What have we proved?
\includegraphics[max width=\textwidth, center]{2023_11_30_c996b59107099c56822bg-06}

\section*{Factorisation of simple quadratics}
A simple quadratic expression is an expression of the form \(x^{2}+b x+c\), where \(b\) and \(c\) are given numbers. When we expand \((x+3)(x+4)\), we obtain a simple quadratic.

\[
\begin{gathered}
x(x+4)+3(x+4)=x^{2}+4 x+3 x+12 \\
=x^{2}+7 x+12
\end{gathered}
\]

We want to develop a method for reversing this process.

In the expansion \((x+3)(x+4)=x^{2}+7 x+12\), notice that the coefficient of \(x\) is \(3+4=7\). The term that is independent of \(x\), the constant term, is \(3 \times 4=12\). This suggests a method of factorising.

In general, when we expand \((x+p)(x+q)\), we obtain

\[
x^{2}+p x+q x+p q=x^{2}+(p+q) x+p q
\]

The coefficient of \(x\) is the sum of \(p\) and \(q\), and the constant term is the product of \(p\) and \(q\).

\section*{Factorisation of simple quadratics}
To factorise a simple quadratic, look for two numbers that add to give the coefficient of \(x\), and that multiply together to give the constant term.

For example, to factorise \(x^{2}+8 x+15\), we look for two numbers that multiply to give 15 and add to give 8 . Of the pairs that multiply to give \(15(15 \times 1,5 \times 3,-5 \times(-3)\) and \(-15 \times(-1))\), only 5 and 3 add to give 8 .

Therefore, \(x^{2}+8 x+15=(x+3)(x+5)\)

The result can be checked by expanding \((x+3)(x+5)\).

\section*{Example 7}
Factorise \(x^{2}-3 x-18\).

\section*{Solution}
We are looking for two numbers that multiply to give -18 and add to give -3 . The numbers -6 and 3 satisfy both conditions.

\(x^{2}-3 x-18=(x-6)(x+3)\)

Consider factorising \(x^{2}-36\).

Since the constant term is -36 and the coefficient of \(x\) is 0 , we are looking for two numbers that multiply to give -36 and add to give 0 . The numbers -6 and 6 satisfy both conditions.

Thus \(x^{2}-36=(x-6)(x+6)\).

Note: It is not really sensible to do the example in this way - it is best to recognise \(x^{2}-36\) as a difference of squares. However, it does show that the method of factorisation in this section is consistent with the earlier technique.

\section*{Example 8}
Factorise \(x^{2}+8 x+16\)

\section*{Solution}
We are looking for two numbers that multiply to give 16 and add to give 8 . The numbers 4 and 4 satisfy both conditions.

Thus \(x^{2}+8 x+16=(x+4)(x+4)\)

\[
=(x+4)^{2}
\]

\section*{Exercise 4C}
1 Factorise these quadratric expressions.
a \(x^{2}+5 x+6\)
b \(x^{2}+11 x+18\)
c \(x^{2}+7 x+10\)
d \(x^{2}+11 x+30\)
e \(x^{2}+9 x+14\)
f \(x^{2}+19 x+90\)
g \(x^{2}+9 x+20\)
h \(x^{2}+7 x+12\)
i \(x^{2}+12 x+32\)
j \(x^{2}+13 x+40\)
\(\mathbf{k} x^{2}+20 x+75\)
l \(x^{2}+28 x+27\)
\(\mathbf{m} x^{2}+15 x+56\)
n \(x^{2}+18 x+56\)

2 Factorise these quadratric expressions.
a \(x^{2}-5 x+6\)
b \(x^{2}-14 x+33\)
c \(x^{2}-17 x+30\)
d \(x^{2}-13 x+42\)
e \(x^{2}-9 x+14\)
f \(x^{2}-47 x+90\)
g \(x^{2}-15 x+44\)
h \(x^{2}-25 x+100\)
i \(x^{2}-18 x+80\)
j \(x^{2}-21 x+80\)
k \(x^{2}-14 x+40\)
l \(x^{2}-11 x+24\)
\(\mathbf{m} x^{2}-30 x+56\)
n \(x^{2}-14 x+24\)

3 Factorise these quadratric expressions.
a \(x^{2}+x-6\)
b \(x^{2}-8 x-33\)
c \(x^{2}+x-30\)
d \(x^{2}-19 x-42\)
e \(x^{2}-5 x-14\)
f \(x^{2}-9 x-90\)
g \(x^{2}-7 x-44\)
h \(x^{2}+15 x-100\)
i \(x^{2}+2 x-80\)
j \(x^{2}-7 x-60\)
k \(x^{2}+3 x-40\)
l \(x^{2}-10 x-24\)
m \(x^{2}+4 x-21\)
n \(x^{2}+2 x-15\)
o \(x^{2}-x+56\)
p \(x^{2}+5 x-24\)

4 Factorise these quadratic expressions.
a \(x^{2}-3 x+2\)
b \(x^{2}+8 x+12\)
c \(x^{2}-3 x-10\)
d \(x^{2}+11 x+30\)
e \(x^{2}-5 x-14\)
f \(x^{2}-9 x-90\)
g \(x^{2}-5 x+4\)
h \(x^{2}-7 x-18\)
i \(x^{2}-x-12\)
j \(x^{2}-11 x+28\)
k \(x^{2}+3 x-10\)
l \(x^{2}+x-90\)

Example 85 Factorise these quadratic expressions.
a \(x^{2}+6 x+9\)
b \(x^{2}+14 x+49\)
c \(x^{2}-10 x+25\)
d \(x^{2}-18 x+81\)
e \(x^{2}+10 x+25\)
f \(x^{2}+12 x+36\)
g \(x^{2}+30 x+225\)
h \(x^{2}-16 x+64\)
i \(x^{2}-20 x+100\)
j \(x^{2}-8 x+16\)

\section*{1 Factorisation using perfect squares}
In Chapter 4, Example 8, we used the methods discussed previously to show that \(x^{2}+8 x+16=(x+4)^{2}\). This form of quadratic factorisation is called a perfect square.

We recall from Chapter 1 that a perfect square is an expression such as \((x+3)^{2},(x-5)^{2}\) or \((2 x+7)^{2}\).

The expansion of a perfect square has a special form. For example:

\[
\begin{aligned}
(x+3)^{2} & =(x+3)(x+3) \\
& =x^{2}+6 x+9 \\
& =x^{2}+2 \times(3 x)+3^{2}
\end{aligned}
\]

Note that the constant term 9 is the square of half the coefficient of \(x\).

The quadratic \(x^{2}+10 x+25\) is a perfect square since the "constant term is equal to the square of half of the coefficient of \(x\) '. So:

\[
x^{2}+10 x+25=(x+5)^{2}
\]

We recognise a perfect square such as this in the following way: the constant term is the square of half of the coefficient of \(x\). For example:

\[
\begin{array}{ll}
x^{2}+12 x+36=(x+6)^{2} & x^{2}+7 x+\frac{49}{4}=\left(x+\frac{7}{2}\right)^{2} \\
x^{2}-14 x+49=(x-7)^{2} & x^{2}-9 x+\frac{81}{4}=\left(x-\frac{9}{2}\right)^{2}
\end{array}
\]

\section*{Factorising using the perfect square identities}
\begin{itemize}
  \item In general, \(a^{2}+2 a b+b^{2}=(a+b)^{2}\)
  \item Similarly, \(a^{2}-2 a b+b^{2}=(a-b)^{2}\)
\end{itemize}

When a quadratic expression has the form of a perfect square, factorisation can occur immediately by application of the relevant identity.

\section*{Example 9}
Factorise:
a \(x^{2}+8 x+16\)
b \(x^{2}-10 x+25\)
c \(x^{2}+11 x+\frac{121}{4}\)

\section*{Solution}
a \(x^{2}+8 x+16=x^{2}+2 \times 4 x+4^{2}\)

\[
=(x+4)^{2}
\]

c \(x^{2}+11 x+\frac{121}{4}=x^{2}+2 \times \frac{11}{2} x+\left(\frac{11}{2}\right)^{2}\)

\[
=\left(x+\frac{11}{2}\right)^{2}
\]

b \(x^{2}-10 x+25=x^{2}-2 \times 5 x+5^{2}\)

\[
=(x-5)^{2}
\]

\section*{Exercise 4D}
Example 9 1 Factorise using the appropriate perfect square identity.
a \(x^{2}+12 x+36\)
b \(x^{2}-8 x+16\)
c \(x^{2}+10 x+25\)
d \(a^{2}-4 a+4\)
e \(m^{2}-26 m+169\)
f \(a^{2}+28 a+196\)
g \(x^{2}-9 x+\frac{81}{4}\)
h \(x^{2}+13 x+\frac{169}{4}\)
i \(x^{2}-11 x+\frac{121}{4}\)

2 Copy and complete:
a \(x^{2}+8 x+16=(x+\)
b \(x^{2}-10 x+\)
\(=(x-5)^{2}\)
c \(x^{2}-\ldots+\ldots=(x-9)^{2}\)
d \(x^{2}+\)
e \(x^{2}-9 x+\)
\(=(x-\)
f \(x^{2}-\frac{5 x}{2}+\)
\(+\_=\left(x+\frac{7}{2}\right)^{2}\)
)\(^{2}\)

3 Identify the simple quadratic expression that cannot be factorised as a perfect square.
a \(\mathbf{i} x^{2}+4 x+4\)
ii \(x^{2}-6 x+12\)
iii \(x^{2}-12 x+36\)
iv \(x^{2}-10 x+25\)
b i \(x^{2}+6 x+9\)
ii \(x^{2}+5 x+\frac{25}{4}\)
iii \(x^{2}-8 x-16\)
iv \(x^{2}-14 x+49\)
c i \(x^{2}+\frac{2 x}{3}+\frac{1}{9}\)
ii \(x^{2}-3 x+\frac{9}{4}\)
iii \(x^{2}-\frac{5 x}{3}+\frac{25}{36}\)
iv \(x^{2}+\frac{7 x}{4}+\frac{49}{16}\)
d \(\mathbf{i} x^{2}-\frac{11 x}{2}+\frac{121}{16}\)
ii \(x^{2}-\frac{4 x}{5}+\frac{4}{25}\)
iii \(x^{2}-\frac{3 x}{2}-\frac{9}{16}\)
iv \(x^{2}-\frac{9 x}{2}+\frac{81}{16}\)

4 A brick company provides rectangular and square pavers.

a Draw a diagram to show how two different square pavers of side lengths \(a\) and \(b\) respectively, and two identical rectangular pavers with dimensions \(a \times b\), can be arranged into a square.

b How many of each type of paver enables you to pave a square area of side length \(a+3 b\) ? Draw a diagram to illustrate how this can be done.

\section*{\(4 E\) Quadratics with common factors}
Sometimes a common factor can be taken out of a quadratic expression so that the expression inside the brackets becomes a simple quadratic that can be factorised.

\section*{Example 10}
Factorise:
a \(3 x^{2}+9 x+6\)
b \(6 x^{2}-54\)
c \(-x^{2}-x+2\)

\section*{Solution}
a \(3 x^{2}+9 x+6=3\left(x^{2}+3 x+2\right) \quad\) (Take out the common factor.)

\[
=3(3 x+2)(3 x+1)
\]

b \(6 x^{2}-54=6\left(x^{2}-9\right)\)

\[
=6(x+3)(x-3)
\]

c \(-x^{2}-x+2=-\left(x^{2}+x-2\right)\)

(Factor -1 from each term.)

\[
=-(x+2)(x-1)
\]

\section*{Exercise 4E}
1 Factorise:
a \(2 x^{2}+14 x+24\)
b \(3 x^{2}+24 x+36\)
c \(3 x^{2}-27 x+24\)
d \(4 x^{2}-24 x+36\)
e \(7 x^{2}+14 x+7\)
f \(5 x^{2}-5 x-30\)
g \(4 x^{2}-4 x+48\)
h \(2 x^{2}-18 x+36\)
i \(5 x^{2}+40 x+35\)
j \(3 x^{2}+9 x-120\)
k \(3 x^{2}-3 x-90\)
l \(5 x^{2}+60 x+180\)
m \(2 x^{2}-4 x-96\)
n \(5 x^{2}+65 x+180\)
o \(3 x^{2}+30 x-72\)
p \(3 x^{2}-18 x+27\)
q \(5 x^{2}-20 x+20\)
r \(3 x^{2}-24 x+36\)

2 Factorise:
a \(4 x^{2}-16\)
b \(2 x^{2}-18\)
c \(3 x^{2}-48\)
d \(3 a^{2}-27\)
e \(6 x^{2}-600\)
f \(3 a^{2}-27 b^{2}\)
g \(27 x^{2}-3 y^{2}\)
h \(45-5 b^{2}\)
i \(12-3 m^{2}\)
j \(128-2 x^{2}\)
k \(\frac{1}{2} a^{2}-2 b^{2}\)
l \(27 x^{2}-\frac{1}{3} y^{2}\)
\(\mathbf{m} \frac{1}{4} a^{2}-9\)
n \(\frac{1}{5} x^{2}-20\)
o \(\frac{1}{4} x^{2}-y^{2}\)

3 Factorise:

a \(-x^{2}-8 x-12\)

b \(12-11 x-x^{2}\)

c \(7-6 x-x^{2}\)

d \(9+8 x-x^{2}\)

e \(-x^{2}-4 x-4\)

f \(-x^{2}-14 x-45\)

g \(-x^{2}+3 x+40\)

h \(42+x-x^{2}\)

i \(22 x-x^{2}-40\)

j \(11 x-x^{2}-24\)

k \(-3 x^{2}-30 x+72\)

l \(-56-x^{2}-15 x\)

\(\mathbf{m}-16 x-63-x^{2}\)

n \(-x^{2}-35+12 x\)

o \(7 x+18-x^{2}\)

\section*{Review exercise}
1 Factorise:
a \(4 x+16\)
b \(7 x-21\)
c \(6 a-9\)
d \(4 a b+7 a\)
e \(6 p q-11 p\)
f \(5 m n-10 n\)
g \(4 u v-8 v\)
h \(a^{2}+9 a\)
i \(4 m^{2} n-12 m n\)
j \(a^{2} b-4 a b^{2}\)
k \(3 p q-6 p^{2}\)
l \(6 p^{3} q-18 p q\)

2 Factorise:
a \(x^{2}-9\)
b \(x^{2}-16\)
c \(9 a^{2}-25\)
d \(16 m^{2}-1\)
e \(9-4 b^{2}\)
f \(100-81 b^{2}\)
g \(16 x^{2}-y^{2}\)
h \(2 m^{2}-50\)
i \(3 a^{2}-27\)
j \(1-36 b^{2}\)
k \(4 y^{2}-\frac{1}{4}\)
l \(p^{2} q^{2}-1\)

3 Factorise:

a \(x^{2}+8 x+12\)

b \(x^{2}+9 x+18\)

c \(x^{2}+11 x+30\)

d \(x^{2}-11 x+24\)

e \(x^{2}-10 x+24\)

f \(x^{2}-14 x+24\)

g \(x^{2}-25 x+24\)

h \(x^{2}+x-20\)

i \(x^{2}-2 x-48\)

j \(x^{2}-4 x-12\)

k \(x^{2}+3 x-40\)

l \(x^{2}-7 x-8\)

m \(x^{2}-x-132\)

n \(a^{2}+19 a+60\)

o \(x^{2}-50 x+96\)

4 Factorise:
a \(a^{2}-22 a+121\)
b \(m^{2}-14 m+49\)
c \(s^{2}+8 s+16\)
d \(a^{2}+24 a+144\)
e \(a^{2}-12 a+36\)
f \(z^{2}-40 z+400\)
g \(x^{2}+5 x+\frac{25}{4}\)
h \(y^{2}-\frac{2 y}{3}+\frac{1}{9}\)
i \(a^{2}+\frac{3 a}{2}+\frac{9}{16}\)

5 Factorise:
a \(2 x^{2}+18 x+40\)
b \(3 x^{2}-30 x+63\)
c \(5 x^{2}-50 x+120\)
d \(2 x^{2}+8 x-90\)
e \(3 x^{2}-6 x-105\)
f \(2 x^{2}-6 x-260\)

6 Factorise:
a \(25 a^{2}-16 b^{2}\)
b \(a^{2}+14 a+49\)
c \(a^{2}-a-20\)
d \(1-36 m^{2}\)
e \(4-9 x^{2} y^{2}\)
f \(\frac{1}{9}-\frac{a^{2}}{25}\)
g \(m^{2}-\frac{1}{4}\)
h \(x^{3}-49 x y^{2}\)
i \(3 a^{2}-75\)
j \(b^{2}-20 b+96\)
k \(n^{2}-31 n+150\)
l \(m^{2}+20 m+91\)
\(\mathbf{m} a^{2}-7 a b-98 b^{2}\)
n \(m^{2}-4 m-165\)
o \(x^{2}+3 x y-4 y^{2}\)
p \(20 m^{2} n-5 n^{3}\)
q \(a^{2}-2 a-63\)
r \(5 q^{2}-5 p q-30 p^{2}\)
s \(x^{2}-3 x-130\)
t \(42-x-x^{2}\)
u \(x^{2}+7 x-18\)

\section*{Challenge exercise}
1 Factorise:
a \(x^{4}-1\)
b \(x^{4}-16\)
c \(x^{2}-3\)
d \(x^{2}-5\)
e \(x^{2}-2 \sqrt{2} x+2\)
f \(x^{2}+2 \sqrt{2} x+2\)
g \(x^{2}+x+\frac{1}{4}\)
h \(x^{2}+3 x+\frac{9}{4}\)
i \(x^{2}-x+\frac{1}{4}\)
j \(x^{2}-3 x+\frac{9}{4}\)
k \(x^{2}-x y-2 y^{2}\)
l \(x^{2}+x y-2 y^{2}\)

2 Simplify:

a \(\frac{x^{2}+x-2}{x^{2}-x-20} \times \frac{x^{2}+5 x+4}{x^{2}-x} \div\left(\frac{x^{2}+3 x+2}{x^{2}-2 x-15} \times \frac{x+3}{x^{2}}\right)\)

b \(\frac{x^{2}-64}{x^{2}+24 x+128} \times \frac{x^{2}+12 x-64}{x^{2}-16} \div \frac{x^{2}-16 x+64}{x^{2}-10 x+16}\)

c \(\frac{x^{2}-18 x+80}{x^{2}-5 x-50} \times \frac{x^{2}-6 x-7}{x^{2}-15 x+56} \div \frac{x-1}{x+5}\)

3 Factorise:
a \((x+2)^{2}-8 x\)
\(\mathbf{b}(x-3)^{2}+12 x\)
c \((x+a)^{2}-4 a x\)
d \((x-a)^{2}+4 a x\)

4 Simplify \(\left(\frac{a}{b}+\frac{c}{d}\right) \div\left(\frac{a}{b}-\frac{c}{d}\right)\).

5 Factorise:
a \(\left(x^{2}+1\right)^{2}-4\)
b \(\left(x^{2}-2\right)^{2}-4\)
c \(x^{2}+4 x+4-y^{2}\)
d \(x^{2}+8 x+16-a^{2}\)
e \(m^{2}-2 m+1-n^{2}\)
f \(p^{2}-5 p+\frac{25}{4}-q^{2}\)

6 a By adding and subtracting \(4 x^{2}\), factorise \(x^{4}+4\).

b Factorise \(x^{4}+4 a^{4}\).

7 A swimming pool is designed in an L-shape with dimensions in metres as shown.

The pool is enlarged or reduced depending on the value of \(x\).

a Find, in terms of \(x\), the length of:
i \(A F\)
ii \(C D\)

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_30_c996b59107099c56822bg-14}
\end{center}

b Show that the perimeter is equal to \((10 x+50) \mathrm{m}\).

c What is the perimeter if \(x=3\) ?

d Find the area of the swimming pool in terms of \(x\). Expand and simplify your answer.

e It is decided that a square swimming pool would be a better use of space.

i By factorising your answer to part \(\mathbf{d}\), find the dimensions, in terms of \(x\), of a square swimming pool with the same area as the L-shaped swimming pool.

ii What is the perimeter, in terms of \(x\), of this square swimming pool?

8 For positive whole numbers \(a\) and \(b\), prove that:
a if \(\frac{a}{b}<1\) then \(\frac{a+1}{b+1}>\frac{a}{b}\)
b if \(\frac{a}{b}>1\) then \(\frac{a+1}{b+1}<\frac{a}{b}\)

9 A right-angled triangle has a hypotenuse of length \(b \mathrm{~cm}\) and one other side of length \(a \mathrm{~cm}\). If \(b-a=1\), find the length of the third side in terms of \(a\) and \(b\).

10 Factorise \(\left(1+\frac{y^{2}+z^{2}-x^{2}}{2 y z}\right) \div\left(1-\frac{x^{2}+y^{2}-z^{2}}{2 x y}\right)\).


\end{document}
|723035582|kitty|/Applications/kitty.app|0|
Number and Algebra

The distributive law can be used to rewrite a product involving brackets as an expression without brackets. For instance, the product \(3(a+2)\) can be rewritten as \(3 a+6\); this is called the expanded form of the expression, and the process is called expansion.

The process of writing an algebraic expression as a product of two or more algebraic factors is called factorisation. Factorisation is the reverse process to expansion. For example, we can write \(3 a+6\) as \(3(a+2)\). This is called the factorised form.

In this chapter we will look at how to factorise expressions such as \(x^{2}+7 x+12\).

We recall that such an expression is called a quadratic.

\section*{Factorisation using common factors}
If each term in the algebraic expression to be factorised contains a common factor, then this common factor is a factor of the entire expression. To find the other factor, divide each term by the common factor. The common factor is placed outside brackets. For this reason the procedure is sometimes called 'taking the common factor outside the brackets'.

\section*{Example 1}
Factorise \(4 a+12\)

\section*{Solution}
4 is a common factor of \(4 a\) and 12.

Thus \(4 a+12=4(a+3)\)

Notice that the answer can be checked by expanding \(4(a+3)\).

In general, take out as many common factors as possible. The common factors may involve both numbers and pronumerals. It may be easier if you first take out the common factor of the numbers and then the common factor of the pronumerals.

\section*{Example 2}
\(\begin{array}{ll}\text { Factorise: } \mathbf{a} 12 x^{2}+3 x & \text { b } 36 a b-27 a\end{array}\)

Solution
a \(12 x^{2}+3 x=3\left(4 x^{2}+x\right)\)
b \(36 a b-27 a=9(4 a b-3 a)\)
\(=3 x(4 x+1)\)
\(=9 a(4 b-3)\)

Note: It is more efficient to take out all the common factors in one step, as shown in the following examples.

\section*{Example 3}
Factorise:
a \(3 x+9\)
b \(2 x^{2}+4 x\)
c \(-7 a^{2}-49\)
d \(7 a^{2}+63 a b\)

Solution

a \(3 x+9=3(x+3)\)

b \(2 x^{2}+4 x=2 x(x+2)\)

c \(-7 a^{2}-49=-7 \times a^{2}+(-7) \times 7\)

d \(7 a^{2}+63 a b=7 a(a+9 b)\)

\(=-7\left(a^{2}+7\right)\)

\(7\left(-a^{2}-7\right)\) is also correct.

\section*{Example 4}
Factorise:
a \(5 p q^{2}+10 p^{2} q+25 p^{2} q^{2}\)
b \(16 a b+10 b^{2}-2 a^{2} b\)

Solution

a \(5 p q^{2}+10 p^{2} q+25 p^{2} q^{2}=5 p q(q+2 p+5 p q)\)

b \(16 a b+10 b^{2}-2 a^{2} b=2 b\left(8 a+5 b-a^{2}\right)\)

\section*{Exercise 4A}
1 Complete each factorisation.
a \(12 x=12 \times \square\)
b \(24 a=12 \times \square\)
c \(36 a b=9 a \times\)
d \(15 a c=5 c \times \square\)
e \(y^{2}=y \times \square\)
f \(6 y^{2}=3 y \times \square\)
g \(24 a^{2}=6 a \times \square\)
h \(-6 b^{2}=2 b \times \square\)
i \(8 a^{2} b=2 a b \times\)
j \(4 x^{2} y=2 x \times \square\)
k \(12 m^{2} n=3 m n \times \square\)
l \(25 a^{2} b^{2}=5 a b \times \square\)

2 Fill in the blanks by finding the missing factors.
a \(12 a+18=6 \times\)
b \(15 p-10=5 \times\)
c \(20 m n-15 n=5 \times \square\)
d \(20 m n-15 n=5 n \times \square\)
e \(a^{2}+4 a=(a+4) \times \square\)
f \(b^{2}-10 b=b \times \square\)
g \(6 y z^{2}-18 y z=3 z \times \square\)
h \(6 y z^{2}-18 y z=y z \times \square\)
i \(6 y z^{2}-18 y z=-3 y z \times \square\)
j \(6 y z^{2}-18 y z=6 y z \times \square\)

3 Factorise:
a \(6 x+24\)
b \(5 a+15\)
c \(a c+5 c\)
d \(a^{2}+a\)
e \(y^{2}+x y\)
f \(4 x+24\)
g \(7 a-63\)
h \(9 a+36\)
i \(y^{2}-3 y\)
\(\mathbf{j}-14 a-21\)
\(\mathbf{k}-6 y-9\)
l \(-4-12 b\)

4 Factorise:
a \(4 a b+16 a\)
b \(12 a^{2}+8 a\)
c \(18 m^{2} n+9 m n^{2}\)
d \(15 a^{2} b^{2}+10 a b^{2}\)
e \(4 a^{2}+6 a\)
f \(8 a^{2}+12 a b\)

5 Factorise:
a \(3 b-6 b^{2}\)
b \(4 x^{2}-6 x y\)
c \(9 m n-12 m^{2} n\)
d \(18 y-9 y^{2}\)
e \(4 a-6 a b^{2}\)
f \(6 x y-4 x^{2}\)
g \(14 m n^{2}-21 m^{2} n\)
h \(6 p q^{2}-21 q p^{2}\)
i \(10 a b^{2}-25 a^{2} b\)

6 Factorise:
\(\mathbf{a}-10 b^{2}+5 b\)
\(\mathbf{b}-16 a^{2} b-8 a b\)
c \(-x^{2} y-3 x y\)
d \(-4 p q+16 p^{2}\)
e \(-5 x^{2} y+30 x\)
f \(18 p^{2}-4 p q\)
\(\mathbf{g}-8 a^{2} b^{2}-2 a b\)
h \(12 x y^{2}-3 x^{2} y\)
i \(-25 m^{2} n^{2}-10 m n^{2}\)

7 In each part, an expression for the area of the rectangle has been given. Find an expression for the missing side length.

a

\(2 a+3\)

Area \(=8 a+12\)

c

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_30_c996b59107099c56822bg-04(1)}
\end{center}

e \(6 a\)

\[
\text { Area }=12 a^{2}+6 a b
\]

b

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_30_c996b59107099c56822bg-04}
\end{center}

d

\(a\)

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_30_c996b59107099c56822bg-04(2)}
\end{center}

\(\mathbf{f}\) \(3 q^{2}-2\)

Area \(=24 p^{2} q^{2}-16 p^{2}\)

Example 4 8 Factorise:

a \(4 a^{2} b-2 a b+8 a b^{2} \quad\) b \(4 m^{2} n-4 m n+16 n^{2}\)

c \(7 a b+14 a^{2}+21 b\)

d \(2 m^{2}+4 m n+6 n\)

e \(5 a^{2} b+3 a b+4 a b^{2}\)

f \(6 a+8 a b+10 a b^{2}\)

g \(5 p^{2} q^{2}+10 p q^{2}+15 p^{2} q\)

h \(5 \ell^{2}-15 \ell m-20 m^{2}\)

\section*{1 B Factorisation using the difference of two squares}
You will recall from Section \(1 \mathrm{G}\) the important identity

\[
(a+b)(a-b)=a^{2}-b^{2},
\]

which is called the difference of two squares. We can now use this result the other way around to factorise an expression that is the difference of two squares.

That is:

\[
a^{2}-b^{2}=(a+b)(a-b)
\]

That is, the factors of \(a^{2}-b^{2}\) are \(a+b\) and \(a-b\).

\section*{Example 5}
Factorise:
a \(x^{2}-9\)
b \(25-y^{2}\)
c \(4 x^{2}-9\)

\section*{Solution}
a \(x^{2}-9=x^{2}-3^{2}\)

\[
=(x+3)(x-3)
\]

Check the answer by expanding \((x+3)(x-3)\) to see that \(x^{2}-9\) is obtained.

b \(25-y^{2}=5^{2}-y^{2}\)

\[
=(5+y)(5-y)
\]

c \(4 x^{2}-9=(2 x)^{2}-3^{2}\)

\[
=(2 x+3)(2 x-3)
\]

\section*{Example 6}
Factorise:
a \(3 a^{2}-27\)
b \(-16+9 x^{2}\)

\section*{Solution}
a \(3 a^{2}-27=3\left(a^{2}-9\right)\)
b \(-16+9 x^{2}=9 x^{2}-16\)
\(=3\left(a^{2}-3^{2}\right)\)
\(=(3 x)^{2}-4^{2}\)
\(=3(a+3)(a-3)\)
\(=(3 x-4)(3 x+4)\)

Factorisation using the difference of two squares identity

\(a^{2}-b^{2}=(a+b)(a-b)\)
|723035650|kitty|/Applications/kitty.app|0|
d \(d^{2}-400\)
|723036599|kitty|/Applications/kitty.app|0|
1 Factorise:
a \(x^{2}-16\)
b \(x^{2}-49\)
c \(a^{2}-121\)
e \((2 x)^{2}-25\)
f \((3 x)^{2}-16\)
g \((4 x)^{2}-1\)
h \((5 m)^{2}-9\)
i \(9 x^{2}-4\)
j \(16 y^{2}-49\)
k \(100 a^{2}-49 b^{2}\)
l \(64 m^{2}-81 p^{2}\)
m \(1-4 a^{2}\)
n \(9-16 y^{2}\)
o \(25 a^{2}-100 b^{2}\)
p \(-9+x^{2}\)
|723036615|kitty|/Applications/kitty.app|0|
Example 6a 2 Factorise:
|723036619|kitty|/Applications/kitty.app|0|
a \(3 x^{2}-48\)
|723036622|kitty|/Applications/kitty.app|0|
a \(3 x^{2}-48\)
e \(10 x^{2}-1000\)
|723036624|kitty|/Applications/kitty.app|0|
a \(3 x^{2}-48\)
e \(10 x^{2}-1000\)
i \(3-12 b^{2}\)
|723036627|kitty|/Applications/kitty.app|0|
a \(3 x^{2}-48\)
e \(10 x^{2}-1000\)
i \(3-12 b^{2}\)
m \(45 m^{2}-125 n^{2}\)
|723036630|kitty|/Applications/kitty.app|0|
p \(-200 p^{2}+32 q^{2}\)
|723036657|kitty|/Applications/kitty.app|0|
l \(16 x^{2}-100 y^{2}\)
p \(-200 p^{2}+32 q^{2}\)
|723036659|kitty|/Applications/kitty.app|0|
h \(12 m^{2}-75\)
l \(16 x^{2}-100 y^{2}\)
p \(-200 p^{2}+32 q^{2}\)
|723036662|kitty|/Applications/kitty.app|0|
d \(6 x^{2}-24\)
h \(12 m^{2}-75\)
l \(16 x^{2}-100 y^{2}\)
p \(-200 p^{2}+32 q^{2}\)
|723036665|kitty|/Applications/kitty.app|0|
b \(4 x^{2}-100\)
c \(5 x^{2}-45\)
f \(7 x^{2}-63\)
g \(8 x^{2}-50\)
j \(20-5 y^{2}\)
k \(27 a^{2}-12 b^{2}\)
n \(27 a^{2}-192 l^{2}\)
o \(-8 x^{2}+32 y^{2}\)
|723036685|kitty|/Applications/kitty.app|0|
a \(-25+x^{2}\)
|723036699|kitty|/Applications/kitty.app|0|
a \(-25+x^{2}\)
d \(-81 x^{2}+16\)
|723036701|kitty|/Applications/kitty.app|0|
a \(-25+x^{2}\)
d \(-81 x^{2}+16\)
g \(-12 x^{2}+27\)
|723036704|kitty|/Applications/kitty.app|0|
c \(-9 x^{2}+4\)
|723036716|kitty|/Applications/kitty.app|0|
c \(-9 x^{2}+4\)
f \(-18+50 x^{2}\)
|723036717|kitty|/Applications/kitty.app|0|
c \(-9 x^{2}+4\)
f \(-18+50 x^{2}\)
i \(-175+28 x^{2}\)
|723036720|kitty|/Applications/kitty.app|0|
b \(-4+9 x^{2}\)
e \(-100 x^{2}+9\)
h \(-36 x^{2}+400\)
|723036728|kitty|/Applications/kitty.app|0|
a \(23^{2}-7^{2}\)
|723036760|kitty|/Applications/kitty.app|0|
a \(23^{2}-7^{2}\)
d \(94^{2}-6^{2}\)
|723036763|kitty|/Applications/kitty.app|0|
a \(23^{2}-7^{2}\)
d \(94^{2}-6^{2}\)
g \(11.3^{2}-8.7^{2}\)
|723036765|kitty|/Applications/kitty.app|0|
4 Use the factorisation of the difference of two squares to evaluate the following. One has been done for you.

\[
\begin{aligned}
17^{2}-3^{2} & =(17+3)(17-3) \\
& =20 \times 14 \\
& =280
\end{aligned}
\]
a \(23^{2}-7^{2}\)
d \(94^{2}-6^{2}\)
g \(11.3^{2}-8.7^{2}\)
|723036770|kitty|/Applications/kitty.app|0|
b \(23^{2}-3^{2}\)
|723036780|kitty|/Applications/kitty.app|0|
b \(23^{2}-3^{2}\)
e \(1.8^{2}-0.2^{2}\)
|723036782|kitty|/Applications/kitty.app|0|
b \(23^{2}-3^{2}\)
e \(1.8^{2}-0.2^{2}\)
h \(92.6^{2}-7.4^{2}\)
|723036783|kitty|/Applications/kitty.app|0|
c \(36^{2}-6^{2}\)
f \(28^{2}-2.2^{2}\)
i \(3.214^{2}-2.214^{2}\)
|723036791|kitty|/Applications/kitty.app|0|
Example 6b 3 Factorise:
|723036799|kitty|/Applications/kitty.app|0|
v \(8^{2}-7^{2}\)
|723036808|kitty|/Applications/kitty.app|0|
vi \(9^{2}-8^{2}\)
|723036813|kitty|/Applications/kitty.app|0|
vii \(10^{2}-9^{2}\)
|723036819|kitty|/Applications/kitty.app|0|
iv \(7^{2}-6^{2}\)
viii \(101^{2}-100^{2}\)
|723036822|kitty|/Applications/kitty.app|0|
ii \(5^{2}-4^{2}\)
vi \(9^{2}-8^{2}\)
iii \(6^{2}-5^{2}\)
vii \(10^{2}-9^{2}\)
|723036832|kitty|/Applications/kitty.app|0|
i \(4^{2}-3^{2}\)
v \(8^{2}-7^{2}\)
|723036837|kitty|/Applications/kitty.app|0|
5 a Evaluate:

b What do you notice?

c Use the factorisation of \((n+1)^{2}-n^{2}\) to prove the result of part \(\mathbf{b}\).
|723036856|kitty|/Applications/kitty.app|0|
6 The following leads you through the geometrical proof that \(a^{2}-b^{2}=(a-b)(a+b)\).

In the diagram opposite, square \(E C F G\), of side length \(b\), is cut out of square \(A B C D\), which has side length \(a\).

a What is the area of hexagon \(A B E G F D\) ?

b What is the length of \(B E\) ?

c What is the length of \(D F\) ?

d The rectangle \(H G F D\) is moved so that \(D F\) is placed on top of \(B E\) (see diagram).

i What is the area of the large shaded rectangle?

ii What have we proved?
\includegraphics[max width=\textwidth, center]{2023_11_30_c996b59107099c56822bg-06}
|723036874|kitty|/Applications/kitty.app|0|
\section*{Example 7}
Factorise \(x^{2}-3 x-18\).

\section*{Solution}
We are looking for two numbers that multiply to give -18 and add to give -3 . The numbers -6 and 3 satisfy both conditions.

\(x^{2}-3 x-18=(x-6)(x+3)\)

Consider factorising \(x^{2}-36\).

Since the constant term is -36 and the coefficient of \(x\) is 0 , we are looking for two numbers that multiply to give -36 and add to give 0 . The numbers -6 and 6 satisfy both conditions.

Thus \(x^{2}-36=(x-6)(x+6)\).

Note: It is not really sensible to do the example in this way - it is best to recognise \(x^{2}-36\) as a difference of squares. However, it does show that the method of factorisation in this section is consistent with the earlier technique.

\section*{Example 8}
Factorise \(x^{2}+8 x+16\)

\section*{Solution}
We are looking for two numbers that multiply to give 16 and add to give 8 . The numbers 4 and 4 satisfy both conditions.

Thus \(x^{2}+8 x+16=(x+4)(x+4)\)

\[
=(x+4)^{2}
\]
|723037078|kitty|/Applications/kitty.app|0|
a \(x^{2}+5 x+6\)
|723037140|kitty|/Applications/kitty.app|0|
a \(x^{2}+5 x+6\)
c \(x^{2}+7 x+10\)
|723037147|kitty|/Applications/kitty.app|0|
a \(x^{2}+5 x+6\)
c \(x^{2}+7 x+10\)
e \(x^{2}+9 x+14\)
g \(x^{2}+9 x+20\)
|723037153|kitty|/Applications/kitty.app|0|
a \(x^{2}+5 x+6\)
c \(x^{2}+7 x+10\)
e \(x^{2}+9 x+14\)
g \(x^{2}+9 x+20\)
i \(x^{2}+12 x+32\)
|723037160|kitty|/Applications/kitty.app|0|
a \(x^{2}+5 x+6\)
c \(x^{2}+7 x+10\)
e \(x^{2}+9 x+14\)
g \(x^{2}+9 x+20\)
i \(x^{2}+12 x+32\)
\(\mathbf{k} x^{2}+20 x+75\)
|723037166|kitty|/Applications/kitty.app|0|
a \(x^{2}+5 x+6\)
c \(x^{2}+7 x+10\)
e \(x^{2}+9 x+14\)
g \(x^{2}+9 x+20\)
i \(x^{2}+12 x+32\)
\(\mathbf{k} x^{2}+20 x+75\)
\(\mathbf{m} x^{2}+15 x+56\)
|723037171|kitty|/Applications/kitty.app|0|
b \(x^{2}+11 x+18\)
d \(x^{2}+11 x+30\)
f \(x^{2}+19 x+90\)
h \(x^{2}+7 x+12\)
j \(x^{2}+13 x+40\)
l \(x^{2}+28 x+27\)
n \(x^{2}+18 x+56\)
|723037212|kitty|/Applications/kitty.app|0|
\section*{Exercise 4C}
1 Factorise these quadratric expressions.
b \(x^{2}+11 x+18\)
d \(x^{2}+11 x+30\)
f \(x^{2}+19 x+90\)
h \(x^{2}+7 x+12\)
j \(x^{2}+13 x+40\)
l \(x^{2}+28 x+27\)
n \(x^{2}+18 x+56\)
|723037219|kitty|/Applications/kitty.app|0|
a \(x^{2}-5 x+6\)
|723037241|kitty|/Applications/kitty.app|0|
a \(x^{2}-5 x+6\)
c \(x^{2}-17 x+30\)
|723037242|kitty|/Applications/kitty.app|0|
a \(x^{2}-5 x+6\)
c \(x^{2}-17 x+30\)
e \(x^{2}-9 x+14\)
|723037244|kitty|/Applications/kitty.app|0|
a \(x^{2}-5 x+6\)
c \(x^{2}-17 x+30\)
e \(x^{2}-9 x+14\)
g \(x^{2}-15 x+44\)
|723037246|kitty|/Applications/kitty.app|0|
a \(x^{2}-5 x+6\)
c \(x^{2}-17 x+30\)
e \(x^{2}-9 x+14\)
g \(x^{2}-15 x+44\)
i \(x^{2}-18 x+80\)
|723037250|kitty|/Applications/kitty.app|0|
a \(x^{2}-5 x+6\)
c \(x^{2}-17 x+30\)
e \(x^{2}-9 x+14\)
g \(x^{2}-15 x+44\)
i \(x^{2}-18 x+80\)
k \(x^{2}-14 x+40\)
|723037253|kitty|/Applications/kitty.app|0|
a \(x^{2}-5 x+6\)
c \(x^{2}-17 x+30\)
e \(x^{2}-9 x+14\)
g \(x^{2}-15 x+44\)
i \(x^{2}-18 x+80\)
k \(x^{2}-14 x+40\)
\(\mathbf{m} x^{2}-30 x+56\)
|723037258|kitty|/Applications/kitty.app|0|
b \(x^{2}-14 x+33\)
d \(x^{2}-13 x+42\)
f \(x^{2}-47 x+90\)
h \(x^{2}-25 x+100\)
j \(x^{2}-21 x+80\)
l \(x^{2}-11 x+24\)
n \(x^{2}-14 x+24\)
|723037280|kitty|/Applications/kitty.app|0|
a \(x^{2}+x-6\)
|723037327|kitty|/Applications/kitty.app|0|
a \(x^{2}+x-6\)
c \(x^{2}+x-30\)
|723037329|kitty|/Applications/kitty.app|0|
a \(x^{2}+x-6\)
c \(x^{2}+x-30\)
e \(x^{2}-5 x-14\)
|723037331|kitty|/Applications/kitty.app|0|
a \(x^{2}+x-6\)
c \(x^{2}+x-30\)
e \(x^{2}-5 x-14\)
g \(x^{2}-7 x-44\)
|723037334|kitty|/Applications/kitty.app|0|
a \(x^{2}+x-6\)
c \(x^{2}+x-30\)
e \(x^{2}-5 x-14\)
g \(x^{2}-7 x-44\)
i \(x^{2}+2 x-80\)
|723037336|kitty|/Applications/kitty.app|0|
a \(x^{2}+x-6\)
c \(x^{2}+x-30\)
e \(x^{2}-5 x-14\)
g \(x^{2}-7 x-44\)
i \(x^{2}+2 x-80\)
k \(x^{2}+3 x-40\)
|723037339|kitty|/Applications/kitty.app|0|
a \(x^{2}+x-6\)
c \(x^{2}+x-30\)
e \(x^{2}-5 x-14\)
g \(x^{2}-7 x-44\)
i \(x^{2}+2 x-80\)
k \(x^{2}+3 x-40\)
m \(x^{2}+4 x-21\)
|723037343|kitty|/Applications/kitty.app|0|
a \(x^{2}+x-6\)
c \(x^{2}+x-30\)
e \(x^{2}-5 x-14\)
g \(x^{2}-7 x-44\)
i \(x^{2}+2 x-80\)
k \(x^{2}+3 x-40\)
m \(x^{2}+4 x-21\)
o \(x^{2}-x+56\)
|723037345|kitty|/Applications/kitty.app|0|
b \(x^{2}-8 x-33\)
d \(x^{2}-19 x-42\)
f \(x^{2}-9 x-90\)
h \(x^{2}+15 x-100\)
j \(x^{2}-7 x-60\)
l \(x^{2}-10 x-24\)
n \(x^{2}+2 x-15\)
p \(x^{2}+5 x-24\)
|723037350|kitty|/Applications/kitty.app|0|
a \(x^{2}-3 x+2\)
|723037363|kitty|/Applications/kitty.app|0|
a \(x^{2}-3 x+2\)
c \(x^{2}-3 x-10\)
|723037364|kitty|/Applications/kitty.app|0|
a \(x^{2}-3 x+2\)
c \(x^{2}-3 x-10\)
e \(x^{2}-5 x-14\)
|723037366|kitty|/Applications/kitty.app|0|
a \(x^{2}-3 x+2\)
c \(x^{2}-3 x-10\)
e \(x^{2}-5 x-14\)
g \(x^{2}-5 x+4\)
|723037368|kitty|/Applications/kitty.app|0|
a \(x^{2}-3 x+2\)
c \(x^{2}-3 x-10\)
e \(x^{2}-5 x-14\)
g \(x^{2}-5 x+4\)
i \(x^{2}-x-12\)
|723037371|kitty|/Applications/kitty.app|0|
a \(x^{2}-3 x+2\)
c \(x^{2}-3 x-10\)
e \(x^{2}-5 x-14\)
g \(x^{2}-5 x+4\)
i \(x^{2}-x-12\)
k \(x^{2}+3 x-10\)
|723037375|kitty|/Applications/kitty.app|0|
b \(x^{2}+8 x+12\)
d \(x^{2}+11 x+30\)
f \(x^{2}-9 x-90\)
h \(x^{2}-7 x-18\)
j \(x^{2}-11 x+28\)
l \(x^{2}+x-90\)
|723037382|kitty|/Applications/kitty.app|0|
a \(x^{2}+6 x+9\)
|723037389|kitty|/Applications/kitty.app|0|
a \(x^{2}+6 x+9\)
c \(x^{2}-10 x+25\)
|723037390|kitty|/Applications/kitty.app|0|
a \(x^{2}+6 x+9\)
c \(x^{2}-10 x+25\)
e \(x^{2}+10 x+25\)
|723037392|kitty|/Applications/kitty.app|0|
a \(x^{2}+6 x+9\)
c \(x^{2}-10 x+25\)
e \(x^{2}+10 x+25\)
g \(x^{2}+30 x+225\)
|723037394|kitty|/Applications/kitty.app|0|
a \(x^{2}+6 x+9\)
c \(x^{2}-10 x+25\)
e \(x^{2}+10 x+25\)
g \(x^{2}+30 x+225\)
i \(x^{2}-20 x+100\)
|723037395|kitty|/Applications/kitty.app|0|
b \(x^{2}+14 x+49\)
d \(x^{2}-18 x+81\)
f \(x^{2}+12 x+36\)
h \(x^{2}-16 x+64\)
j \(x^{2}-8 x+16\)
|723037402|kitty|/Applications/kitty.app|0|
The expansion of a perfect square has a special form. For example:

\[
\begin{aligned}
(x+3)^{2} & =(x+3)(x+3) \\
& =x^{2}+6 x+9 \\
& =x^{2}+2 \times(3 x)+3^{2}
\end{aligned}
\]
|723037620|kitty|/Applications/kitty.app|0|
\section*{Factorising using the perfect square identities}
\begin{itemize}
  \item In general, \(a^{2}+2 a b+b^{2}=(a+b)^{2}\)
  \item Similarly, \(a^{2}-2 a b+b^{2}=(a-b)^{2}\)
\end{itemize}
|723037659|kitty|/Applications/kitty.app|0|
\section*{Example 9}
Factorise:
a \(x^{2}+8 x+16\)
b \(x^{2}-10 x+25\)
c \(x^{2}+11 x+\frac{121}{4}\)

\section*{Solution}
a \(x^{2}+8 x+16=x^{2}+2 \times 4 x+4^{2}\)

\[
=(x+4)^{2}
\]

c \(x^{2}+11 x+\frac{121}{4}=x^{2}+2 \times \frac{11}{2} x+\left(\frac{11}{2}\right)^{2}\)

\[
=\left(x+\frac{11}{2}\right)^{2}
\]

b \(x^{2}-10 x+25=x^{2}-2 \times 5 x+5^{2}\)

\[
=(x-5)^{2}
\]
|723037681|kitty|/Applications/kitty.app|0|
a \(x^{2}+12 x+36\)
|723037698|kitty|/Applications/kitty.app|0|
a \(x^{2}+12 x+36\)
d \(a^{2}-4 a+4\)
|723037701|kitty|/Applications/kitty.app|0|
a \(x^{2}+12 x+36\)
d \(a^{2}-4 a+4\)
g \(x^{2}-9 x+\frac{81}{4}\)
|723037704|kitty|/Applications/kitty.app|0|
a \(x^{2}+8 x+16=(x+\)
|723037711|kitty|/Applications/kitty.app|0|
a \(x^{2}+8 x+16=(x+\)
c \(x^{2}-\ldots+\ldots=(x-9)^{2}\)
|723037712|kitty|/Applications/kitty.app|0|
3 Identify the simple quadratic expression that cannot be factorised as a perfect square.
c i \(x^{2}+\frac{2 x}{3}+\frac{1}{9}\)
ii \(x^{2}-3 x+\frac{9}{4}\)
iii \(x^{2}-\frac{5 x}{3}+\frac{25}{36}\)
iv \(x^{2}+\frac{7 x}{4}+\frac{49}{16}\)
d \(\mathbf{i} x^{2}-\frac{11 x}{2}+\frac{121}{16}\)
ii \(x^{2}-\frac{4 x}{5}+\frac{4}{25}\)
iii \(x^{2}-\frac{3 x}{2}-\frac{9}{16}\)
iv \(x^{2}-\frac{9 x}{2}+\frac{81}{16}\)
|723037766|kitty|/Applications/kitty.app|0|
b \(x^{2}-8 x+16\)
|723037782|kitty|/Applications/kitty.app|0|
b \(x^{2}-8 x+16\)
e \(m^{2}-26 m+169\)
|723037784|kitty|/Applications/kitty.app|0|
b \(x^{2}-8 x+16\)
e \(m^{2}-26 m+169\)
h \(x^{2}+13 x+\frac{169}{4}\)
|723037786|kitty|/Applications/kitty.app|0|
3 Identify the simple quadratic expression that cannot be factorised as a perfect square.
a \(\mathbf{i} x^{2}+4 x+4\)
ii \(x^{2}-6 x+12\)
iii \(x^{2}-12 x+36\)
iv \(x^{2}-10 x+25\)
b i \(x^{2}+6 x+9\)
ii \(x^{2}+5 x+\frac{25}{4}\)
iii \(x^{2}-8 x-16\)
iv \(x^{2}-14 x+49\)
|723037805|kitty|/Applications/kitty.app|0|
c \(x^{2}+10 x+25\)
f \(a^{2}+28 a+196\)
i \(x^{2}-11 x+\frac{121}{4}\)
|723037817|kitty|/Applications/kitty.app|0|
2 Copy and complete:
b \(x^{2}-10 x+\)
\(=(x-5)^{2}\)
d \(x^{2}+\)
e \(x^{2}-9 x+\)
\(=(x-\)
f \(x^{2}-\frac{5 x}{2}+\)
\(+\_=\left(x+\frac{7}{2}\right)^{2}\)
)\(^{2}\)
|723037877|kitty|/Applications/kitty.app|0|

4 A brick company provides rectangular and square pavers.

a Draw a diagram to show how two different square pavers of side lengths \(a\) and \(b\) respectively, and two identical rectangular pavers with dimensions \(a \times b\), can be arranged into a square.

b How many of each type of paver enables you to pave a square area of side length \(a+3 b\) ? Draw a diagram to illustrate how this can be done.
|723037881|kitty|/Applications/kitty.app|0|
\section*{Example 10}
Factorise:
a \(3 x^{2}+9 x+6\)
b \(6 x^{2}-54\)
c \(-x^{2}-x+2\)

\section*{Solution}
a \(3 x^{2}+9 x+6=3\left(x^{2}+3 x+2\right) \quad\) (Take out the common factor.)

\[
=3(3 x+2)(3 x+1)
\]

b \(6 x^{2}-54=6\left(x^{2}-9\right)\)

\[
=6(x+3)(x-3)
\]

c \(-x^{2}-x+2=-\left(x^{2}+x-2\right)\)

(Factor -1 from each term.)

\[
=-(x+2)(x-1)
\]
|723038066|kitty|/Applications/kitty.app|0|
a \(2 x^{2}+14 x+24\)
|723038074|kitty|/Applications/kitty.app|0|
a \(2 x^{2}+14 x+24\)
d \(4 x^{2}-24 x+36\)
|723038076|kitty|/Applications/kitty.app|0|
a \(2 x^{2}+14 x+24\)
d \(4 x^{2}-24 x+36\)
g \(4 x^{2}-4 x+48\)
|723038081|kitty|/Applications/kitty.app|0|
a \(2 x^{2}+14 x+24\)
d \(4 x^{2}-24 x+36\)
g \(4 x^{2}-4 x+48\)
j \(3 x^{2}+9 x-120\)
|723038085|kitty|/Applications/kitty.app|0|
a \(2 x^{2}+14 x+24\)
d \(4 x^{2}-24 x+36\)
g \(4 x^{2}-4 x+48\)
j \(3 x^{2}+9 x-120\)
m \(2 x^{2}-4 x-96\)
|723038088|kitty|/Applications/kitty.app|0|
b \(3 x^{2}+24 x+36\)
|723038099|kitty|/Applications/kitty.app|0|
b \(3 x^{2}+24 x+36\)
e \(7 x^{2}+14 x+7\)
|723038101|kitty|/Applications/kitty.app|0|
b \(3 x^{2}+24 x+36\)
e \(7 x^{2}+14 x+7\)
h \(2 x^{2}-18 x+36\)
|723038104|kitty|/Applications/kitty.app|0|
b \(3 x^{2}+24 x+36\)
e \(7 x^{2}+14 x+7\)
h \(2 x^{2}-18 x+36\)
k \(3 x^{2}-3 x-90\)
|723038107|kitty|/Applications/kitty.app|0|
b \(3 x^{2}+24 x+36\)
e \(7 x^{2}+14 x+7\)
h \(2 x^{2}-18 x+36\)
k \(3 x^{2}-3 x-90\)
n \(5 x^{2}+65 x+180\)
|723038112|kitty|/Applications/kitty.app|0|
b \(3 x^{2}+24 x+36\)
e \(7 x^{2}+14 x+7\)
h \(2 x^{2}-18 x+36\)
k \(3 x^{2}-3 x-90\)
n \(5 x^{2}+65 x+180\)
q \(5 x^{2}-20 x+20\)
|723038115|kitty|/Applications/kitty.app|0|
a \(2 x^{2}+14 x+24\)
d \(4 x^{2}-24 x+36\)
g \(4 x^{2}-4 x+48\)
j \(3 x^{2}+9 x-120\)
m \(2 x^{2}-4 x-96\)
p \(3 x^{2}-18 x+27\)
|723038125|kitty|/Applications/kitty.app|0|
c \(3 x^{2}-27 x+24\)
f \(5 x^{2}-5 x-30\)
i \(5 x^{2}+40 x+35\)
l \(5 x^{2}+60 x+180\)
o \(3 x^{2}+30 x-72\)
r \(3 x^{2}-24 x+36\)
|723038130|kitty|/Applications/kitty.app|0|
\section*{Exercise 4E}
1 Factorise:
c \(3 x^{2}-27 x+24\)
f \(5 x^{2}-5 x-30\)
i \(5 x^{2}+40 x+35\)
l \(5 x^{2}+60 x+180\)
o \(3 x^{2}+30 x-72\)
r \(3 x^{2}-24 x+36\)
|723038134|kitty|/Applications/kitty.app|0|
a \(4 x^{2}-16\)
|723038143|kitty|/Applications/kitty.app|0|
a \(4 x^{2}-16\)
d \(3 a^{2}-27\)
|723038145|kitty|/Applications/kitty.app|0|
a \(4 x^{2}-16\)
d \(3 a^{2}-27\)
g \(27 x^{2}-3 y^{2}\)
|723038148|kitty|/Applications/kitty.app|0|
a \(4 x^{2}-16\)
d \(3 a^{2}-27\)
g \(27 x^{2}-3 y^{2}\)
j \(128-2 x^{2}\)
|723038154|kitty|/Applications/kitty.app|0|
a \(4 x^{2}-16\)
d \(3 a^{2}-27\)
g \(27 x^{2}-3 y^{2}\)
j \(128-2 x^{2}\)
\(\mathbf{m} \frac{1}{4} a^{2}-9\)
|723038157|kitty|/Applications/kitty.app|0|
b \(2 x^{2}-18\)
|723038175|kitty|/Applications/kitty.app|0|
b \(2 x^{2}-18\)
e \(6 x^{2}-600\)
|723038176|kitty|/Applications/kitty.app|0|
b \(2 x^{2}-18\)
e \(6 x^{2}-600\)
h \(45-5 b^{2}\)
|723038179|kitty|/Applications/kitty.app|0|
b \(2 x^{2}-18\)
e \(6 x^{2}-600\)
h \(45-5 b^{2}\)
k \(\frac{1}{2} a^{2}-2 b^{2}\)
|723038182|kitty|/Applications/kitty.app|0|
b \(2 x^{2}-18\)
e \(6 x^{2}-600\)
h \(45-5 b^{2}\)
k \(\frac{1}{2} a^{2}-2 b^{2}\)
n \(\frac{1}{5} x^{2}-20\)
|723038186|kitty|/Applications/kitty.app|0|
c \(3 x^{2}-48\)
f \(3 a^{2}-27 b^{2}\)
i \(12-3 m^{2}\)
l \(27 x^{2}-\frac{1}{3} y^{2}\)
o \(\frac{1}{4} x^{2}-y^{2}\)
|723038200|kitty|/Applications/kitty.app|0|
a \(-x^{2}-8 x-12\)
|723038212|kitty|/Applications/kitty.app|0|
a \(-x^{2}-8 x-12\)
d \(9+8 x-x^{2}\)
|723038214|kitty|/Applications/kitty.app|0|
a \(-x^{2}-8 x-12\)
d \(9+8 x-x^{2}\)
g \(-x^{2}+3 x+40\)
|723038218|kitty|/Applications/kitty.app|0|
a \(-x^{2}-8 x-12\)
d \(9+8 x-x^{2}\)
g \(-x^{2}+3 x+40\)
j \(11 x-x^{2}-24\)
|723038223|kitty|/Applications/kitty.app|0|
a \(-x^{2}-8 x-12\)
d \(9+8 x-x^{2}\)
g \(-x^{2}+3 x+40\)
j \(11 x-x^{2}-24\)
m \(-16 x-63-x^{2}\)
|723038235|kitty|/Applications/kitty.app|0|
b \(12-11 x-x^{2}\)
|723038246|kitty|/Applications/kitty.app|0|
b \(12-11 x-x^{2}\)
e \(-x^{2}-4 x-4\)
|723038248|kitty|/Applications/kitty.app|0|
b \(12-11 x-x^{2}\)
e \(-x^{2}-4 x-4\)
h \(42+x-x^{2}\)
|723038252|kitty|/Applications/kitty.app|0|
b \(12-11 x-x^{2}\)
e \(-x^{2}-4 x-4\)
h \(42+x-x^{2}\)
k \(-3 x^{2}-30 x+72\)
|723038256|kitty|/Applications/kitty.app|0|
b \(12-11 x-x^{2}\)
e \(-x^{2}-4 x-4\)
h \(42+x-x^{2}\)
k \(-3 x^{2}-30 x+72\)
n \(-x^{2}-35+12 x\)
|723038258|kitty|/Applications/kitty.app|0|

c \(7-6 x-x^{2}\)



f \(-x^{2}-14 x-45\)


i \(22 x-x^{2}-40\)



l \(-56-x^{2}-15 x\)



o \(7 x+18-x^{2}\)
|723038267|kitty|/Applications/kitty.app|0|
                \draw (0,0) -- (3,0) -- (3,2) -- (0,2) -- cycle;
                \node at (1.5, 2) [above] {$2a + 3$};
                \node at (1.5, 1) {Area = $8a + 12$};
|723038410|kitty|/Applications/kitty.app|0|
            \begin{tikzpicture} 
                \draw (0,0) -- (3,0) -- (3,2) -- (0,2) -- cycle;
                \node at (1.5, 2) [above] {$2a + 3$};
                \node at (1.5, 1) {Area = $8a + 12$};
            \end{tikzpicture}
|723038419|kitty|/Applications/kitty.app|0|
 \begin{tikzpicture} \end{tikzpicture}|723038425|kitty|/Applications/kitty.app|0|
            \begin{tikzpicture} \end{tikzpicture}
|723038427|kitty|/Applications/kitty.app|0|
\begin{questions}
\question[3] In each part, an expression for the area of the rectangle has been given. Find an expression for the missing side length.
    \begin{subparts}\begin{multicols}{3}
        \part 
            \begin{tikzpicture} 
                \draw (0,0) -- (3,0) -- (3,2) -- (0,2) -- cycle;
                \node at (1.5, 2) [above] {$2a + 3$};
                \node at (1.5, 1) {Area = $8a + 12$};
            \end{tikzpicture}
            \begin{solutionordottedlines}[1in]
                $4$
            \end{solutionordottedlines}
        \part 
            \begin{tikzpicture} 
                \draw (0,0) -- (3,0) -- (3,2) -- (0,2) -- cycle;
                \node at (1.5, 2) [above] {$2a + 3$};
                \node at (1.5, 1) {Area = $8a + 12$};
            \end{tikzpicture}
            \begin{solutionordottedlines}[1in]
                $2b+3$
            \end{solutionordottedlines}
        \part
            \begin{tikzpicture} 
                \draw (0,0) -- (3,0) -- (3,2) -- (0,2) -- cycle;
                \node at (1.5, 2) [above] {$2a + 3$};
                \node at (1.5, 1) {Area = $8a + 12$};
            \end{tikzpicture}
            \begin{solutionordottedlines}[1in]
                $2a + b$
            \end{solutionordottedlines}
    \end{multicols}\end{subparts}
\end{questions}
|723038500|kitty|/Applications/kitty.app|0|
\graphicspath{ {./images/} }
|723038735|kitty|/Applications/kitty.app|0|
\randomword{square}%
|723039980|kitty|/Applications/kitty.app|0|
 Factorisation using common factors|723040025|kitty|/Applications/kitty.app|0|
examples
|723040135|kitty|/Applications/kitty.app|0|
\section*{Example 1}
Factorise \(4 a+12\)

\section*{Solution}
4 is a common factor of \(4 a\) and 12.

Thus \(4 a+12=4(a+3)\)
|723040138|kitty|/Applications/kitty.app|0|
Factorise: a \(12 x^{2}+3 x\)
|723040164|kitty|/Applications/kitty.app|0|
b \(36 a b-27 a\)
|723040187|kitty|/Applications/kitty.app|0|
a \(12 x^{2}+3 x=3\left(4 x^{2}+x\right)\)
|723040207|kitty|/Applications/kitty.app|0|
b \(36 a b-27 a=9(4 a b-3 a)\)
\(=3 x(4 x+1)\)
\(=9 a(4 b-3)\)
|723040215|kitty|/Applications/kitty.app|0|

\section*{Solution}

|723040237|kitty|/Applications/kitty.app|0|
exercises
|723040327|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\section*{Example 2}
|723040348|kitty|/Applications/kitty.app|0|
a \(3 x+9\)
b \(2 x^{2}+4 x\)
c \(-7 a^{2}-49\)
d \(7 a^{2}+63 a b\)
|723040367|kitty|/Applications/kitty.app|0|
\section*{Example 3}
Factorise:

|723040396|kitty|/Applications/kitty.app|0|
Solution
|723040397|kitty|/Applications/kitty.app|0|
a \(3 x+9=3(x+3)\)
|723040398|kitty|/Applications/kitty.app|0|
            a \(3 x+9=3(x+3)\)
|723040424|kitty|/Applications/kitty.app|0|
b \(2 x^{2}+4 x=2 x(x+2)\)
|723040430|kitty|/Applications/kitty.app|0|
c \(-7 a^{2}-49=-7 \times a^{2}+(-7) \times 7\)

|723040447|kitty|/Applications/kitty.app|0|
d \(7 a^{2}+63 a b=7 a(a+9 b)\)

\(=-7\left(a^{2}+7\right)\)

\(7\left(-a^{2}-7\right)\) is also correct.
|723040457|kitty|/Applications/kitty.app|0|
a \(5 p q^{2}+10 p^{2} q+25 p^{2} q^{2}\)
b \(16 a b+10 b^{2}-2 a^{2} b\)
|723040471|kitty|/Applications/kitty.app|0|
a \(5 p q^{2}+10 p^{2} q+25 p^{2} q^{2}=5 p q(q+2 p+5 p q)\)
|723040491|kitty|/Applications/kitty.app|0|
b \(16 a b+10 b^{2}-2 a^{2} b=2 b\left(8 a+5 b-a^{2}\right)\)
|723040501|kitty|/Applications/kitty.app|0|

\section*{Example 4}
Factorise:

\section*{Solution}


|723040510|kitty|/Applications/kitty.app|0|
Note: It is more efficient to take out all the common factors in one step, as shown in the following examples.
|723040511|kitty|/Applications/kitty.app|0|
    \end{}
|723040552|kitty|/Applications/kitty.app|0|
 3 Factorise:|723040695|kitty|/Applications/kitty.app|0|
        \question[6]
        \begin{parts}
            \part 
|723040780|kitty|/Applications/kitty.app|0|
a \(3 b-6 b^{2}\)
b \(4 x^{2}-6 x y\)
d \(18 y-9 y^{2}\)
e \(4 a-6 a b^{2}\)
g \(14 m n^{2}-21 m^{2} n\)
h \(6 p q^{2}-21 q p^{2}\)
|723040784|kitty|/Applications/kitty.app|0|
a \(-10 b^{2}+5 b\)
d \(-4 p q+16 p^{2}\)
b \(-16 a^{2} b-8 a b\)
g \(-8 a^{2} b^{2}-2 a b\)
e \(-5 x^{2} y+30 x\)
h \(12 x y^{2}-3 x^{2} y\)
|723040795|kitty|/Applications/kitty.app|0|
            \part \(6 y z^{2}-18 y z=-3 y z \times \fillin[]\)
|723041019|kitty|/Applications/kitty.app|0|
\fillin[]|723041053|kitty|/Applications/kitty.app|0|
=\fillin[]|723041057|kitty|/Applications/kitty.app|0|
                    \draw (0,0) -- (4,0) -- (4,2) -- (0,2) -- cycle;
                    \node at (2, 2) [above] {$2a + 3$};
                    \node at (2, 1) {Area = $8a + 12$};
|723041198|kitty|/Applications/kitty.app|0|
                    \draw (0,0) -- (4,0) -- (4,2) -- (0,2) -- cycle;
                    \node at (2, 2) [above] {$5$};
                    \node at (2, 1) {Area = $10b + 15$};
|723041217|kitty|/Applications/kitty.app|0|
                    \draw (0,0) -- (3,0) -- (3,2) -- (0,2) -- cycle;
                    \node at (1.5, 2) [above] {$2a + 3$};
                    \node at (1.5, 1) {Area = $8a + 12$};
|723041221|kitty|/Applications/kitty.app|0|
                \begin{solutionordottedlines}[1in]
|723041270|kitty|/Applications/kitty.app|0|
\section*{\(\triangle \mathrm{B}\) Factorisation using the difference of two squares}
You will recall from Section \(1 \mathrm{G}\) the important identity
|723041624|kitty|/Applications/kitty.app|0|
which is called the difference of two squares. We can now use this result the other way around to factorise an expression that is the difference of two squares.
|723041634|kitty|/Applications/kitty.app|0|
That is:|723041637|kitty|/Applications/kitty.app|0|
That is, the factors of \(a^{2}-b^{2}\) are \(a+b\) and \(a-b\).
|723041642|kitty|/Applications/kitty.app|0|
\section*{Solution}
a \(x^{2}-9=x^{2}-3^{2}\)

\[
=(x+3)(x-3)
\]
|723041777|kitty|/Applications/kitty.app|0|
Check the answer by expanding \((x+3)(x-3)\) to see that \(x^{2}-9\) is obtained.
|723041812|kitty|/Applications/kitty.app|0|
\(25-y^{2}=5^{2}-y^{2}\)

\[
=(5+y)(5-y)
\]
|723041833|kitty|/Applications/kitty.app|0|
\(4 x^{2}-9=(2 x)^{2}-3^{2}\)

\[
=(2 x+3)(2 x-3)
\]
|723041845|kitty|/Applications/kitty.app|0|
            \part 
            \begin{solutionordottedlines}[1in]
            \end{solutionordottedlines}
|723041870|kitty|/Applications/kitty.app|0|
a \(3 a^{2}-27\)
|723041876|kitty|/Applications/kitty.app|0|
b \(-16+9 x^{2}\)
|723041883|kitty|/Applications/kitty.app|0|
a \(3 a^{2}-27=3\left(a^{2}-9\right)\)
|723041889|kitty|/Applications/kitty.app|0|
\(-16+9 x^{2}=9 x^{2}-16\)
\(=3\left(a^{2}-3^{2}\right)\)
\(=(3 x)^{2}-4^{2}\)
\(=3(a+3)(a-3)\)
\(=(3 x-4)(3 x+4)\)
|723041905|kitty|/Applications/kitty.app|0|
\section*{Example 6}
Factorise:



|723041920|kitty|/Applications/kitty.app|0|
a \(x^{2}-16\)
e \((2 x)^{2}-25\)
i \(9 x^{2}-4\)
m \(1-4 a^{2}\)

a \(3 x^{2}-48\)
e \(10 x^{2}-1000\)
i \(3-12 b^{2}\)
m \(45 m^{2}-125 n^{2}\)

a \(-25+x^{2}\)
d \(-81 x^{2}+16\)
g \(-12 x^{2}+27\)
|723041940|kitty|/Applications/kitty.app|0|
4 Use the factorisation of the difference of two squares to evaluate the following. One has been done for you.

\[
\begin{aligned}
17^{2}-3^{2} & =(17+3)(17-3) \\
& =20 \times 14 \\
& =280
\end{aligned}
\]
a \(23^{2}-7^{2}\)
d \(94^{2}-6^{2}\)
g \(11.3^{2}-8.7^{2}\)


i \(4^{2}-3^{2}\)
v \(8^{2}-7^{2}\)
|723042006|kitty|/Applications/kitty.app|0|
\[
\begin{aligned}
17^{2}-3^{2} & =(17+3)(17-3) \\
& =20 \times 14 \\
& =280
\end{aligned}
\]
|723042039|kitty|/Applications/kitty.app|0|

\[
a^{2}-b^{2}=(a+b)(a-b)
\]

|723042091|kitty|/Applications/kitty.app|0|
Factorisation using the difference of two squares identity
|723042092|kitty|/Applications/kitty.app|0|
            \part \(-12 x^{2}+27\)
            \begin{solutionordottedlines}[1in]
            \end{solutionordottedlines}
|723042234|kitty|/Applications/kitty.app|0|
c \(a^{2}-121\)

g \((4 x)^{2}-1\)

k \(100 a^{2}-49 b^{2}\)

o \(25 a^{2}-100 b^{2}\) d \(d^{2}-400\)
|723042343|kitty|/Applications/kitty.app|0|
b \(4 x^{2}-100\)
c \(5 x^{2}-45\)
f \(7 x^{2}-63\)
g \(8 x^{2}-50\)
j \(20-5 y^{2}\)
k \(27 a^{2}-12 b^{2}\)
n \(27 a^{2}-192 l^{2}\)
o \(-8 x^{2}+32 y^{2}\)

b \(-4+9 x^{2}\)
e \(-100 x^{2}+9\)
h \(-36 x^{2}+400\)

b \(23^{2}-3^{2}\)
e \(1.8^{2}-0.2^{2}\)
h \(92.6^{2}-7.4^{2}\)


ii \(5^{2}-4^{2}\)
vi \(9^{2}-8^{2}\)
iii \(6^{2}-5^{2}\)
vii \(10^{2}-9^{2}\)
|723042363|kitty|/Applications/kitty.app|0|
            \part  \(5^{2}-4^{2}\)
            \part  \(9^{2}-8^{2}\)
            \part \(6^{2}-5^{2}\)
            \part \(10^{2}-9^{2}\)
|723042382|kitty|/Applications/kitty.app|0|
    \question[1]
    \begin{boxdef}
        Quadratic: \fillin[a polynomial whose highest power is 2]
    \end{boxdef}
|723042998|kitty|/Applications/kitty.app|0|
Quadratic|723043001|kitty|/Applications/kitty.app|0|
\begin{boxdef}
    \
\end{boxdef}
|723043014|kitty|/Applications/kitty.app|0|
fillin|723043328|kitty|/Applications/kitty.app|0|
answerline|723043371|kitty|/Applications/kitty.app|0|
\begin{solutionordottedlines}[1cm]the terminology for a specific topic\end{solutionordottedlines}
|723043553|kitty|/Applications/kitty.app|0|
a polynomial whose highest power is 2|723043562|kitty|/Applications/kitty.app|0|
\begin{solutionordottedlines}[1cm]a polynomial whose highest power is 2\end{solutionordottedlines}
|723043578|kitty|/Applications/kitty.app|0|
the number before the variable|723043588|kitty|/Applications/kitty.app|0|
\fillin[]
|723043593|kitty|/Applications/kitty.app|0|
\(x^{2}-3 x-18=(x-6)(x+3)\)

}|723043790|kitty|/Applications/kitty.app|0|
\section*{Example 7}
Factorise \(x^{2}-3 x-18\).

\section*{Solution}
We are looking for two numbers that multiply to give -18 and add to give -3 . The numbers -6 and 3 satisfy both conditions.

|723043810|kitty|/Applications/kitty.app|0|

Consider factorising \(x^{2}-36\).

Since the constant term is -36 and the coefficient of \(x\) is 0 , we are looking for two numbers that multiply to give -36 and add to give 0 . The numbers -6 and 6 satisfy both conditions.

Thus \(x^{2}-36=(x-6)(x+6)\).

Note: It is not really sensible to do the example in this way - it is best to recognise \(x^{2}-36\) as a difference of squares. However, it does show that the method of factorisation in this section is consistent with the earlier technique.
|723043816|kitty|/Applications/kitty.app|0|
            \part ection*{Solution}
|723043844|kitty|/Applications/kitty.app|0|
a \(x^{2}+5 x+6\)
c \(x^{2}+7 x+10\)
e \(x^{2}+9 x+14\)
|723043891|kitty|/Applications/kitty.app|0|
            \part \(x^{2}-10 x+25\)
            \begin{solutionordottedlines}[1in]
            \end{solutionordottedlines}
            \part \(x^{2}+10 x+25\)
            \begin{solutionordottedlines}[1in]
            \end{solutionordottedlines}
            \part \(x^{2}+30 x+225\)
            \begin{solutionordottedlines}[1in]
            \end{solutionordottedlines}
            \part \(x^{2}-20 x+100\)
            \begin{solutionordottedlines}[1in]
            \end{solutionordottedlines}
|723044152|kitty|/Applications/kitty.app|0|
\newmdenv|723044514|kitty|/Applications/kitty.app|0|
\newmdenv[nobreak=true]{myframe}|723044578|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\newmdenv[nobreak=true]{myframe}
|723044602|kitty|/Applications/kitty.app|0|
myframe|723044615|kitty|/Applications/kitty.app|0|
nobreak=true|723044658|kitty|/Applications/kitty.app|0|
\newmdenv[]{examplebox}
|723044754|kitty|/Applications/kitty.app|0|
  skipabove=\baselineskip,
  skipbelow=\baselineskip,|723044862|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
        \vspace*{-1cm}
|723079429|kitty|/Applications/kitty.app|0|
false|723079465|kitty|/Applications/kitty.app|0|
,nobreak=true|723079484|kitty|/Applications/kitty.app|0|
            \part \(16 a b+10 b^{2}-2 a^{2} b\)
            \begin{solutionordottedlines}[1in]
                \(16 a b+10 b^{2}-2 a^{2} b=2 b\left(8 a+5 b-a^{2}\right)\)
            \end{solutionordottedlines}
|723079587|kitty|/Applications/kitty.app|0|
            \begin{solutionordottedlines}[1in]
                \(16 a b+10 b^{2}-2 a^{2} b=2 b\left(8 a+5 b-a^{2}\right)\)
            \end{solutionordottedlines}
|723079678|kitty|/Applications/kitty.app|0|
            \part \(2 x^{2}+4 x\)
            \begin{solutionordottedlines}[1in]
                \(2 x^{2}+4 x=2 x(x+2)\)
            \end{solutionordottedlines}
|723079750|kitty|/Applications/kitty.app|0|
            \begin{solutionordottedlines}[1in]
                \(2 x^{2}+4 x=2 x(x+2)\)
            \end{solutionordottedlines}
|723079758|kitty|/Applications/kitty.app|0|
        \end{multicols}\end{parts}
|723080247|kitty|/Applications/kitty.app|0|
            \
|723080267|kitty|/Applications/kitty.app|0|
        \newpage
|723081014|kitty|/Applications/kitty.app|0|
        \end{multicols}\end{parts}
        \begin{parts}\begin{multicols}{2}
            \setcounter{partno}{8}
|723081104|kitty|/Applications/kitty.app|0|
        \end{multicols}\end{parts}
        \begin{parts}\begin{multicols}{2}
            \setcounter{partno}{12}
|723081225|kitty|/Applications/kitty.app|0|
\section*{Factorising using the perfect square identities}
|723081420|kitty|/Applications/kitty.app|0|
itemize|723081506|kitty|/Applications/kitty.app|0|
\begin{center}
|723081513|kitty|/Applications/kitty.app|0|
\end{itemize}
|723081527|kitty|/Applications/kitty.app|0|
    \[
  a^{2}+2 a b+b^{2}=(a+b)^{2}

  a^{2}-2 a b+b^{2}=(a-b)^{2}
    \]
|723081558|kitty|/Applications/kitty.app|0|
math|723081592|kitty|/Applications/kitty.app|0|
Factorise:
|723081698|kitty|/Applications/kitty.app|0|
a \(x^{2}+8 x+16\)
b \(x^{2}-10 x+25\)
c \(x^{2}+11 x+\frac{121}{4}\)
|723081699|kitty|/Applications/kitty.app|0|
a \(x^{2}+8 x+16=x^{2}+2 \times 4 x+4^{2}\)

\[
=(x+4)^{2}
\]
|723081732|kitty|/Applications/kitty.app|0|
b \(x^{2}-10 x+25=x^{2}-2 \times 5 x+5^{2}\)

\[
=(x-5)^{2}
\]
|723081746|kitty|/Applications/kitty.app|0|
c \(x^{2}+11 x+\frac{121}{4}=x^{2}+2 \times \frac{11}{2} x+\left(\frac{11}{2}\right)^{2}\)

\[
=\left(x+\frac{11}{2}\right)^{2}
\]
|723081756|kitty|/Applications/kitty.app|0|
\section*{Example 9}
|723081770|kitty|/Applications/kitty.app|0|
3 Identify the simple quadratic expression that cannot be factorised as a perfect square.
|723081782|kitty|/Applications/kitty.app|0|
a \(\mathbf{i} x^{2}+4 x+4\)
ii \(x^{2}-6 x+12\)
iii \(x^{2}-12 x+36\)
iv \(x^{2}-10 x+25\)
b i \(x^{2}+6 x+9\)
ii \(x^{2}+5 x+\frac{25}{4}\)
iii \(x^{2}-8 x-16\)
iv \(x^{2}-14 x+49\)
|723081808|kitty|/Applications/kitty.app|0|
4 A brick company provides rectangular and square pavers.

a Draw a diagram to show how two different square pavers of side lengths \(a\) and \(b\) respectively, and two identical rectangular pavers with dimensions \(a \times b\), can be arranged into a square.

b How many of each type of paver enables you to pave a square area of side length \(a+3 b\) ? Draw a diagram to illustrate how this can be done.
|723081907|kitty|/Applications/kitty.app|0|

        \
|723081929|kitty|/Applications/kitty.app|0|
            \part \(8^{2}-7^{2}\)
            \begin{solutionordottedlines}[1in]
            \end{solutionordottedlines}
|723081963|kitty|/Applications/kitty.app|0|
            \part \(x^{2}+8 x+16=(x+\)
|723082085|kitty|/Applications/kitty.app|0|
            \begin{solutionordottedlines}[0.5in]
            \end{solutionordottedlines}
|723082270|kitty|/Applications/kitty.app|0|
a \(3 x^{2}+9 x+6=3\left(x^{2}+3 x+2\right) \quad\) (Take out the common factor.)

\[
=3(3 x+2)(3 x+1)
\]
|723082613|kitty|/Applications/kitty.app|0|
b \(6 x^{2}-54=6\left(x^{2}-9\right)\)

\[
=6(x+3)(x-3)
\]
|723082621|kitty|/Applications/kitty.app|0|
c \(-x^{2}-x+2=-\left(x^{2}+x-2\right)\)

(Factor -1 from each term.)

\[
=-(x+2)(x-1)
\]
|723082631|kitty|/Applications/kitty.app|0|
a \(2 x^{2}+14 x+24\)
d \(4 x^{2}-24 x+36\)
g \(4 x^{2}-4 x+48\)
j \(3 x^{2}+9 x-120\)
m \(2 x^{2}-4 x-96\)
p \(3 x^{2}-18 x+27\)
a \(4 x^{2}-16\)
d \(3 a^{2}-27\)
g \(27 x^{2}-3 y^{2}\)
j \(128-2 x^{2}\)
m \(\frac{1}{4} a^{2}-9\)
a \(-x^{2}-8 x-12\)
d \(9+8 x-x^{2}\)
g \(-x^{2}+3 x+40\)
j \(11 x-x^{2}-24\)
m \(-16 x-63-x^{2}\)
|723082657|kitty|/Applications/kitty.app|0|
            \begin{solutionordottedlines}[2in]
            \end{solutionordottedlines}
|723082680|kitty|/Applications/kitty.app|0|
\newmdenv[backgroundcolor=exampleboxcolor]{examplebox}
|723082845|kitty|/Applications/kitty.app|0|
%\newmdenv[linecolor=exerciseboxcolor,linewidth=10pt]{exercisebox}
|723082866|kitty|/Applications/kitty.app|0|
% now sparing PEN eds printers
|723082871|kitty|/Applications/kitty.app|0|
\newmdenv[backgroundcolor=exerciseboxcolor]{exercisebox}
|723082873|kitty|/Applications/kitty.app|0|
b \(3 x^{2}+24 x+36\)
e \(7 x^{2}+14 x+7\)
h \(2 x^{2}-18 x+36\)
k \(3 x^{2}-3 x-90\)
n \(5 x^{2}+65 x+180\)
q \(5 x^{2}-20 x+20\)
b \(2 x^{2}-18\)
e \(6 x^{2}-600\)
h \(45-5 b^{2}\)
k \(\frac{1}{2} a^{2}-2 b^{2}\)
n \(\frac{1}{5} x^{2}-20\)
b \(12-11 x-x^{2}\)
e \(-x^{2}-4 x-4\)
h \(42+x-x^{2}\)
k \(-3 x^{2}-30 x+72\)
n \(-x^{2}-35+12 x\)
|723082941|kitty|/Applications/kitty.app|0|
Exercise 4A|723083176|kitty|/Applications/kitty.app|0|
\section*{Exercise 4B}
|723083192|kitty|/Applications/kitty.app|0|
\subsection{Difference of Two Squares}
|723083202|kitty|/Applications/kitty.app|0|
\section*{Exercise 4C}
|723083208|kitty|/Applications/kitty.app|0|
Difference of Two Squares|723083209|kitty|/Applications/kitty.app|0|
\section*{Exercise 4D}
|723083222|kitty|/Applications/kitty.app|0|
\section*{Exercise 4E}
|723083235|kitty|/Applications/kitty.app|0|
b \(4 m^{2} n-4 m n+16 n^{2}\)
d \(2 m^{2}+4 m n+6 n\)
f \(6 a+8 a b+10 a b^{2}\)
h \(5 \ell^{2}-15 \ell m-20 m^{2}\)
|723083472|kitty|/Applications/kitty.app|0|
                \begin{tikzpicture} 
                    \draw (0,0) -- (4,0) -- (4,2) -- (0,2) -- cycle;
                    \node at (2, 2) [above] {$2a + 3$};
                    \node at (2, 1) {Area = $8a + 12$};
                \end{tikzpicture}
                \begin{solutionordottedlines}[1in]
                    $4$
                \end{solutionordottedlines}
|723083637|kitty|/Applications/kitty.app|0|
        \begin{parts}\begin{multicols}{3}
            \part 
                \begin{tikzpicture} 
                    \draw (0,0) -- (4,0) -- (4,2) -- (0,2) -- cycle;
                    \node at (2, 2) [above] {$2a + 3$};
                    \node at (2, 1) {Area = $8a + 12$};
                \end{tikzpicture}
                \begin{solutionordottedlines}[1in]
                    $4$
                \end{solutionordottedlines}
            \part 
                \begin{tikzpicture} 
                    \draw (0,0) -- (4,0) -- (4,2) -- (0,2) -- cycle;
                    \node at (2, 2) [above] {$5$};
                    \node at (2, 1) {Area = $10b + 15$};
                \end{tikzpicture}
                \begin{solutionordottedlines}[1in]
                    $2b+3$
                \end{solutionordottedlines}
            \part
                \begin{tikzpicture} 
                    \draw (0,0) -- (4,0) -- (4,2) -- (0,2) -- cycle;
                    \node at (2, 2) [above] {$6a$};
                    \node at (2, 1) {Area = $12a^2 + 6ab$};
                \end{tikzpicture}
                \begin{solutionordottedlines}[1in]
                    $2a + b$
                \end{solutionordottedlines}
        \end{multicols}\end{parts}
|723083656|kitty|/Applications/kitty.app|0|
        \begin{parts}\begin{multicols}{3}
            \part 
                \begin{tikzpicture} 
                    \draw (0,0) -- (3,0) -- (3,2) -- (0,2) -- cycle;
                    \node at (1.5, 2) [above] {$3a$};
                    \node at (1.5, 1) {Area = $9a+6ab$};
                \end{tikzpicture}
                \begin{solutionordottedlines}[1in]
                    $3+2b$
                \end{solutionordottedlines}
            \part \begin{tikzpicture} \end{tikzpicture}
                \begin{solutionordottedlines}[1in]
                    $2+a$
                \end{solutionordottedlines}
            \part \begin{tikzpicture} \end{tikzpicture}
                \begin{solutionordottedlines}[1in]
                    $8p^2$
                \end{solutionordottedlines}
        \end{multicols}\end{parts}
|723083668|kitty|/Applications/kitty.app|0|
2a + 3|723083679|kitty|/Applications/kitty.app|0|
*/
|723083888|kitty|/Applications/kitty.app|0|
d \(d^{2}-400\)
h \((5 m)^{2}-9\)
l \(64 m^{2}-81 p^{2}\)
p \(-9+x^{2}\)
d \(6 x^{2}-24\)
h \(12 m^{2}-75\)
l \(16 x^{2}-100 y^{2}\)
p \(-200 p^{2}+32 q^{2}\)
c \(-9 x^{2}+4\)
f \(-18+50 x^{2}\)
i \(-175+28 x^{2}\)
c \(36^{2}-6^{2}\)
f \(28^{2}-2.2^{2}\)
i \(3.214^{2}-2.214^{2}\)
iv \(7^{2}-6^{2}\)
viii \(101^{2}-100^{2}\)

i|723083961|kitty|/Applications/kitty.app|0|
1 Factorise these quadratric expressions.
|723083973|kitty|/Applications/kitty.app|0|
l \(x^{2}-11 x+24\)
|723083991|kitty|/Applications/kitty.app|0|
j \(x^{2}-11 x+28\)
|723083994|kitty|/Applications/kitty.app|0|
        \part \(x^{2}-9 x-90\)
        \part \(x^{2}+15 x-100\)
        \part \(x^{2}-7 x-60\)
        \part \(x^{2}-10 x-24\)
        \part \(x^{2}+2 x-15\)
        \part \(x^{2}+5 x-24\)
        \part \(x^{2}+8 x+12\)
        \part \(x^{2}+11 x+30\)
        \part \(x^{2}-9 x-90\)
        \part \(x^{2}-7 x-18\)
        \part \(x^{2}+x-90\)
        \part \(x^{2}+14 x+49\)
        \part \(x^{2}-18 x+81\)
        \part \(x^{2}+12 x+36\)
        \part \(x^{2}-16 x+64\)
        \part \(x^{2}-8 x+16\)
|723084174|kitty|/Applications/kitty.app|0|
    \question[16] Factorise:
|723084217|kitty|/Applications/kitty.app|0|
        \begin{solutionordottedlines}[1cm]
        \end{solutionordottedlines}
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\begin{center}
\gradetable[h][pages]
\end{center}|723084941|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
pages|723085167|kitty|/Applications/kitty.app|0|
\begin{center}
\gradetable[h][questions]
\end{center}
|723085214|kitty|/Applications/kitty.app|0|

So much happens on our dates that I want to include a portion of prose before the gallery so that at least I can organise my memories.

# Date 1: Mirac, Eastwood (Korean side!)

No real sparks. "If you want it all then why you half way in?". Aj feels like Kiyo just left him. Worried there would not be a second date. 

Drank a bit, she didn't let me hold her hand, handled a dead body in my closet well. Felt like this is a girl that could handle the other skeletons too.

---

# Date 2: Burgers on Broadway, Summer Hill

Aj comes crying to Kiyo, she comforts him on not becoming Prince of UNSW Ultimate. Also apologises for a non-classy first date and wants a second. Aj internally glad. AK roll on the grass of Ashfield park for a bit - become best friends. Sit on the bench for a bit, feeling like puzzle pieces.

Timelapse.

Aj breaks up with Kiyo. Aj not good enough for Kiyo. 

One day later.

Aj&Kiyo too greedy, they find a way to communicate.

A whispers to K: come training?
Kiyo comes to UNSW Ultimate Frisbee training. Aj shaves; the mark of reformation. Aj&Kiyo go on date. The eat burgers in Summer Hill, Norman's name still stuck in both their throats.

More chemistry this time. No drinking, just messy burgers.

---

# Date 3: Cricket Pitch, Ashfield

A is getting close with K and now knows her schedule. He waits outside and elopes w K on a picnic date when she was ready.

{{< image-gallery gallery_dir="album/23-11-11-cricket-pitch" >}}

We forgot matches for the candle and got bitten by bugs. We drank and danced and cuddled and kissed. We forgot to tell the rest of the world where we went; we were on the cricket pitch.

---

# Date 4: Sinchon, West Ryde

OMg, perhaps my best meal. Yes I think let's call it that. We were trying to do our Malatang date which was overdue, but Aj had stolen a piece of art from uni and had some books to donate at Vinnies so we stopped at West Ryde first. 

AK were also fighting, so they went for a walk (something about a different boy, I forget his name). We didn't kiss, but made up and ended up looking for food in West Ryde. We found Sinchon, a Korean Fusion restaurant and ordered Tofu soup, Pork on a hotplate and A KIMCHI PANCAKE (or was it seafood??). Yeah I think it was seafood, we got a Kimchi pancake on the first date.

Kiyo sat on the right, Aj on the left. We ate and fed each other too. We felt like best friends. We scribbled on parchment for a bit. The food was awesome.

More chemistry.

---

# Date 5: Master Hot Pot, Eastwood (Chinese side!)

Oh boy, we got busted holding hands. It was only for 15 SECONDs! Ugh, anyways it was a cute date. OMG I think she kissed me after this one!!! muahaha. We ate on the Chinese side this time! Things are coming full circle I suppose. Kiyo eats so much, I love it. We walked to the park afterwards and started the evil exes list :D. We listed them all and put prices on all of them... except norman >:(. 
{{< image-gallery gallery_dir="album/23-11-19-malatang" >}}

---

# Date 6: Coffee Date

We finish scott pilgrim! Kiyo sees the house. She's not allowed upstairs. We drink coffee. Go eat sushi for breafast!! aj pays, heh. Kiyo buys a puzzle; to her own demise. We play with the puzzle and with the pets. Aj&Kiyo snooze on the bus. Aj get snapchat 👻️.

---

# Date 7: Unit 7

Date 7 happens at Unit 7, stupid fateful universe. Oh this is a big adventure. It starts at 2:39am on Monday with "Swollen Eyes" and ends ...(kiyo ) 

Monday
- 3:46am Vomit
- 6am aj sad
- 9am aj exam
- 1pm aj calls k. he wants to help.
- 3pm k kills norman.
- 6pm k links with aj
- 7pm unicorns frisbee. a little chardonnay. elbows to the eye, kiyo cares for aj.
- 8pm aj abducts kiyo from second frisbee game
- 9pm thai: kiyo pays
- 10pm aj does a scott pilgrim, brings knives to old house.
- 11pm more chardonnay, trespassing, and dinner by the pool.
- 12am secrets spilled

Tuesday
- 1am we get back to herm
- 2am photos & kisses & naps
- 5am stranger danger. angry tenants with important things to do in the early morning.
- 6am 38 Spurway Street
- 7am coffee + breakfast v1
- 8am puzzle, friends, naps
- 10am smacking sticks, giggles on a sofa
- 12am breakfast v2, more friends
- 2pm K leaves, gives A a bye-kiss 🥰️
- 3pm Aj sleeps for 16hours
- 4pm K does chores like a crazy person
- 6pm K sleeps (aj hopes)

{{< image-gallery gallery_dir="album/23-11-28-unit7" >}}
|723249705|kitty|/Applications/kitty.app|0|
https://www.youtube.com/watch?v=-LnEkcNeuF8|723250368|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
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layout: image-gallery
|723251088|kitty|/Applications/kitty.app|0|
| Ex # | Name   | Value | Challenge          |
|------|--------|-------|---------------------|
| 1.   | Pie    | $80   | Aerial             |
| 2.   | Kaito  | $38   | Div 1              |
| 3.   | Tobias | $9    | New Car            |
| 4.   | Eric   | $50   | Half Marathon      |
| 5.   | Johann | $60   | 1x Degree          |
| 6.   | Norman | $103  | Earn $3 by busking |
| 7.   | AJ     | $340  |                     |
|723251277|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
|---------------------------------|
| Ex # | Name | Value | Challenge |
|---------------------------------|
| 1.   | Pie | $80 | Aerial |
| 2.   | Kaito | $38 | Div 1 |
| 3.   | Tobias | $9 | New Car |
| 4.   | Eric | $50 | Half Marathon |
| 5.   | Johann | $60 | 1x Degree |
| 6.   | Norman | $103 | Earn $3 by busking |
| 7.   | AJ | $340 |  |
|---------------------------------|
|723251281|kitty|/Applications/kitty.app|0|
Image: 1440x900 (19.8 MB)|723252881|Brave Browser Beta|/Applications/Brave Browser Beta.app|1|56cfb9c2adbdd5945892005770f4b6f0817380c4.tiff
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:|723511581|kitty|/Applications/kitty.app|0|
supdog
|723530094|kitty|/Applications/kitty.app|0|
test2

|723530122|kitty|/Applications/kitty.app|0|
howdy
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example1.mp4
|723530569|kitty|/Applications/kitty.app|0|
 output.dat

|723530574|kitty|/Applications/kitty.app|0|
Screen Recording 2023-12-06 at 3.41.49 pm
Screen Recording 2023-12-06 at 3.40.06 pm|723530873|Finder|/System/Library/CoreServices/Finder.app|0|
https://drive.google.com/drive/folders/1t-mWkF9cAh7UTljFhS9Zw5g3cTPNy5Rm?usp=sharing|723531375|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
dev/disk8 (external, physical):
   #:                       TYPE NAME                    SIZE       IDENTIFIER
   0:     FDisk_partition_scheme                        *63.9 GB    disk8
   1:               Windows_NTFS ⁨⁩                        63.8 GB    disk8s1

|723619201|kitty|/Applications/kitty.app|0|
#include <stdio.h>

int add(int a, int b) {
    return a + b;
}

int subtract(int a, int b) {
    return a - b;
}

int do_op(/* what goes here */, int a, int b) {
    return func(a, b);
} 

int main() {
    do_op(add, 7, 5);
    do_op(subtract, 7, 5);
    return 0;
}
|724041512|kitty|/Applications/kitty.app|0|
XLv7eTnD5fk#jmCHG%|724053693|kitty|/Applications/kitty.app|0|
#include <stdio.h>

int add(int a, int b) {
    return a + b;
}

int subtract(int a, int b) {
    return a - b;
}

int do_op(int (*func)(int, int), int a, int b) {
    return func(a, b);
} 

int main() {
    printf("%d\n", do_op(add, 7, 5));
    printf("%d\n", do_op(subtract, 7, 5));
    return 0;
}
|724051423|kitty|/Applications/kitty.app|0|
    int 
|724051461|kitty|/Applications/kitty.app|0|
int increment_by_7(int a){
    return a + 7;
}
|724051965|kitty|/Applications/kitty.app|0|
    increment_by_7(a);
|724051972|kitty|/Applications/kitty.app|0|
    printf("deref pa: %d\n", *pa);

/
/
/
/|724052256|kitty|/Applications/kitty.app|0|
pa: 0x30e4b236c
|724052892|kitty|/Applications/kitty.app|0|
pa: 0x30e4b236c
va: 0x30e4b236c

/
/
/
/|724052946|kitty|/Applications/kitty.app|0|
    printf("deref pa: %d\n", *pa);
    printf("deref va: %d\n", *va);
|724052970|kitty|/Applications/kitty.app|0|
int|724053009|kitty|/Applications/kitty.app|0|
CPU: Apple M1
|724129105|kitty|/Applications/kitty.app|0|
laptop|724129334|kitty|/Applications/kitty.app|0|
 gas|724129561|kitty|/Applications/kitty.app|0|
incomes:
kiyo:
    - keep alive
        - rent, groceries, bills (++electricity, water)
aj:
    - feel alive
        - trip to japan



|724129565|kitty|/Applications/kitty.app|0|
security find-generic-password -wa Man2.4G
|724130054|kitty|/Applications/kitty.app|0|
Manish123c!!|724130068|kitty|/Applications/kitty.app|0|
where is the function f(z)=\frac{z^2+4iz+1}{2-cos(z)} holomorphic?|724650337|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
how would I solve \int(\frac{e^{i\theta}{7+4\cos{\theta}}}) using complex methods|724653954|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
# Tuyo

Rodrigo Amarante


1. Soy el fuego que arde tu piel
2. Soy el agua que mata tu sed
3. El castillo, la torre yo soy
4. La espada que guarda el caudal
5. Tú, el aire que respiro yo
6. Y la luz de la luna en el mar
7. La garganta que ansío mojar
8. Que temo ahogar de amor
9. Y cuáles deseos me vas a dar?
10. Oh, oh, oh
11. Dices tu, mi tesoro basta con mirarlo
12. Y tuyo será, y tuyo será


1. I am the fire that burns your skin
2. I am the water that kills your thirst
3. Of the castle, I am the tower
4. The sword that guards the treasure
5. You, the air that I breathe
6. And the light of the moon on the sea
7. The throat that I long to wet
8. But I'm afraid I'll drown in love
9. And which desires will you give me?
10. Oh, oh, oh
11. You say: my treasure enough is enough to look at
12. and yours it will be, yours it will be

la = feminine
el = masculine
soy = am
tu = you
arde = burns
piel = skin
fuego = fire
sed = thirst
mata = kill
yo = I
garganta = throat
ansio = longing
mojar = wet
ahogar = drown
temo = fear
de = of
Y = and
cuales = which
deseos = wishes
me = I
vas a dar = will you give
vas = go
dices tu = you say
basta = enough
con = with
mirarlo = looking
sera = be
mi = my
tesoro = treasure

|724661369|kitty|/Applications/kitty.app|0|
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DSC00151.JPG|724662190|Finder|/System/Library/CoreServices/Finder.app|0|
take it out on us." - not me.|724662869|kitty|/Applications/kitty.app|0|
 , aj gave them 90 coins for the night. |724663325|kitty|/Applications/kitty.app|0|
drink break,|724663500|kitty|/Applications/kitty.app|0|
{{< image-gallery gallery_dir="album/23-11-28-unit7" >}}
|724663551|kitty|/Applications/kitty.app|0|
- [.] [COMP6771](COMP6771/index.md)
|724729535|kitty|/Applications/kitty.app|0|
- [ ] 
|724729583|kitty|/Applications/kitty.app|0|
    - [ ] [COMP3411](COMP3411/index.md)
|724729593|kitty|/Applications/kitty.app|0|
- [.] [COMP4121](COMP4121/index.md)
|724729606|kitty|/Applications/kitty.app|0|
: 1683857830:0;bw get template item | jq ".name=\"macquarie oneid\" | .login=$(bw get template item.login | jq '.username="mqg747572" | .password="*GCyRMaH5n6iLjeb65"')" | bw encode | bw create item
|724730030|kitty|/Applications/kitty.app|0|
\section{Introduction}
    \input{1-intro}
\section{Constructing Expressions}
    \input{2-expressions}
\section{Solving simple linear equations}
    \input{3-linear-simple}
\section{Equations with brackets}
    \input{4-brackets}
\section{Linear Equations involving fractions}
    \input{5-fractions}
\section{Outro}
    \input{6-outro}
\newpage
\section{Homework}
    \input{0-homework}
|724757716|kitty|/Applications/kitty.app|0|
    \input{2-expressions}
|724757723|kitty|/Applications/kitty.app|0|
    \input{3-linear-simple}
|724757724|kitty|/Applications/kitty.app|0|
    \input{4-brackets}
|724757725|kitty|/Applications/kitty.app|0|
    \input{5-fractions}
|724757725|kitty|/Applications/kitty.app|0|
    \input{6-outro}
|724757726|kitty|/Applications/kitty.app|0|
\section{Outro}
|724757729|kitty|/Applications/kitty.app|0|
\section{Homework}
|724757729|kitty|/Applications/kitty.app|0|
    \input{0-homework}
|724757730|kitty|/Applications/kitty.app|0|
\section{Introduction}
|724757736|kitty|/Applications/kitty.app|0|
\section*{Linear equations and inequalities}
We have seen that many real-world problems can be converted to equations. We have also seen how to solve simple equations. In this chapter, we consider a wider variety of equations and use them to solve practical problems.

Equations arise naturally when solving problems. In fact, a lot of problem-solving relies on us being able to translate a given word or real world problem into an equation, or equations, solve the equation(s) and relate the solution to the original problem. Turning a complicated problem into an equation enables us to understand and solve difficult problems. Before we reach this stage, we need to have at hand a collection of techniques for solving equations.

In earlier times, people used a number of ad hoc methods for solving equations. Only since the development of modern algebra have standard procedures and notations been introduced that enable us to solve equations quickly and efficiently.

Two examples of linear expressions are \(2 x+3\) and \(x+7\). The expression \(2 x+3\) is called linear because the graph of \(\mathrm{y}=2 x+3\) is a straight line, as we saw in Year 8 . We begin with some revision examples and exercises.

\section*{Example 1}
Brian is \(6 \mathrm{~cm}\) taller than Geoff. Represent this information algebraically.

\section*{Solution}
Since we do not know how tall Geoff is, we call his height \(x \mathrm{~cm}\).

Then Brian is \((x+6) \mathrm{cm}\) tall.

\section*{Example 2}
The length of a rectangle is \(3 \mathrm{~m}\) more than its width. The width of the rectangle is \(w \mathrm{~m}\).

a Express the length of the rectangle in terms of \(w\).

b Express the perimeter of the rectangle in terms of \(w\).
|724758066|kitty|/Applications/kitty.app|0|
\section*{Solution}
a Length of rectangle \(=(w+3) \mathrm{m}\)

b Perimeter \(=w+w+(w+3)+(w+3)\)

\[
=(4 w+6) \mathrm{m}
\]

\section*{Exercise 5A}
1 Fiona is \(5 \mathrm{~cm}\) taller than Tristan. Tristan's height is \(h \mathrm{~cm}\). What is Fiona's height?

2 Seuret weighs \(2 \mathrm{~kg}\) less than his older sister Vivian. If Vivian weighs \(w \mathrm{~kg}\), what is Seuret's weight?
|724758073|kitty|/Applications/kitty.app|0|
3 When Andriana's age is doubled, the number is 3 more than Helen's age. If Andriana's age is \(x\) years, what is Helen's age?

4 The length of a rectangle is 5 metres more than its width.

a If \(w\) metres is the width of the rectangle, express the length of the rectangle in terms of \(w\). b If \(\ell\) metres is the length of the rectangle, express the width of the rectangle in terms of \(\ell\).

5 In a competition, Deeksha scored 18 points more than Greta and Deirdre scored 5 points less than twice the number of points Greta scored. If \(a\) is the number of points Greta scored:

a express the number of points Deeksha scored in terms of \(a\)

b express the number of points Deirdre scored in terms of \(a\)
|724758266|kitty|/Applications/kitty.app|0|
6 The length of a rectangular paddock is \(20 \mathrm{~m}\) less than three times its width. If the width of the paddock is \(x \mathrm{~m}\), express the length of the paddock in terms of \(x\).

7 In a triathlon race, Luca ran at an average speed 5 times his average swimming speed. Also, when his average running speed was multiplied by 4 , this number was 3 less than his average speed for the cycling leg.

If \(x \mathrm{~km} / \mathrm{h}\) is Luca's average swimming speed, find expressions in terms of \(x\) for his:

a average running speed

b average cycling speed
|724758278|kitty|/Applications/kitty.app|0|
8 Match each of the following mathematical expressions with its corresponding English expression.
a \(4+2 x \quad\) i \(\quad\) Six less than four times \(x\)
b \(x-5\) ii Three times one more than \(x\)
c \(2 x-4\) iii Two less than one-quarter of \(x\)
d \(3(x+1)\) iv One-quarter of two less than \(x\)

\includegraphics[max width=\textwidth, center]{2023_12_09_f9189d42272205e0a381g-03}
f \(\frac{x}{4}-2 \quad\) vi \(\quad\) Six more than half of \(x\)
g \(\frac{x-2}{4} \quad\) vii One more than three times \(x\)
h \(x+6\) viii Five less than \(x\)
i \(3 x+1 \quad\) ix \(\quad\) Six more than \(x\)
\(\mathbf{j} \frac{x}{2}+6 \quad \mathbf{x} \quad\) Four more than twice \(x\)

|724758351|kitty|/Applications/kitty.app|0|
\includegraphics[max width=\textwidth, center]{2023_12_09_f9189d42272205e0a381g-03}
|724758373|kitty|/Applications/kitty.app|0|
9 Gemma is \(6 \mathrm{~cm}\) shorter than Gavin and \(4 \mathrm{~cm}\) taller than Brent. If \(x \mathrm{~cm}\) represents Gemma's height, express:

a Gavin's height in terms of \(x\)

b Brent's height in terms of \(x\)

|724758424|kitty|/Applications/kitty.app|0|
\section*{Solving simple linear equations}
A statement such as \(x+7=11\) is called an equation and we may solve the equation to find a value for \(x\) that makes the statement true. In this case, the solution is \(x=4\).

\section*{Reading equations}
The equation \(2 x+4=10\) can be read as 'two times a number plus 4 is equal to 10 '.

The instruction 'Solve the equation \(2 x+4=10\) ' can also be read as 'A number is multiplied by 2 and 4 is then added. The result is 10 . Find the number.'

In this case, the number is 3 . The solution is \(x=3\).

\section*{Equivalent equations}
Consider these equations:

|724758776|kitty|/Applications/kitty.app|0|
\[
\begin{aligned}
& 2 x+7=11 \\
& 2 x+9=13
\end{aligned}
\]

Equation (2) is obtained from equation (1) by adding 2 to each side of the equation.

So equation (1) is obtained from equation (2) by subtracting 2 from each side of the equation.

Equations (1) and (2) are said to be equivalent equations.

Equation (3), below, is obtained from equation (1) by subtracting 7 from each side of the equation.

\[
\begin{aligned}
2 x & =4 \\
x & =2
\end{aligned}
\]

Equation (4) is obtained from equation (3) by dividing each side of the equation by 2. You can obtain equation (3) from equation (4) by multiplying each side of the equation by 2 .

All of the above equations are equivalent.
|724758796|kitty|/Applications/kitty.app|0|
\section*{Example 3}
Solve:
a \(3 x+5=20\)
b \(5 x-7=18\)
c \(3-2 x=15\)
d \(-3 p=\frac{2}{5}\)

\section*{Solution}
a \(3 x+5=20\)

\(3 x=15 \quad\) (Subtract 5 from both sides.)

\(x=5 \quad\) (Divide both sides by 3 .)

b \(5 x-7=18\)

\(5 x=25 \quad\) (Add 7 to both sides.)

\(x=5 \quad\) (Divide both sides by 5.)

c \(3-2 x=15\)

\(-2 x=12 \quad\) (Subtract 3 from both sides.)

\(x=-6 \quad\) (Divide both sides by -2 .)

d \(-3 p=\frac{2}{5}\)

\[
p=-\frac{2}{15} \quad \text { (Divide both sides by }-3 \text {.) }
\]
|724758813|kitty|/Applications/kitty.app|0|
\section*{Example 4}
Solve:

a \(3 x+7=2 x+13 \quad\) b \(5 a-21=14-2 a\)

\section*{Solution}
a

\[
3 x+7=2 x+13
\]

\[
\begin{aligned}
3 x+7-2 x & =2 x+13-2 x & & \text { (Subtract } 2 x \text { from both sides.) } \\
x+7 & =13 & & \\
x & =6 & & \text { (Subtract } 7 \text { from both sides.) }
\end{aligned}
\]

b \(5 a-21=14-2 a\)

\[
\begin{aligned}
7 a-21 & =14 & & \text { (Add } 2 a \text { to both sides.) } \\
7 a & =35 & & \text { (Add } 21 \text { to both sides.) } \\
a & =5 & & \text { (Divide both sides by 7.) }
\end{aligned}
\]
|724758835|kitty|/Applications/kitty.app|0|
a \(a+2=5\)
|724758859|kitty|/Applications/kitty.app|0|
a \(a+2=5\)
d \(d-15=3\)
|724758861|kitty|/Applications/kitty.app|0|
a \(a+2=5\)
d \(d-15=3\)
g \(6 d=42\)
|724758864|kitty|/Applications/kitty.app|0|
a \(a+2=5\)
d \(d-15=3\)
g \(6 d=42\)
j \(9 q=-27\)
|724758868|kitty|/Applications/kitty.app|0|
a \(a+2=5\)
d \(d-15=3\)
g \(6 d=42\)
j \(9 q=-27\)
m \(x-6=5 \frac{1}{3}\)
|724758871|kitty|/Applications/kitty.app|0|
c \(c-6=11\)
|724758884|kitty|/Applications/kitty.app|0|
f \(3 b=9\)
|724758885|kitty|/Applications/kitty.app|0|
i \(-2 n=8\)
|724758887|kitty|/Applications/kitty.app|0|
l \(3 a=\frac{2}{3}\)
|724758888|kitty|/Applications/kitty.app|0|
o \(2 x=-\frac{5}{6}\)
|724758889|kitty|/Applications/kitty.app|0|
b \(b+7=19\)
e \(2 a=6\)
h \(-3 m=6\)
k \(b+7=29 \frac{1}{2}\)
n \(-4 y=\frac{8}{9}\)
|724758912|kitty|/Applications/kitty.app|0|
1 Solve these equations.
|724758931|kitty|/Applications/kitty.app|0|
a \(2 a+5=7\)
|724759058|kitty|/Applications/kitty.app|0|
a \(2 a+5=7\)
d \(5 d-7=23\)
|724759061|kitty|/Applications/kitty.app|0|
a \(2 a+5=7\)
d \(5 d-7=23\)
g \(5 h+21=11\)
|724759065|kitty|/Applications/kitty.app|0|
a \(2 a+5=7\)
d \(5 d-7=23\)
g \(5 h+21=11\)
j \(3 a-16=-31\)
|724759068|kitty|/Applications/kitty.app|0|
a \(2 a+5=7\)
d \(5 d-7=23\)
g \(5 h+21=11\)
j \(3 a-16=-31\)
m \(2 x+11=7 \frac{1}{4}\)
|724759072|kitty|/Applications/kitty.app|0|
c \(3 c-1=20\)
|724759081|kitty|/Applications/kitty.app|0|
f \(3 g+17=5\)
|724759082|kitty|/Applications/kitty.app|0|
i \(4 a+23=-9\)
|724759083|kitty|/Applications/kitty.app|0|
l \(2 b+5=7 \frac{1}{2}\)
|724759085|kitty|/Applications/kitty.app|0|
o \(-2 b+4=9 \frac{1}{4}\)
|724759086|kitty|/Applications/kitty.app|0|
b \(3 b+4=19\)
e \(4 f-3=13\)
h \(6 a+17=-1\)
k \(7 b-17=-66\)
n \(4 m-9=-13 \frac{4}{5}\)
|724759088|kitty|/Applications/kitty.app|0|
a \(2-3 a=8\)
|724759102|kitty|/Applications/kitty.app|0|
a \(2-3 a=8\)
d \(3-5 d=-22\)
|724759111|kitty|/Applications/kitty.app|0|
b \(3-4 b=15\)
|724759125|kitty|/Applications/kitty.app|0|
b \(3-4 b=15\)
e \(-6-7 e=15\)
|724759126|kitty|/Applications/kitty.app|0|
2 Solve these equations.

3 Solve these equations.
c \(5-2 c=-13\)
f \(-4-3 f=14\)
|724759133|kitty|/Applications/kitty.app|0|
a \(5 x+5=3 x+1\)
|724759140|kitty|/Applications/kitty.app|0|
a \(5 x+5=3 x+1\)
d \(5 x-6=x-2\)
|724759142|kitty|/Applications/kitty.app|0|
a \(5 x+5=3 x+1\)
d \(5 x-6=x-2\)
g \(4 x-3=18-3 x\)
|724759145|kitty|/Applications/kitty.app|0|
b \(7 x+15=2 x+20\)
|724759152|kitty|/Applications/kitty.app|0|
b \(7 x+15=2 x+20\)
e \(4 x+7=x-2\)
|724759153|kitty|/Applications/kitty.app|0|
b \(7 x+15=2 x+20\)
e \(4 x+7=x-2\)
h \(2 x-3=7-x\)
|724759155|kitty|/Applications/kitty.app|0|
4 Solve these equations for \(x\) and check your answers.
c \(9 x-7=7 x+3\)
f \(3 x+1=9-x\)
i \(8-3 x=2 x-7\)
|724759285|kitty|/Applications/kitty.app|0|
\section*{Equations with brackets}
In this section, we look at solving equations in which brackets are involved. In previous work in this area, you have always expanded the brackets first. In some examples, we do not do this.

|724759287|kitty|/Applications/kitty.app|0|
Solve:
a \(3(x+2)=21\)
b \(2(3-x)=12\)

\section*{Solution}
a \(3(x+2)=21\)

\[
\begin{aligned}
x+2 & =7 \quad \text { (Divide both sides by 3.) } \\
x & =5
\end{aligned}
\]

b \(2(3-x)=12\)

\(3-x=6 \quad\) (Divide both sides by 2.)

\(-x=3\)

\[
x=-3
\]
|724759293|kitty|/Applications/kitty.app|0|
\section*{Example 6}
Solve \(3(x+5)=31\).

\section*{Solution}
\section*{Method 1}
\(3(x+5)=31\)

\(3 x+15=31\)

\(3 x=16\)

\(x=\frac{16}{3}\)

\(x=5 \frac{1}{3}\)

\section*{Method 2}
\[
\begin{array}{r}
3(x+5)=31 \\
x+5=\frac{31}{3} \\
x=\frac{16}{3}
\end{array}
\]
|724759320|kitty|/Applications/kitty.app|0|
\section*{Example 7}
Solve:
a \(2(x+1)+4(x+3)=26\)
b \(3(a+5)=2(a+6)\)
c \(3(y-3)-2(y-4)=4\)

\section*{Solution}
a \(2(x+1)+4(x+3)=26\)

\(2 x+2+4 x+12=26\)

\(6 x+14=26\)

(Expand the brackets.)

\(6 x=12\)

\(x=2\)

b \(3(a+5)=2(a+6)\)

\(3 a+15=2 a+12\)

\(a+15=12\)

\(a=-3\)

(Subtract \(2 a\) from both sides.)

c \(3(y-3)-2(y-4)=4\)

\(3 y-9-2 y+8=4\)

\(y-1=4\)

(Expand both sets of brackets,

being careful with the signs.)

\(y=5\)
|724759348|kitty|/Applications/kitty.app|0|
b \(3(x-2)=15\)
c \(4(x-1)=12\)
|724759361|kitty|/Applications/kitty.app|0|
b \(3(x-2)=15\)
c \(4(x-1)=12\)
e \(3(5-x)=9\)
f \(4(8-x)=36\)
|724759364|kitty|/Applications/kitty.app|0|
1 Solve for \(x\).
a \(2(x+3)=8\)
d \(5(x+1)=10\)
g \(-3(2 x-6)=12\)
|724759371|kitty|/Applications/kitty.app|0|
c \(4(x-1)=12\)
|724759382|kitty|/Applications/kitty.app|0|
f \(4(8-x)=36\)
|724759382|kitty|/Applications/kitty.app|0|
i \(-2(7 x+6)=86\)
|724759383|kitty|/Applications/kitty.app|0|
b \(3(x-2)=15\)
e \(3(5-x)=9\)
h \(-4(3 x-7)=-8\)
|724759384|kitty|/Applications/kitty.app|0|
a \(2(x+3)=15\)
|724760106|kitty|/Applications/kitty.app|0|
a \(2(x+3)=15\)
d \(4(7-x)=7\)
|724760108|kitty|/Applications/kitty.app|0|
a \(2(x+3)=15\)
d \(4(7-x)=7\)
g \(5(x-3)=\frac{2}{3}\)
|724760110|kitty|/Applications/kitty.app|0|
c \(7(3-x)=20\)
|724760115|kitty|/Applications/kitty.app|0|
f \(-3(2 x-1)=2\)
|724760116|kitty|/Applications/kitty.app|0|
i \(-3(5 x+2)=\frac{1}{4}\)
|724760117|kitty|/Applications/kitty.app|0|
b \(5(x-2)=16\)
e \(-2(x-5)=7\)
h \(2(2 x-6)=\frac{2}{6}\)
|724760118|kitty|/Applications/kitty.app|0|
b \(4(b-1)+3(b+2)=30\)
|724760131|kitty|/Applications/kitty.app|0|
b \(4(b-1)+3(b+2)=30\)
d \(4(2 d+1)-5(d-2)=17\)
|724760133|kitty|/Applications/kitty.app|0|
b \(4(b-1)+3(b+2)=30\)
d \(4(2 d+1)-5(d-2)=17\)
f \(5(2 \mathrm{y}-3)-3(\mathrm{y}-5)=21\)
|724760137|kitty|/Applications/kitty.app|0|
b \(4(b-1)+3(b+2)=30\)
d \(4(2 d+1)-5(d-2)=17\)
f \(5(2 \mathrm{y}-3)-3(\mathrm{y}-5)=21\)
h \(5(2 a-1)=2(3 a+2)\)
|724760140|kitty|/Applications/kitty.app|0|
b \(4(b-1)+3(b+2)=30\)
d \(4(2 d+1)-5(d-2)=17\)
f \(5(2 \mathrm{y}-3)-3(\mathrm{y}-5)=21\)
h \(5(2 a-1)=2(3 a+2)\)
j \(-5(x+3)-4(x+1)=17\)
|724760143|kitty|/Applications/kitty.app|0|
b \(4(b-1)+3(b+2)=30\)
d \(4(2 d+1)-5(d-2)=17\)
f \(5(2 \mathrm{y}-3)-3(\mathrm{y}-5)=21\)
h \(5(2 a-1)=2(3 a+2)\)
j \(-5(x+3)-4(x+1)=17\)
l \(\frac{1}{2}(4 x+1)+2(x-2)=13\)
|724760145|kitty|/Applications/kitty.app|0|
a \(2(a+1)+4(a+2)=22\)
c \(5(c+2)-2(c+1)=17\)
e \(2(x+3)-3(x-4)=20\)
g \(5(a+3)=3(2 a+1)\)
i \(-2(x+4)+3(x-2)=16\)
k \(\frac{1}{2}(2 x+5)+6(x-2)=4 \frac{1}{2}\)
|724760156|kitty|/Applications/kitty.app|0|
\section*{Exercise 5C}

2 Solve for \(x\).

3 Solve:
|724760174|kitty|/Applications/kitty.app|0|
When there are fractions in equations, the standard procedure is to remove the fractions by multiplying both sides of the equation by an appropriate whole number. The next step is to remove the brackets.
|724760184|kitty|/Applications/kitty.app|0|
Solve:
a \(2 x+\frac{1}{2}=\frac{2}{3}\)
b \(3 x-\frac{1}{4}=\frac{4}{5}\)

\section*{Solution}
a

\[
2 x+\frac{1}{2}=\frac{2}{3}
\]

\[
6\left(2 x+\frac{1}{2}\right)=6 \times \frac{2}{3} \quad \begin{gathered}
\text { (Multiply both sides by } 6, \text { the lowest common multiple of the } \\
\text { denominators. })
\end{gathered}
\]

\[
12 x+3=4
\]

\[
\begin{aligned}
12 x & =1 \\
x & =\frac{1}{12}
\end{aligned}
\]

b \(\quad 3 x-\frac{1}{4}=\frac{4}{5}\)

\[
\begin{aligned}
20\left(3 x-\frac{1}{4}\right) & =20 \times \frac{4}{5} \\
60 x-5 & =16 \\
60 x & =21 \quad \text { (Multiply both sides by 20.) } \\
x & =\frac{21}{60} \\
& =\frac{7}{20}
\end{aligned}
\]
|724760205|kitty|/Applications/kitty.app|0|
Solve:
a \(\frac{2 x}{3}+\frac{1}{5}=4\)
b \(10-\frac{a+3}{4}=6\)

\section*{Solution}
a

\[
\begin{aligned}
& \frac{2 x}{3}+\frac{1}{5}=4 \\
& \text { b } \quad 10-\frac{a+3}{4}=6 \\
& 15\left(\frac{2 x}{3}+\frac{1}{5}\right)=15 \times 4 \\
& 15 \times \frac{2 x}{3}+15 \times \frac{1}{5}=15 \times 4 \\
& 10 x+3=60 \\
& 10 x=57 \\
& 4 \times 10-4 \times \frac{a+3}{4}=4 \times 6 \\
& 40-(a+3)=24 \\
& 37-a=24 \\
& -a=-13 \\
& x=\frac{57}{10} \text { or } x=5 \frac{7}{10} \\
& a=13
\end{aligned}
\]

|724760248|kitty|/Applications/kitty.app|0|
Solve \(2.1 x+3.5=9.4\)

\section*{Solution}
\[
\begin{aligned}
2.1 x+3.5 & =9.4 \\
21 x+35 & =94 \quad \text { (Multiply both sides by 10.) } \\
21 x & =59 \\
x & =\frac{59}{21}
\end{aligned}
\]
|724760271|kitty|/Applications/kitty.app|0|
Solve \(\frac{2 x}{3}-3=x+\frac{3}{4}\)

\section*{Solution}
\[
\begin{aligned}
\frac{2 x}{3}-3 & =x+\frac{3}{4} \\
8 x-36 & =12 x+9 \quad \text { (Multiply both sides by 12.) } \\
-4 x & =45 \\
x & =-\frac{45}{4}
\end{aligned}
\]
|724760279|kitty|/Applications/kitty.app|0|
Solve \(\frac{a+5}{4}=\frac{a+3}{3}\)

\section*{Solution}
\[
\begin{aligned}
\frac{a+5}{4} & =\frac{a+3}{3} \\
3(a+5) & =4(a+3) \quad \text { (Multiply both sides by 12.) } \\
3 a+15 & =4 a+12 \\
15 & =a+12 \\
3 & =a \\
a & =3
\end{aligned}
\]
|724760289|kitty|/Applications/kitty.app|0|
\section*{Solving linear equations}
To solve linear equations:

\begin{itemize}
  \item Remove all fractions by multiplying both sides of the equation by the lowest common multiple of the denominators.
  \item Remove all brackets.
  \item Collect like terms and solve the equation.
\end{itemize}
|724760309|kitty|/Applications/kitty.app|0|
a \(2 x-\frac{1}{2}=\frac{1}{4}\)
|724760319|kitty|/Applications/kitty.app|0|
a \(2 x-\frac{1}{2}=\frac{1}{4}\)
d \(\frac{7}{2}-3 y=\frac{2}{3}\)
|724760321|kitty|/Applications/kitty.app|0|
a \(2 x-\frac{1}{2}=\frac{1}{4}\)
d \(\frac{7}{2}-3 y=\frac{2}{3}\)
g \(-2 x+\frac{1}{3}=\frac{1}{5}\)
|724760324|kitty|/Applications/kitty.app|0|
b \(3 x+\frac{3}{2}=\frac{5}{3}\)
|724760335|kitty|/Applications/kitty.app|0|
b \(3 x+\frac{3}{2}=\frac{5}{3}\)
e \(\frac{2}{3}+4 x=\frac{2}{3}\)
|724760336|kitty|/Applications/kitty.app|0|
b \(3 x+\frac{3}{2}=\frac{5}{3}\)
e \(\frac{2}{3}+4 x=\frac{2}{3}\)
h \(\frac{2}{3}+3 x=\frac{1}{5}\)
|724760337|kitty|/Applications/kitty.app|0|
a \(\frac{a}{3}+5=3\)
|724760347|kitty|/Applications/kitty.app|0|
a \(\frac{a}{3}+5=3\)
d \(\frac{3 b}{7}+6=2\)
|724760349|kitty|/Applications/kitty.app|0|
a \(\frac{a}{3}+5=3\)
d \(\frac{3 b}{7}+6=2\)
g \(\frac{2}{3}(m-3)=1\)
|724760351|kitty|/Applications/kitty.app|0|
b \(\frac{3 a}{4}-\frac{4}{5}=\frac{2}{3}\)
|724760357|kitty|/Applications/kitty.app|0|
b \(\frac{3 a}{4}-\frac{4}{5}=\frac{2}{3}\)
e \(2-\frac{x}{3}=6\)
|724760359|kitty|/Applications/kitty.app|0|
b \(\frac{3 a}{4}-\frac{4}{5}=\frac{2}{3}\)
e \(2-\frac{x}{3}=6\)
h \(3\left(\frac{m}{5}+2\right)=2\)
|724760360|kitty|/Applications/kitty.app|0|
a \(\frac{2 y+1}{3}+4=7\)
|724760373|kitty|/Applications/kitty.app|0|
a \(\frac{2 y+1}{3}+4=7\)
d \(\frac{3 y-1}{2}+2=9\)
|724760374|kitty|/Applications/kitty.app|0|
a \(\frac{2 y+1}{3}+4=7\)
d \(\frac{3 y-1}{2}+2=9\)
g \(2+\frac{2 x-1}{5}=5\)
|724760376|kitty|/Applications/kitty.app|0|
c \(\frac{5 p-2}{4}-1=6\)
|724760381|kitty|/Applications/kitty.app|0|
f \(\frac{4 a+3}{5}-2=1\)
|724760382|kitty|/Applications/kitty.app|0|
i \(1-\frac{5 y-3}{4}=-2\)
|724760382|kitty|/Applications/kitty.app|0|
b \(\frac{2 x+5}{3}=9\)
e \(\frac{2 x-3}{5}=3\)
h \(18-\frac{7 x+2}{3}=8\)
|724760384|kitty|/Applications/kitty.app|0|
Example 9b 3 Solve:

Example 10
|724760391|kitty|/Applications/kitty.app|0|
a \(e+1.8=2.9\)
|724760393|kitty|/Applications/kitty.app|0|
a \(e+1.8=2.9\)
d \(1.2 u=15.6\)
|724760395|kitty|/Applications/kitty.app|0|
a \(e+1.8=2.9\)
d \(1.2 u=15.6\)
g \(1.2 x+4=10\)
|724760396|kitty|/Applications/kitty.app|0|
b \(f+3.6=7.5\)
|724760406|kitty|/Applications/kitty.app|0|
b \(f+3.6=7.5\)
e \(3.6 r=9\)
|724760408|kitty|/Applications/kitty.app|0|
b \(f+3.6=7.5\)
e \(3.6 r=9\)
h \(3.8 x-7=8.2\)
|724760409|kitty|/Applications/kitty.app|0|
a \(\frac{3 x}{4}-1=\frac{x}{2}+3\)
|724760419|kitty|/Applications/kitty.app|0|
a \(\frac{3 x}{4}-1=\frac{x}{2}+3\)
d \(\frac{x}{3}-4=6-\frac{2 x}{5}\)
|724760421|kitty|/Applications/kitty.app|0|
b \(\frac{x}{3}+2=\frac{4 x}{3}+3\)
|724760429|kitty|/Applications/kitty.app|0|
b \(\frac{x}{3}+2=\frac{4 x}{3}+3\)
e \(\frac{5}{3}-\frac{x}{2}=\frac{3 x}{4}+\frac{7}{6}\)
|724760430|kitty|/Applications/kitty.app|0|
Example 11 5 Solve these equations for \(x\) and check your answers.
c \(\frac{5 x}{6}-3=7-\frac{x}{3}\)
f \(\frac{11}{12}-\frac{5 x}{6}=\frac{3 x}{4}-\frac{2}{3}\)
|724760437|kitty|/Applications/kitty.app|0|
a \(1.6 x+10=0.9 x+12\)
|724760439|kitty|/Applications/kitty.app|0|
a \(1.6 x+10=0.9 x+12\)
d \(4.8-1.3 x=23+1.3 x\)
|724760441|kitty|/Applications/kitty.app|0|
b \(5.9 x-7=2.4 x+35\)
|724760446|kitty|/Applications/kitty.app|0|
b \(5.9 x-7=2.4 x+35\)
e \(1.5 x+3.9=6.7-0.5 x\)
|724760447|kitty|/Applications/kitty.app|0|
b \(\frac{a+1}{3}=\frac{2 a-1}{7}\)
|724760472|kitty|/Applications/kitty.app|0|
b \(\frac{a+1}{3}=\frac{2 a-1}{7}\)
d \(\frac{a+1}{2}+1=\frac{a-1}{5}\)
|724760474|kitty|/Applications/kitty.app|0|
b \(\frac{a+1}{3}=\frac{2 a-1}{7}\)
d \(\frac{a+1}{2}+1=\frac{a-1}{5}\)
f \(\frac{2 a+1}{2}+\frac{a}{3}=4\)
|724760477|kitty|/Applications/kitty.app|0|
b \(\frac{a+1}{3}=\frac{2 a-1}{7}\)
d \(\frac{a+1}{2}+1=\frac{a-1}{5}\)
f \(\frac{2 a+1}{2}+\frac{a}{3}=4\)
h \(\frac{2 a-1}{3}-2=\frac{a+3}{4}\)
|724760479|kitty|/Applications/kitty.app|0|
b \(\frac{a+1}{3}=\frac{2 a-1}{7}\)
d \(\frac{a+1}{2}+1=\frac{a-1}{5}\)
f \(\frac{2 a+1}{2}+\frac{a}{3}=4\)
h \(\frac{2 a-1}{3}-2=\frac{a+3}{4}\)
j \(\frac{a}{2}+\frac{a-1}{3}=\frac{a+1}{4}\)
|724760482|kitty|/Applications/kitty.app|0|
a \(\frac{a+3}{2}=\frac{a+1}{5}\)
c \(\frac{2 a+1}{3}=\frac{3 a+1}{4}\)
e \(\frac{3 a+2}{4}+2=a\)
g \(\frac{3 a-2}{4}=\frac{a-5}{2}\)
i \(\frac{4 a-1}{3}+a=2\)
|724760492|kitty|/Applications/kitty.app|0|
a \(5 a+9=24\)
|724760506|kitty|/Applications/kitty.app|0|
a \(5 a+9=24\)
c \(\frac{c}{3}-2=4\)
|724760508|kitty|/Applications/kitty.app|0|
a \(5 a+9=24\)
c \(\frac{c}{3}-2=4\)
e \(\frac{4 e}{3}+\frac{1}{2}=2\)
|724760520|kitty|/Applications/kitty.app|0|
a \(5 a+9=24\)
c \(\frac{c}{3}-2=4\)
e \(\frac{4 e}{3}+\frac{1}{2}=2\)
g \(2 g+5=7 g-6\)
|724760524|kitty|/Applications/kitty.app|0|
a \(5 a+9=24\)
c \(\frac{c}{3}-2=4\)
e \(\frac{4 e}{3}+\frac{1}{2}=2\)
g \(2 g+5=7 g-6\)
i \(2(i-1)=5(i+6)\)
|724760530|kitty|/Applications/kitty.app|0|
a \(5 a+9=24\)
c \(\frac{c}{3}-2=4\)
e \(\frac{4 e}{3}+\frac{1}{2}=2\)
g \(2 g+5=7 g-6\)
i \(2(i-1)=5(i+6)\)
k \(2(k+1)-3(k-2)=7\)
|724760532|kitty|/Applications/kitty.app|0|
a \(5 a+9=24\)
c \(\frac{c}{3}-2=4\)
e \(\frac{4 e}{3}+\frac{1}{2}=2\)
g \(2 g+5=7 g-6\)
i \(2(i-1)=5(i+6)\)
k \(2(k+1)-3(k-2)=7\)
\(\mathbf{m} \frac{m+1}{3}=\frac{m-2}{5}\)
|724760537|kitty|/Applications/kitty.app|0|
a \(5 a+9=24\)
c \(\frac{c}{3}-2=4\)
e \(\frac{4 e}{3}+\frac{1}{2}=2\)
g \(2 g+5=7 g-6\)
i \(2(i-1)=5(i+6)\)
k \(2(k+1)-3(k-2)=7\)
\(\mathbf{m} \frac{m+1}{3}=\frac{m-2}{5}\)
o \(\frac{2 q-2}{5}+1=4 q\)
|724760540|kitty|/Applications/kitty.app|0|
b \(3 b-7=32\)
d \(\frac{d}{2}+6=3\)
f \(\frac{2 f}{3}-1=\frac{3}{7}\)
h \(4 h-2=5-3 h\)
j \(3(j+2)=2(2 j-1)\)
l \(4(\ell-1)+3(\ell+2)=8\)
n \(\frac{2 n-1}{3}=\frac{4 n+1}{5}\)
\(\mathbf{p} \frac{2 r+1}{3}+2=\frac{3 r-1}{4}\)
|724760546|kitty|/Applications/kitty.app|0|
loginzlib2vrak5zzpcocc3ouizykn6k5qecgj2tzlnab5wcbqhembyd.onion|724841219|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
1592575641|724806725|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
http://ja.bookszlibb74ugqojhzhg2a63w5i2atv5bqarulgczawnbmsb6s6qead.onion/dl/689520/d4ef19|724841197|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
The web server reported a gateway time-out error.|724808088|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
file:///Users/aayushbajaj/Downloads/The%20Complete%20Idiots%20Guide%20to%20Playing%20Piano%20(Brad%20Hill)%20(Z-Library).pdf|724809658|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
2018 Greatest Pop & Movie Hits Songbook for Piano:|724810145|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
147064049X|724810171|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Class|724827265|kitty|/Applications/kitty.app|0|
    \hline
    \large{\boxtimes} & 7 & N/A & \textbf{Class Quiz 1} & November '23 & 100 & \large{\boxtimes}\\
    \hline
|724827463|kitty|/Applications/kitty.app|0|
    \hline
    \large{\boxtimes} & 7 & N/A & \textbf{Topic Test 1} & November '23 & 100 & \large{\boxtimes}\\
    \hline
|724827499|kitty|/Applications/kitty.app|0|
    \large{\boxtimes} & 6 & CH:4A-E & Factorisation & November '23 & 100 & \large{\boxtimes}\\
|724827528|kitty|/Applications/kitty.app|0|
    \large{\Square} & 14 & CH:8A-E & Index Laws & December '23 & 100 & \large{\Square}\\
|724827545|kitty|/Applications/kitty.app|0|
    \hline
    \large{\Square} & 11 & N/A & \textbf{Topic Test 3} & December '23 & 100 & \large{\Square}\\
    \hline
|724827551|kitty|/Applications/kitty.app|0|
    \hline
    \large{\Square} & 11 & N/A & \textbf{Class Quiz 2} & December '23 & 100 & \large{\Square}\\
    \hline
|724827557|kitty|/Applications/kitty.app|0|
    \hline
    \large{\Square} & 11 & N/A & \textbf{Topic Test 4} & December '23 & 100 & \large{\Square}\\
    \hline
|724827561|kitty|/Applications/kitty.app|0|
    \hline
    \large{\Square} & 22 & N/A & \textbf{Class Quiz 4} & January '24 & 100 & \large{\Square}\\
    \hline
|724827611|kitty|/Applications/kitty.app|0|
    \hline
    \large{\Square} & 11 & N/A & \textbf{Topic Test 5} & December '23 & 100 & \large{\Square}\\
    \hline
|724827614|kitty|/Applications/kitty.app|0|
Class Quiz 5|724827637|kitty|/Applications/kitty.app|0|
    \hline
    \large{\Square} & 28 & N/A & \textbf{Topic Test 8} & January '24 & 100 & \large{\Square}\\
    \hline
|724827647|kitty|/Applications/kitty.app|0|
    \hline
    \large{\Square} & 28 & N/A & \textbf{Topic Test 9} & January '24 & 100 & \large{\Square}\\
    \hline
|724827671|kitty|/Applications/kitty.app|0|
    \hline
    \large{\Square} & 34 & N/A & \textbf{Class Quiz 6} & January '24 & 100 & \large{\Square}\\
    \hline
|724827681|kitty|/Applications/kitty.app|0|
    \hline
    \large{\Square} & 28 & N/A & \textbf{Topic Test 10} & January '24 & 100 & \large{\Square}\\
    \hline
|724827699|kitty|/Applications/kitty.app|0|
i have a latex table with a column for numbering the rows, how can I make this part automatic as opposed to numbering them 1...n? because sometimes I swap two rows and mess up the numberings|724827849|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\newcommand{\rownumber}{\stepcounter{rownum}\arabic{rownum}}|724827889|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
10|724827920|kitty|/Applications/kitty.app|0|
14|724827921|kitty|/Applications/kitty.app|0|
12|724827925|kitty|/Applications/kitty.app|0|
13|724827926|kitty|/Applications/kitty.app|0|
15|724827930|kitty|/Applications/kitty.app|0|
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21|724827937|kitty|/Applications/kitty.app|0|
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25|724827945|kitty|/Applications/kitty.app|0|
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30|724827952|kitty|/Applications/kitty.app|0|
31|724827955|kitty|/Applications/kitty.app|0|
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39|724827969|kitty|/Applications/kitty.app|0|
    \large{\Square} & 40 & N/A & Spare Booklet 1 & February '24 & 100 & \large{\Square}\\
    \large{\Square} & 41 & N/A & Spare Booklet 2 & February '24 & 100 & \large{\Square}\\
    \large{\Square} & 42 & N/A & Spare Booklet 3 & February '24 & 100 & \large{\Square}\\
    \large{\Square} & 43 & N/A & Spare Booklet 4 & February '24 & 100 & \large{\Square}\\
    \large{\Square} & 44 & N/A & Spare Booklet 5 & February '24 & 100 & \large{\Square}\\
    \large{\Square} & 45 & N/A & Spare Booklet 6 & February '24 & 100 & \large{\Square}\\
    \large{\Square} & 46 & N/A & Spare Booklet 7 & February '24 & 100 & \large{\Square}\\
    \large{\Square} & 47 & N/A & Spare Booklet 8 & February '24 & 100 & \large{\Square}\\
    \large{\Square} & 48 & N/A & Spare Booklet 9 & February '24 & 100 & \large{\Square}\\
    \large{\Square} & 49 & N/A & Spare Booklet 10 & February '24 & 100 & \large{\Square}\\
|724828610|kitty|/Applications/kitty.app|0|
\Square|724828272|kitty|/Applications/kitty.app|0|
    \large{\Square} & \rownumber & N/A & \textbf{Topic Test 6} & January '24 & 100 & \large{\Square}\\
|724828394|kitty|/Applications/kitty.app|0|
    \large{\Square} & \rownumber & N/A & \textbf{Half Yearly} & December '23 & 100 & \large{\Square}\\
|724828396|kitty|/Applications/kitty.app|0|
    \hline
    \large{\Square} & \rownumber & N/A & \textbf{Topic Test 5} & January '24 & 100 & \large{\Square}\\
    \hline
|724828467|kitty|/Applications/kitty.app|0|
    \hline
    \large{\Square} & \rownumber & N/A & \textbf{Topic Test 6} & January '24 & 100 & \large{\Square}\\
    \hline
|724828471|kitty|/Applications/kitty.app|0|
    \large{\Square} & \rownumber & CH:16 & Measurement \textbf{Workshop} & January '24 & 100 & \large{\Square}\\
|724828496|kitty|/Applications/kitty.app|0|
January|724828589|kitty|/Applications/kitty.app|0|
        \large{\Square} & 40 & N/A & Spare Booklet 1 & February '24 & 100 & \large{\Square}\\
|724828613|kitty|/Applications/kitty.app|0|
    \large{\Square} & 45 & N/A & Spare Booklet 6 & February '24 & 100 & \large{\Square}\\
    \large{\Square} & 46 & N/A & Spare Booklet 7 & February '24 & 100 & \large{\Square}\\
    \large{\Square} & 47 & N/A & Spare Booklet 8 & February '24 & 100 & \large{\Square}\\
    \large{\Square} & 48 & N/A & Spare Booklet 9 & February '24 & 100 & \large{\Square}\\
    \large{\Square} & 49 & N/A & Spare Booklet 10 & February '24 & 100 & \large{\Square}\\
|724828622|kitty|/Applications/kitty.app|0|
    \hline
|724828623|kitty|/Applications/kitty.app|0|
41|724828628|kitty|/Applications/kitty.app|0|
42|724828632|kitty|/Applications/kitty.app|0|
43|724828633|kitty|/Applications/kitty.app|0|
44|724828634|kitty|/Applications/kitty.app|0|
Nidhi Bajaj <nidhi.bajaj@det.nsw.edu.au>|724832188|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
nidhi.bajaj@det.nsw.edu.au|729063414|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
cp6X9JYw!RQJZ9c53C|724840975|kitty|/Applications/kitty.app|0|
Tutors should have handed out an extra quiz for each student.|724839294|kitty|/Applications/kitty.app|0|
quiz|724839320|kitty|/Applications/kitty.app|0|
Everyone|724839326|kitty|/Applications/kitty.app|0|
realise |724839393|kitty|/Applications/kitty.app|0|
 over the past 4 weeks you have covered the following topics: 1 - Algebra techniques, 2 - Pythagoras' Theorem and Surds, 3 - Consumer Arithmetic and most recently: 4 - Factorisation.\\|724839421|kitty|/Applications/kitty.app|0|
 So, we have planned half-yearly and yearly exams containing mixed questions for you. Yay!|724839491|kitty|/Applications/kitty.app|0|
1-intro|724839585|kitty|/Applications/kitty.app|0|
Algebra|724839586|kitty|/Applications/kitty.app|0|
2-algebra|724839587|kitty|/Applications/kitty.app|0|
Pythagoras' Theorem and Surds|724839587|kitty|/Applications/kitty.app|0|
3-pythag|724839588|kitty|/Applications/kitty.app|0|
Consumer Arithmetic|724839589|kitty|/Applications/kitty.app|0|
4-consumer|724839590|kitty|/Applications/kitty.app|0|
Factorisation|724839591|kitty|/Applications/kitty.app|0|
5-factorisation|724839593|kitty|/Applications/kitty.app|0|
\section{}
    \input{}
\section{}
    \input{}
\section{}
    \input{}
\section{}
    \input{}
|724839598|kitty|/Applications/kitty.app|0|
Questions|724839621|kitty|/Applications/kitty.app|0|
Cambridge Maths Stage 4 NSW Year 7 2ed|724840760|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
継続する|724841359|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\section{}
    \input{}
|724844945|kitty|/Applications/kitty.app|0|
\randomword{quadratic}%
\randomword{perfect square}%
\randomword{factorisation}%
\randomword{expansion}%
|724845094|kitty|/Applications/kitty.app|0|
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{shadows.blur}

\newcommand{\fuzzytext}[1]{
    \begin{tikzpicture}
        \node[blur shadow={shadow blur steps=5}] {#1};
    \end{tikzpicture}
}

\begin{document}
\fuzzytext{Fuzzy Text}
\end{document}|724845468|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{shadows.blur}

\newcommand{\fuzzytext}[1]{
    \begin{tikzpicture}
        \node[blur shadow={shadow blur steps=5}] {#1};
    \end{tikzpicture}
}

\begin{document}
\fuzzytext{Fuzzy Text}
\end{document}
|724845581|kitty|/Applications/kitty.app|0|
\newcommand{\fuzzyfont}[1]{%
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|724845611|kitty|/Applications/kitty.app|0|
Three children earn weekly pocket money. Andrew earns \(\$ 2\) more than Gina, and Katya earns twice the amount Gina earns. The total of the weekly pocket money is \(\$ 22\).

a How much money does Gina earn?

b How much money do Andrew and Katya earn?

\section*{Solution}
a Let \(\$ m\) be the amount of pocket money Gina earns in a week.

Then Andrew earns \(\$(m+2)\) and Katya earns \(\$(2 m)\),

so \(\quad m+(m+2)+2 m=22\) (The total weekly pocket money is \(\$ 22\).)

\[
4 m+2=22
\]

\(4 m=20\)

\[
m=5
\]

So Gina earns \(\$ 5\) per week.

b Andrew earns \(\$ 7\) and Katya earns \(\$ 10\) per week.
|724845734|kitty|/Applications/kitty.app|0|
\section*{Example 13}
Three children earn weekly pocket money. Andrew earns \(\$ 2\) more than Gina, and Katya earns twice the amount Gina earns. The total of the weekly pocket money is \(\$ 22\).

a How much money does Gina earn?

b How much money do Andrew and Katya earn?

\section*{Solution}
a Let \(\$ m\) be the amount of pocket money Gina earns in a week.

Then Andrew earns \(\$(m+2)\) and Katya earns \(\$(2 m)\),

so \(\quad m+(m+2)+2 m=22\) (The total weekly pocket money is \(\$ 22\).)

\[
4 m+2=22
\]

\(4 m=20\)

\[
m=5
\]

So Gina earns \(\$ 5\) per week.

b Andrew earns \(\$ 7\) and Katya earns \(\$ 10\) per week.
|724845755|kitty|/Applications/kitty.app|0|
Ali and Jasmine each have a number of swap cards. Jasmine has 25 more cards than Ali, and in total the two children have 149 cards.

a How many cards does Ali have?

b How many cards does Jasmine have?

\section*{Solution}
a Let \(x\) be the number of cards Ali has.

Jasmine has \((x+25)\) cards.

Total number of cards is 149 ,

\[
\begin{aligned}
& \text { so } x+(x+25)=149 \\
& 2 x+25=149 \\
& 2 x=124 \\
& x=62
\end{aligned}
\]

So Ali has 62 cards.

b Jasmine has \(62+25=87\) cards.

\section*{Harder examples involving rates}
Speed is one of the most familiar rates. In problems involving speed, we use the relationship:

average speed \(=\frac{\text { distance travelled }}{\text { time taken }}\)

or

\[
\text { time taken }=\frac{\text { distance travelled }}{\text { average speed }}
\]

or distance travelled \(=\) average speed \(\times\) time taken

|724845763|kitty|/Applications/kitty.app|0|
Ping and Anna compete in a handicap sprint race. Anna starts the race \(10 \mathrm{~m}\) ahead of Ping. Ping runs at an average speed that is \(20 \%\) faster than Anna's average speed. The two sprinters will be level in the race after 9 seconds. Find the average speed of:
a Anna
b Ping
|724845797|kitty|/Applications/kitty.app|0|
\section*{Example 14}
\section*{Example 15}

\section*{Solution}
a The diagram below represents the situation.

\begin{center}
\includegraphics[max width=\textwidth]{2023_12_09_f9189d42272205e0a381g-14}
\end{center}

Let \(x\) be Anna's average speed, measured in \(\mathrm{m} / \mathrm{s}\).

Ping's average speed is \(120 \%\) of \(x=\frac{120 x}{100}\)

\[
=\frac{6 x}{5} \mathrm{~m} / \mathrm{s}
\]

We use distance \(=\) average speed \(\times\) time taken

After 9 seconds, Anna has run a distance of \(9 x \mathrm{~m}\) and Ping has run a

distance of \(\frac{6 x}{5} \times 9=\frac{54 x}{5} \mathrm{~m}\).

Anna started \(10 \mathrm{~m}\) in front of Ping. So when they are level,

\[
\begin{aligned}
9 x+10 & =\frac{54 x}{5} \\
45 x+50 & =54 x \\
50 & =9 x \\
x & =\frac{50}{9} \\
x & =5 \frac{5}{9}
\end{aligned}
\]

So Anna runs at an average speed of \(5 \frac{5}{9} \mathrm{~m} / \mathrm{s}\).

b Ping runs at \(\frac{6 x}{5}=\frac{6}{5} \times \frac{50}{9}=6 \frac{2}{3} \mathrm{~m} / \mathrm{s}\).
|724845815|kitty|/Applications/kitty.app|0|
For a training run, a triathlete covers \(50 \mathrm{~km}\) in \(4 \frac{1}{4}\) hours. She runs part of the way at a speed of \(10 \mathrm{~km} / \mathrm{h}\), cycles part of the way at a speed of \(40 \mathrm{~km} / \mathrm{h}\) and swims the remaining distance at a speed of \(2 \frac{1}{2} \mathrm{~km} / \mathrm{h}\). The athlete runs for twice the time it takes to complete the cycle leg. How long did she take to complete the cycle leg?

\section*{Solution}
Let \(t\) hours be the time for the cycle leg.

Then \(2 t\) hours is the time for the running leg and \(\left(4 \frac{1}{4}-t-2 t\right)\) hours is the time for the swim leg.

Now, distance of run + distance of cycle + distance of swim \(=\) total distance,

so \(10 \times 2 t+40 \times t+2 \frac{1}{2} \times\left(4 \frac{1}{4}-t-2 t\right)=50\)

\[
\begin{aligned}
20 t+40 t+\frac{5}{2}\left(\frac{17}{4}-3 t\right) & =50 \\
60 t+\frac{85}{8}-\frac{15 t}{2} & =50 \\
480 t+85-60 t & =400 \\
420 t & =315 \\
t & =\frac{315}{420} \\
t & =\frac{3}{4}
\end{aligned}
\]

The athlete takes \(\frac{3}{4}\) hour, or 45 minutes, to complete the cycle leg.
|724845823|kitty|/Applications/kitty.app|0|
1 Jacques thinks of a number \(x\). When he adds 17 to his number, the result is 32 . What is the value of \(x\) ?

2 When 16 is added to twice Simone's age, the answer is 44 . How old is Simone?

3 When 14 is added to half of Suzette's weight in kilograms, the result is 42 . How much does Suzette weigh?
|724845837|kitty|/Applications/kitty.app|0|
4 Yolan buys 8 pens and receives 80 cents change from \(\$ 20.00\). How much does a pen cost, assuming each pen costs the same amount?

5 If the sum of \(2 p\) and 19 is the same as the sum of \(4 p\) and 11 , find the value of \(p\).

6 If the sum of half of \(q\) and 6 is equal to the sum of one-third of \(q\) and 2, find the value of \(q\).
|724845843|kitty|/Applications/kitty.app|0|
7 Derek is presently 20 years older than his daughter, Alana.

a If \(x\) represents Alana's present age, express each of the following in terms of \(x\).

i Derek's present age ii Alana's age in 12 years' time

iii Derek's age in 12 years' time

b If Derek's age 12 years from now is twice Alana's age 12 years from now, find their present ages.
|724845859|kitty|/Applications/kitty.app|0|
8 Alan, Brendan and Calum each have a number of plastic toys from a fast food store. Brendan has 5 more toys than Alan, and Calum has twice as many toys as Alan.

a If \(x\) represents the number of toys Alan has, express each of the following in terms of \(x\) :

i the number of toys Brendan has

ii the number of toys Calum has

iii the total number of toys the three boys have

b If the boys have 37 toys in total, determine how many toys each boy has.
|724845869|kitty|/Applications/kitty.app|0|
definitionbox|724846439|kitty|/Applications/kitty.app|0|
\begin{defbox}Expression: \end{defbox}
|724846452|kitty|/Applications/kitty.app|0|
Expression|724846454|kitty|/Applications/kitty.app|0|
dottedlinesorsolution|724846808|kitty|/Applications/kitty.app|0|
        \begin{solutionordottedlines}
|724846814|kitty|/Applications/kitty.app|0|
        \begin{solutionordottedlines}
        \end{solutionordottedlines}
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English|724847382|kitty|/Applications/kitty.app|0|
\usepackage{xcolor}
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\definecolor{Blue}{rgb}{0.0, 0.0, 1.0} % Standard RGB blue
|724848044|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
2E3093|724848141|ColorSlurp|/Applications/ColorSlurp.app|0|
\definecolor{myColor}{rgb}{0.18, 0.19, 0.58}
|724848241|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\definecolor{Blue}{rgb}{0.0, 0.0, 1.0} % Standard RGB blue|724848240|kitty|/Applications/kitty.app|0|
myColor|724848245|kitty|/Applications/kitty.app|0|
    2E3093
|724848247|kitty|/Applications/kitty.app|0|
    \definecolor{Blue}{rgb}{0.18, 0.19, 0.58}
|724848315|kitty|/Applications/kitty.app|0|
\documentclass{article}
\usepackage{tikz}

\newcommand{\fuzzyfont}[1]{%
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}

\begin{document}
\fuzzyfont{Fuzzy Font}
\end{document}
|724848675|kitty|/Applications/kitty.app|0|
\documentclass{article}
\usepackage{calligra}
\usepackage[T1]{fontenc}

\begin{document}
{\calligra\Large \&}
\end{document}
|724848739|kitty|/Applications/kitty.app|0|
\documentclass{article}
\usepackage{newcent}
\begin{document}
{\fontfamily{pnc}\selectfont\itshape\Large \&}
\end{document}
|724848746|kitty|/Applications/kitty.app|0|
{\calligra\Large \&}
|724848858|kitty|/Applications/kitty.app|0|
\usepackage{calligra}
|724848878|kitty|/Applications/kitty.app|0|
\usepackage[T1]{fontenc}
|724848896|kitty|/Applications/kitty.app|0|
L|724848961|kitty|/Applications/kitty.app|0|
\large|724849083|kitty|/Applications/kitty.app|0|
\begin{boxdef}
    \large{\textbf{Equation:}}
    \begin{solutionordottedlines}[1in]
    \end{solutionordottedlines}
\end{boxdef}
|724849184|kitty|/Applications/kitty.app|0|
what is an equation?

what does it mean for 2 equations to be equal?
|724849280|kitty|/Applications/kitty.app|0|
examplebox|724849423|kitty|/Applications/kitty.app|0|
exaboxplebox|724849430|kitty|/Applications/kitty.app|0|
a \(3 x+5=20\)

\(3 x=15 \quad\) (Subtract 5 from both sides.)

\(x=5 \quad\) (Divide both sides by 3 .)
|724849505|kitty|/Applications/kitty.app|0|
b \(5 x-7=18\)

\(5 x=25 \quad\) (Add 7 to both sides.)

\(x=5 \quad\) (Divide both sides by 5.)
|724849511|kitty|/Applications/kitty.app|0|
c \(3-2 x=15\)

\(-2 x=12 \quad\) (Subtract 3 from both sides.)

\(x=-6 \quad\) (Divide both sides by -2 .)
|724849518|kitty|/Applications/kitty.app|0|
d \(-3 p=\frac{2}{5}\)

\[
p=-\frac{2}{15} \quad \text { (Divide both sides by }-3 \text {.) }
\]
|724849523|kitty|/Applications/kitty.app|0|

\section*{Solution}



|724849557|kitty|/Applications/kitty.app|0|

\section*{Example 4}
|724849560|kitty|/Applications/kitty.app|0|
a
|724849603|kitty|/Applications/kitty.app|0|
\[
3 x+7=2 x+13
\]

\[
\begin{aligned}
3 x+7-2 x & =2 x+13-2 x & & \text { (Subtract } 2 x \text { from both sides.) } \\
x+7 & =13 & & \\
x & =6 & & \text { (Subtract } 7 \text { from both sides.) }
\end{aligned}
\]
|724849605|kitty|/Applications/kitty.app|0|
\(5 a-21=14-2 a\)

\[
\begin{aligned}
7 a-21 & =14 & & \text { (Add } 2 a \text { to both sides.) } \\
7 a & =35 & & \text { (Add } 21 \text { to both sides.) } \\
a & =5 & & \text { (Divide both sides by 7.) }
\end{aligned}
\]
|724849616|kitty|/Applications/kitty.app|0|
exercisebox|724849649|kitty|/Applications/kitty.app|0|
exercises:

|724849656|kitty|/Applications/kitty.app|0|
m \(x-6=5 \frac{1}{3}\)
|724849678|kitty|/Applications/kitty.app|0|
g \(6 d=42\)
|724849701|kitty|/Applications/kitty.app|0|
            \begin{solutionordottedlines}[1in]
|724849808|kitty|/Applications/kitty.app|0|
See the following examples:|724849847|kitty|/Applications/kitty.app|0|
Solve:
a \(3(x+2)=21\)
b \(2(3-x)=12\)
|724850601|kitty|/Applications/kitty.app|0|
    \begin{enumerate}
        \item 
|724850641|kitty|/Applications/kitty.app|0|
    \begin{enumerate}[(a)]
        \item 
|724850703|kitty|/Applications/kitty.app|0|
a \(3(x+2)=21\)

\[
\begin{aligned}
x+2 & =7 \quad \text { (Divide both sides by 3.) } \\
x & =5
\end{aligned}
\]
|724850706|kitty|/Applications/kitty.app|0|
            a \(3(x+2)=21\)

|724850714|kitty|/Applications/kitty.app|0|
\(3-x=6 \quad\) (Divide both sides by 2.)

\(-x=3\)

\[
x=-3
\]
|724850721|kitty|/Applications/kitty.app|0|

b \(2(3-x)=12\)

|724850729|kitty|/Applications/kitty.app|0|
\section*{Method 1}
\(3(x+5)=31\)

\(3 x+15=31\)

\(3 x=16\)

\(x=\frac{16}{3}\)

\(x=5 \frac{1}{3}\)

\section*{Method 2}
\[
\begin{array}{r}
3(x+5)=31 \\
x+5=\frac{31}{3} \\
x=\frac{16}{3}
\end{array}
\]
|724850777|kitty|/Applications/kitty.app|0|
a \(2(x+1)+4(x+3)=26\)

\(2 x+2+4 x+12=26\)

\(6 x+14=26\)
|724850860|kitty|/Applications/kitty.app|0|
a \(2(x+1)+4(x+3)=26\)

\(2 x+2+4 x+12=26\)

\(6 x+14=26\)

(Expand the brackets.)

\(6 x=12\)

\(x=2\)
|724850863|kitty|/Applications/kitty.app|0|
b \(3(a+5)=2(a+6)\)

\(3 a+15=2 a+12\)

\(a+15=12\)

\(a=-3\)

(Subtract \(2 a\) from both sides.)
|724850875|kitty|/Applications/kitty.app|0|
c \(3(y-3)-2(y-4)=4\)

\(3 y-9-2 y+8=4\)

\(y-1=4\)

(Expand both sets of brackets,

being careful with the signs.)

\(y=5\)
|724850890|kitty|/Applications/kitty.app|0|
        \question[]
|724850899|kitty|/Applications/kitty.app|0|
 1 Solve for \(x\).|724850931|kitty|/Applications/kitty.app|0|
(|724851045|kitty|/Applications/kitty.app|0|
 Solve:|724851190|kitty|/Applications/kitty.app|0|
a

|724851220|kitty|/Applications/kitty.app|0|
\[
2 x+\frac{1}{2}=\frac{2}{3}
\]

\[
6\left(2 x+\frac{1}{2}\right)=6 \times \frac{2}{3} \quad \begin{gathered}
\text { (Multiply both sides by } 6, \text { the lowest common multiple of the } \\
\text { denominators. })
\end{gathered}
\]

\[
12 x+3=4
\]

\[
\begin{aligned}
12 x & =1 \\
x & =\frac{1}{12}
\end{aligned}
\]
|724851224|kitty|/Applications/kitty.app|0|
\(\quad 3 x-\frac{1}{4}=\frac{4}{5}\)

\[
\begin{aligned}
20\left(3 x-\frac{1}{4}\right) & =20 \times \frac{4}{5} \\
60 x-5 & =16 \\
60 x & =21 \quad \text { (Multiply both sides by 20.) } \\
x & =\frac{21}{60} \\
& =\frac{7}{20}
\end{aligned}
\]
|724851232|kitty|/Applications/kitty.app|0|
\section*{Solution}
a
|724851286|kitty|/Applications/kitty.app|0|
\[
\begin{aligned}
& \frac{2 x}{3}+\frac{1}{5}=4 \\
& \text { b } \quad 10-\frac{a+3}{4}=6 \\
& 15\left(\frac{2 x}{3}+\frac{1}{5}\right)=15 \times 4 \\
& 15 \times \frac{2 x}{3}+15 \times \frac{1}{5}=15 \times 4 \\
& 10 x+3=60 \\
& 10 x=57 \\
& 4 \times 10-4 \times \frac{a+3}{4}=4 \times 6 \\
& 40-(a+3)=24 \\
& 37-a=24 \\
& -a=-13 \\
& x=\frac{57}{10} \text { or } x=5 \frac{7}{10} \\
& a=13
\end{aligned}
\]
|724851354|kitty|/Applications/kitty.app|0|
            \begin{solutionordottedlines}[1in]
            \end{solutionordottedlines}
|724851367|kitty|/Applications/kitty.app|0|
\[
\begin{aligned}
2.1 x+3.5 & =9.4 \\
21 x+35 & =94 \quad \text { (Multiply both sides by 10.) } \\
21 x & =59 \\
x & =\frac{59}{21}
\end{aligned}
\]
|724851372|kitty|/Applications/kitty.app|0|
{}|724851389|kitty|/Applications/kitty.app|0|
        \begi
|724851392|kitty|/Applications/kitty.app|0|
\[
\begin{aligned}
\frac{2 x}{3}-3 & =x+\frac{3}{4} \\
8 x-36 & =12 x+9 \quad \text { (Multiply both sides by 12.) } \\
-4 x & =45 \\
x & =-\frac{45}{4}
\end{aligned}
\]
|724851409|kitty|/Applications/kitty.app|0|
\section*{Solution}


|724851416|kitty|/Applications/kitty.app|0|
\[
\begin{aligned}
\frac{a+5}{4} & =\frac{a+3}{3} \\
3(a+5) & =4(a+3) \quad \text { (Multiply both sides by 12.) } \\
3 a+15 & =4 a+12 \\
15 & =a+12 \\
3 & =a \\
a & =3
\end{aligned}
\]
|724851435|kitty|/Applications/kitty.app|0|
\section*{Solution}

|724851442|kitty|/Applications/kitty.app|0|

\section*{Solving linear equations}
To solve linear equations:
|724851501|kitty|/Applications/kitty.app|0|
        \question[7]
|724851655|kitty|/Applications/kitty.app|0|
When there are fractions in equations, the standard procedure is to remove the fractions by multiplying both sides of the equation by an appropriate whole number. The next step is to remove the brackets.

\begin{examplebox}
    \begin{questions}
        \question[2] Solve:
        \begin{parts}
            \part \(2 x+\frac{1}{2}=\frac{2}{3}\)
            \begin{solutionordottedlines}[1in]
                \[
                2 x+\frac{1}{2}=\frac{2}{3}
                \]

                \[
                6\left(2 x+\frac{1}{2}\right)=6 \times \frac{2}{3} \quad \begin{gathered}
                \text { (Multiply both sides by } 6, \text { the lowest common multiple of the } \\
                \text { denominators. })
                \end{gathered}
                \]

                \[
                12 x+3=4
                \]

                \[
                \begin{aligned}
                12 x & =1 \\
                x & =\frac{1}{12}
                \end{aligned}
                \]
            \end{solutionordottedlines}
            \part \(3 x-\frac{1}{4}=\frac{4}{5}\)
            \begin{solutionordottedlines}[1in]
                \(\quad 3 x-\frac{1}{4}=\frac{4}{5}\)

                \[
                \begin{aligned}
                20\left(3 x-\frac{1}{4}\right) & =20 \times \frac{4}{5} \\
                60 x-5 & =16 \\
                60 x & =21 \quad \text { (Multiply both sides by 20.) } \\
                x & =\frac{21}{60} \\
                & =\frac{7}{20}
                \end{aligned}
                \]
            \end{solutionordottedlines}
        \end{parts}
        \question[2] Solve:
        \begin{parts}
            \part \(\frac{2 x}{3}+\frac{1}{5}=4\)
            \begin{solutionordottedlines}[1in]
            \end{solutionordottedlines}
            \part \(10-\frac{a+3}{4}=6\)
            \begin{solutionordottedlines}[1in]
            \end{solutionordottedlines}
        \end{parts}
        \question[1] Solve \(2.1 x+3.5=9.4\)
        \begin{solutionordottedlines}[1in]
            \[
            \begin{aligned}
            2.1 x+3.5 & =9.4 \\
            21 x+35 & =94 \quad \text { (Multiply both sides by 10.) } \\
            21 x & =59 \\
            x & =\frac{59}{21}
            \end{aligned}
            \]
        \end{solutionordottedlines}
        \question[1] Solve \(\frac{2 x}{3}-3=x+\frac{3}{4}\)
        \begin{solutionordottedlines}[1in]
            \[
            \begin{aligned}
            \frac{2 x}{3}-3 & =x+\frac{3}{4} \\
            8 x-36 & =12 x+9 \quad \text { (Multiply both sides by 12.) } \\
            -4 x & =45 \\
            x & =-\frac{45}{4}
            \end{aligned}
            \]
        \end{solutionordottedlines}
        \question[1] Solve \(\frac{a+5}{4}=\frac{a+3}{3}\)
        \begin{solutionordottedlines}[1in]
            \[
            \begin{aligned}
            \frac{a+5}{4} & =\frac{a+3}{3} \\
            3(a+5) & =4(a+3) \quad \text { (Multiply both sides by 12.) } \\
            3 a+15 & =4 a+12 \\
            15 & =a+12 \\
            3 & =a \\
            a & =3
            \end{aligned}
            \]
        \end{solutionordottedlines}

\fbox{\parbox{\textwidth}
\large{\textbf{Crash Course:}}
\begin{itemize}
  \item Remove all fractions by multiplying both sides of the equation by the lowest common multiple of the denominators.
  \item Remove all brackets.
  \item Collect like terms and solve the equation.
\end{itemize}
}

\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \question[9] Solve for the unknown pronumeral:
        \begin{parts}
            \part \(2 x-\frac{1}{2}=\frac{1}{4}\)
            \part \(\frac{7}{2}-3 y=\frac{2}{3}\)
            \part \(-2 x+\frac{1}{3}=\frac{1}{5}\)
            \part \(\frac{a}{3}+5=3\)
            \part \(\frac{3 b}{7}+6=2\)
            \part \(\frac{2}{3}(m-3)=1\)
            \part \(\frac{2 y+1}{3}+4=7\)
            \part \(\frac{3 y-1}{2}+2=9\)
            \part \(2+\frac{2 x-1}{5}=5\)
        \end{parts}
        \begin{parts}
            \part \(e+1.8=2.9\)
            \part \(1.2 u=15.6\)
            \part \(1.2 x+4=10\)
            \part \(\frac{3 x}{4}-1=\frac{x}{2}+3\)
            \part \(\frac{x}{3}-4=6-\frac{2 x}{5}\)
            \part \(1.6 x+10=0.9 x+12\)
            \part \(4.8-1.3 x=23+1.3 x\)
        \end{parts}
        \begin{parts}
            \part \(\frac{a+3}{2}=\frac{a+1}{5}\)
            \part \(\frac{2 a+1}{3}=\frac{3 a+1}{4}\)
            \part \(\frac{3 a+2}{4}+2=a\)
            \part \(\frac{3 a-2}{4}=\frac{a-5}{2}\)
            \part \(\frac{4 a-1}{3}+a=2\)
            \part \(5 a+9=24\)
            \part \(\frac{c}{3}-2=4\)
            \part \(\frac{4 e}{3}+\frac{1}{2}=2\)
            \part \(2 g+5=7 g-6\)
            \part \(2(i-1)=5(i+6)\)
            \part \(2(k+1)-3(k-2)=7\)
            \part \(\frac{m+1}{3}=\frac{m-2}{5}\)
            \part \(\frac{2 q-2}{5}+1=4 q\)
        \end{parts}
    \end{questions}
\end{exercisebox}
|724851788|kitty|/Applications/kitty.app|0|
\fbox{\parbox{\textwidth}
\large{\textbf{Crash Course:}}
\begin{itemize}
  \item Remove all fractions by multiplying both sides of the equation by the lowest common multiple of the denominators.
  \item Remove all brackets.
  \item Collect like terms and solve the equation.
\end{itemize}
}

\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \question[9] Solve for the unknown pronumeral:
        \begin{parts}
            \part \(2 x-\frac{1}{2}=\frac{1}{4}\)
            \part \(\frac{7}{2}-3 y=\frac{2}{3}\)
            \part \(-2 x+\frac{1}{3}=\frac{1}{5}\)
            \part \(\frac{a}{3}+5=3\)
            \part \(\frac{3 b}{7}+6=2\)
            \part \(\frac{2}{3}(m-3)=1\)
            \part \(\frac{2 y+1}{3}+4=7\)
            \part \(\frac{3 y-1}{2}+2=9\)
            \part \(2+\frac{2 x-1}{5}=5\)
        \end{parts}
        \begin{parts}
            \part \(e+1.8=2.9\)
            \part \(1.2 u=15.6\)
            \part \(1.2 x+4=10\)
            \part \(\frac{3 x}{4}-1=\frac{x}{2}+3\)
            \part \(\frac{x}{3}-4=6-\frac{2 x}{5}\)
            \part \(1.6 x+10=0.9 x+12\)
            \part \(4.8-1.3 x=23+1.3 x\)
        \end{parts}
        \begin{parts}
            \part \(\frac{a+3}{2}=\frac{a+1}{5}\)
            \part \(\frac{2 a+1}{3}=\frac{3 a+1}{4}\)
            \part \(\frac{3 a+2}{4}+2=a\)
            \part \(\frac{3 a-2}{4}=\frac{a-5}{2}\)
            \part \(\frac{4 a-1}{3}+a=2\)
            \part \(5 a+9=24\)
            \part \(\frac{c}{3}-2=4\)
            \part \(\frac{4 e}{3}+\frac{1}{2}=2\)
            \part \(2 g+5=7 g-6\)
            \part \(2(i-1)=5(i+6)\)
            \part \(2(k+1)-3(k-2)=7\)
            \part \(\frac{m+1}{3}=\frac{m-2}{5}\)
            \part \(\frac{2 q-2}{5}+1=4 q\)
        \end{parts}
    \end{questions}
\end{exercisebox}
|724851840|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \question[9] Solve for the unknown pronumeral:
        \begin{parts}
            \part \(2 x-\frac{1}{2}=\frac{1}{4}\)
            \part \(\frac{7}{2}-3 y=\frac{2}{3}\)
            \part \(-2 x+\frac{1}{3}=\frac{1}{5}\)
            \part \(\frac{a}{3}+5=3\)
            \part \(\frac{3 b}{7}+6=2\)
            \part \(\frac{2}{3}(m-3)=1\)
            \part \(\frac{2 y+1}{3}+4=7\)
            \part \(\frac{3 y-1}{2}+2=9\)
            \part \(2+\frac{2 x-1}{5}=5\)
        \end{parts}
        \begin{parts}
            \part \(e+1.8=2.9\)
            \part \(1.2 u=15.6\)
            \part \(1.2 x+4=10\)
            \part \(\frac{3 x}{4}-1=\frac{x}{2}+3\)
            \part \(\frac{x}{3}-4=6-\frac{2 x}{5}\)
            \part \(1.6 x+10=0.9 x+12\)
            \part \(4.8-1.3 x=23+1.3 x\)
        \end{parts}
        \begin{parts}
            \part \(\frac{a+3}{2}=\frac{a+1}{5}\)
            \part \(\frac{2 a+1}{3}=\frac{3 a+1}{4}\)
            \part \(\frac{3 a+2}{4}+2=a\)
            \part \(\frac{3 a-2}{4}=\frac{a-5}{2}\)
            \part \(\frac{4 a-1}{3}+a=2\)
            \part \(5 a+9=24\)
            \part \(\frac{c}{3}-2=4\)
            \part \(\frac{4 e}{3}+\frac{1}{2}=2\)
            \part \(2 g+5=7 g-6\)
            \part \(2(i-1)=5(i+6)\)
            \part \(2(k+1)-3(k-2)=7\)
            \part \(\frac{m+1}{3}=\frac{m-2}{5}\)
            \part \(\frac{2 q-2}{5}+1=4 q\)
        \end{parts}
    \end{questions}
\end{exercisebox}
|724851846|kitty|/Applications/kitty.app|0|
\fbox{\parbox{\textwidth}
\large{\textbf{Crash Course:}}
\begin{itemize}
  \item Remove all fractions by multiplying both sides of the equation by the lowest common multiple of the denominators.
  \item Remove all brackets.
  \item Collect like terms and solve the equation.
\end{itemize}
}
|724851885|kitty|/Applications/kitty.app|0|
        \end{parts}
        \begin{parts}
|724852094|kitty|/Applications/kitty.app|0|
        \part \(-3 m=6\)
|724852376|kitty|/Applications/kitty.app|0|
        \part \(4(b-1)+3(b+2)=30\)
        \part \(4(2 d+1)-5(d-2)=17\)
        \part \(5(2 \mathrm{y}-3)-3(\mathrm{y}-5)=21\)
        \part \(5(2 a-1)=2(3 a+2)\)
        \part \(-5(x+3)-4(x+1)=17\)
        \part \(\frac{1}{2}(4 x+1)+2(x-2)=13\)
|724852422|kitty|/Applications/kitty.app|0|
        \part \(2(2 x-6)=\frac{2}{6}\)



\subsection{Linear Equations involving fractions}

|724852424|kitty|/Applications/kitty.app|0|
    \part \(4(\ell-1)+3(\ell+2)=8\)
|724852533|kitty|/Applications/kitty.app|0|
begin|724852713|kitty|/Applications/kitty.app|0|
Image: 772x228 (2.7 MB)|724886326|Brave Browser Beta|/Applications/Brave Browser Beta.app|1|01ae6fc6f8d90a6515471f91c0e431cc72f9c28d.tiff
the \numquestions the \numparts the \numsubparts the \numsubsubparts
|724888006|kitty|/Applications/kitty.app|0|
\renewcommand{\question}[1][]{%
  \ifx\relax#1\relax
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    \addtocounter{totalpoints}{#1}%
    \oldquestion[#1]
  \fi
}|724888063|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
    \thetotalpoints
|724888180|kitty|/Applications/kitty.app|0|
    \droptotalpoints
|724888193|kitty|/Applications/kitty.app|0|
    \setcounter{totalpoints}{0}
    \input{1-intro}
|724888229|kitty|/Applications/kitty.app|0|
\usepackage{etoolbox}|724888272|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
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See the LaTeX manual or LaTeX Companion for explanation.
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l.73 \renewcommand{\question}
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Try typing  <return>  to proceed.
If that doesn't work, type  X <return>  to quit.
|724888328|kitty|/Applications/kitty.app|0|
\renewcommand{\question}[1][]{%
  \ifx\relax#1\relax
    % if no optional argument is given, just use the original question command
    \oldquestion
  \else
    % if optional argument is given (points), add to total points counter
    \addtocounter{totalpoints}{#1}%
    \oldquestion[#1]
  \fi
}
|724888341|kitty|/Applications/kitty.app|0|
\let\oldquestion\question
|724888352|kitty|/Applications/kitty.app|0|
thetotalpoints|724888380|kitty|/Applications/kitty.app|0|
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\renewcommand{\question}[1][]{%
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    \addtocounter{totalpoints}{#1}%
    \oldquestion[#1]
  \fi
}
|724888392|kitty|/Applications/kitty.app|0|
\let\oldquestion\question

\renewcommand{\question}[1][]{%
  \ifx\relax#1\relax
    % if no optional argument is given, just use the original question command
    \oldquestion
  \else
    % if optional argument is given (points), add to total points counter
    \addtocounter{totalpoints}{#1}%
    \oldquestion[#1]
  \fi
}

|724888425|kitty|/Applications/kitty.app|0|
totalpoints|724888434|kitty|/Applications/kitty.app|0|
\newcounter{sectionpoints}

\let\oldquestion\question

\renewcommand{\question}[1][]{%
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}
|724888687|kitty|/Applications/kitty.app|0|
\let\oldquestion\question

\renewcommand{\question}[1][]{%
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    % if no optional argument is given, just use the original question command
    \oldquestion
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    % if optional argument is given (points), add to total points counter
|724888749|kitty|/Applications/kitty.app|0|
    \oldquestion[#1]
  \fi
}
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\newcommand{\questionpoints}[2][]{
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  \fi
}
|724889070|kitty|/Applications/kitty.app|0|
    \input{1-intro}
|724889308|kitty|/Applications/kitty.app|0|
\begin{center}
    \multirowgradetable{2}[questions]
\end{center}
|724889315|kitty|/Applications/kitty.app|0|
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Question}}
    & \ref{q1}-\ref{q10}
    & \ref{q11}-\ref{q15}
    & \ref{q16}-\ref{q19}
    & \ref{q20}-\ref{q21}
    & \ref{q22}-\ref{q26}
    & \ref{q27}-\ref{q30}
    & \ref{q31}-\ref{q33}
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    & $\dfrac{}{10}$ 
    & $\dfrac{}{12}$  
    & $\dfrac{}{16}$ 
    & $\dfrac{}{11}$ 
    & $\dfrac{}{18}$ 
    & $\dfrac{}{17}$
    & $\dfrac{}{16}$
    & $\dfrac{}{100}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
|724889350|kitty|/Applications/kitty.app|0|
q1-q10|724889383|kitty|/Applications/kitty.app|0|
Question|724889390|kitty|/Applications/kitty.app|0|
11|724889670|kitty|/Applications/kitty.app|0|
18|724889671|kitty|/Applications/kitty.app|0|
17|724889672|kitty|/Applications/kitty.app|0|
16|724889673|kitty|/Applications/kitty.app|0|
100|724889674|kitty|/Applications/kitty.app|0|
\newcounter{sec1marks}
|724889705|kitty|/Applications/kitty.app|0|
\newcounter{sec6marks}
|724889713|kitty|/Applications/kitty.app|0|
    \setcounter{sectionpoints}{0}
|724889758|kitty|/Applications/kitty.app|0|
\let\oldquestion\question

|724889765|kitty|/Applications/kitty.app|0|
sectionpoints|724889820|kitty|/Applications/kitty.app|0|
thesectionpoints|724889824|kitty|/Applications/kitty.app|0|
    \thesectionm
|724889829|kitty|/Applications/kitty.app|0|
    \thesectionpoints
|724889836|kitty|/Applications/kitty.app|0|
\setcounter{sectionpoints}{0}
|724889850|kitty|/Applications/kitty.app|0|
    \setcounter{sec1marks}{secmarks}
|724889919|kitty|/Applications/kitty.app|0|
\foreach \i in {1,2,...,7} {
  \addtocounter{totalmarks}{\value{sec\i marks}}
}|724890170|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\setcounter{totalmarks}{for i in {1..7} ; do totalmarks + sec{$i}marks; done}
|724890175|kitty|/Applications/kitty.app|0|
    & $\dfrac{}{\thesec1marks}$ 
|724890275|kitty|/Applications/kitty.app|0|
\thesec2marks|724890456|kitty|/Applications/kitty.app|0|
\thesec3marks|724890464|kitty|/Applications/kitty.app|0|
\thesec4marks|724890471|kitty|/Applications/kitty.app|0|
\thessec4marksec5marks|724890481|kitty|/Applications/kitty.app|0|
\thesec6marks|724890487|kitty|/Applications/kitty.app|0|
\thesec7marks|724890492|kitty|/Applications/kitty.app|0|
friends|725090767|kitty|/Applications/kitty.app|0|
collated|725091053|kitty|/Applications/kitty.app|0|
corresponding|725091161|kitty|/Applications/kitty.app|0|
found|725091268|kitty|/Applications/kitty.app|0|

occasionally fall under the domain of a computer scientist
|725091600|kitty|/Applications/kitty.app|0|
As the years progress, I expect this challenge to first warp further towards Pure Mathematics and Algorithmic Networking, and then later to Organic Chemistry and Theoretical Physics.
|725091836|kitty|/Applications/kitty.app|0|
scrartcl|725091956|kitty|/Applications/kitty.app|0|
\noindent\fbox{
    \parbox{\textwidth}{Welcome! Today is the $26^{th}$ of December, and it is my birthday :D.\\\\Today we are going to be playing a game called \textit{21 Problems}. This game consists of 21 \textbf{mathematical} problems and whoever has the highest score by midnight will be the winner!
}
}

\bigbreak
\dotfill
|725092010|kitty|/Applications/kitty.app|0|
\f
|725092017|kitty|/Applications/kitty.app|0|
\marks{3}|725092306|kitty|/Applications/kitty.app|0|
\marks{4}|725092308|kitty|/Applications/kitty.app|0|
marks{4} |725092335|kitty|/Applications/kitty.app|0|
\marks{2}|725092338|kitty|/Applications/kitty.app|0|
    \item Submission fi
|725092652|kitty|/Applications/kitty.app|0|
\begin{tikzpicture}
  % Button background
  \node[draw, rounded corners=8pt, fill=blue!30, inner sep=10pt] (button) {\textbf{Click Me!}};

  % Text label
  \node[text=white, align=center] at (button) {Visit Website};

  % Define the URL
  \def\myurl{https://www.example.com}

  % Add a link
  \node[anchor=center] (link) at (button.center) {\hyperlink{\myurl}{\phantom{\rule{\widthof{Visit Website}}{\heightof{Click Me!}}}}};
\end{tikzpicture}
|725092762|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
https://www.example.com|725092771|kitty|/Applications/kitty.app|0|
Click Me!|725092798|kitty|/Applications/kitty.app|0|
\usepackage{hyperref}
\usepackage{tikz}
\usetikzlibrary{calc}|725092818|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\usepackage{tikz}
|725092823|kitty|/Applications/kitty.app|0|
<argument> \widthof 
         {Visit Website}
l.49 ...widthof{Visit Website}}{\heightof{Submit}}}}
                                                  };

}|725092922|kitty|/Applications/kitty.app|0|
  \node[anchor=center] (link) at (button.center) {\hyperlink{\myurl}{\phantom{Visit Website}}};
|725092946|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
  \node[anchor=center] (link) at (button.center) {\hyperlink{\myurl}{\phantom{Visit Website}}{\heightof{Submit}}};
|725092951|kitty|/Applications/kitty.app|0|
Visit Website|725092987|kitty|/Applications/kitty.app|0|
  \node[text=white, align=center] at (button) {Submit};
|725093008|kitty|/Applications/kitty.app|0|
{questions}|725093172|kitty|/Applications/kitty.app|0|
\renewcommand{\marks}[1]{\hfill{}\textit{(#1 marks)}}
\renewcommand{\mark}{\hfill{}\textit{(1 marks)}}
|725093344|kitty|/Applications/kitty.app|0|
\marksnotpoints
|725093363|kitty|/Applications/kitty.app|0|
\pointsinrightmargin
|725093492|kitty|/Applications/kitty.app|0|
\usepackage[top=20mm,bottom=20mm,right=10mm,left=10mm]{geometry}
|725093508|kitty|/Applications/kitty.app|0|
\vspace{-1in}
|725093964|kitty|/Applications/kitty.app|0|
\vspace*{-2cm}|725094023|kitty|/Applications/kitty.app|0|
    \begin{solutionordottedlines}[1in]
    \end{solutionordottedlines}
|725094771|kitty|/Applications/kitty.app|0|
solutionordottsolutionorboxedlines|725094893|kitty|/Applications/kitty.app|0|
Fourier decompose Earthquake.|725095034|kitty|/Applications/kitty.app|0|
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{positioning, arrows.meta}

\begin{document}
\begin{tikzpicture}[
    node distance=2cm,
    >=Stealth,
    node/.style={circle, draw, minimum size=1.5cm},
    arrow/.style={->, shorten >=1pt, >=Stealth}
]

% Create nodes
\node[node] (A) {1};
\node[node, below right=of A] (B) {2};
\node[node, below=of A] (C) {3};
\node[node, below left=of A] (D) {4};
\node[node, right=of B] (E) {5};
\node[node, left=of D] (F) {6};
\node[node, below right=of B] (G) {7};
\node[node, below left=of D] (H) {8};
\node[node, below=of C] (I) {9};
\node[node, below=of I] (J) {10};

% Create directed edges with labels
\draw[arrow] (A) -- (B) node[midway, above] {1};
\draw[arrow] (A) -- (C) node[midway, above] {2};
\draw[arrow] (A) -- (D) node[midway, above] {3};
\draw[arrow] (B) -- (E) node[midway, above] {4};
\draw[arrow] (D) -- (F) node[midway, above] {5};
\draw[arrow] (B) -- (G) node[midway, above] {6};
\draw[arrow] (D) -- (H) node[midway, above] {7};
\draw[arrow] (C) -- (I) node[midway, above] {8};
\draw[arrow] (I) -- (J) node[midway, above] {9};
\draw[arrow] (E) -- (J) node[midway, above] {10};
\end{tikzpicture}
\end{document}
|725100041|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\documentclass{article}
\usepackage{palatino}
\begin{document}
{\fontfamily{ppl}\selectfont\itshape\Large \&}
\end{document}
|725100063|kitty|/Applications/kitty.app|0|
\includegraphics[width=\textwidth]{2023_12_24_f3a4aee8e898d0afdf24g-001}
|725100487|kitty|/Applications/kitty.app|0|
\texttt{https://cdn.mathpix.com/cropped/2023_12_24_f3a4aee8e898d0afdf24g-196.jpg?height=730\&width=1088\&top_left_y=217\&top_left_x=84}
|725100613|kitty|/Applications/kitty.app|0|
\texttt{https://cdn.mathpix.com/cropped/2023_12_24_f3a4aee8e898d0afdf24g-198.jpg?height=730&width=1086&top_left_y=217&top_left_x=86}
|725100615|kitty|/Applications/kitty.app|0|
\usepackage{tikz}
\usetikzlibrary{positioning, arrows.meta}

\begin{document}
\begin{tikzpicture}[
    node distance=2cm,
    >=Stealth,
    node/.style={circle, draw, minimum size=1.5cm},
    arrow/.style={->, shorten >=1pt, >=Stealth}
]

% Create nodes
\node[node] (A) {1};
\node[node, below right=of A] (B) {2};
\node[node, below=of A] (C) {3};
\node[node, below left=of A] (D) {4};
\node[node, right=of B] (E) {5};
\node[node, left=of D] (F) {6};
\node[node, below right=of B] (G) {7};
\node[node, below left=of D] (H) {8};
\node[node, below=of C] (I) {9};
\node[node, below=of I] (J) {10};

% Create directed edges with labels
\draw[arrow] (A) -- (B) node[midway, above] {1};
\draw[arrow] (A) -- (C) node[midway, above] {2};
\draw[arrow] (A) -- (D) node[midway, above] {3};
\draw[arrow] (B) -- (E) node[midway, above] {4};
\draw[arrow] (D) -- (F) node[midway, above] {5};
\draw[arrow] (B) -- (G) node[midway, above] {6};
\draw[arrow] (D) -- (H) node[midway, above] {7};
\draw[arrow] (C) -- (I) node[midway, above] {8};
\draw[arrow] (I) -- (J) node[midway, above] {9};
\draw[arrow] (E) -- (J) node[midway, above] {10};
\end{tikzpicture}
\end{document}
|725101009|kitty|/Applications/kitty.app|0|
\documentclass{article}
|725101010|kitty|/Applications/kitty.app|0|
\documentclass[tikz,border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{arrows.meta, positioning, arrows}

\begin{document}
\begin{tikzpicture}[>=Stealth, node distance=2cm, on grid, auto]
   \node (s) {$s$};
   \node (v1) [above right=of s] {$v_1$};
   \node (v2) [below right=of s] {$v_2$};
   \node (v3) [above right=of v2] {$v_3$};
   \node (v4) [below right=of v3] {$v_4$};
   \node (t) [below right=of v1] {$t$};

   \path[->]
    (s) edge node {16} (v1)
    (s) edge node {13} (v2)
    (s) edge node {10} (v3)
    (v1) edge node {12} (v3)
    (v2) edge node {14} (v4)
    (v3) edge node {20} (t)
    (v4) edge node {4} (t)
    (v1) edge node {4} (v2)
    (v2) edge node {9} (v3)
    (v3) edge node {7} (v4);

\end{tikzpicture}
\end{document}
|725101057|kitty|/Applications/kitty.app|0|
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{arrows.meta, positioning}

\begin{document}
\begin{tikzpicture}[>=Stealth, node distance=3cm and 3cm, on grid, auto]

   % Nodes
   \node (s) {s};
   \node (v1) [above right=of s] {$v_1$};
   \node (v2) [below right=of s] {$v_2$};
   \node (v3) [above right=of v2] {$v_3$};
   \node (v4) [below right=of v2] {$v_4$};
   \node (t) [above right=of v4] {t};

   % Edges
   \path[->]
   (s) edge node [sloped, above] {16} (v1)
   (s) edge node [sloped, below] {13} (v2)
   (v1) edge node [sloped, above] {12} (v3)
   (v2) edge node [sloped, above] {14} (v4)
   (v3) edge node [sloped, above] {20} (t)
   (v4) edge node [sloped, below] {4} (t)
   (v1) edge node [sloped, below] {4} (v2)
   (v1) edge node [sloped, below] {10} (v3)
   (v2) edge node [sloped, below] {9} (v3)
   (v3) edge node [sloped, above] {7} (v4);
   
\end{tikzpicture}
\end{document}
|725101337|kitty|/Applications/kitty.app|0|
Image: 860x493 (6.5 MB)|725101379|Brave Browser Beta|/Applications/Brave Browser Beta.app|1|1d2269c738c13826d77fc6b47c4301e1a0552da4.tiff
\documentclass{standalone}
\usepackage{tkz-graph}

\begin{document}
\begin{tikzpicture}[node distance=5cm]
   \SetGraphUnit{3}
   \GraphInit[vstyle=Normal]
   \SetVertexNormal[Shape=circle,MinSize=20pt,FillColor=white]
   \SetUpEdge[lw=1.5pt,color=black]
   \Vertices{circle}{s,v1,v3,v4,v2,t}
   \Edges[label=16](s,v1)
   \Edges[label=12](v1,v3)
   \Edges[label=20](v3,t)
   \Edges[label=4](v4,t)
   \Edges[label=14](v2,v4)
   \Edges[label=13](s,v2)
   \Edges[label=10](s,v3)
   \Edges[label=9](v2,v3)
   \Edges[label=7](v3,v4)
   \Edges[label=4](v1,v2)
\end{tikzpicture}
\end{document}
|725101392|kitty|/Applications/kitty.app|0|
\documentclass{article}
\usepackage{tkz-graph}

\begin{document}
\begin{tikzpicture}[scale=1,transform shape]
  % Nodes
  \Vertex[x=0,y=0]{s}
  \Vertex[x=2,y=2]{v1}
  \Vertex[x=2,y=-2]{v2}
  \Vertex[x=5,y=2]{v3}
  \Vertex[x=5,y=-2]{v4}
  \Vertex[x=7,y=0]{t}

  % Edges
  \tikzset{EdgeStyle/.style={->}}
  \Edge[label=16](s)(v1)
  \Edge[label=13](s)(v2)
  \Edge[label=12](v1)(v3)
  \Edge[label=4](v1)(v2)
  \Edge[label=10](s)(v3)
  \Edge[label=9](v2)(v3)
  \Edge[label=14](v2)(v4)
  \Edge[label=7](v3)(v4)
  \Edge[label=20](v3)(t)
  \Edge[label=4](v4)(t)
\end{tikzpicture}
\end{document}
|725101433|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\documentclass{article}
\usepackage{tkz-graph}

\begin{document}
\begin{tikzpicture}
  \SetGraphUnit{2}
  \GraphInit[vstyle=Simple]
  \SetVertexSimple[MinSize=5pt]
  \Vertex{S}
  \NOEA(S){v1} \SOEA(S){v2}
  \NOEA(v2){v3} \SOEA(v3){v4}
  \NOEA(v1){t}

  \tikzset{EdgeStyle/.append style = {->}}
  \Edge[label=16](S)(v1)
  \Edge[label=13](S)(v2)
  \Edge[label=10](S)(v3)
  \Edge[label=12](v1)(v3)
  \Edge[label=4](v1)(v2)
  \Edge[label=9](v2)(v3)
  \Edge[label=14](v2)(v4)
  \Edge[label=7](v3)(v4)
  \Edge[label=20](v3)(t)
  \Edge[label=4](v4)(t)
\end{tikzpicture}
\end{document}
|725101435|kitty|/Applications/kitty.app|0|
  \tikzset{EdgeStyle/.append style = {bend left = 50}}
|725101574|kitty|/Applications/kitty.app|0|
  \Edge[label=4](v1)(v2)
  \Edge[label=10](v2)(v1)
|725101579|kitty|/Applications/kitty.app|0|
  \Vertex[x=0,y=0, L={$s$}]{s}
  \Vertex[x=2,y=2, L={$v_1$}]{v1}
  \Vertex[x=2,y=-2, L={$v_2$}]{v2}
  \Vertex[x=5,y=2, L={$v_3$}]{v3}
  \Vertex[x=5,y=-2, L={$v_4$}]{v4}
  \Vertex[x=7,y=0, L={$t$}]{t}|725101720|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
  \Vertex[x=0,y=0]{s}
  \Vertex[x=2,y=2]{v1}
  \Vertex[x=2,y=-2]{v2}
  \Vertex[x=5,y=2]{v3}
  \Vertex[x=5,y=-2]{v4}
  \Vertex[x=7,y=0]{t}
|725101729|kitty|/Applications/kitty.app|0|
\vspace{1cm}
|725101967|kitty|/Applications/kitty.app|0|
\begin{tikzpicture}[scale=1,transform shape]
  % Nodes
  \Vertex[x=0,y=0, L={$s$}]{s}
  \Vertex[x=2,y=2, L={$v_1$}]{v1}
  \Vertex[x=2,y=-2, L={$v_2$}]{v2}
  \Vertex[x=5,y=2, L={$v_3$}]{v3}
  \Vertex[x=5,y=-2, L={$v_4$}]{v4}
  \Vertex[x=7,y=0, L={$t$}]{t}

  % Edges
  \tikzset{EdgeStyle/.style={->}}
  \Edge[label=16](s)(v1)
  \Edge[label=13](s)(v2)
  \Edge[label=12](v1)(v3)
  \Edge[label=9](v2)(v3)
  \Edge[label=14](v2)(v4)
  \Edge[label=7](v3)(v4)
  \Edge[label=20](v3)(t)
  \Edge[label=4](v4)(t)
  \tikzset{EdgeStyle/.append style = {bend left = 15}}
  \Edge[label=4](v1)(v2)
  \Edge[label=10](v2)(v1)
\end{tikzpicture}
|725102014|kitty|/Applications/kitty.app|0|
\usepackage{tkz-graph}
|725102036|kitty|/Applications/kitty.app|0|
\usepackage{tikz}
\usetikzlibrary{calc}
|725102052|kitty|/Applications/kitty.app|0|
\renewenvironment{solutionordottedlines}{}{}|725102113|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Latex[length=3mm, width=2mm]|725102292|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
EdgeStyle/.append style = {->, >=Latex, line width=1pt}|725102360|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\usetikzlibrary{arrows.meta} |725102382|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
    \question[2] What does the sum $1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - ...$ converge to?
|725103415|kitty|/Applications/kitty.app|0|
        \captionof*{figure}{Q4. Sugar Cube}
|725108670|kitty|/Applications/kitty.app|0|
Q4. Sugar Cube|725108714|kitty|/Applications/kitty.app|0|
    \question[3] How many permutations of the rubiks cube exist? Show full working.
|725109013|kitty|/Applications/kitty.app|0|
43,252,003,274,489,856,000|725109108|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Urqh zdv qrw exlow lq d gdb|725109281|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Cipher: Rome was not built in a day.|725109303|kitty|/Applications/kitty.app|0|
    \question[2] Negate the following statement and reexpress it as an equivalent positive one.
    \textsc{Everyone who is majoring in math has a friend who needs help with his or her homework.}
|725111247|kitty|/Applications/kitty.app|0|
There is at least one math major who has no friends needing help with their homework|725111266|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
196,418|725111740|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\begin{tikzpicture}
\begin{axis}[
    axis lines = left,
    xlabel = \( x \),
    ylabel = {\( y \)},
]

% Add the plot of the function
\addplot [
    domain=0:4,
    samples=100,
    color=blue,
    fill=blue, 
    fill opacity=0.2
] {x^2} \closedcycle;

% Highlight the section from 0 to 4
\addplot [
    domain=-1:5, 
    samples=100, 
    color=red,
] {x^2};|725111674|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\usepackage{pgfplots}|725111707|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\renewenvironment{solutionordottedlines}[1][]
  {% begin code
   \def\@tempa{#1}%
   \expandafter\comment
  }
  {% end code
   \expandafter\endcomment
  }
\makeatother|725111815|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\renewenvironment{solutionordottedlines}[1][]{}{}
|725111828|kitty|/Applications/kitty.app|0|
\renewenvironment{solutionorbox}[1][]{}{}
|725111829|kitty|/Applications/kitty.app|0|
\renewenvironment{solutionordottedlines}[1][]
  {% begin code
   \def\@tempa{#1}%
   \expandafter\comment
  }
  {% end code
   \expandafter\endcomment
  }
\makeatother
|725111831|kitty|/Applications/kitty.app|0|
IP address question|725111923|kitty|/Applications/kitty.app|0|
\begin{tikzpicture}
\begin{axis}[
    axis lines = left,
    xlabel = \( x \),
    ylabel = {\( y \)},
]

% Plot the function y = x^2
\addplot [
    domain=-1:5, 
    samples=100, 
    color=red,
] {x^2};

% Highlight the section from 0 to 4 in blue
\addplot [
    domain=0:4,
    samples=100,
    color=blue,
    thick,
] {x^2};

\end{axis}
\end{tikzpicture}|725111961|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
        \begin{tikzpicture}
            \begin{axis}[
                axis lines = left,
                xlabel = \( x \),
                ylabel = {\( y \)},
            ]

            % Add the plot of the function
            \addplot [
                domain=0:4,
                samples=100,
                color=blue,
                fill=blue, 
                fill opacity=0.2
            ] {x^2} \closedcycle;

            % Highlight the section from 0 to 4
            \addplot [
                domain=-1:5, 
                samples=100, 
                color=red,
            ] {x^2};
            \end{axis}
        \end{tikzpicture}
|725111967|kitty|/Applications/kitty.app|0|
    \begin{minipage}{\linewidth}
    \end{minipage}
|725111995|kitty|/Applications/kitty.app|0|
\begin{tikzpicture}
    \begin{axis}[
        axis lines=middle,
        xmin=-0.5, xmax=3.5,
        ymin=-0.5, ymax=1.5,
        xlabel=$x$,
        ylabel=$f(x)$,
        xtick={0, 1, 2, 3},
        ytick={0, 1},
        yticklabels={0, 1},
        clip=false
    ]
    % Drawing the function, but only showing dots at rational points
    \addplot[only marks, mark=*, mark options={scale=0.5}, color=blue] coordinates {
        (0, 1)
        (1, 1)
        (2, 1)
        (3, 1)
        % Add more points as needed to represent rationals
    };
    \node at (axis cs:1.5, 1.3) {Rationals};
    \node at (axis cs:2.5, 0.3) {Irrationals};
    \end{axis}
\end{tikzpicture}|725112875|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\begin{tikzpicture}
    \begin{axis}[
        axis lines=middle,
        width=15cm, height=15cm,     % size of the image
        grid = both,
        grid style={dashed, gray!30},
        enlargelimits=true,
        xmin=-1,     % start the diagram at this x-coordinate
        xmax= 1,     % end   the diagram at this x-coordinate
        ymin= 0,     % start the diagram at this y-coordinate
        ymax= 1,     % end   the diagram at this y-coordinate
        /pgfplots/xtick={-1,-0.8,...,1}, % make steps of length 0.2
        /pgfplots/ytick={0,0.1,...,1},   % make steps of length 0.1
        axis background/.style={fill=white},
        ylabel=y,
        xlabel=x,]
      \addplot[domain=-1:1, ultra thick,samples=100,blue]  {1};
      \label{plot one}
      \addplot[domain=-1:1, ultra thick,samples=100,red] {0};
      \label{plot two}
      \node [draw,fill=white] at (rel axis cs: 0.8,0.8) {\shortstack[l]{
      $f(x) =
    \left\lbrace\begin{array}{@{}l@{}l@{}l@{}}
       \tikz[baseline=-0.5ex]\node{\ref{plot one}}; \phantom{1cm}& 1 &  \text{ if } x \in \mathbb{Q}\\
       \tikz[baseline=-0.5ex]\node{\ref{plot two}}; & 0 & \text{ if } x \in \mathbb{R} \setminus \mathbb{Q}
    \end{array}\right.
$}};
    \end{axis}
\end{tikzpicture}|725112992|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
        \begin{tikzpicture}[scale=0.5]
        \begin{axis}[
            axis lines=middle,
            xmin=-0.5, xmax=3.5,
            ymin=-0.5, ymax=1.5,
            xlabel=$x$,
            ylabel=$f(x)$,
            xtick={0, 1, 2, 3},
            ytick={0, 1},
            yticklabels={0, 1},
            clip=false
        ]
        % Drawing the function, but only showing dots at rational points
        \addplot[only marks, mark=*, mark options={scale=0.5}, color=blue] coordinates {
            (0, 1)
            (1, 1)
            (2, 1)
            (3, 1)
            % Add more points as needed to represent rationals
        };
        \node at (axis cs:1.5, 1.3) {Rationals};
        \node at (axis cs:2.5, 0.3) {Irrationals};
        \end{axis}
    \end{tikzpicture}
|725112995|kitty|/Applications/kitty.app|0|
\usepackage{amssymb,amsmath}|725113007|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\begin{tikzpicture}
    \begin{axis}[
        axis lines=middle,
        xmin=0, xmax=1,
        ymin=0, ymax=2,
        xlabel=$x$,
        ylabel=$f(x)$,
        ytick={1},
        yticklabels={1},
        clip=false,
        small
    ]
    
    % Plot rational points
    \foreach \p in {1,...,100}{
        \pgfmathsetmacro{\q}{int(rand*10)}
        \pgfmathsetmacro{\rational}{\p/(\q+1)}
        \addplot[only marks, mark=*, mark options={scale=0.3}, color=blue] coordinates {(\rational, 1)};
    }
    
    % Plot irrational points
    \foreach \p in {1,...,100}{
        \pgfmathsetmacro{\irrational}{\p + sqrt(2)}
        \addplot[only marks, mark=*, mark options={scale=0.3}, color=red] coordinates {(\irrational, 0)};
    }

    \end{axis}
\end{tikzpicture}|725113065|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\begin{tikzpicture}
    \begin{axis}[
        axis lines=middle,
        width=15cm, height=15cm,     % size of the image
        grid = both,
        grid style={dashed, gray!30},
        enlargelimits=true,
        xmin=-1,     % start the diagram at this x-coordinate
        xmax= 1,     % end   the diagram at this x-coordinate
        ymin= 0,     % start the diagram at this y-coordinate
        ymax= 1,     % end   the diagram at this y-coordinate
        /pgfplots/xtick={-1,-0.8,...,1}, % make steps of length 0.2
        /pgfplots/ytick={0,0.1,...,1},   % make steps of length 0.1
        axis background/.style={fill=white},
        ylabel=y,
        xlabel=x,]
      \addplot[domain=-1:1, ultra thick,samples=100,blue]  {1};
      \label{plot one}
      \addplot[domain=-1:1, ultra thick,samples=100,red] {0};
      \label{plot two}
      \node [draw,fill=white] at (rel axis cs: 0.8,0.8) {\shortstack[l]{
      $f(x) =
    \left\lbrace\begin{array}{@{}l@{}l@{}l@{}}
       \tikz[baseline=-0.5ex]\node{\ref{plot one}}; \phantom{1cm}& 1 &  \text{ if } x \in \mathbb{Q}\\
       \tikz[baseline=-0.5ex]\node{\ref{plot two}}; & 0 & \text{ if } x \in \mathbb{R} \setminus \mathbb{Q}
    \end{array}\right.
$}};
    \end{axis}
\end{tikzpicture}
|725113075|kitty|/Applications/kitty.app|0|
! Package PGF Math Error: You've asked me to divide `2' by `0.0', but I cannot 
divide any number by `0.0' (in '2/(-1+1)').
|725113100|kitty|/Applications/kitty.app|0|
\begin{tikzpicture}
    \begin{axis}[
        axis lines=middle,
        xmin=0, xmax=1,
        ymin=0, ymax=2,
        xlabel=$x$,
        ylabel=$f(x)$,
        ytick={1},
        yticklabels={1},
        clip=false,
        small
    ]
    
    % Plot rational points
    \foreach \p in {1,...,100}{
        \pgfmathsetmacro{\q}{int(rand*10 + 10)}
        \pgfmathsetmacro{\rational}{\p/(\q+1)}
        \addplot[only marks, mark=*, mark options={scale=0.3}, color=blue] coordinates {(\rational, 1)};
    }
    
    % Plot irrational points
    \foreach \p in {1,...,100}{
        \pgfmathsetmacro{\irrational}{mod(\p + sqrt(2), 1)} % keep the value within [0,1]
        \addplot[only marks, mark=*, mark options={scale=0.3}, color=red] coordinates {(\irrational, 0)};
    }

    \end{axis}
\end{tikzpicture}|725113136|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\begin{tikzpicture}
    \begin{axis}[
        axis lines=middle,
        xmin=0, xmax=1,
        ymin=0, ymax=2,
        xlabel=$x$,
        ylabel=$f(x)$,
        ytick={1},
        yticklabels={1},
        clip=false,
        small
    ]
    
    % Plot rational points
    \foreach \p in {1,...,100}{
        \pgfmathsetmacro{\q}{int(rand*10)}
        \pgfmathsetmacro{\rational}{\p/(\q+1)}
        \addplot[only marks, mark=*, mark options={scale=0.3}, color=blue] coordinates {(\rational, 1)};
    }
    
    % Plot irrational points
    \foreach \p in {1,...,100}{
        \pgfmathsetmacro{\irrational}{\p + sqrt(2)}
        \addplot[only marks, mark=*, mark options={scale=0.3}, color=red] coordinates {(\irrational, 0)};
    }

    \end{axis}
\end{tikzpicture}
|725113140|kitty|/Applications/kitty.app|0|
\begin{tikzpicture}
    \begin{axis}[
        axis lines=middle,
        xmin=0, xmax=1,
        ymin=0, ymax=1.5,
        xlabel=$x$,
        ylabel=$f(x)$,
        ytick={0, 1},
        yticklabels={0, 1},
        clip=false,
        small
    ]
    
    % Plot rational points
    \foreach \p in {1,...,100}{
        \pgfmathsetmacro{\q}{int(rand*10 + 10)}
        \pgfmathsetmacro{\rational}{mod(\p/\q, 1)}
        \addplot[only marks, mark=*, mark options={scale=0.3}, color=blue] coordinates {(\rational, 1)};
    }
    
    % Plot irrational points
    \foreach \i in {1,...,100}{
        \pgfmathsetmacro{\irrational}{mod(\i*sqrt(2), 1)}
        \addplot[only marks, mark=*, mark options={scale=0.3}, color=red] coordinates {(\irrational, 0)};
    }

    \end{axis}
\end{tikzpicture}|725113249|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\begin{tikzpicture}
    \begin{axis}[
        axis lines=middle,
        xmin=0, xmax=1,
        ymin=0, ymax=2,
        xlabel=$x$,
        ylabel=$f(x)$,
        ytick={1},
        yticklabels={1},
        clip=false,
        small
    ]
    
    % Plot rational points
    \foreach \p in {1,...,100}{
        \pgfmathsetmacro{\q}{int(rand*10 + 10)}
        \pgfmathsetmacro{\rational}{\p/(\q+1)}
        \addplot[only marks, mark=*, mark options={scale=0.3}, color=blue] coordinates {(\rational, 1)};
    }
    
    % Plot irrational points
    \foreach \p in {1,...,100}{
        \pgfmathsetmacro{\irrational}{mod(\p + sqrt(2), 1)} % keep the value within [0,1]
        \addplot[only marks, mark=*, mark options={scale=0.3}, color=red] coordinates {(\irrational, 0)};
    }

    \end{axis}
\end{tikzpicture}
|725113255|kitty|/Applications/kitty.app|0|
! Package PGF Math Error: You've asked me to divide `20' by `0', but I cannot d
ivide any number by `0' (in 'mod(20/0, 1)').

|725113291|kitty|/Applications/kitty.app|0|
\begin{tikzpicture}
    \begin{axis}[
        axis lines=middle,
        xmin=0, xmax=1,
        ymin=0, ymax=1.5,
        xlabel=$x$,
        ylabel=$f(x)$,
        ytick={0, 1},
        yticklabels={0, 1},
        clip=false,
        small
    ]
    
    % Define a fixed non-zero denominator for rational numbers
    \def\denominator{10}
    
    % Plot rational points
    \foreach \p in {1,...,100}{
        \pgfmathsetmacro{\rational}{mod(\p/\denominator, 1)}
        \addplot[only marks, mark=*, mark options={scale=0.3}, color=blue] coordinates {(\rational, 1)};
    }
    
    % Plot irrational points
    \foreach \i in {1,...,100}{
        \pgfmathsetmacro{\irrational}{mod(\i*sqrt(2), 1)}
        \addplot[only marks, mark=*, mark options={scale=0.3}, color=red] coordinates {(\irrational, 0)};
    }

    \end{axis}
\end{tikzpicture}|725113330|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\begin{tikzpicture}
    \begin{axis}[
        axis lines=middle,
        xmin=0, xmax=1,
        ymin=0, ymax=1.5,
        xlabel=$x$,
        ylabel=$f(x)$,
        ytick={0, 1},
        yticklabels={0, 1},
        clip=false,
        small
    ]
    
    % Plot rational points
    \foreach \p in {1,...,100}{
        \pgfmathsetmacro{\q}{int(rand*10 + 10)}
        \pgfmathsetmacro{\rational}{mod(\p/\q, 1)}
        \addplot[only marks, mark=*, mark options={scale=0.3}, color=blue] coordinates {(\rational, 1)};
    }
    
    % Plot irrational points
    \foreach \i in {1,...,100}{
        \pgfmathsetmacro{\irrational}{mod(\i*sqrt(2), 1)}
        \addplot[only marks, mark=*, mark options={scale=0.3}, color=red] coordinates {(\irrational, 0)};
    }

    \end{axis}
\end{tikzpicture}
|725113334|kitty|/Applications/kitty.app|0|
\pgfplotsset{compat=1.16}|725113397|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
        \begin{tikzpicture}[scale=0.7,transform shape]
|725113442|kitty|/Applications/kitty.app|0|
        \begin{axis}[
            axis lines = left,
            xlabel = \( x \),
            ylabel = {\( y \)},
        ]
|725113591|kitty|/Applications/kitty.app|0|
ylabel style={rotate=90},|725113614|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
[scale=0.7, transform shape]|725113668|kitty|/Applications/kitty.app|0|
textwidth}|725113687|kitty|/Applications/kitty.app|0|
                small
|725113742|kitty|/Applications/kitty.app|0|
                clip=false,
|725113778|kitty|/Applications/kitty.app|0|
    \begin{figure}
|725113808|kitty|/Applications/kitty.app|0|
    \begin{minipage}{\linewidth}
|725113809|kitty|/Applications/kitty.app|0|
    \end{minipage}
|725113811|kitty|/Applications/kitty.app|0|
\DeclareMathOperator{\Dirichlet}{D}

\begin{document}

The Dirichlet function is defined as:
\[
\Dirichlet(x) =
\begin{cases} 
1 & \text{if } x \text{ is rational}, \\
0 & \text{if } x \text{ is irrational}.
\end{cases}
\]|725114020|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\begin{document}
|725114032|kitty|/Applications/kitty.app|0|
The Dirichlet function is defined as:
\[
\Dirichlet(x) =
\begin{cases} 
1 & \text{if } x \text{ is rational}, \\
0 & \text{if } x \text{ is irrational}.
\end{cases}
\]
|725114035|kitty|/Applications/kitty.app|0|
a
l|725114981|kitty|/Applications/kitty.app|0|
\setlength{\rightpointsmargin}{1cm}
|725115064|kitty|/Applications/kitty.app|0|
\begin{tikzpicture}
  % Button background
  \node[draw, rounded corners=8pt, fill=blue!30, inner sep=10pt] (button) {\textbf{Submit}};

  % Text label

  % Define the URL
  \def\myurl{https://abaj.io/bday/problems/upload}

  % Add a link
    \node[anchor=center] (link) at (button.center) {\hyperlink{\myurl}{\phantom{Submit}}};

\end{tikzpicture}
|725115532|kitty|/Applications/kitty.app|0|
\renewenvironment{solutionordottedlines}[1][]
  {% begin code
   \def\@tempa{#1}%
   \expandafter\comment
  }
  {% end code
   \expandafter\endcomment
  }
\makeatother
\renewenvironment{solutionorbox}[1][]
  {% begin code
   \def\@tempa{#1}%
   \expandafter\comment
  }
  {% end code
   \expandafter\endcomment
  }
\makeatother
|725115709|kitty|/Applications/kitty.app|0|
curl -fsSL https://raw.githubusercontent.com/filebrowser/get/master/get.sh | bash
filebrowser -r /path/to/your/files|725116587|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
    listen 80;
    server_name abaj.io;

    location /bday/problems/upload {
        proxy_pass http://localhost:8080;
        proxy_set_header Host $host;
        proxy_set_header X-Real-IP $remote_addr;
        proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for;
        proxy_set_header X-Forwarded-Proto $scheme;
    }|725119604|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
---
title: "File Upload"
---

<form action="/upload" method="post" enctype="multipart/form-data">
    <input type="file" name="file" id="file">
    <input type="submit" value="Upload File" name="submit">
</form>|725117773|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
---
title: "File Upload"
---

<form action="/upload" method="post" enctype="multipart/form-data">
    <input type="file" name="file" id="file">
    <input type="submit" value="Upload File" name="submit">
</form>
|725118296|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Some introductory text here.

<form action="https://yourbackend.com/upload" method="post" enctype="multipart/form-data">
    <label for="fileUpload">Choose file:</label>
    <input type="file" id="fileUpload" name="fileUpload">
    <button type="submit">Upload</button>
</form>

More content here.|725118757|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
npm install express multer
|725118893|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
const express = require('express');
const multer = require('multer');
const app = express();

const upload = multer({ dest: 'uploads/' });

app.post('/upload', upload.single('fileUpload'), (req, res) => {
    console.log(req.file);
    res.send('File uploaded successfully.');
});

app.listen(3000, () => {
    console.log('Server started on port 3000');
});|725118908|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
[markup]
  [markup.goldmark]
    [markup.goldmark.renderer]
      unsafe = true|725119184|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
https://abaj.io/bday/problems/upload/|725119273|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
const express = require('express');
const multer = require('multer');
const app = express();

const storage = multer.diskStorage({
    destination: function (req, file, cb) {
        cb(null, 'uploads/');
    },
    filename: function (req, file, cb) {
        cb(null, file.originalname); // Use the original file name
    }
});

const upload = multer({ storage: storage });

app.post('/upload', upload.single('fileUpload'), (req, res) => {
    console.log(req.file);
    res.send('File uploaded successfully.');
});

app.listen(3000, () => {
    console.log('Server started on port 3000');
});
|725120059|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Image: 535x464 (973.9 KB)|725145337|Brave Browser Beta|/Applications/Brave Browser Beta.app|1|dd1519a73b398375aa696c0e3976cefa3df81e83.tiff
\begin{document}
\begin{tikzpicture}
  % Define the triangle
  \draw[fill=pink!50] (0,0) -- (4,0) -- (2,3.5) -- cycle;
  
  % Add labels
  \node at (2,1.2) {Speed};
  \node at (2,2.3) {Distance};
  \node at (2,0.4) {Time};
  
  % Add the lines
  \draw (2,0.85) -- (2,1.55);
\end{tikzpicture}|725145383|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\documentclass{article}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
    % Triangle
    \draw[thick] (0,0) -- (5,0) -- (2.5,4.33) -- cycle;
    
    % Labels for corners
    \node at (2.5,4.7) {Distance};
    \node at (2.5,-0.5) {Speed};
    \node at (5.5,2) {Time};
    
    % Lines to split the triangle
    \draw[thick] (2.5,1.44) -- (2.5,0);
    \draw[thick] (1.25,0.72) -- (3.75,0.72);
    
    % Color the triangle
    \fill[pink!40] (0,0) -- (5,0) -- (2.5,4.33) -- cycle;
\end{tikzpicture}
\end{document}
|725145428|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\documentclass[tikz,border=10pt]{standalone}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
  % Define the triangle
  \draw[fill=pink!50] (0,0) -- (4,0) -- (2,3.5) -- cycle;
  
  % Add labels
  \node at (2,1.2) {Speed};
  \node at (2,2.3) {Distance};
  \node at (2,0.4) {Time};
  
  % Add the lines
  \draw (2,0.85) -- (2,1.55);
\end{tikzpicture}
\end{document}
|725145431|kitty|/Applications/kitty.app|0|
    \fill[pink!40] (0,0) -- (5,0) -- (2.5,4.33) -- cycle;
|725145446|kitty|/Applications/kitty.app|0|

\documentclass{article}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
    % Triangle
    \draw[thick] (0,0) -- (5,0) -- (2.5,4.33) -- cycle;
    
    % Labels for corners
    \node at (2.5,4.7) {Distance};
    \node at (2.5,-0.5) {Speed};
    \node at (5.5,2) {Time};
    
    % Lines to split the triangle
    \draw[thick] (2.5,1.44) -- (2.5,0);
    \draw[thick] (1.25,0.72) -- (3.75,0.72);
    
    % Color the triangle
\end{tikzpicture}
\end{document}
|725145485|kitty|/Applications/kitty.app|0|
\documentclass[tikz,border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{arrows.meta}

\begin{document}
\begin{tikzpicture}
  % Draw the number line
  \draw[latex-latex] (-3,0) -- (3,0); % The line
  \foreach \x in  {-2,...,2} % The ticks and their labels
    \draw (\x,0.1) -- (\x,-0.1) node[below] {\x};

  % Highlight the inequality x < 2
  \draw[very thick, blue, -{Stealth[scale=1.5]}] 
    (-2.8,0) -- (1.9,0); % The arrow on the line
  \filldraw[fill=white, draw=blue, very thick] (2,0) circle (2pt); % Open circle at x = 2
\end{tikzpicture}
\end{document}
|725158958|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\documentclass[tikz,border=10pt]{standalone}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}[scale=1, every node/.style={scale=1}]
  % Define nodes
  \node (A) at (0,0) {};
  \node (B) at (4,0) {};
  \node (C) at (2,3.464) {};

  % Draw triangle
  \filldraw[fill=pink!50] (A) -- (B) -- (C) -- cycle;

  % Draw dividing lines
  \draw (A) -- node[below] {Speed} (B);
  \draw (A) -- node[above left] {Time} (C);
  \draw (B) -- node[above right] {Distance} (C);

  % Label vertices (optional, not present in the original image)
  % \node at (A) [left] {A};
  % \node at (B) [right] {B};
  % \node at (C) [above] {C};
\end{tikzpicture}
\end{document}
|725158967|kitty|/Applications/kitty.app|0|
(-2.8,0)|725159166|kitty|/Applications/kitty.app|0|
-- |725159174|kitty|/Applications/kitty.app|0|
@students.mq.edu.au|725274306|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Greetings academics,

Today is the 26th of December, 2023, and it is my 22nd birthday. In celebration of life after yet another lap of the Sun, I have compiled 22 interesting problems that I have encountered this year.

The nature of these problems (at least presently) are largely mathematical and occasionally of a computer science flavour.

There is prize money for the recipients of 1st and 2nd place: \$100 and \$50 respectively.

The rules of the game are enclosed within the PDF. 

Last years Problem Set can be found [here](abaj.io/bday/problems/21.pdf), and the list of previous winners [here](abaj.io/bday/winners). Finally, if you'd prefer to print and have dotted lines to scribble on, that PDF can be found here: abaj.io/bday/problems/22-lined.pdf.

頑張って.



|725275523|kitty|/Applications/kitty.app|0|
頑張ってね|725274423|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
頑張って|725274426|kitty|/Applications/kitty.app|0|
Sent|725274543|kitty|/Applications/kitty.app|0|
 awoo5332@gmail.com, aaravbajaj08@gmail.com, kiiroi.kirin1@gmail.com, sudarakapieris@gmail.com,
INBOX      130/ 173│        astertheking5@gmail.com, erick.rajan@gmail.com, usij.gnos@gmail.com, masonwong1999@gmail.com,
                   │        sarro.aghrel@students.mq.edu.au, prannavindra@gmail.com
|725275305|kitty|/Applications/kitty.app|0|
To: awoo5332@gmail.com, aaravbajaj08@gmail.com, kiiroi.kirin1@gmail.com, sudarakapieris@gmail.com,
        astertheking5@gmail.com, erick.rajan@gmail.com, usij.gnos@gmail.com, masonwong1999@gmail.com,
        sarro.aghrel@students.mq.edu.au, prannavindra@gmail.com
|725274848|kitty|/Applications/kitty.app|0|
🙄|725343847|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Windows_FAT_32 ⁨record⁩                  31.7 GB    disk4s1

|725511521|kitty|/Applications/kitty.app|0|
 disk4s1
|725511534|kitty|/Applications/kitty.app|0|
Windows_FAT_32 ⁨|725511545|kitty|/Applications/kitty.app|0|
    Windows_FAT_32 ⁨rec|725511548|kitty|/Applications/kitty.app|0|
i ran dd on my microsd card but it couldn't write over 2gb|725523973|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
i ran dd on my microsd card but it couldn't write over 2gb. i think the sd card is corrupted, the disk identifier is 0x00000000 and formatting it doesn't remove the files. neither does rewriting the partition table with fdisk.|725524007|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
b1b8b499-26bf-41a8-9d2e-acfd00413f09|725529448|kitty|/Applications/kitty.app|0|
Q7P7NKy#b9*wKgsxMxJ2qhXR|725529457|kitty|/Applications/kitty.app|0|
cfa35ee1-c666-45cd-b5ab-ac1c0122e920|725529466|kitty|/Applications/kitty.app|0|
Wp2&rHSO4Rr0T#Om|725529476|kitty|/Applications/kitty.app|0|
eNrCU6AhWA99RCHF|725596722|kitty|/Applications/kitty.app|0|
https://my.gov.au/en/myaccount/dashboard|725531534|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
https://my.gov.au/auth-callback|725531530|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
QQ239939|725596767|Safari|/Applications/Safari.app|0|
2848937761|725596943|kitty|/Applications/kitty.app|0|
 209 020 525A|725597082|Safari|/Applications/Safari.app|0|
ECBrxt9c59SwTR4n|725599114|kitty|/Applications/kitty.app|0|
How to Read and Why by Harold Bloom|725617630|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
好い子|725618069|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
い|725618035|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
client_max_body_size 100M;
|725709612|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
%|725766511|kitty|/Applications/kitty.app|0|
What do we need to be careful of?
\mdots{2}
|725766537|kitty|/Applications/kitty.app|0|
\definecolor{Blue}{rgb}{0.18, 0.19, 0.58}
|725767081|kitty|/Applications/kitty.app|0|
  \rput[l](1cm,0.5ex){  % Place text
    %\textcolor{Blue}{\parbox[m]{0.66cm}{\includegraphics[width=0.66cm]{./include/books8_Blue.eps}} \textbf{\defboxtext}}%
    \color{Blue} \textbf{\defboxtext}
  }
  \psline[linewidth=3pt,linecolor=Blue](0.22,0)(0.22,-\defboxheight)
|725767117|kitty|/Applications/kitty.app|0|
  \rput[l](1cm,0.5ex){  % Place text
    %\textcolor{Blue}{\parbox[m]{0.66cm}{\includegraphics[width=0.66cm]{./include/books8_Blue.eps}} \textbf{\defboxtext}}%
    \color{blue} \textbf{\defboxtext}
  }
  \psline[linewidth=3pt,linecolor=blue](0.22,0)(0.22,-\defboxheight)
|725767124|kitty|/Applications/kitty.app|0|
\section{Homework}
\input{9-homework}
|725769921|kitty|/Applications/kitty.app|0|
    \setcounter{sec8marks}{\thesecmarks}
|725769928|kitty|/Applications/kitty.app|0|
    & $\dfrac{}{\arabic{sec8marks}}$
|725770039|kitty|/Applications/kitty.app|0|
\usepackage[top=20mm,bottom=20mm,right=20mm,left=20mm]{geometry}
\usepackage{pgffor}
\usepackage{setspace}
\usepackage{mathtools}
\usepackage{enumerate}
\usepackage{multicol}
\usepackage{cancel}
\usepackage{fontspec}
|725770080|kitty|/Applications/kitty.app|0|
\setcounter{tocdepth}{1}
\setlength\parindent{0pt}
|725770145|kitty|/Applications/kitty.app|0|
\marksnotpoints
\pointsinrightmargin
\boxedpoints
\setlength{\rightpointsmargin}{1cm}


\newcommand{\randomword}[1]{%
  \pgfmathsetmacro{\angle}{random(-45,45)}% Generate a random angle between -45 and 45
  \rotatebox{\angle}{#1}%
}
|725770172|kitty|/Applications/kitty.app|0|
\usepackage{background}
|725770219|kitty|/Applications/kitty.app|0|
\section{The End}
\hspace{4.5cm}
\input{10-outro}
|725770274|kitty|/Applications/kitty.app|0|
\mdots{3}
|725770362|kitty|/Applications/kitty.app|0|
\mdots{4}
|725770363|kitty|/Applications/kitty.app|0|
    \begin{qe}
|725770409|kitty|/Applications/kitty.app|0|
    \begin{solutionordottedlines}[1cm]
    \end{solutio}
|725770471|kitty|/Applications/kitty.app|0|
\begin{questions}
|725770477|kitty|/Applications/kitty.app|0|
cfill|725770533|kitty|/Applications/kitty.app|0|
fill|725770539|kitty|/Applications/kitty.app|0|
 \mdots{1}|725770624|kitty|/Applications/kitty.app|0|
            \begin{parts}
|725770663|kitty|/Applications/kitty.app|0|
            \question[5]
|725770668|kitty|/Applications/kitty.app|0|
p1art|725770676|kitty|/Applications/kitty.app|0|
            \end{parts}
|725770697|kitty|/Applications/kitty.app|0|
    \end{enumerate}
|725770713|kitty|/Applications/kitty.app|0|
Which of the following are pairs of like terms?
|725770857|kitty|/Applications/kitty.app|0|
        \begin{enumerate}[(a)]

item|725771033|kitty|/Applications/kitty.app|0|
                        \begin{solutionordottedlines}[1in]
                        \end{solutionordottedlines}
|725771194|kitty|/Applications/kitty.app|0|
                        \mdots{3}
|725771197|kitty|/Applications/kitty.app|0|
            \newpage
|725771228|kitty|/Applications/kitty.app|0|
                        \begin{solutionordottedlines}[1cm]
                        \end{solutionordottedlines}
|725771270|kitty|/Applications/kitty.app|0|
                        \begin{solutionordottedlines}[2cm]
                        \end{solutionordottedlines}
|725771278|kitty|/Applications/kitty.app|0|
                        \mdots{2}
|725771288|kitty|/Applications/kitty.app|0|
\fillin[something][0.5\textwidth]|725771317|kitty|/Applications/kitty.app|0|
 \fillin[something][0.5\textwidth]|725771339|kitty|/Applications/kitty.app|0|
enumerate
\dotfilL
\dotfilL
\dotfilL|725771432|kitty|/Applications/kitty.app|0|
multicols
\dotfilL
\dotfilL
\dotfilL
\dotfilL|725771511|kitty|/Applications/kitty.app|0|
item
\dotfilL
\dotfilL
\dotfilL
\dotfilL
\dotfilL
\dotfilL|725771545|kitty|/Applications/kitty.app|0|
            \end{enumerate}
|725771579|kitty|/Applications/kitty.app|0|
enumerate
item
item
item
\dotfilL
\dotfilL
\dotfilL|725771683|kitty|/Applications/kitty.app|0|
    \begin{enumerate}
|725771712|kitty|/Applications/kitty.app|0|
        \begin{multicols}{2}

item
item
item
item
item
item|725771752|kitty|/Applications/kitty.app|0|
enumerate
{2}|725771820|kitty|/Applications/kitty.app|0|
    \begin{enumerate}[(a)]
|725771862|kitty|/Applications/kitty.app|0|
[(a)]
item
item
item
item
item
item
item
item
item|725771886|kitty|/Applications/kitty.app|0|
    \end{multicols}

\mdots{2}|725771906|kitty|/Applications/kitty.app|0|
)\mdots{2}|725771910|kitty|/Applications/kitty.app|0|
enumerate
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item
item|725772113|kitty|/Applications/kitty.app|0|
sub|725772286|kitty|/Applications/kitty.app|0|
Please attempt every question in your exercise books!
|725772423|kitty|/Applications/kitty.app|0|
DSCN9437.JPG|725792066|Finder|/System/Library/CoreServices/Finder.app|0|
uYEwGumpf#8|=!|725792416|kitty|/Applications/kitty.app|0|
Image: 622x121 (1.2 MB)|725793009|Brave Browser Beta|/Applications/Brave Browser Beta.app|1|578d9301d5a51705cb434e5fe01f1e4f4e842703.tiff
\usetikzlibrary{decorations.text, shadows}
\newfontfamily\looney[]{That's Font Folks!}
\definecolor{darkblueOuter}{RGB}{1,11,23}
\definecolor{darkblueInner}{RGB}{1,18,37}
|725887201|kitty|/Applications/kitty.app|0|
% that's all folks
|725887202|kitty|/Applications/kitty.app|0|
%
|725887202|kitty|/Applications/kitty.app|0|
\fancyhf{}
|725887204|kitty|/Applications/kitty.app|0|
\usepackage{fancyhdr}
|725887206|kitty|/Applications/kitty.app|0|
article|725887211|kitty|/Applications/kitty.app|0|
\setcounter{tocdepth}{1}
|725887214|kitty|/Applications/kitty.app|0|
\thispagestyle{fancy}
|725887225|kitty|/Applications/kitty.app|0|
\section{The End}
\hspace{4.5cm}
|725887232|kitty|/Applications/kitty.app|0|
    \input{8-outro}
|725887233|kitty|/Applications/kitty.app|0|
\backgroundsetup{placement = top, hshift=0cm, vshift=-\textwidth*0.70, scale = 0.75, contents ={\ifnum\value{page}=1 \includegraphics[width=\textwidth]{img/logo-cover.png}\else \fi}}
|725887264|kitty|/Applications/kitty.app|0|
    & $\dfrac{}{\arabic{sec6marks}}$
|725887377|kitty|/Applications/kitty.app|0|
    & $\$
|725887380|kitty|/Applications/kitty.app|0|
\usepackage{pgffor}
|725887404|kitty|/Applications/kitty.app|0|
\usepackage{setspace}
\usepackage{mathtools}
\usepackage{enumerate}
\usepackage{multicol}
\usepackage{cancel}
\usepackage{fontspec}
|725887408|kitty|/Applications/kitty.app|0|
\usepackage[top=15mm,bottom=20mm,right=20mm,left=20mm]{geometry}
|725887415|kitty|/Applications/kitty.app|0|
 $ \coloneqq $ \dotfill|725887569|kitty|/Applications/kitty.app|0|
    \begi}
|725887571|kitty|/Applications/kitty.app|0|
    \begin{solutionordottedlines}[2cm]
    \end{solutionordottedlines}
|725887573|kitty|/Applications/kitty.app|0|
mdots{3}|725887642|kitty|/Applications/kitty.app|0|
\,|725887655|kitty|/Applications/kitty.app|0|
item\,|725887667|kitty|/Applications/kitty.app|0|
\begin{solutionordottedlines}[1in]
\end{solutionordottedlines}
|725887694|kitty|/Applications/kitty.app|0|
\mdots{5}|725887762|kitty|/Applications/kitty.app|0|
item \,|725887863|kitty|/Applications/kitty.app|0|
] \,|725887869|kitty|/Applications/kitty.app|0|
 \,|725887877|kitty|/Applications/kitty.app|0|
                    \begin{solutionordottedlines}[1in]
                    \end{solutionordottedlines}
|725887904|kitty|/Applications/kitty.app|0|
                    \mdots{3}
|725887906|kitty|/Applications/kitty.app|0|
\fillin[][\textwidth]|725888137|kitty|/Applications/kitty.app|0|
enumerate
textwidth
textwidth|725888172|kitty|/Applications/kitty.app|0|
-1cm|725888196|kitty|/Applications/kitty.app|0|
\dimexpr\linewidth-2cm\relax|725888379|kitty|/Applications/kitty.app|0|
\usepackage{linegoal}|725888654|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
linefill|725888661|kitty|/Applications/kitty.app|0|
\l1ininewidth|725888783|kitty|/Applications/kitty.app|0|
\linewidth|725888788|kitty|/Applications/kitty.app|0|
1in|725888812|kitty|/Applications/kitty.app|0|
                \begin{solutionordottedlines}[1in]
                \end{solutionordottedlines}
|725888846|kitty|/Applications/kitty.app|0|
            \mdots{4}
|725888866|kitty|/Applications/kitty.app|0|
            \begin{solutionordottedlines}[1.5in]
            \end{solutionordottedlines}
|725888873|kitty|/Applications/kitty.app|0|
            \mdots{5}
|725888876|kitty|/Applications/kitty.app|0|
\dotfill|725889151|kitty|/Applications/kitty.app|0|
\newlength{\mylen}
\uselengthunit{mm}

\newsavebox{\mybox}
\newcommand{\measureddotfill}{
  \savebox{\mybox}{\dotfill}
  \setlength{\mylen}{\wd\mybox}
  \usebox{\mybox}
  \printlength{\mylen}
}|725889389|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\newlength{\mylen}
\uselengthunit{mm}

\newsavebox{\mybox}
\newcommand{\measureddotfill}{
  \savebox{\mybox}{\dotfill}
  \setlength{\mylen}{\wd\mybox}
  \usebox{\mybox}
  \printlength{\mylen}
}
|725889649|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\usepackage{printlen}|725889738|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\measureddotfill{}|725889840|kitty|/Applications/kitty.app|0|
\newsavebox{\mybox}
\newcommand{\measureddotfill}{
  \savebox{\mybox}{\dotfill}
  \setlength{\mylen}{\wd\mybox}
  \usebox{\mybox}
  \printlength{\mylen}
}


\begin{examplebox}
    \section*{Examples}
    Try to simplify the following:
    \begin{multicols}{2}
        \begin{questions}
            \question[1] $\sqrt{18}=\fillin[][\dimexpr\mylen]$\measureddotfill{}
            \question[1] $\sqrt{108}=\fillin[][1in]$
            \question[1] $\sqrt{45}=\fillin[][1in]$
            \question[1] $\sqrt{32}=\fillin[][1in]$
        \end{questions}
    \end{multicols}
\end{examplebox}
|725889874|kitty|/Applications/kitty.app|0|
\newlength{\mylen}
\uselengthunit{mm}

\newsavebox{\mybox}
\newcommand{\measureddotfill}{
  \savebox{\mybox}{\dotfill}
  \setlength{\mylen}{\wd\mybox}
  \usebox{\mybox}
  \printlength{\mylen}
}


\begin{examplebox}
    \section*{Examples}
    Try to simplify the following:
    \begin{multicols}{2}
        \begin{questions}
            \question[1] $\sqrt{18}=\fillin[][\dimexpr\mylen]$\measureddotfill{}
|725889880|kitty|/Applications/kitty.app|0|
\newlength{\mylen}
\uselengthunit{mm}

\newsavebox{\mybox}
\newcommand{\measureddotfill}{
  \savebox{\mybox}{\dotfill}
  \setlength{\mylen}{\wd\mybox}
  \usebox{\mybox}
  \printlength{\mylen}
}


\begin{examplebox}
    \section*{Examples}
    Try to simplify the following:
    \begin{multicols}{2}
        \begin{questions}
            \question[1] $\sqrt{18}=\measureddotfill{}\fillin[][\dimexpr\mylen]$
|725889888|kitty|/Applications/kitty.app|0|
\newlength{\mylen}
\uselengthunit{mm}

\newsavebox{\mybox}
\newcommand{\measureddotfill}{
  \savebox{\mybox}{\dotfill}
  \setlength{\mylen}{\wd\mybox}
  \usebox{\mybox}
  \printlength{\mylen}
}

|725890022|kitty|/Applications/kitty.app|0|
\fillin[][1in]|725890039|kitty|/Applications/kitty.app|0|
o
l|725890045|kitty|/Applications/kitty.app|0|
y|725890048|kitty|/Applications/kitty.app|0|
\fbox{\parbox{\hfill}{text}}|725890215|kitty|/Applications/kitty.app|0|
hfill|725890241|kitty|/Applications/kitty.app|0|
\fbox{\parbox{\hfill}{}\hspace*{\fill}}|725890288|kitty|/Applications/kitty.app|0|
question[1]
item
item|725890430|kitty|/Applications/kitty.app|0|
it1em|725890431|kitty|/Applications/kitty.app|0|
            \begin{}
|725890437|kitty|/Applications/kitty.app|0|
            \question[1] $4\sqrt{7} + 5\sqrt{7}=$
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \question[1] $6\sqrt{7}-2\sqrt{7}=$
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \question[1] $2\sqrt{2}+5\sqrt{2}-3\sqrt{2}=$
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \question[1] $3\sqrt{5}+2\sqrt{7}-\sqrt{5}+4\sqrt{7}=$
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \question[1] $\sqrt{27}+2\sqrt{5}+\sqrt{20}-2\sqrt{3}=$
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \question[1] $\frac{\sqrt{3}}{2}+3\sqrt{3}=$
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \question[1] $\frac{\sqrt{8}}{3}-\frac{\sqrt{2}}{5}=$
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \question[1] $6\sqrt{2} + 4\sqrt{2}=$
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \question[1] $10\sqrt{7}-5\sqrt{7}=$
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \question[1] $3\sqrt{11}-5\sqrt{11}=$
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \question[1] $2\sqrt{2}+3\sqrt{2}+\sqrt{2}=$
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \question[1] $10\sqrt{5}-\sqrt{5}-6\sqrt{5}=$
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \question[1] $\sqrt{7}-8\sqrt{7}+5\sqrt{7}=$
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
|725890913|kitty|/Applications/kitty.app|0|
s/\$\$\([^$]*\)\$\$/\\[\1\\]/g|725891212|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\usepackage{linegoal}
\usepackage{printlen}
|725891139|kitty|/Applications/kitty.app|0|
            \question[1] \[\frac{\sqrt{8}}{3}-\frac{\sqrt{2}}{5}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
|725891497|kitty|/Applications/kitty.app|0|
question[1|725892056|kitty|/Applications/kitty.app|0|
        \end{multicols}\end{questions}
        \begin{questions}\begin{multicols}{2}
|725892075|kitty|/Applications/kitty.app|0|
            \part \[\frac{\sqrt{8}}{3}-\frac{\sqrt{2}}{5}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
|725892115|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \section*{Exercises}
        \begin{questions}
            \question[8]
            \begin{parts}\begin{multicols}{2}
            \part \[4\sqrt{7} + 5\sqrt{7}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[6\sqrt{7}-2\sqrt{7}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[2\sqrt{2}+5\sqrt{2}-3\sqrt{2}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[3\sqrt{5}+2\sqrt{7}-\sqrt{5}+4\sqrt{7}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\sqrt{27}+2\sqrt{5}+\sqrt{20}-2\sqrt{3}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{\sqrt{3}}{2}+3\sqrt{3}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[6\sqrt{2} + 4\sqrt{2}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[10\sqrt{7}-5\sqrt{7}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[3\sqrt{11}-5\sqrt{11}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[2\sqrt{2}+3\sqrt{2}+\sqrt{2}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[10\sqrt{5}-\sqrt{5}-6\sqrt{5}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\sqrt{7}-8\sqrt{7}+5\sqrt{7}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \end{questions}
\end{exercisebox}
|725892206|kitty|/Applications/kitty.app|0|
i1tem|725892366|kitty|/Applications/kitty.app|0|
1]|725892369|kitty|/Applications/kitty.app|0|
question[1]|725892371|kitty|/Applications/kitty.app|0|
item
item
item
item
item
item
|725892385|kitty|/Applications/kitty.app|0|
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
|725892395|kitty|/Applications/kitty.app|0|
\mdots{2|725892411|kitty|/Applications/kitty.app|0|
\mdots{2}|725892442|kitty|/Applications/kitty.app|0|
i
l|725892449|kitty|/Applications/kitty.app|0|
Please attempt every question in your exercise books!

|725892554|kitty|/Applications/kitty.app|0|
qusetion|725892598|kitty|/Applications/kitty.app|0|
part
item
item
item
item
item
item
item
item|725892700|kitty|/Applications/kitty.app|0|
        \part $\sqrt{2}\times\sqrt{6}$
|725892725|kitty|/Applications/kitty.app|0|
\end{questions}
|725892745|kitty|/Applications/kitty.app|0|
\begin{enumerate}[1)]

item
item|725892761|kitty|/Applications/kitty.app|0|
[1)]|725892795|kitty|/Applications/kitty.app|0|
            \item $\sqrt{7}-8\sqrt{7}+5\sqrt{7}$

item
item
item
item
item
item
item
item
item
item|725892824|kitty|/Applications/kitty.app|0|
        \item $6\sqrt{3}\div2\sqrt{3}$

item
item
item
item
item|725892856|kitty|/Applications/kitty.app|0|
\end{enumerate}
\begin{enumerate}[1)]

item
item
item
item
item
item
item
item
item
item|725892886|kitty|/Applications/kitty.app|0|
\end{enumerate}
\begin{enumerate}[1)]
|725892890|kitty|/Applications/kitty.app|0|
enumerate
question|725892994|kitty|/Applications/kitty.app|0|
[1]
[1]
[1]
|725893000|kitty|/Applications/kitty.app|0|
        \end{parts}
|725893173|kitty|/Applications/kitty.app|0|
[1]
[1]
|725893226|kitty|/Applications/kitty.app|0|
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
|725893337|kitty|/Applications/kitty.app|0|
[|725893592|kitty|/Applications/kitty.app|0|
s/\%V.\{-}/\=ROT13(submatch(0))/g|726054093|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
:'<,'>s/\%V.\{-}/\=ROT13(submatch(0))/g|726053327|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
<details>
  <summary>Click to expand!</summary>
  <p>This is the content that was hidden and is now visible!</p>
</details>|726053504|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
function! ROT13(str)
  let result = ''
  for i in range(len(a:str))
    let c = a:str[i]
    if c >= 'a' && c <= 'z'
      " Shift lowercase letter by 13
      let result .= nr2char(((char2nr(c) - char2nr('a') + 13) % 26) + char2nr('a'))
    elseif c >= 'A' && c <= 'Z'
      " Shift uppercase letter by 13
      let result .= nr2char(((char2nr(c) - char2nr('A') + 13) % 26) + char2nr('A'))
    else
      " Non-alphabetic characters remain unchanged
      let result .= c
    endif
  endfor
  return result
endfunction
|726053814|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
:set encoding=utf-8
|726054031|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
:s/.*/\=ROT13(submatch(0))/|726054253|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\usepackage[top=15mm,bottom=20mm,right=20mm,left=20mm]{geometry}
\usepackage{fancyhdr}
\usepackage{pgffor}
\usepackage{setspace}
\usepackage{mathtools}
\usepackage{enumerate}
\usepackage{multicol}
\usepackage{cancel}
\usepackage{fontspec}
\usepackage{background}
\usepackage{hyperref}
\usepackage{xcolor}
\usepackage{wasysym}


\usepackage{rotating}
|726061569|kitty|/Applications/kitty.app|0|
\usetikzlibrary{decorations.text, shadows}
\newfontfamily\looney[]{That's Font Folks!}
\definecolor{darkblueOuter}{RGB}{1,11,23}
\definecolor{darkblueInner}{RGB}{1,18,37}
%
\fancypagestyle{pen}{
    \fancyhf{} % clear all header and footer fields
    \fancyfoot[CE,CO]{\thepage}
    \fancyfoot[RO]{\small \copyright{} PEN Education 2023}
  %\renewcommand{\headrulewidth}{0pt} % remove the header rule
  %\renewcommand{\footrulewidth}{0pt} % remove the footer rule
}
|726061591|kitty|/Applications/kitty.app|0|

\pagestyle{pen}
|726061593|kitty|/Applications/kitty.app|0|

\setcounter{tocdepth}{1}
|726061595|kitty|/Applications/kitty.app|0|

\setlength\parindent{0pt}
|726061597|kitty|/Applications/kitty.app|0|
\today|726061599|kitty|/Applications/kitty.app|0|
    \setcounter{}
|726061623|kitty|/Applications/kitty.app|0|
\pagestyle{pen}
|726061633|kitty|/Applications/kitty.app|0|
    \setcounter{sec1marks}{\thesecmarks}
|726061636|kitty|/Applications/kitty.app|0|
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    & 1
    & 2
    & 3
    & 4
    & 5
    & 6
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    & $\dfrac{}{\arabic{sec1marks}}$ 
    & $\dfrac{}{\arabic{sec2marks}}$  
    & $\dfrac{}{\arabic{sec3marks}}$ 
    & $\dfrac{}{\arabic{sec4marks}}$ 
    & $\dfrac{}{\arabic{sec5marks}}$ 
    & $\dfrac{}{\arabic{sec6marks}}$
    & $\dfrac{}{\arabic{sec7marks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
|726642653|kitty|/Applications/kitty.app|0|
\foreach \i in {1,2,...,10} {
  \addtocounter{totalmarks}{\value{sec\i marks}}
}
|726061666|kitty|/Applications/kitty.app|0|
    & $\dfrac{}{\arabic{sec7marks}}$
|726061681|kitty|/Applications/kitty.app|0|
|>{\centering}X|726061691|kitty|/Applications/kitty.app|0|

\fancyfoot[L]{
    \begin{turn}{180}$\frac{9}{11}$\end{turn}
}
|726062484|kitty|/Applications/kitty.app|0|

\newpage
\fancyfoot{}
|726062502|kitty|/Applications/kitty.app|0|
Answer: bottom left|726062651|kitty|/Applications/kitty.app|0|
Warm-up|726062797|kitty|/Applications/kitty.app|0|
{2}|726062831|kitty|/Applications/kitty.app|0|
                    \begin{solutionor}
|726062857|kitty|/Applications/kitty.app|0|
=\dotfill|726063075|kitty|/Applications/kitty.app|0|
        \begin{doublespace}
|726063139|kitty|/Applications/kitty.app|0|

            \mdots{2}
|726063149|kitty|/Applications/kitty.app|0|
:%s/\\(\(.\{-}\)\\)/\\[\1\\]/g|726063265|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
questions|726063354|kitty|/Applications/kitty.app|0|
    \begin{multicols}{2}

item|726063391|kitty|/Applications/kitty.app|0|
        \item[] Express the following with a \emph{rational} denominator
|726063540|kitty|/Applications/kitty.app|0|
            \mdots{2}
|726063586|kitty|/Applications/kitty.app|0|
            mdots{2}
|726063597|kitty|/Applications/kitty.app|0|
                \begin{multicols}{2}

item
item|726063635|kitty|/Applications/kitty.app|0|
            \end{pa}
|726063639|kitty|/Applications/kitty.app|0|
                    \begin{solutionordottedlines}[4cm]
                    \end{solutionordottedlines}
|726063703|kitty|/Applications/kitty.app|0|
                \end{multicols}
|726063721|kitty|/Applications/kitty.app|0|
                \mdots{2}
|726063794|kitty|/Applications/kitty.app|0|
supart|726063843|kitty|/Applications/kitty.app|0|
    \end{parts}
|726063850|kitty|/Applications/kitty.app|0|
\
|726064276|kitty|/Applications/kitty.app|0|
\mdots{5}
|726064277|kitty|/Applications/kitty.app|0|
\begin{solutionordottedlines}[2in]
\end{solutionordottedlines}
|726064293|kitty|/Applications/kitty.app|0|
    \mdots{4}
|726064295|kitty|/Applications/kitty.app|0|
[(a)]|726064318|kitty|/Applications/kitty.app|0|
isubparttem|726064330|kitty|/Applications/kitty.app|0|
    \begin{solutionordottedlines}[2in]
    \end{solutionordottedlines}
|726064376|kitty|/Applications/kitty.app|0|
                            \begin{solutionordottedlines}[2cm]
                            \end{solutionordottedlines}
|726064418|kitty|/Applications/kitty.app|0|
            The rectangular prism in the diagram has a length of \(12 \mathrm{~cm}\), a width of \(5 \mathrm{~cm}\) and a height of \(6 \mathrm{~cm}\).
|726064569|kitty|/Applications/kitty.app|0|
            \begin{solutionordottedlines}[3cm]
            \end{solutionordottedlines}
|726064640|kitty|/Applications/kitty.app|0|
            \mdots{3}
|726064642|kitty|/Applications/kitty.app|0|
    \begin{multicols}{2}
|726064735|kitty|/Applications/kitty.app|0|
    \end{multicols}
|726064750|kitty|/Applications/kitty.app|0|
parts
item|726064761|kitty|/Applications/kitty.app|0|
                \begin{solutionordottedlines}[3cm]
                \end{solutionordottedlines}
|726064769|kitty|/Applications/kitty.app|0|
                \mdots{3}
|726064780|kitty|/Applications/kitty.app|0|
Worked Examples|726064801|kitty|/Applications/kitty.app|0|
        \begin{enumerate}

item
item
item
item
item
item
item
item
item|726064847|kitty|/Applications/kitty.app|0|
            \[] Which of these numbers are irrational?
|726064879|kitty|/Applications/kitty.app|0|
                \begin{enumerate}

item
item|726064886|kitty|/Applications/kitty.app|0|
dotfill
\)
dotfill|726064958|kitty|/Applications/kitty.app|0|
        \begin{onehalfspacing}
|726065004|kitty|/Applications/kitty.app|0|
        \end{onehalfspacing}
|726065056|kitty|/Applications/kitty.app|0|
\end{parts}|726065159|kitty|/Applications/kitty.app|0|
            \item If \(x=\sqrt{2}-1\) and \(y=\sqrt{2}+1\), find:
|726065170|kitty|/Applications/kitty.app|0|
\(|726065213|kitty|/Applications/kitty.app|0|
    \item Simplify:
|726065230|kitty|/Applications/kitty.app|0|
parts|726065320|kitty|/Applications/kitty.app|0|
part|726065326|kitty|/Applications/kitty.app|0|
                \begin{solutionordottedlines}[3in]
                \end{solutionordottedlines}
|726065385|kitty|/Applications/kitty.app|0|
subparts
subpart|726065430|kitty|/Applications/kitty.app|0|
subparts|726065437|kitty|/Applications/kitty.app|0|
        \end{multicols}
|726065469|kitty|/Applications/kitty.app|0|
enumerate
item
item
item
item
item
item|726065486|kitty|/Applications/kitty.app|0|
enumerate
item|726065522|kitty|/Applications/kitty.app|0|
multicols|726065523|kitty|/Applications/kitty.app|0|
        \end{enumerate}
|726065532|kitty|/Applications/kitty.app|0|
            \item \(\frac{5}{(\sqrt{7}-\sqrt{2})^{2}}\)
|726065602|kitty|/Applications/kitty.app|0|
            \item \(\frac{2}{6-3 \sqrt{3}}-\frac{1}{2 \sqrt{3}+3}\)
|726065605|kitty|/Applications/kitty.app|0|
        \begin{enumerate}

item
item
item
item|726065623|kitty|/Applications/kitty.app|0|
            \end{multicols}
|726065625|kitty|/Applications/kitty.app|0|
item|726065652|kitty|/Applications/kitty.app|0|
enumerate|726065655|kitty|/Applications/kitty.app|0|
OPENAI_API_KEY=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|726138481|kitty|/Applications/kitty.app|0|
O|726138507|kitty|/Applications/kitty.app|0|
A|726138508|kitty|/Applications/kitty.app|0|
P|726138508|kitty|/Applications/kitty.app|0|
I|726138508|kitty|/Applications/kitty.app|0|
_|726138508|kitty|/Applications/kitty.app|0|
K|726138508|kitty|/Applications/kitty.app|0|
Y|726138509|kitty|/Applications/kitty.app|0|
for chap in 3-new.tex 4-individual.tex 5-group.tex; do\
  { cat "$chap" | sgpt "take this tex file and insert fully worked mathematical solutions inside all of the fillin and solutionordottedlines environments." | tee "$chap.gpt"; }\
done\
|726138699|kitty|/Applications/kitty.app|0|
2022
|726139485|kitty|/Applications/kitty.app|0|
Computer Science + Actuary
|726140150|kitty|/Applications/kitty.app|0|
B Acc. B It maj Data Science.

500k not in a corporate world. must own business
this degree is an enabler. -> 200k to play with your business ideas

Banking, Finance.

~~Bored: Compsci~~

Don't focus too much on R.

**Do well in uni.** You get internships. Big and important companies let you work for them; these companies are the ones that will pay you big bucks when you finish your degree

UNSW Math Degree Computer Science.
|726140639|kitty|/Applications/kitty.app|0|
 What does the word \emph{algebra} mean?                                                                        
 \begin{solutionordottedlines}                                                                                  
 Algebra is a branch of mathematics dealing with symbols and the rules for manipulating those symbols; it is a  
 unifying thread of almost all of mathematics.                                                                  
 \end{solutionordottedlines}                                                                                    
                                                                                                                
 Where does the word \textbf{algebra} come from?                                                                
 \begin{solutionordottedlines}                                                                                  
 The word "algebra" comes from the Arabic word "al-jabr" which means "reunion of broken parts," and it was used 
 in the title of a book by the mathematician Al-Khwarizmi.                                                      
 \end{solutionordottedlines}                                                                                    
                                                                                                                
 What is an example of \textit{algebra}?                                                                        
 \begin{solutionordottedlines}                                                                                  
 An example of algebra is solving for \( x \) in the equation \( 2x + 3 = 7 \). The solution would be \( x = 2  
 \).                                                                                                            
 \end{solutionordottedlines}                                                                                    
                                                                                                                
 What can I do with \textsc{algebra}?                                                                           
 \begin{solutionordottedlines}                                                                                  
 With algebra, you can solve equations, work with unknowns, model real-world situations, and understand how     
 different mathematical symbols and functions relate to one another.                                            
 \end{solutionordottedlines}                                                                                    
|726140865|kitty|/Applications/kitty.app|0|
What does the word \emph{algebra} mean?
\begin{solutionordottedlines}[1in]
\end{solutionordottedlines}

Where does the word \textbf{algebra} come from?
\begin{solutionordottedlines}[1in]
\end{solutionordottedlines}

What is an example of \textit{algebra}?
\begin{solutionordottedlines}[1in]
\end{solutionordottedlines}

What can I do with \textsc{algebra}?
\begin{solutionordottedlines}[1in]
\end{solutionordottedlines}
|726140870|kitty|/Applications/kitty.app|0|
 Algebra is a branch of mathematics dealing with symbols and the rules for manipulating those symbols; it is a unifying thread of almost all of mathematics.                                                                  
|726140893|kitty|/Applications/kitty.app|0|
 The word "algebra" comes from the Arabic word "al-jabr" which means "reunion of broken parts," and it was used in the title of a book by the mathematician Al-Khwarizmi.                                                      
|726140898|kitty|/Applications/kitty.app|0|
 An example of algebra is solving for \( x \) in the equation \( 2x + 3 = 7 \). The solution would be \( x = 2  
|726140904|kitty|/Applications/kitty.app|0|
 With algebra, you can solve equations, work with unknowns, model real-world situations, and understand how different mathematical symbols and functions relate to one another.                                            
|726140924|kitty|/Applications/kitty.app|0|
         A symbol used to represent a number in algebraic expressions and equations.                            
|726140975|kitty|/Applications/kitty.app|0|
         A specific number represented by a symbol or an expression.                                            
|726140980|kitty|/Applications/kitty.app|0|
 \begin{examplebox}                                                                                             
     \subsection{Examples}                                                                                      
     \begin{doublespace}                                                                                        
         \begin{questions}                                                                                      
             \Question[1] Evaluate $ 2x $ when $ x = 3 $                                                        
                 \begin{solutionordottedlines}[1cm]                                                             
                     $2 \times 3 = 6$                                                                           
                 \end{solutionordottedlines}                                                                    
             \Question[1] Evaluate $ 5a + 2b $ when $ a = 2 $ and $ b = -3 $                                    
                 \begin{solutionordottedlines}[1cm]                                                             
                     $5 \times 2 + 2 \times (-3) = 10 - 6 = 4$                                                  
                 \end{solutionordottedlines}                                                                    
             \Question[1] Evaluate $ 2p(3q-2) $ when $ p = 1 $ and $ q = -2 $                                   
                 \begin{solutionordottedlines}[1cm]                                                             
                     $2 \times 1 \times (3 \times (-2) - 2) = 2 \times (-6 - 2) = 2 \times (-8) = -16$          
                 \end{solutionordottedlines}                                                                    
             \Question[1] Evaluate $ 7m - 4n $ when $ m = -3 $ and $ n = -2 $                                   
                 \begin{solutionordottedlines}[1cm]                                                             
                     $7 \times (-3) - 4 \times (-2) = -21 + 8 = -13$                                            
                 \end{solutionordottedlines}                                                                    
             \Question[1] Evaluate $ a + 2b - 3c $ when $ a = 3, b = -5, c = -2 $                               
                 \begin{solutionordottedlines}[1cm]                                                             
                     $3 + 2 \times (-5) - 3 \times (-2) = 3 - 10 + 6 = -1$                                      
                 \end{solutionordottedlines}                                                                    
         \end{questions}                                                                                        
     \end{doublespace}                                                                                          
 \end{examplebox}                                                                                               
|726140988|kitty|/Applications/kitty.app|0|
\begin{examplebox}
    \subsection{Examples}
    \begin{doublespace}
        \begin{questions}
            \Question[1] Evaluate $ 2x $ when $ x = 3 $
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \Question[1] Evaluate $ 5a + 2b $ when $ a = 2 $ and $ b = -3 $
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \Question[1] Evaluate $ 2p(3q-2) $ when $ p = 1 $ and $ q = -2 $
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \Question[1] Evaluate $ 7m - 4n $ when $ m = -3 $ and $ n = -2 $
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \Question[1] Evaluate $ a + 2b - 3c $ when $ a = 3, b = -5, c = -2 $
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
        \end{questions}
    \end{doublespace}
\end{examplebox}
|726140992|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                            
     \subsection{Exercises}                                                                                     
     \begin{questions}                                                                                          
         \Question[2] Evaluate \(2x-3y\) when:                                                                  
                 \begin{parts}                                                                                  
                 \begin{multicols}{2}                                                                           
                     \part \(x = \frac{2}{5}, y = -\frac{1}{4}\)                                                
                         \begin{solutionordottedlines}[1cm]                                                     
                             $2 \times \frac{2}{5} - 3 \times \left(-\frac{1}{4}\right) = \frac{4}{5} +         
 \frac{3}{4} = \frac{16}{20} + \frac{15}{20} = \frac{31}{20} = 1\frac{11}{20}$                                  
                         \end{solutionordottedlines}                                                            
                     \part \(x = \frac{1}{3}, y = \frac{1}{6}\)                                                 
                         \begin{solutionordottedlines}[1cm]                                                     
                             $2 \times \frac{1}{3} - 3 \times \frac{1}{6} = \frac{2}{3} - \frac{3}{6} =         
 \frac{2}{3} - \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6}$                                           
                         \end{solutionordottedlines}                                                            
                 \end{multicols}                                                                                
                 \end{parts}                                                                                    
         \Question[2] Evaluate \(p^2 -2q\) when:                                                                
             \begin{parts}                                                                                      
                 \begin{multicols}{2}                                                                           
                 \part \(p=-7, q = 2\)                                                                          
                         \begin{solutionordottedlines}[1cm]                                                     
                             $(-7)^2 - 2 \times 2 = 49 - 4 = 45$                                                
                         \end{solutionordottedlines}                                                            
                 \part \(p = -\frac{1}{3}, q = \frac{5}{6}\)                                                    
                         \begin{solutionordottedlines}[1cm]                                                     
                             $\left(-\frac{1}{3}\right)^2 - 2 \times \frac{5}{6} = \frac{1}{9} - \frac{10}{6} = 
 \frac{1}{9} - \frac{15}{9} = -\frac{14}{9} = -1\frac{5}{9}$                                                    
                         \end{solutionordottedlines}                                                            
                 \end{multicols}                                                                                
             \end{parts}                                                                                        
     \end{questions}                                                                                            
 \end{exercisebox}                                                                                              
|726141004|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises}
    \begin{questions}
        \Question[2] Evaluate \(2x-3y\) when:
                \begin{parts}
                \begin{multicols}{2}
                    \part \(x = \frac{2}{5}, y = -\frac{1}{4}\)
                        \mdots{5}
                    \part \(x = \frac{1}{3}, y = \frac{1}{6}\)
                        \mdots{5}
                \end{multicols}
                \end{parts}
        \Question[2] Evaluate \(p^2 -2q\) when:
            \begin{parts}
                \begin{multicols}{2}
                \part \(p=-7, q = 2\)
                        \mdots{5}
                \part \(p = -\frac{1}{3}, q = \frac{5}{6}\)
                        \mdots{5}
                \end{multicols}
            \end{parts}
    \end{questions}
\end{exercisebox}
|726141007|kitty|/Applications/kitty.app|0|
 \begin{examplebox}                                                                                             
     \subsection{Examples}                                                                                      
     \begin{questions}                                                                                          
         \Question[3] Which of the following are pairs of like terms?                                           
         \begin{parts}\begin{multicols}{3}                                                                      
             \part \(3x, 2x\) \hfill Yes                                                                        
             \part \(3m, 2c\) \hfill No                                                                         
             \part \(3x^2, 3x\) \hfill No                                                                       
             \part \(2x^2y, 3yx^2\) \hfill Yes                                                                  
             \part \(2mn, 3nm\) \hfill Yes                                                                      
             \part \(5y^2, 6y^2x\) \hfill No                                                                    
         \end{multicols}\end{parts}                                                                             
         \Question[6] Simplify each expression if possible:                                                     
         \begin{parts}\begin{multicols}{2}                                                                      
             \part \(4a + 7a = \)                                                                               
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(11a\)                                                                                        
                 \end{solutionordottedlines}                                                                    
             \part \(3x^2y + 4x^2 -2x^2y = \)                                                                   
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(x^2y + 4x^2\)                                                                                
                 \end{solutionordottedlines}                                                                    
             \part \(5m + 6n = \)                                                                               
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(5m + 6n\)                                                                                    
                 \end{solutionordottedlines}                                                                    
             \part \(9b + 2c - 3b + 6c = \)                                                                     
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(6b + 8c\)                                                                                    
                 \end{solutionordottedlines}                                                                    
             \part \(3z+5yx-z-6xy=\)                                                                            
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(2z - yx\)                                                                                    
                 \end{solutionordottedlines}                                                                    
             \part \(6x^3-4x^2+5x^3\)                                                                           
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(11x^3 - 4x^2\)                                                                               
                 \end{solutionordottedlines}                                                                    
         \end{multicols}\end{parts}                                                                             
     \end{questions}                                                                                            
 \end{examplebox}                                                                                               
|726141016|kitty|/Applications/kitty.app|0|
\begin{examplebox}
    \subsection{Examples}
    \begin{questions}
        \Question[3] Which of the following are pairs of like terms?
        \begin{parts}\begin{multicols}{3}
            \part \(3x, 2x\)
            \part \(3m, 2c\)
            \part \(3x^2, 3x\)
            \part \(2x^2y, 3yx^2\)
            \part \(2mn, 3nm\)
            \part \(5y^2, 6y^2x\)
        \end{multicols}\end{parts}
        \Question[6] Simplify each expression if possible:
        \begin{parts}\begin{multicols}{2}
            \part \(4a + 7a = \)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \part \(3x^2y + 4x^2 -2x^2y = \)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \part \(5m + 6n = \)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \part \(9b + 2c - 3b + 6c = \)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \part \(3z+5yx-z-6xy=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \part \(6x^3-4x^2+5x^3\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \end{questions}
\end{examplebox}
|726141026|kitty|/Applications/kitty.app|0|
hfill Yes                                                                  |726141056|kitty|/Applications/kitty.app|0|
                 \begin{solutionordottedlines}[1cm]                                                             
|726141075|kitty|/Applications/kitty.app|0|
\begin{solutionordottedlines}[1in]                                                             
\end{solutionordottedlines}
|726141092|kitty|/Applications/kitty.app|0|
\mdots{2}
|726141093|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                            
     \subsection{Exercises}                                                                                     
     \begin{questions}                                                                                          
         \Question[2] Simplify:                                                                                 
             \begin{multicols}{2}                                                                               
                 \begin{parts}                                                                                  
                     \part \(\frac{x}{2} + \frac{x}{3}\)                                                        
                         \begin{solutionordottedlines}[1in]                                                     
                         \(\frac{5x}{6}\)                                                                       
                         \end{solutionordottedlines}                                                            
                     \part \(\frac{3x}{4} - \frac{2x}{5}\)                                                      
                         \begin{solutionordottedlines}[1in]                                                     
                         \(\frac{7x}{20}\)                                                                      
                         \end{solutionordottedlines}                                                            
                 \end{parts}                                                                                    
             \end{multicols}                                                                                    
         \Question[4] Which of the following are pairs of like terms?                                           
             \begin{multicols}{2}                                                                               
                 \begin{parts}                                                                                  
                     \part \(12m, 5m\) \hfill Yes                                                               
                     \part \(-6a, 7b\) \hfill No                                                                
                     \part \(6ab, -7b\) \hfill No                                                               
                     \part \(6x^2, -7x^2\) \hfill Yes                                                           
                 \end{parts}                                                                                    
             \end{multicols}                                                                                    
         \Question[3] Simplify each expression by collecting like terms.                                        
             \begin{multicols}{3}                                                                               
                 \begin{parts}                                                                                  
                     \part \(8b + 3b\)                                                                          
                         \begin{solutionordottedlines}[1cm]                                                     
                         \(11b\)                                                                                
                         \end{solutionordottedlines}                                                            
                     \part \(6x^2 + 4x^2\)\dotfill                                                              
                         \begin{solutionordottedlines}[1cm]                                                     
                         \(10x^2\)                                                                              
                         \end{solutionordottedlines}                                                            
                     \part \(7f-3f+9f\)\dotfill                                                                 
                         \begin{solutionordottedlines}[1cm]                                                     
                         \(13f\)                                                                                
                         \end{solutionordottedlines}                                                            
                 \end{parts}                                                                                    
             \end{multicols}                                                                                    
         \Question[3] Fill in the missing term.                                                                 
             \begin{multicols}{3}                                                                               
                 \begin{parts}                                                                                  
                     \part \(8mn + \dotfill = 12mn\)                                                            
                         \begin{solutionordottedlines}[1cm]                                                     
                         \(4mn\)                                                                                
                         \end{solutionordottedlines}                                                            
                     \part \(6m^2 - \dotfill = m^2\)                                                            
                         \begin{solutionordottedlines}[1cm]                                                     
                         \(5m^2\)                                                                               
                         \end{solutionordottedlines}                                                            
                     \part \(-7a^2b + \dotfill = a^2b\)                                                         
                         \begin{solutionordottedlines}[1cm]                                                     
                         \(8a^2b\)                                                                              
                         \end{solutionordottedlines}                                                            
                 \end{parts}                                                                                    
             \end{multicols}                                                                                    
         \Question[6] Simplify by collecting like terms.                                                        
             \begin{multicols}{2}                                                                               
                 \begin{parts}                                                                                  
                     \part \(8p + 6 + 3p - 2=\)\dotfill                                                         
                         \begin{solutionordottedlines}[1cm]                                                     
                         \(11p + 4\)                                                                            
                         \end{solutionordottedlines}                                                            
                     \part \(10ab+11b-12b+3ab=\)\dotfill                                                        
                         \begin{solutionordottedlines}[1cm]                                                     
                         \(13ab - b\)                                                                           
                         \end{solutionordottedlines}                                                            
                     \part \(4p^2-3p-8p-3p^2=\)\dotfill                                                         
                         \begin{solutionordottedlines}[1cm]                                                     
                         \(p^2 - 11p\)                                                                          
                         \end{solutionordottedlines}                                                            
                     \part \(-4x^2+3x^2-3y-7y=\)\dotfill                                                        
                         \begin{solutionordottedlines}[1cm]                                                     
                         \(-x^2 - 10y\)                                                                         
                         \end{solutionordottedlines}                                                            
                     \part \(7x^3+6x^2-4y^3-x^2=\)\dotfill                                                      
                         \begin{solutionordottedlines}[1cm]                                                     
                         \(7x^3 + 5x^2 - 4y^3\)                                                                 
                         \end{solutionordottedlines}                                                            
                     \part \(-3ab^2+4a^2b-5ab^2+a^2b=\)\dotfill                                                 
                         \begin{solutionordottedlines}[1cm]                                                     
                         \(-8ab^2 + 5a^2b\)                                                                     
                         \end{solutionordottedlines}                                                            
                 \end{parts}                                                                                    
             \end{multicols}                                                                                    
         \Question[6] Simplify:                                                                                 
             \begin{multicols}{3}                                                                               
                 \begin{parts}                                                                                  
                     \part \(\frac{c}{6} + \frac{c}{7}\)                                                        
                         \begin{solutionordottedlines}[2cm]                                                     
                         \(\frac{13c}{42}\)                                                                     
                         \end{solutionordottedlines}                                                            
                     \part \(\frac{x}{7} - \frac{x}{8}\)                                                        
                         \begin{solutionordottedlines}[2cm]                                                     
                         \(\frac{x}{56}\)                                                                       
                         \end{solutionordottedlines}                                                            
                     \part \(c - \frac{c}{7}\)                                                                  
                         \begin{solutionordottedlines}[2cm]                                                     
                         \(\frac{6c}{7}\)                                                                       
                         \end{solutionordottedlines}                                                            
                     \part \(\frac{2x}{3} + \frac{x}{4}\)                                                       
                         \begin{solutionordottedlines}[2cm]                                                     
                         \(\frac{11x}{12}\)                                                                     
                         \end{solutionordottedlines}                                                            
                     \part \(\frac{5x}{3} + \frac{x}{2}\)                                                       
                         \begin{solutionordottedlines}[2cm]                                                     
                         \(\frac{19x}{6}\)                                                                      
                         \end{solutionordottedlines}                                                            
                     \part \(\frac{5x}{11} - \frac{2x}{3}\)                                                     
                         \begin{solutionordottedlines}[2cm]                                                     
                         \(-\frac{31x}{33}\)                                                                    
                         \end{solutionordottedlines}                                                            
                 \end{parts}                                                                                    
             \end{multicols}                                                                                    
     \end{questions}                                                                                            
 \end{exercisebox}                                                                                              
|726141109|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises}
    \begin{questions}
        \Question[2] Simplify:
            \begin{multicols}{2}
                \begin{parts}
                    \part \(\frac{x}{2} + \frac{x}{3}\)
                        \begin{solutionordottedlines}[1in]
                        \end{solutionordottedlines}
                    \part \(\frac{3x}{4} - \frac{2x}{5}\)
                        \begin{solutionordottedlines}[1in]
                        \end{solutionordottedlines}
                \end{parts}
            \end{multicols}
        \Question[4] Which of the following are pairs of like terms?
            \begin{multicols}{2}
                \begin{parts}
                    \part \(12m, 5m\)
                    \part \(-6a, 7b\)
                    \part \(6ab, -7b\)
                    \part \(6x^2, -7x^2\)
                \end{parts}
            \end{multicols}
        \Question[3] Simplify each expression by collecting like terms.
            \begin{multicols}{3}
                \begin{parts}
                    \part \(8b + 3b\)
                        \begin{solutionordottedlines}[1cm]
                        \end{solutionordottedlines}
                    \part \(6x^2 + 4x^2\)\dotfill
                        \begin{solutionordottedlines}[1cm]
                        \end{solutionordottedlines}
                    \part \(7f-3f+9f\)\dotfill
                        \begin{solutionordottedlines}[1cm]
                        \end{solutionordottedlines}
                \end{parts}
            \end{multicols}
        \Question[3] Fill in the missing term.
            \begin{multicols}{3}
                \begin{parts}
                    \part \(8mn + \dotfill = 12mn\)
                    \part \(6m^2 - \dotfill = m^2\)
                    \part \(-7a^2b + \dotfill = a^2b\)
                \end{parts}
            \end{multicols}
        \Question[6] Simplify by collecting like terms.
            \begin{multicols}{2}
                \begin{parts}
                    \part \(8p + 6 + 3p - 2=\)\dotfill
                    \part \(10ab+11b-12b+3ab=\)\dotfill
                    \part \(4p^2-3p-8p-3p^2=\)\dotfill
                    \part \(-4x^2+3x^2-3y-7y=\)\dotfill
                    \part \(7x^3+6x^2-4y^3-x^2=\)\dotfill
                    \part \(-3ab^2+4a^2b-5ab^2+a^2b=\)\dotfill
                \end{parts}
            \end{multicols}
        \Question[6] Simplify:
            \begin{multicols}{3}
                \begin{parts}
                    \part \(\frac{c}{6} + \frac{c}{7}\)
                        \begin{solutionordottedlines}[2cm]
                        \end{solutionordottedlines}
                    \part \(\frac{x}{7} - \frac{x}{8}\)
                        \begin{solutionordottedlines}[2cm]
                        \end{solutionordottedlines}
                    \part \(c - \frac{c}{7}\)
                        \begin{solutionordottedlines}[2cm]
                        \end{solutionordottedlines}
                    \part \(\frac{2x}{3} + \frac{x}{4}\)
                        \begin{solutionordottedlines}[2cm]
                        \end{solutionordottedlines}
                    \part \(\frac{5x}{3} + \frac{x}{2}\)
                        \begin{solutionordottedlines}[2cm]
                        \end{solutionordottedlines}
                    \part \(\frac{5x}{11} - \frac{2x}{3}\)
                        \begin{solutionordottedlines}[2cm]
                        \end{solutionordottedlines}
                \end{parts}
            \end{multicols}
    \end{questions}
\end{exercisebox}
|726141113|kitty|/Applications/kitty.app|0|
 \begin{examplebox}                                                                                             
     \subsection{Examples}                                                                                      
     \begin{questions}                                                                                          
         \Question[6] Multiplications                                                                           
             \begin{multicols}{3}                                                                               
             \begin{parts}                                                                                      
                 \part \(4 \times 3a = 12a\)                                                                    
                 \begin{solutionordottedlines}[1cm]                                                             
                 12a                                                                                            
                 \end{solutionordottedlines}                                                                    
                 \part \(2d \times 5e = 10de\)                                                                  
                 \begin{solutionordottedlines}[1cm]                                                             
                 10de                                                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \(4m \times 5m = 20m^2\)                                                                 
                 \begin{solutionordottedlines}[1cm]                                                             
                 20m^2                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(3p \times 2pq = 6p^2q\)                                                                
                 \begin{solutionordottedlines}[1cm]                                                             
                 6p^2q                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(3x \times (-6) = -18x\)                                                                
                 \begin{solutionordottedlines}[1cm]                                                             
                 -18x                                                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \(-5ab \times -3bc = 15abc^2\)                                                           
                 \begin{solutionordottedlines}[1cm]                                                             
                 15abc^2                                                                                        
                 \end{solutionordottedlines}                                                                    
             \end{parts}                                                                                        
             \end{multicols}                                                                                    
         \Question[6] Divisions                                                                                 
             \begin{multicols}{3}                                                                               
             \begin{parts}                                                                                      
                 \part \(24x \div 6 = 4x\)                                                                      
                 \begin{solutionordottedlines}[1cm]                                                             
                 4x                                                                                             
                 \end{solutionordottedlines}                                                                    
                 \part \(36a\div 4 = 9a\)                                                                       
                 \begin{solutionordottedlines}[1cm]                                                             
                 9a                                                                                             
                 \end{solutionordottedlines}                                                                    
                 \part \(-18x^2 \div (-3) = 6x^2\)                                                              
                 \begin{solutionordottedlines}[1cm]                                                             
                 6x^2                                                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \(\frac{15a}{3} = 5a\)                                                                   
                 \begin{solutionordottedlines}[1cm]                                                             
                 5a                                                                                             
                 \end{solutionordottedlines}                                                                    
                 \part \(\frac{12x}{21} = \frac{4x}{7}\)                                                        
                 \begin{solutionordottedlines}[1cm]                                                             
                 \frac{4x}{7}                                                                                   
                 \end{solutionordottedlines}                                                                    
                 \part \(\frac{-24xy}{6y} = -4x\)                                                               
                 \begin{solutionordottedlines}[1cm]                                                             
                 -4x                                                                                            
                 \end{solutionordottedlines}                                                                    
             \end{parts}                                                                                        
             \end{multicols}                                                                                    
     \end{questions}                                                                                            
 \end{examplebox}                                                                                               
|726141204|kitty|/Applications/kitty.app|0|
\begin{examplebox}
    \subsection{Examples}
    \begin{questions}
        \Question[6] Multiplications
            \begin{multicols}{3}
            \begin{parts}
                \part \(4 \times 3a = \)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(2d \times 5e =\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(4m \times 5m = \)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(3p \times 2pq = \)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(3x \times (-6) = \)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(-5ab \times -3bc\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \end{parts}
            \end{multicols}
        \Question[6] Divisions
            \begin{multicols}{3}
            \begin{parts}
                \part \(24x \div 6 = \)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(36a\div 4\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(-18x^2 \div (-3)\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(\frac{15a}{3}\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(\frac{12x}{21}\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(\frac{-24xy}{6y}\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \end{parts}
            \end{multicols}
    \end{questions}
\end{examplebox}
|726141208|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                            
     \subsection{Exercises}                                                                                     
     \begin{questions}                                                                                          
         \Question[6] Rewrite as a single fraction:                                                             
             \begin{multicols}{3}                                                                               
             \begin{doublespace}                                                                                
             \begin{parts}                                                                                      
                 \part \(\frac{2a}{5} \times \frac{a}{4}=\frac{2a^2}{20}=\frac{a^2}{10}\)                       
                 \begin{solutionordottedlines}[1cm]                                                             
                 \frac{a^2}{10}                                                                                 
                 \end{solutionordottedlines}                                                                    
                 \part \(\frac{3x}{7} \times \frac{5y}{12}=\frac{15xy}{84}=\frac{5xy}{28}\)                     
                 \begin{solutionordottedlines}[1cm]                                                             
                 \frac{5xy}{28}                                                                                 
                 \end{solutionordottedlines}                                                                    
                 \part \(\frac{4p}{q} \times \frac{3}{2p}=\frac{12p}{2pq}=\frac{6}{q}\)                         
                 \begin{solutionordottedlines}[1cm]                                                             
                 \frac{6}{q}                                                                                    
                 \end{solutionordottedlines}                                                                    
                 \part \(\frac{15}{x} \times \frac{2}{3x}=\frac{30}{3x^2}=\frac{10}{x^2}\)                      
                 \begin{solutionordottedlines}[1cm]                                                             
                 \frac{10}{x^2}                                                                                 
                 \end{solutionordottedlines}                                                                    
                 \part \(\frac{2x}{3} \div \frac{3x}{5}=\frac{2x}{3} \times \frac{5}{3x}=\frac{10}{9}\)         
                 \begin{solutionordottedlines}[1cm]                                                             
                 \frac{10}{9}                                                                                   
                 \end{solutionordottedlines}                                                                    
                 \part \(\frac{6a}{7b} \div \frac{2ab}{3}=\frac{6a}{7b} \times                                  
 \frac{3}{2ab}=\frac{18}{14b^2}=\frac{9}{7b^2}\)                                                                
                 \begin{solutionordottedlines}[1cm]                                                             
                 \frac{9}{7b^2}                                                                                 
                 \end{solutionordottedlines}                                                                    
             \end{parts}                                                                                        
             \end{doublespace}                                                                                  
             \end{multicols}                                                                                    
         \Question[6] Simplify                                                                                  
             \begin{multicols}{3}                                                                               
             \begin{parts}                                                                                      
                 \part \(5c \times 2d = 10cd\)                                                                  
                 \begin{solutionordottedlines}[1cm]                                                             
                 10cd                                                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \(-6l \times (-5m)= 30lm\)                                                               
                 \begin{solutionordottedlines}[1cm]                                                             
                 30lm                                                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \(-2m \times (-4m)= 8m^2\)                                                               
                 \begin{solutionordottedlines}[1cm]                                                             
                 8m^2                                                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \(24a^2 \div 8 = 3a^2\)                                                                  
                 \begin{solutionordottedlines}[1cm]                                                             
                 3a^2                                                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \(7 \times 15p \div 21 = 5p\)                                                            
                 \begin{solutionordottedlines}[1cm]                                                             
                 5p                                                                                             
                 \end{solutionordottedlines}                                                                    
                 \part \(18y \div 6 \times 2= 6y\)                                                              
                 \begin{solutionordottedlines}[1cm]                                                             
                 6y                                                                                             
                 \end{solutionordottedlines}                                                                    
             \end{parts}                                                                                        
             \end{multicols}                                                                                    
         \Question[8] Simplify by first cancelling out common factors:                                          
             \begin{multicols}{4}                                                                               
             \begin{doublespace}                                                                                
             \begin{parts}                                                                                      
                 \part \(\frac{14p}{21} =\frac{2p}{3}\)                                                         
                 \begin{solutionordottedlines}[1cm]                                                             
                 \frac{2p}{3}                                                                                   
                 \end{solutionordottedlines}                                                                    
                 \part \(\frac{22x^2}{33} =\frac{2x^2}{3}\)                                                     
                 \begin{solutionordottedlines}[1cm]                                                             
                 \frac{2x^2}{3}                                                                                 
                 \end{solutionordottedlines}                                                                    
                 \part \(\frac{2xy}{6xy}=\frac{1}{3}\)                                                          
                 \begin{solutionordottedlines}[1cm]                                                             
                 \frac{1}{3}                                                                                    
                 \end{solutionordottedlines}                                                                    
                 \part \(\frac{-4xy}{8x} =\frac{-y}{2}\)                                                        
                 \begin{solutionordottedlines}[1cm]                                                             
                 \frac{-y}{2}                                                                                   
                 \end{solutionordottedlines}                                                                    
                 \part \(\frac{2y}{5} \times \frac{y}{4} =\frac{2y^2}{20}=\frac{y^2}{10}\)                      
                 \begin{solutionordottedlines}[1cm]                                                             
                 \frac{y^2}{10}                                                                                 
                 \end{solutionordottedlines}                                                                    
                 \part \(\frac{p}{6q} \times \frac{9p}{4q} =\frac{9p^2}{24q^2}=\frac{3p^2}{8q^2}\)              
                 \begin{solutionordottedlines}[1cm]                                                             
                 \frac{3p^2}{8q^2}                                                                              
                 \end{solutionordottedlines}                                                                    
                 \part \(\frac{2yz}{5xy} \times \frac{3xy}{4yz} =\frac{6xyz}{20xyz}=\frac{3}{10}\)              
                 \begin{solutionordottedlines}[1cm]                                                             
                 \frac{3}{10}                                                                                   
                 \end{solutionordottedlines}                                                                    
                 \part \(\frac{2y}{5} \div \frac{y}{4} =\frac{2y}{5} \times \frac{4}{y}=\frac{8}{5}\)           
                 \begin{solutionordottedlines}[1cm]                                                             
                 \frac{8}{5}                                                                                    
                 \end{solutionordottedlines}                                                                    
             \end{parts}                                                                                        
             \end{doublespace}                                                                                  
             \end{multicols}                                                                                    
     \end{questions}                                                                                            
 \end{exercisebox}                                                                                              
|726141221|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises}
    \begin{questions}
        \Question[6] Rewrite as a single fraction:
            \begin{multicols}{3}
            \begin{doublespace}
            \begin{parts}
                \part \(\frac{2a}{5} \times \frac{a}{4}=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(\frac{3x}{7} \times \frac{5y}{12}=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(\frac{4p}{q} \times \frac{3}{2p}=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(\frac{15}{x} \times \frac{2}{3x}=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(\frac{2x}{3} \div \frac{3x}{5}=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(\frac{6a}{7b} \div \frac{2ab}{3}=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \end{parts}
            \end{doublespace}
            \end{multicols}
        \Question[6] Simplify
            \begin{multicols}{3}
            \begin{parts}
                \part \(5c \times 2d =\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(-6l \times (-5m)=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(-2m \times (-4m)=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(24a^2 \div 8 = \)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(7 \times 15p \div 21 = \)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(18y \div 6 \times 2= \)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \end{parts}
            \end{multicols}
        \Question[8] Simplify by first cancelling out common factors:
            \begin{multicols}{4}
            \begin{doublespace}
            \begin{parts}
                \part \(\frac{14p}{21} =\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(\frac{22x^2}{33} =\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(\frac{2xy}{6xy}=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(\frac{-4xy}{8x} =\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(\frac{2y}{5} \times \frac{y}{4} =\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(\frac{p}{6q} \times \frac{9p}{4q} =\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(\frac{2yz}{5xy} \times \frac{3xy}{4yz} =\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(\frac{2y}{5} \div \frac{y}{4} =\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \end{parts}
            \end{doublespace}
            \end{multicols}
    \end{questions}
\end{exercisebox}
|726141225|kitty|/Applications/kitty.app|0|
 \begin{examplebox}                                                                                             
     \subsection{Examples}                                                                                      
     \begin{questions}                                                                                          
         \Question[3] Expand:                                                                                   
             \begin{multicols}{3}                                                                               
             \begin{parts}                                                                                      
                 \part \(2(a+3)=\)                                                                              
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(2a+6\)                                                                                       
                 \end{solutionordottedlines}                                                                    
                 \part \(3(x-2)=\)                                                                              
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(3x-6\)                                                                                       
                 \end{solutionordottedlines}                                                                    
                 \part \(4(2m-7)=\)                                                                             
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(8m-28\)                                                                                      
                 \end{solutionordottedlines}                                                                    
             \end{parts}                                                                                        
             \end{multicols}                                                                                    
         \Question[3] Now try:                                                                                  
             \begin{multicols}{3}                                                                               
             \begin{parts}                                                                                      
                 \part \(5(a+1)+6=\)                                                                            
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(5a+5+6\)                                                                                     
                 \(5a+11\)                                                                                      
                 \end{solutionordottedlines}                                                                    
                 \part \(4(2b-1)+7=\)                                                                           
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(8b-4+7\)                                                                                     
                 \(8b+3\)                                                                                       
                 \end{solutionordottedlines}                                                                    
                 \part \(6(d+5)-3d=\)                                                                           
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(6d+30-3d\)                                                                                   
                 \(3d+30\)                                                                                      
                 \end{solutionordottedlines}                                                                    
             \end{parts}                                                                                        
             \end{multicols}                                                                                    
         \Question[2] Can you handle some more terms?                                                           
             \begin{multicols}{2}                                                                               
             \begin{parts}                                                                                      
                 \part \(2(b+5)+3(b+2)=\)                                                                       
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(2b+10+3b+6\)                                                                                 
                 \(5b+16\)                                                                                      
                 \end{solutionordottedlines}                                                                    
                 \part \(3(x-2)-2(x+1)=\)                                                                       
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(3x-6-2x-2\)                                                                                  
                 \(x-8\)                                                                                        
                 \end{solutionordottedlines}                                                                    
             \end{parts}                                                                                        
             \end{multicols}                                                                                    
     \end{questions}                                                                                            
 \end{examplebox}                                                                                               
|726141251|kitty|/Applications/kitty.app|0|
\begin{examplebox}
    \subsection{Examples}
    \begin{questions}
        \Question[3] Expand:
            \begin{multicols}{3}
            \begin{parts}
                \part \(2(a+3)=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(3(x-2)=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(4(2m-7)=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \end{parts}
            \end{multicols}
        \Question[3] Now try:
            \begin{multicols}{3}
            \begin{parts}
                \part \(5(a+1)+6=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(4(2b-1)+7=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(6(d+5)-3d=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \end{parts}
            \end{multicols}
        \Question[2] Can you handle some more terms?
            \begin{multicols}{2}
            \begin{parts}
                \part \(2(b+5)+3(b+2)=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \(3(x-2)-2(x+1)=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \end{parts}
            \end{multicols}
    \end{questions}
\end{examplebox}
|726141255|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                            
     \subsection{Exercises}                                                                                     
     \begin{questions}                                                                                          
         \Question[10] Have a go at these ones yourselves:                                                      
         \begin{parts}                                                                                          
         \begin{multicols}{2}                                                                                   
             \part \(\frac{3}{5}(6x+\frac{7}{3})=\)\dotfill                                                     
             \( \frac{18}{5}x + \frac{7}{5} \)                                                                  
             \part \(\frac{4}{3}(6x+11)+\frac{2}{3}=\)\dotfill                                                  
             \( 8x + \frac{44}{3} + \frac{2}{3} = 8x + \frac{46}{3} \)                                          
             \part \(-12(4y-5)=\)\dotfill                                                                       
             \( -48y + 60 \)                                                                                    
             \part \(\frac{2}{3}(12p+6)=\)\dotfill                                                              
             \( 8p + 4 \)                                                                                       
             \part \(-\frac{1}{2}(10d-6)=\)\dotfill                                                             
             \( -5d + 3 \)                                                                                      
             \part \(-\frac{4}{5}(25m-100)=\)\dotfill                                                           
             \( -20m + 80 \)                                                                                    
             \part \(\frac{3}{5}(\frac{x}{6}+\frac{1}{3})=\)\dotfill                                            
             \( \frac{1}{10}x + \frac{1}{5} \)                                                                  
             \part \(-\frac{3}{5}(\frac{a}{3}-\frac{2}{3})=\)\dotfill                                           
             \( -\frac{1}{5}a + \frac{2}{5} \)                                                                  
             \part \(c(c-5)=\)\dotfill                                                                          
             \( c^2 - 5c \)                                                                                     
             \part \(2i(5i+7)=\)\dotfill                                                                        
             \( 10i^2 + 14i \)                                                                                  
         \end{multicols}                                                                                        
         \end{parts}                                                                                            
     \end{questions}                                                                                            
 \end{exercisebox}                                                                                              
|726141259|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises}
    \begin{questions}
        \Question[10] Have a go at these ones yourselves:
        \begin{parts}
        \begin{multicols}{2}
            \part \(\frac{3}{5}(6x+\frac{7}{3})=\)\dotfill
            \part \(\frac{4}{3}(6x+11)+\frac{2}{3}=\)\dotfill
            \part \(-12(4y-5)=\)\dotfill
            \part \(\frac{2}{3}(12p+6)=\)\dotfill
            \part \(-\frac{1}{2}(10d-6)=\)\dotfill
            \part \(-\frac{4}{5}(25m-100)=\)\dotfill
            \part \(\frac{3}{5}(\frac{x}{6}+\frac{1}{3})=\)\dotfill
            \part \(-\frac{3}{5}(\frac{a}{3}-\frac{2}{3})=\)\dotfill
            \part \(c(c-5)=\)\dotfill
            \part \(2i(5i+7)=\)\dotfill
        \end{multicols}
        \end{parts}
    \end{questions}
\end{exercisebox}
|726141264|kitty|/Applications/kitty.app|0|
 \begin{solutionordottedlines}[2cm]                                                                             
 The prefix \textit{\textbf{bi}} means two.                                                                     
|726141282|kitty|/Applications/kitty.app|0|
 \begin{solutionordottedlines}[2cm]                                                                             
|726141285|kitty|/Applications/kitty.app|0|
dotfill|726141297|kitty|/Applications/kitty.app|0|
 \begin{examplebox}                                                                                             
     \subsection{Examples}                                                                                      
     \begin{questions}                                                                                          
         \Question[4] Expand the following:                                                                     
             \begin{multicols}{2}                                                                               
             \begin{parts}                                                                                      
                 \part \((x+4)(x+5)=\)                                                                          
                 \begin{solutionordottedlines}[1cm]                                                             
                 $x^2 + 5x + 4x + 20 = x^2 + 9x + 20$                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \((x+3)(x-2)=\)                                                                          
                 \begin{solutionordottedlines}[1cm]                                                             
                 $x^2 - 2x + 3x - 6 = x^2 + x - 6$                                                              
                 \end{solutionordottedlines}                                                                    
                 \part \((x-4)(x-3)=\)                                                                          
                 \begin{solutionordottedlines}[1cm]                                                             
                 $x^2 - 3x - 4x + 12 = x^2 - 7x + 12$                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \((2y+1)(3y-4)=\)                                                                        
                 \begin{solutionordottedlines}[1cm]                                                             
                 $6y^2 - 8y + 3y - 4 = 6y^2 - 5y - 4$                                                           
                 \end{solutionordottedlines}                                                                    
             \end{parts}                                                                                        
             \end{multicols}                                                                                    
     \end{questions}                                                                                            
 \end{examplebox}                                                                                               
|726141393|kitty|/Applications/kitty.app|0|
\begin{examplebox}
    \subsection{Examples}
    \begin{questions}
        \Question[4] Expand the following:
            \begin{multicols}{2}
            \begin{parts}
                \part \((x+4)(x+5)=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \((x+3)(x-2)=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \((x-4)(x-3)=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
                \part \((2y+1)(3y-4)=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \end{parts}
            \end{multicols}
    \end{questions}
\end{examplebox}
|726141398|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                            
     \subsection{Exercises}                                                                                     
     \begin{questions}                                                                                          
         \question[]                                                                                            
         \begin{parts}\begin{multicols}{2}                                                                      
         \part \((a+3)(a+9)=\)                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                 
             $a^2 + 9a + 3a + 27 = a^2 + 12a + 27$                                                              
             \end{solutionordottedlines}                                                                        
         \part \((a+8)(9+a)=\)                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                 
             $a^2 + 9a + 8a + 72 = a^2 + 17a + 72$                                                              
             \end{solutionordottedlines}                                                                        
         \part \((p-6)(p+4)=\)                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                 
             $p^2 + 4p - 6p - 24 = p^2 - 2p - 24$                                                               
             \end{solutionordottedlines}                                                                        
         \part \((x+3)(x-8)=\)                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                 
             $x^2 - 8x + 3x - 24 = x^2 - 5x - 24$                                                               
             \end{solutionordottedlines}                                                                        
         \part \((x+7)(x-4)=\)                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                 
             $x^2 - 4x + 7x - 28 = x^2 + 3x - 28$                                                               
             \end{solutionordottedlines}                                                                        
         \part \((5x+1)(x+2)=\)                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                 
             $5x^2 + 10x + x + 2 = 5x^2 + 11x + 2$                                                              
             \end{solutionordottedlines}                                                                        
         \part \((4m+3)(2m-1)=\)                                                                                
             \begin{solutionordottedlines}[2cm]                                                                 
             $8m^2 - 4m + 6m - 3 = 8m^2 + 2m - 3$                                                               
             \end{solutionordottedlines}                                                                        
         \part \((2x-7)(3x-1)=\)                                                                                
             \begin{solutionordottedlines}[2cm]                                                                 
             $6x^2 - 2x - 21x + 7 = 6x^2 - 23x + 7$                                                             
             \end{solutionordottedlines}                                                                        
         \part \((2b+3)(4b-2)=\)                                                                                
             \begin{solutionordottedlines}[2cm]                                                                 
             $8b^2 - 4b + 12b - 6 = 8b^2 + 8b - 6$                                                              
             \end{solutionordottedlines}                                                                        
         \part \((4c+d)(2c-3d)=\)                                                                               
             \begin{solutionordottedlines}[2cm]                                                                 
             $8c^2 - 12cd + 2cd - 3d^2 = 8c^2 - 10cd - 3d^2$                                                    
             \end{solutionordottedlines}                                                                        
         \part \((3x-y)(2x+5y)=\)                                                                               
             \begin{solutionordottedlines}[2cm]                                                                 
             $6x^2 + 15xy - 2xy - 5y^2 = 6x^2 + 13xy - 5y^2$                                                    
             \end{solutionordottedlines}                                                                        
         \part \((2p-5q)(3q-2p)=\)                                                                              
             \begin{solutionordottedlines}[2cm]                                                                 
             $-4p^2 + 10pq + 6pq - 15q^2 = -4p^2 + 16pq - 15q^2$                                                
             \end{solutionordottedlines}                                                                        
         \end{multicols}\end{parts}                                                                             
     \end{questions}                                                                                            
 \end{exercisebox}                                                                                              
|726141404|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises}
    \begin{questions}
        \question[]
        \begin{parts}\begin{multicols}{2}
        \part \((a+3)(a+9)=\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \((a+8)(9+a)=\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \((p-6)(p+4)=\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \((x+3)(x-8)=\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \((x+7)(x-4)=\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \((5x+1)(x+2)=\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \((4m+3)(2m-1)=\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \((2x-7)(3x-1)=\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \((2b+3)(4b-2)=\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \((4c+d)(2c-3d)=\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \((3x-y)(2x+5y)=\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \((2p-5q)(3q-2p)=\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \end{questions}
\end{exercisebox}
|726141410|kitty|/Applications/kitty.app|0|
 \begin{examplebox}                                                                                             
     \subsection{Examples}                                                                                      
     Let's only do a few examples this time. We'll come back and do more practise after covering                
 \emph{differences of two squares.}                                                                             
     \begin{doublespace}                                                                                        
     \begin{questions}                                                                                          
         \question[]                                                                                            
         \begin{parts}                                                                                          
             \part \((x-5)^2= \)                                                                                
                 \begin{solutionordottedlines}[1cm]                                                             
                 $x^2 - 10x + 25$                                                                               
                 \end{solutionordottedlines}                                                                    
             \part \((x+7)^2=\)                                                                                 
                 \begin{solutionordottedlines}[1cm]                                                             
                 $x^2 + 14x + 49$                                                                               
                 \end{solutionordottedlines}                                                                    
             \part \((3x-1)^2=\)                                                                                
                 \begin{solutionordottedlines}[1cm]                                                             
                 $9x^2 - 6x + 1$                                                                                
                 \end{solutionordottedlines}                                                                    
         \end{parts}                                                                                            
     \end{questions}                                                                                            
     \end{doublespace}                                                                                          
 \end{examplebox}                                                                                               
|726141443|kitty|/Applications/kitty.app|0|
\begin{examplebox}
    \subsection{Examples}
    Let's only do a few examples this time. We'll come back and do more practise after covering \emph{differences of two squares.}
    \begin{doublespace}
    \begin{questions}
        \question[]
        \begin{parts}
            \part \((x-5)^2= \)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \part \((x+7)^2=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \part \((3x-1)^2=\)
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
        \end{parts}
    \end{questions}
    \end{doublespace}
\end{examplebox}
|726141450|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                            
     \subsection{Exercises}                                                                                     
     \begin{questions}                                                                                          
         \Question[3] Let's practise:                                                                           
             \begin{parts}                                                                                      
                 \part \((x-5)(x+5)=\)\fillin[x^2 - 25]{2}                                                      
                 \part \((3x-4)(3x+4)=\)\fillin[9x^2 - 16]{2}                                                   
                 \part \((a+b)(a-b)=\)\fillin[a^2 - b^2]{2}                                                     
             \end{parts}                                                                                        
         \Question[4] Now back to perfect squares:                                                              
             \begin{multicols}{2}                                                                               
             \begin{parts}                                                                                      
                 \part \((x+1)^2=\)\fillin[x^2 + 2x + 1]{2}                                                     
                 \part \((x+5)^2=\)\fillin[x^2 + 10x + 25]{2}                                                   
                 \part \((2+x)^2=\)\fillin[x^2 + 4x + 4]{2}                                                     
                 \part \((x+20)^2=\)\fillin[x^2 + 40x + 400]{2}                                                 
             \end{parts}                                                                                        
             \end{multicols}                                                                                    
         \Question[6] Try a mix now:                                                                            
             \begin{multicols}{2}                                                                               
             \begin{parts}                                                                                      
                 \part \((3x-2)(3x+2)=\)\fillin[9x^2 - 4]{3}                                                    
                 \part \((3a-4b)^2=\)\fillin[9a^2 - 24ab + 16b^2]{2}                                            
                 \part \((2x+3y)^2=\)\fillin[4x^2 + 12xy + 9y^2]{2}                                             
                 \part \((5a+2b)(5a-2b)=\)\fillin[25a^2 - 4b^2]{2}                                              
                 \part \((\frac{x}{2}+3)^2=\)\fillin[\frac{x^2}{4} + 3x + 9]{2}                                 
                 \part \((3c-b)^2=\)\fillin[9c^2 - 6bc + b^2]{2}                                                
             \end{parts}                                                                                        
             \end{multicols}                                                                                    
     \end{questions}                                                                                            
 \end{exercisebox}                                                                                              
|726141468|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises}
    \begin{questions}
        \Question[3] Let's practise:
            \begin{parts}
                \part \((x-5)(x+5)=\)\mdots{2}
                \part \((3x-4)(3x+4)=\)\mdots{2}
                \part \((a+b)(a-b)=\)\mdots{2}
            \end{parts}
        \Question[4] Now back to perfect squares:
            \begin{multicols}{2}
            \begin{parts}
                \part \((x+1)^2=\)\mdots{2}
                \part \((x+5)^2=\)\mdots{2}
                \part \((2+x)^2=\)\mdots{2}
                \part \((x+20)^2=\)\mdots{2}
            \end{parts}
            \end{multicols}
        \Question[6] Try a mix now:
            \begin{multicols}{2}
            \begin{parts}
                \part \((3x-2)(3x+2)=\)\mdots{3}
                \part \((3a-4b)^2=\)\mdots{2}
                \part \((2x+3y)^2=\)\mdots{2}
                \part \((5a+2b)(5a-2b)=\)\mdots{2}
                \part \((\frac{x}{2}+3)^2=\)\mdots{2}
                \part \((3c-b)^2=\)\mdots{2}
            \end{parts}
            \end{multicols}
    \end{questions}
\end{exercisebox}
|726141472|kitty|/Applications/kitty.app|0|
\begin{questions}
    \Question[3] Evaluate \(2m(m-3n)\) when:
        \begin{multicols}{3}
        \begin{parts}
            \part \(m=3,n=5\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(m=-3, n=-2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(m=\frac{1}{3}, n=\frac{1}{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \end{multicols}
    \Question[1] Evaluate \(\frac{p+2q}{3r}\) when \(p = 7, q = -2, r = 2\) 
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \Question[1] Evaluate \(\frac{x+y}{3}\) when \(x = -6, y = -5\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \Question[3] Fill in the missing term:
        \begin{multicols}{3}
        \begin{parts}
            \part \(2a+\dotfill=7a\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(5m^2-\dotfill = -6m^2n\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(-6lm+\dotfill=lm\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \end{multicols}
    \Question[6] Simplify by collecting like terms:
        \begin{multicols}{3}
        \begin{parts}
            \part \(9a^2+5a^2-12a^2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(14a^2d-10a^2d-6a^2d\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(17m^2-14m^2+8m^2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(-4x^2+3x^2-3y-7y\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(7x^3+6x^2-4y^3-x^2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(-3ab^2+4a^2b-5ab^2+a^2b\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \end{multicols}
    \Question[12] Simplify
        \begin{multicols}{4}
        \begin{parts}
            \part \(4a\times 3b\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(-2p \times(-3q)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(27y \div 3\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(3 \times 12t \div 9\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(24x\div 8 \times 3\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(-\frac{12m}{18}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{12ab}{4a}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{3x}{5}\times\frac{2}{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{2}{5a}\times\frac{1}{4a}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{3x}{5}\div\frac{3}{4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{9y}{2}\div 18\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{5p}{6}\div(-\frac{10p}{3})\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \end{multicols}
    \Question[4] Fill in the missing boxes. Each box contains the product of the 2 boxes below it.
        \begin{multicols}{2}
        \begin{centering}
        \begin{parts}
            \part
                \begin{centering}
                \begin{tikzpicture}
                    \draw (0,0) rectangle (1,1);
                    \draw (1,0) rectangle (2,1);
                    \draw (2,0) rectangle (3,1);
                    \draw (0.5,1) rectangle (1.5,2);
                    \draw (1.5,1) rectangle (2.5,2);
                    \draw (1,2) rectangle (2,3);

                    \node at (0.5,0.5) {$2a$};
                    \node at (1.5,0.5) {$4b$};
                    \node at (2.5,0.5) {$5a$};
                \end{tikzpicture}
                \end{centering}
            \part 
                \begin{centering}
                \begin{tikzpicture}
                    \draw (0,0) rectangle (1,1);
                    \draw (1,0) rectangle (2,1);
                    \draw (2,0) rectangle (3,1);
                    \draw (0.5,1) rectangle (1.5,2);
                    \draw (1.5,1) rectangle (2.5,2);
                    \draw (1,2) rectangle (2,3);

                    \node at (1,1.5) {$24a$};
                    \node at (1.5,2.5) {$48a^2b$};
                    \node at (2.5,0.5) {$2b$};
                \end{tikzpicture}
                \end{centering}
        \end{parts}
        \end{centering}
        \end{multicols}
    \Question[8] Expand:
        \begin{multicols}{4}
        \begin{parts}
            \part \(b(b+7)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(4h(5h-7)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(-k(5k-4)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(-4x(3x-5)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(4c(2c-d)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(-3x(2x+5y)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(3p(2-5pq)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(-10b(3a-7b)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \end{multicols}
    \Question[6] Expand and collect like terms
        \begin{multicols}{3}
        \begin{parts}
            \part \(\frac{1}{4}(x+2)+\frac{x}{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{3}{7}(3x+5)+\frac{x}{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(-\frac{1}{2}(3x+2)-\frac{2x}{5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(2p(3p+1)-5(p+1)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(2p(3p+1)-4(2p+1)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(4z(4z-2)-z(z+2)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \end{multicols}
|726141532|kitty|/Applications/kitty.app|0|
 \begin{questions}                                                                                              
     \Question[3] Evaluate \(2m(m-3n)\) when:                                                                   
         \begin{multicols}{3}                                                                                   
         \begin{parts}                                                                                          
             \part \(m=3,n=5\)                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(2 \cdot 3(3-3 \cdot 5) = 6(3-15) = 6(-12) = -72\)                                            
             \end{solutionordottedlines}                                                                        
             \part \(m=-3, n=-2\)                                                                               
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(2 \cdot (-3)(-3-3 \cdot (-2)) = -6(-3+6) = -6 \cdot 3 = -18\)                                
             \end{solutionordottedlines}                                                                        
             \part \(m=\frac{1}{3}, n=\frac{1}{2}\)                                                             
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(2 \cdot \frac{1}{3}\left(\frac{1}{3}-3 \cdot \frac{1}{2}\right) =                            
 \frac{2}{3}\left(\frac{1}{3}-\frac{3}{2}\right) = \frac{2}{3}\left(\frac{2-9}{6}\right) = \frac{2}{3} \cdot    
 \frac{-7}{6} = -\frac{7}{9}\)                                                                                  
             \end{solutionordottedlines}                                                                        
         \end{parts}                                                                                            
         \end{multicols}                                                                                        
     \Question[1] Evaluate \(\frac{p+2q}{3r}\) when \(p = 7, q = -2, r = 2\)                                    
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(\frac{7+2(-2)}{3 \cdot 2} = \frac{7-4}{6} = \frac{3}{6} = \frac{1}{2}\)                      
             \end{solutionordottedlines}                                                                        
     \Question[1] Evaluate \(\frac{x+y}{3}\) when \(x = -6, y = -5\)                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(\frac{-6-5}{3} = \frac{-11}{3}\)                                                             
             \end{solutionordottedlines}                                                                        
     \Question[3] Fill in the missing term:                                                                     
         \begin{multicols}{3}                                                                                   
         \begin{parts}                                                                                          
             \part \(2a+\dotfill=7a\)                                                                           
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(2a+5a=7a\)                                                                                   
             \end{solutionordottedlines}                                                                        
             \part \(5m^2-\dotfill = -6m^2n\)                                                                   
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(5m^2-11m^2n = -6m^2n\)                                                                       
             \end{solutionordottedlines}                                                                        
             \part \(-6lm+\dotfill=lm\)                                                                         
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(-6lm+7lm=lm\)                                                                                
             \end{solutionordottedlines}                                                                        
         \end{parts}                                                                                            
         \end{multicols}                                                                                        
     \Question[6] Simplify by collecting like terms:                                                            
         \begin{multicols}{3}                                                                                   
         \begin{parts}                                                                                          
             \part \(9a^2+5a^2-12a^2\)                                                                          
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(9a^2+5a^2-12a^2 = 2a^2\)                                                                     
             \end{solutionordottedlines}                                                                        
             \part \(14a^2d-10a^2d-6a^2d\)                                                                      
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(14a^2d-10a^2d-6a^2d = -2a^2d\)                                                               
             \end{solutionordottedlines}                                                                        
             \part \(17m^2-14m^2+8m^2\)                                                                         
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(17m^2-14m^2+8m^2 = 11m^2\)                                                                   
             \end{solutionordottedlines}                                                                        
             \part \(-4x^2+3x^2-3y-7y\)                                                                         
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(-4x^2+3x^2-3y-7y = -x^2-10y\)                                                                
             \end{solutionordottedlines}                                                                        
             \part \(7x^3+6x^2-4y^3-x^2\)                                                                       
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(7x^3+6x^2-4y^3-x^2 = 7x^3+5x^2-4y^3\)                                                        
             \end{solutionordottedlines}                                                                        
             \part \(-3ab^2+4a^2b-5ab^2+a^2b\)                                                                  
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(-3ab^2+4a^2b-5ab^2+a^2b = -8ab^2+5a^2b\)                                                     
             \end{solutionordottedlines}                                                                        
         \end{parts}                                                                                            
         \end{multicols}                                                                                        
     \Question[12] Simplify                                                                                     
         \begin{multicols}{4}                                                                                   
         \begin{parts}                                                                                          
             \part \(4a\times 3b\)                                                                              
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(4a \times 3b = 12ab\)                                                                        
             \end{solutionordottedlines}                                                                        
             \part \(-2p \times(-3q)\)                                                                          
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(-2p \times(-3q) = 6pq\)                                                                      
             \end{solutionordottedlines}                                                                        
             \part \(27y \div 3\)                                                                               
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(27y \div 3 = 9y\)                                                                            
             \end{solutionordottedlines}                                                                        
             \part \(3 \times 12t \div 9\)                                                                      
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(3 \times 12t \div 9 = 4t\)                                                                   
             \end{solutionordottedlines}                                                                        
             \part \(24x\div 8 \times 3\)                                                                       
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(24x\div 8 \times 3 = 9x\)                                                                    
             \end{solutionordottedlines}                                                                        
             \part \(-\frac{12m}{18}\)                                                                          
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(-\frac{12m}{18} = -\frac{2}{3}m\)                                                            
             \end{solutionordottedlines}                                                                        
             \part \(\frac{12ab}{4a}\)                                                                          
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(\frac{12ab}{4a} = 3b\)                                                                       
             \end{solutionordottedlines}                                                                        
             \part \(\frac{3x}{5}\times\frac{2}{3}\)                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(\frac{3x}{5}\times\frac{2}{3} = \frac{2x}{5}\)                                               
             \end{solutionordottedlines}                                                                        
             \part \(\frac{2}{5a}\times\frac{1}{4a}\)                                                           
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(\frac{2}{5a}\times\frac{1}{4a} = \frac{2}{20a^2} = \frac{1}{10a^2}\)                         
             \end{solutionordottedlines}                                                                        
             \part \(\frac{3x}{5}\div\frac{3}{4}\)                                                              
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(\frac{3x}{5}\div\frac{3}{4} = \frac{3x}{5} \times \frac{4}{3} = \frac{4x}{5}\)               
             \end{solutionordottedlines}                                                                        
             \part \(\frac{9y}{2}\div 18\)                                                                      
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(\frac{9y}{2}\div 18 = \frac{9y}{2} \times \frac{1}{18} = \frac{y}{4}\)                       
             \end{solutionordottedlines}                                                                        
             \part \(\frac{5p}{6}\div(-\frac{10p}{3})\)                                                         
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(\frac{5p}{6}\div(-\frac{10p}{3}) = \frac{5p}{6} \times \frac{-3}{10p} = -\frac{1}{4}\)       
             \end{solutionordottedlines}                                                                        
         \end{parts}                                                                                            
         \end{multicols}                                                                                        
     \Question[4] Fill in the missing boxes. Each box contains the product of the 2 boxes below it.             
         \begin{multicols}{2}                                                                                   
         \begin{centering}                                                                                      
         \begin{parts}                                                                                          
             \part                                                                                              
                 \begin{centering}                                                                              
                 \begin{tikzpicture}                                                                            
                     \draw (0,0) rectangle (1,1);                                                               
                     \draw (1,0) rectangle (2,1);                                                               
                     \draw (2,0) rectangle (3,1);                                                               
                     \draw (0.5,1) rectangle (1.5,2);                                                           
                     \draw (1.5,1) rectangle (2.5,2);                                                           
                     \draw (1,2) rectangle (2,3);                                                               
                                                                                                                
                     \node at (0.5,0.5) {$2a$};                                                                 
                     \node at (1.5,0.5) {$4b$};                                                                 
                     \node at (2.5,0.5) {$5a$};                                                                 
                     \node at (1,1.5) {$8ab$};                                                                  
                     \node at (2,1.5) {$20ab$};                                                                 
                     \node at (1.5,2.5) {$160a^2b$};                                                            
                 \end{tikzpicture}                                                                              
                 \end{centering}                                                                                
             \part                                                                                              
                 \begin{centering}                                                                              
                 \begin{tikzpicture}                                                                            
                     \draw (0,0) rectangle (1,1);                                                               
                     \draw (1,0) rectangle (2,1);                                                               
                     \draw (2,0) rectangle (3,1);                                                               
                     \draw (0.5,1) rectangle (1.5,2);                                                           
                     \draw (1.5,1) rectangle (2.5,2);                                                           
                     \draw (1,2) rectangle (2,3);                                                               
                                                                                                                
                     \node at (1,1.5) {$24a$};                                                                  
                     \node at (1.5,2.5) {$48a^2b$};                                                             
                     \node at (2.5,0.5) {$2b$};                                                                 
                     \node at (0.5,0.5) {$12a$};                                                                
                     \node at (1.5,0.5) {$2a$};                                                                 
                     \node at (2,1.5) {$4ab$};                                                                  
                 \end{tikzpicture}                                                                              
                 \end{centering}                                                                                
         \end{parts}                                                                                            
         \end{centering}                                                                                        
         \end{multicols}                                                                                        
     \Question[8] Expand:                                                                                       
         \begin{multicols}{4}                                                                                   
         \begin{parts}                                                                                          
             \part \(b(b+7)\)                                                                                   
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(b(b+7) = b^2+7b\)                                                                            
             \end{solutionordottedlines}                                                                        
             \part \(4h(5h-7)\)                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(4h(5h-7) = 20h^2-28h\)                                                                       
             \end{solutionordottedlines}                                                                        
             \part \(-k(5k-4)\)                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(-k(5k-4) = -5k^2+4k\)                                                                        
             \end{solutionordottedlines}                                                                        
             \part \(-4x(3x-5)\)                                                                                
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(-4x(3x-5) = -12x^2+20x\)                                                                     
             \end{solutionordottedlines}                                                                        
             \part \(4c(2c-d)\)                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(4c(2c-d) = 8c^2-4cd\)                                                                        
             \end{solutionordottedlines}                                                                        
             \part \(-3x(2x+5y)\)                                                                               
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(-3x(2x+5y) = -6x^2-15xy\)                                                                    
             \end{solutionordottedlines}                                                                        
             \part \(3p(2-5pq)\)                                                                                
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(3p(2-5pq) = 6p-15p^2q\)                                                                      
             \end{solutionordottedlines}                                                                        
             \part \(-10b(3a-7b)\)                                                                              
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(-10b(3a-7b) = -30ab+70b^2\)                                                                  
             \end{solutionordottedlines}                                                                        
         \end{parts}                                                                                            
         \end{multicols}                                                                                        
     \Question[6] Expand and collect like terms                                                                 
         \begin{multicols}{3}                                                                                   
         \begin{parts}                                                                                          
             \part \(\frac{1}{4}(x+2)+\frac{x}{3}\)                                                             
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(\frac{1}{4}(x+2)+\frac{x}{3} = \frac{1}{4}x+\frac{1}{2}+\frac{x}{3} =                        
 \frac{7}{12}x+\frac{1}{2}\)                                                                                    
             \end{solutionordottedlines}                                                                        
             \part \(\frac{3}{7}(3x+5)+\frac{x}{3}\)                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(\frac{3}{7}(3x+5)+\frac{x}{3} = \frac{9}{7}x+\frac{15}{7}+\frac{x}{3} =                      
 \frac{37}{21}x+\frac{15}{7}\)                                                                                  
             \end{solutionordottedlines}                                                                        
             \part \(-\frac{1}{2}(3x+2)-\frac{2x}{5}\)                                                          
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(-\frac{1}{2}(3x+2)-\frac{2x}{5} = -\frac{3}{2}x-1-\frac{2x}{5} = -\frac{19}{10}x-1\)         
             \end{solutionordottedlines}                                                                        
             \part \(2p(3p+1)-5(p+1)\)                                                                          
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(2p(3p+1)-5(p+1) = 6p^2+2p-5p-5 = 6p^2-3p-5\)                                                 
             \end{solutionordottedlines}                                                                        
             \part \(2p(3p+1)-4(2p+1)\)                                                                         
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(2p(3p+1)-4(2p+1) = 6p^2+2p-8p-4 = 6p^2-6p-4\)                                                
             \end{solutionordottedlines}                                                                        
             \part \(4z(4z-2)-z(z+2)\)                                                                          
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(4z(4z-2)-z(z+2) = 16z^2-8z-z^2-2z = 15z^2-10z\)                                              
             \end{solutionordottedlines}                                                                        
         \end{parts}                                                                                            
         \end{multicols}                                                                                        
|726141572|kitty|/Applications/kitty.app|0|
 \Question[8] Expand:                                                                                                           
         \begin{multicols}{4}                                                                                                   
         \begin{parts}                                                                                                          
             \part \((x-6)(x-4)\)                                                                                               
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(x^2 - 10x + 24\)                                                                                                 
             \end{solutionordottedlines}                                                                                        
             \part \((4x+1)(3x-1)\)                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(12x^2 - 1\)                                                                                                      
             \end{solutionordottedlines}                                                                                        
             \part \((4x+3)(2x-1)\)                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(8x^2 - 4x - 3\)                                                                                                  
             \end{solutionordottedlines}                                                                                        
             \part \((x-4)(2x+5)\)                                                                                              
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(2x^2 + x - 20\)                                                                                                  
             \end{solutionordottedlines}                                                                                        
             \part \((x+3)(x+3)\)                                                                                               
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(x^2 + 6x + 9\)                                                                                                   
             \end{solutionordottedlines}                                                                                        
             \part \((2x-5)(x+3)\)                                                                                              
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(2x^2 + x - 15\)                                                                                                  
             \end{solutionordottedlines}                                                                                        
             \part \((2x+3)(2x+3)\)                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(4x^2 + 12x + 9\)                                                                                                 
             \end{solutionordottedlines}                                                                                        
             \part \((\frac{2b}{3}+2)(\frac{b}{5}-2)\)                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(\frac{2b^2}{15} - \frac{4b}{3} - \frac{2b}{5} + 4\)                                                              
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \end{multicols}                                                                                                        
     \Question[6] Fill in the blanks:                                                                                           
         \begin{multicols}{2}                                                                                                   
         \begin{parts}                                                                                                          
             \part \((x+5)(\fillin[x+3])=x^2+8x+15\)                                                                            
             \part \((x+3)(\fillin[x-5])=x^2-2x-15\)                                                                            
             \part \((3x+4)(\fillin[\frac{1}{3}x-4])=3x^2+x-4\)                                                                 
             \part \((x+\fillin[3])(x+6)=x^2+9x+\fillin[18]\)                                                                   
             \part \((2x+3)(\fillin[\frac{1}{2}x+5])=2x^2+7x+\fillin[15]\)                                                      
             \part \((\fillin[4] x - 3)(\fillin[-3] x 5 \fillin[-1]) = 12x^2-x-6\)                                              
         \end{parts}                                                                                                            
         \end{multicols}                                                                                                        
     \Question[4] Expand                                                                                                        
         \begin{multicols}{4}                                                                                                   
         \begin{parts}                                                                                                          
             \part \((x-7)^2\)                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(x^2 - 14x + 49\)                                                                                                 
             \end{solutionordottedlines}                                                                                        
             \part \((a+8)^8\)                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                 
             This is a binomial expansion that requires the binomial theorem to expand fully.                                   
             \end{solutionordottedlines}                                                                                        
             \part \((9+x)^2\)                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(x^2 + 18x + 81\)                                                                                                 
             \end{solutionordottedlines}                                                                                        
             \part \(x-11)^2\)                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(x^2 - 22x + 121\)                                                                                                
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \end{multicols}                                                                                                        
     \Question[2] Expand                                                                                                        
         \begin{multicols}{2}                                                                                                   
         \begin{parts}                                                                                                          
             \part \((\frac{2x}{5}-1)^2\)                                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(\frac{4x^2}{25} - \frac{4x}{5} + 1\)                                                                             
             \end{solutionordottedlines}                                                                                        
             \part \((\frac{3x}{4}+\frac{2}{3})^2\)                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(\frac{9x^2}{16} + 2x + \frac{4}{9}\)                                                                             
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \end{multicols}                                                                                                        
     \Question[3] Evaluate the following using $ (a+b)^2 = a^2 + 2ab + b^2 $ and $(a-b)^2 = a^2 - 2ab + b^2$.                   
         \begin{multicols}{3}                                                                                                   
         \begin{parts}                                                                                                          
             \part \((1.01)^2\)                                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(1.0201\)                                                                                                         
             \end{solutionordottedlines}                                                                                        
             \part \((0.99)^2\)                                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(0.9801\)                                                                                                         
             \end{solutionordottedlines}                                                                                        
             \part \((4.01)^2\)                                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(16.0801\)                                                                                                        
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \end{multicols}                                                                                                        
     \Question[4] Expand and collect like terms                                                                                 
         \begin{multicols}{2}                                                                                                   
         \begin{parts}                                                                                                          
             \part \((x-2)^2+(x-4)^2\)                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(2x^2 - 12x + 20\)                                                                                                
             \end{solutionordottedlines}                                                                                        
             \part \((2x+5)^2+(2x-5)^2\)                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(8x^2 + 50\)                                                                                                      
             \end{solutionordottedlines}                                                                                        
             \part \(x^2+(x+1)^2+(x+2)^2+(x+3)^2\)                                                                              
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(4x^2 + 20x + 14\)                                                                                                
             \end{solutionordottedlines}                                                                                        
             \part \((\frac{x}{2}+1)^2 + (\frac{x}{2}-1)^2\)                                                                    
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(\frac{x^2}{2} + 2\)                                                                                              
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \end{multicols}                                                                                                        
     \Question[6] Expand                                                                                                        
         \begin{multicols}{2}                                                                                                   
         \begin{parts}                                                                                                          
             \part \((z-7)(z+7)\)                                                                                               
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(z^2 - 49\)                                                                                                       
             \end{solutionordottedlines}                                                                                        
             \part \((10-x)(10+x)\)                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(100 - x^2\)                                                                                                      
             \end{solutionordottedlines}                                                                                        
             \part \((3x-2)(3x+2)\)                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(9x^2 - 4\)                                                                                                       
             \end{solutionordottedlines}                                                                                        
             \part \((\frac{x}{2}+3)(\frac{x}{2}-3)\)                                                                           
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(\frac{x^2}{4} - 9\)                                                                                              
             \end{solutionordottedlines}                                                                                        
             \part \((\frac{x}{3}+\frac{1}{2})(\frac{x}{3}-\frac{1}{2})\)                                                       
             \begin{solutionordottedlines}[2cm]                                                                                 
             \(\frac{x^2}{9} - \frac{1}{4}\)                                                                                    
             \end{solutionordottedlines}                                                                                        
             \part Is $ a^2-2a+1 $ a perfect square expansion or a difference of 2 squares?                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
             Perfect square expansion of $(a-1)^2$.                                                                             
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \end{multicols}                                                                                                        
 \end{questions}                                                                                                                
                                                                                                                                
 \subsection{Challenge Problems}                                                                                                
 \begin{questions}                                                                                                              
     \Question[5]                                                                                                               
         \begin{parts}                                                                                                          
             \part Show that the perimeter of the rectangle is \((4x+6)\)cm                                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
             Perimeter = $2(AB + AD) = 2((x+3) + x) = 2(2x+3) = 4x+6$ cm.                                                       
             \end{solutionordottedlines}                                                                                        
             \part Find the perimeter if $AD = 2\text{cm}$                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                 
             Perimeter = $2(AB + AD) = 2((2+3) + 2) = 2(5+2) = 14$ cm.                                                          
             \end{solutionordottedlines}                                                                                        
             \part Find $x$ if the perimeter = $36\text{cm}$                                                                    
             \begin{solutionordottedlines}[2cm]                                                                                 
             $4x+6 = 36 \Rightarrow 4x = 30 \Rightarrow x = 7.5$ cm.                                                            
             \end{solutionordottedlines}                                                                                        
             \part Find the area of $ABCD$ in terms of $x$                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                 
             Area = $AB \times AD = (x+3) \times x = x^2 + 3x$ cm\(^2\).                                                        
             \end{solutionordottedlines}                                                                                        
             \part Find the area of the rectangle if $AB=6\text{cm}$                                                            
             \begin{solutionordottedlines}[2cm]                                                                                 
             Area = $AB \times AD = 6 \times 2 = 12$ cm\(^2\).                                                                  
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \begin{center}                                                                                                         
         \begin{tikzpicture}                                                                                                    
             \draw (0,0) rectangle (5,2);                                                                                       
             \draw (0,0) rectangle (0.2,0.2);                                                                                   
             \draw (4.8,0.2) rectangle (5,0);                                                                                   
             \draw (0,2) rectangle (0.2,1.8);                                                                                   
             \draw (5,2) rectangle (4.8,1.8);                                                                                   
                                                                                                                                
             \node at (0,0) [below left] {$D$};                                                                                 
             \node at (0,2) [above left] {$A$};                                                                                 
             \node at (5,2) [above right] {$B$};                                                                                
             \node at (5,0) [below right] {$C$};                                                                                
             \node at (0,1) [left] {$x$};                                                                                       
             \node at (2.5,2) [above] {$x+3$};                                                                                  
                                                                                                                                
         \end{tikzpicture}                                                                                                      
         \end{center}                                                                                                           
     \Question[6]  Expand and collect:                                                                                          
         \begin{parts}                                                                                                          
             \part \((x-1)(x^2+x+1)\)                                                                                           
             \begin{solutionordottedlines}[3cm]                                                                                 
             \(x^3 - 1\)                                                                                                        
             \end{solutionordottedlines}                                                                                        
             \part \((x-1)(x^4+x^3+x^2+x+1)\)                                                                                   
             \begin{solutionordottedlines}[3cm]                                                                                 
             \(x^5 - 1\)                                                                                                        
             \end{solutionordottedlines}                                                                                        
             \part What do you expect the result of expanding \((x-1)(x^9+x^8+\dots+1)\) will be?                               
             \begin{solutionordottedlines}[3cm]                                                                                 
             \(x^{10} - 1\)                                                                                                     
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
 \end{questions}                                                                                                                
|726141773|kitty|/Applications/kitty.app|0|
    \Question[8] Expand:
        \begin{multicols}{4}
        \begin{parts}
            \part \((x-6)(x-4)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((4x+1)(3x-1)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((4x+3)(2x-1)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((x-4)(2x+5)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((x+3)(x+3)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((2x-5)(x+3)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((2x+3)(2x+3)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((\frac{2b}{3}+2)(\frac{b}{5}-2)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \end{multicols}
    \Question[6] Fill in the blanks:
        \begin{multicols}{2}
        \begin{parts}
            \part \((x+5)(\fillin[])=x^2+8x+15\)
            \part \((x+3)(\fillin[])=x^2-2x-15\)
            \part \((3x+4)(\fillin[])=3x^2+x-4\)
            \part \((x+\fillin[])(x+6)=x^2+9x+\fillin[]\)
            \part \((2x+3)(\fillin[])=2x^2+7x+\fillin[]\)
            \part \((\fillin[] x - 3)(\fillin[] x 5 \fillin[]) = 12x^2-x-6\)
        \end{parts}
        \end{multicols}
    \Question[4] Expand
        \begin{multicols}{4}
        \begin{parts}
            \part \((x-7)^2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((a+8)^8\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((9+x)^2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(x-11)^2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \end{multicols}
    \Question[2] Expand
        \begin{multicols}{2}
        \begin{parts}
            \part \((\frac{2x}{5}-1)^2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((\frac{3x}{4}+\frac{2}{3})^2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \end{multicols}
    \Question[3] Evaluate the following using $ (a+b)^2 = a^2 + 2ab + b^2 $ and $(a-b)^2 = a^2 - 2ab + b^2$.
        \begin{multicols}{3}
        \begin{parts}
            \part \((1.01)^2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((0.99)^2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((4.01)^2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \end{multicols}
    \Question[4] Expand and collect like terms
        \begin{multicols}{2}
        \begin{parts}
            \part \((x-2)^2+(x-4)^2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((2x+5)^2+(2x-5)^2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(x^2+(x+1)^2+(x+2)^2+(x+3)^2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((\frac{x}{2}+1)^2 + (\frac{x}{2}-1)^2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \end{multicols}
    \Question[6] Expand
        \begin{multicols}{2}
        \begin{parts}
            \part \((z-7)(z+7)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((10-x)(10+x)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((3x-2)(3x+2)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((\frac{x}{2}+3)(\frac{x}{2}-3)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((\frac{x}{3}+\frac{1}{2})(\frac{x}{3}-\frac{1}{2})\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part Is $ a^2-2a+1 $ a perfect square expansion or a difference of 2 squares?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \end{multicols}
\end{questions}

\subsection{Challenge Problems}
\begin{questions}
    \Question[5]
        \begin{parts}
            \part Show that the perimeter of the rectangle is \((4x+6)\)cm
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part Find the perimeter if $AD = 2\text{cm}$
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part Find $x$ if the perimeter = $36\text{cm}$
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part Find the area of $ABCD$ in terms of $x$
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part Find the area of the rectangle if $AB=6\text{cm}$
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \begin{center}
        \begin{tikzpicture}
            \draw (0,0) rectangle (5,2);
            \draw (0,0) rectangle (0.2,0.2);
            \draw (4.8,0.2) rectangle (5,0);
            \draw (0,2) rectangle (0.2,1.8);
            \draw (5,2) rectangle (4.8,1.8);

            \node at (0,0) [below left] {$D$};
            \node at (0,2) [above left] {$A$};
            \node at (5,2) [above right] {$B$};
            \node at (5,0) [below right] {$C$};
            \node at (0,1) [left] {$x$};
            \node at (2.5,2) [above] {$x+3$};
        
        \end{tikzpicture}
        \end{center}
    \Question[6]  Expand and collect:
        \begin{parts}
            \part \((x-1)(x^2+x+1)\)
            \begin{solutionordottedlines}[3cm]
            \end{solutionordottedlines}
            \part \((x-1)(x^4+x^3+x^2+x+1)\)
            \begin{solutionordottedlines}[3cm]
            \end{solutionordottedlines}
            \part What do you expect the result of expanding \((x-1)(x^9+x^8+\dots+1)\) will be?
            \begin{solutionordottedlines}[3cm]
            \end{solutionordottedlines}
        \end{parts}
\end{questions}
|726141787|kitty|/Applications/kitty.app|0|
q|726141925|kitty|/Applications/kitty.app|0|
-18x|726141944|kitty|/Applications/kitty.app|0|
6|726141966|kitty|/Applications/kitty.app|0|
^|726141967|kitty|/Applications/kitty.app|0|
         A statement that has been proven on the basis of previously established statements.                                                 
|726191957|kitty|/Applications/kitty.app|0|
         The longest side of a right-angled triangle, opposite the right angle.                                                              
|726191962|kitty|/Applications/kitty.app|0|
 \begin{examplebox}                                                                                                                          
     \section*{Examples}                                                                                                                     
         \begin{questions}                                                                                                                   
             \Question[1]\,                                                                                                                  
                 \begin{center}                                                                                                              
                     \begin{tikzpicture}[scale=1.5]                                                                                          
                         \draw (0,0) -- (4,0) node[midway,below] {$40$};                                                                     
                         \draw (4,0) -- (0,0.9) node[midway,sloped,above] {$x$};                                                             
                         \draw (0,0.9) -- (0,0) node[midway,left] {$9$};                                                                     
                         \draw (0,0) rectangle (0.2, 0.2);                                                                                   
                     \end{tikzpicture}                                                                                                       
                 \end{center}                                                                                                                
                 \begin{solutionordottedlines}[1in]                                                                                          
                     $x = \sqrt{40^2 + 9^2} = \sqrt{1600 + 81} = \sqrt{1681} = 41$                                                           
                 \end{solutionordottedlines}                                                                                                 
             \Question[1]\,                                                                                                                  
                 \begin{center}                                                                                                              
                     \begin{tikzpicture}[scale=1.5]                                                                                          
                         \draw (0,0) -- (1.6,0) node[midway,below] {$x$};                                                                    
                         \draw (1.6,0) -- (1.6,3) node[midway, right] {$30$};                                                                
                         \draw (1.6,3)--(0,0) node[midway,above,sloped] {$34$};                                                              
                         \draw (1.6,0) rectangle (1.4, 0.2);                                                                                 
                     \end{tikzpicture}                                                                                                       
                 \end{center}                                                                                                                
                 \begin{solutionordottedlines}[1in]                                                                                          
                     $x = \sqrt{34^2 - 30^2} = \sqrt{1156 - 900} = \sqrt{256} = 16$                                                          
                 \end{solutionordottedlines}                                                                                                 
             \Question[1]\,                                                                                                                  
                 \begin{center}                                                                                                              
                     \begin{tikzpicture}[scale=1.5]                                                                                          
                         \draw (0,0) -- (3,0) node[midway,below] {$30$};                                                                     
                         \draw (3,0) -- (0,1.2) node[midway, above] {$x$};                                                                   
                         \draw (0,1.2)--(0,0) node[midway,left] {$12$};                                                                      
                         \draw (0,0) rectangle (0.2, 0.2);                                                                                   
                     \end{tikzpicture}                                                                                                       
                 \end{center}                                                                                                                
                 \begin{solutionordottedlines}[1in]                                                                                          
                     $x = \sqrt{30^2 + 12^2} = \sqrt{900 + 144} = \sqrt{1044} \approx 32.31$                                                 
                 \end{solutionordottedlines}                                                                                                 
         \end{questions}                                                                                                                     
 \end{examplebox}                                                                                                                            
|726192003|kitty|/Applications/kitty.app|0|
\begin{examplebox}
    \section*{Examples}
        \begin{questions}
            \Question[1]\,
                \begin{center}
                    \begin{tikzpicture}[scale=1.5]
                        \draw (0,0) -- (4,0) node[midway,below] {$40$};
                        \draw (4,0) -- (0,0.9) node[midway,sloped,above] {$x$};
                        \draw (0,0.9) -- (0,0) node[midway,left] {$9$};
                        \draw (0,0) rectangle (0.2, 0.2);
                    \end{tikzpicture}
                \end{center}
                \begin{solutionordottedlines}[1in]
                \end{solutionordottedlines}
            \Question[1]\,
                \begin{center}
                    \begin{tikzpicture}[scale=1.5]
                        \draw (0,0) -- (1.6,0) node[midway,below] {$x$};
                        \draw (1.6,0) -- (1.6,3) node[midway, right] {$30$};
                        \draw (1.6,3)--(0,0) node[midway,above,sloped] {$34$};
                        \draw (1.6,0) rectangle (1.4, 0.2);
                    \end{tikzpicture}
                \end{center}
                \begin{solutionordottedlines}[1in]
                \end{solutionordottedlines}
            \Question[1]\,
                \begin{center}
                    \begin{tikzpicture}[scale=1.5]
                        \draw (0,0) -- (3,0) node[midway,below] {$30$};
                        \draw (3,0) -- (0,1.2) node[midway, above] {$x$};
                        \draw (0,1.2)--(0,0) node[midway,left] {$12$};
                        \draw (0,0) rectangle (0.2, 0.2);
                    \end{tikzpicture}
                \end{center}
                \begin{solutionordottedlines}[1in]
                \end{solutionordottedlines}
        \end{questions}
\end{examplebox}
|726192008|kitty|/Applications/kitty.app|0|
     The third example involves finding the hypotenuse $x$ given the two shorter sides, whereas the previous two examples involve finding on 
 of the shorter sides given the hypotenuse and the other shorter side.                                                                       
|726192034|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                                                         
     \section*{Exercises}                                                                                                                    
     \begin{questions}                                                                                                                       
         \question                                                                                                                           
             A door frame has height $1.7$m and width $1$m. Will a square piece of board $2$m by $2$m fit through the doorway?               
             \begin{solutionordottedlines}[1.5in]                                                                                            
                 The diagonal of the door frame is $\sqrt{1.7^2 + 1^2} = \sqrt{2.89 + 1} = \sqrt{3.89} \approx 1.97$m, which is less than th 
 side of the board ($2$m). Therefore, the board will not fit through the doorway.                                                            
             \end{solutionordottedlines}                                                                                                     
                                                                                                                                             
             \begin{multicols}{2}                                                                                                            
                 \Question[1]\,                                                                                                              
                     \begin{center}                                                                                                          
                         \drawTriangle{bottom left}{10}{6}{8}{$6$}{$8$}{$x$}{0.5}                                                            
                     \end{center}                                                                                                            
                     \begin{solutionordottedlines}[1in]                                                                                      
                         $x = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$                                                           
                     \end{solutionordottedlines}                                                                                             
                 \Question[1]\,                                                                                                              
                     \begin{center}                                                                                                          
                         \drawTriangle{top left}{13}{5}{12}{$12$}{$5$}{$x$}{0.5}                                                             
                     \end{center}                                                                                                            
                     \begin{solutionordottedlines}[1in]                                                                                      
                         $x = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$                                                         
                     \end{solutionordottedlines}                                                                                             
                 \Question[1]\,                                                                                                              
                     \begin{center}                                                                                                          
                         \drawTriangle{bottom right}{5}{3}{4}{$3$}{$4$}{$x$}{0.75}                                                           
                     \end{center}                                                                                                            
                     \begin{solutionordottedlines}[1in]                                                                                      
                         $x = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$                                                              
                     \end{solutionordottedlines}                                                                                             
                 \Question[1]\,                                                                                                              
                     \begin{center}                                                                                                          
                         \drawTriangle{top right}{17}{8}{15}{$x$}{$15$}{$17$}{0.4}                                                           
                     \end{center}                                                                                                            
                     \begin{solutionordottedlines}[1in]                                                                                      
                         $x = \sqrt{17^2 - 15^2} = \sqrt{289 - 225} = \sqrt{64} = 8$                                                         
                     \end{solutionordottedlines}                                                                                             
             \end{multicols}                                                                                                                 
     \end{questions}                                                                                                                         
 \end{exercisebox}                                                                                                                           
|726192055|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \section*{Exercises}
    \begin{questions}
        \question 
            A door frame has height $1.7$m and width $1$m. Will a square piece of board $2$m by $2$m fit through the doorway?
            \begin{solutionordottedlines}[1.5in]
            \end{solutionordottedlines}
            
            \begin{multicols}{2}
                \Question[1]\,
                    \begin{center}
                        \drawTriangle{bottom left}{10}{6}{8}{$6$}{$8$}{$x$}{0.5}
                    \end{center}
                    \begin{solutionordottedlines}[1in]
                    \end{solutionordottedlines}
                \Question[1]\,
                    \begin{center}
                        \drawTriangle{top left}{13}{5}{12}{$12$}{$5$}{$x$}{0.5}
                    \end{center}
                    \begin{solutionordottedlines}[1in]
                    \end{solutionordottedlines}
                \Question[1]\,
                    \begin{center}
                        \drawTriangle{bottom right}{5}{3}{4}{$3$}{$4$}{$x$}{0.75}
                    \end{center}
                    \begin{solutionordottedlines}[1in]
                    \end{solutionordottedlines}
                \Question[1]\,
                    \begin{center}
                        \drawTriangle{top right}{17}{8}{15}{$x$}{$15$}{$17$}{0.4}
                    \end{center}
                    \begin{solutionordottedlines}[1in]
                    \end{solutionordottedlines}
            \end{multicols}
    \end{questions}
\end{exercisebox}
|726192067|kitty|/Applications/kitty.app|0|
 \begin{examplebox}                                                                                                                          
     \section*{Examples}                                                                                                                     
     Decide if the triangles with the following side lengths are \textbf{right-angled}.                                                      
     \begin{questions}                                                                                                                       
         \Question[1] 16,30,34                                                                                                               
         \begin{solutionordottedlines}[1in]                                                                                                  
             $16^2 + 30^2 = 256 + 900 = 1156$ and $34^2 = 1156$. Since $16^2 + 30^2 = 34^2$, the triangle is right-angled.                   
         \end{solutionordottedlines}                                                                                                         
         \Question[1] 10,24,26                                                                                                               
         \begin{solutionordottedlines}[1in]                                                                                                  
             $10^2 + 24^2 = 100 + 576 = 676$ and $26^2 = 676$. Since $10^2 + 24^2 = 26^2$, the triangle is right-angled.                     
         \end{solutionordottedlines}                                                                                                         
         \Question[1] 4,6,7                                                                                                                  
         \begin{solutionordottedlines}[1in]                                                                                                  
             $4^2 + 6^2 = 16 + 36 = 52$ and $7^2 = 49$. Since $4^2 + 6^2 \neq 7^2$, the triangle is not right-angled.                        
         \end{solutionordottedlines}                                                                                                         
     \end{questions}                                                                                                                         
 \end{examplebox}                                                                                                                            
|726192101|kitty|/Applications/kitty.app|0|
\begin{examplebox}
    \section*{Examples}
    Decide if the triangles with the following side lengths are \textbf{right-angled}.
    \begin{questions}
        \Question[1] 16,30,34
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \Question[1] 10,24,26
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \Question[1] 4,6,7
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
    \end{questions}
\end{examplebox}
|726192109|kitty|/Applications/kitty.app|0|
 A surd is an irrational number that can be expressed in the form of a root, such as \(\sqrt{2}\), which cannot be simplified to a rational  
|726192131|kitty|/Applications/kitty.app|0|
 A surd is an irrational number that can be expressed in the form of a root, such as \(\sqrt{2}\), which cannot be simplified to a rational number.                                                                                                                                     
|726192135|kitty|/Applications/kitty.app|0|
 \begin{examplebox}                                                                                                                          
     \section*{Examples}                                                                                                                     
     \begin{multicols}{4}                                                                                                                    
         \begin{questions}                                                                                                                   
             \Question[1] \(\sqrt{19}=\)\fillin[4.359]{0.7in}                                                                                
             \Question[1] \(\sqrt{37}=\)\fillin[6.083]{0.7in}                                                                                
             \Question[1] \(\sqrt{161}=\)\fillin[12.688]{0.65in}                                                                             
             \Question[1] \(\sqrt{732}=\)\fillin[27.055]{0.65in}                                                                             
         \end{questions}                                                                                                                     
     \end{multicols}                                                                                                                         
 \end{examplebox}                                                                                                                            
|726192163|kitty|/Applications/kitty.app|0|
\begin{examplebox}
    \section*{Examples}
    \begin{multicols}{4}
        \begin{questions}
            \Question[1] $\sqrt{19}=\fillin[][0.7in]$
            \Question[1] $\sqrt{37}=\fillin[][0.7in]$
            \Question[1] $\sqrt{161}=\fillin[][0.65in]$
            \Question[1] $\sqrt{732}=\fillin[][0.65in]$
        \end{questions}
    \end{multicols}
\end{examplebox}
|726192168|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                                                         
     \section*{Exercises}                                                                                                                    
         \begin{questions}                                                                                                                   
             \begin{multicols}{3}                                                                                                            
                 \Question[1]\,                                                                                                              
                 \drawTriangle{bottom left}{5.83}{3}{5}{$3$}{$5$}{$x$}{0.5}                                                                  
                 \begin{solutionordottedlines}[1in]                                                                                          
                 \( x = \sqrt{5.83^2 - 3^2} = \sqrt{34} \approx 5.831 \)                                                                     
                 \end{solutionordottedlines}                                                                                                 
                 \Question[1]\,                                                                                                              
                 \drawTriangle{top left}{8.944}{8}{4}{$4$}{$8$}{$y$}{0.5}                                                                    
                 \begin{solutionordottedlines}[1in]                                                                                          
                 \( y = \sqrt{8^2 - 4^2} = \sqrt{48} \approx 6.928 \)                                                                        
                 \end{solutionordottedlines}                                                                                                 
                 \Question[1]\,                                                                                                              
                 \drawTriangle{bottom right}{7}{2}{6.708}{$2$}{$x$}{$7$}{0.5}                                                                
                 \begin{solutionordottedlines}[1in]                                                                                          
                 \( x = \sqrt{7^2 - 2^2} = \sqrt{45} \approx 6.708 \)                                                                        
                 \end{solutionordottedlines}                                                                                                 
             \end{multicols}                                                                                                                 
         \Question[2] A door frame has height $1.8$m and width $1$m. Will a square piece of board $2.1$m wide fit through the opening?       
             \begin{solutionordottedlines}[1.5in]                                                                                            
             The diagonal of the door frame is \(\sqrt{1.8^2 + 1^2} = \sqrt{4.24} \approx 2.059\), which is greater than $2.1$m, so the boar 
 will fit.                                                                                                                                   
             \end{solutionordottedlines}                                                                                                     
         \Question[3] A signwriter leans his ladder against a wall so that he can paint a sign. The wall is vertical and the ground in front 
 of the wall is horizontal. The signwriter's ladder is $4$m long.\\If the signwriter wants the top of the ladder to be $3.8$m above the grou 
 when leaning against the wall, how far, correct to $1$ decimal place should the foot of the ladder be placed from the wall?                 
             \begin{solutionordottedlines}[1.5in]                                                                                            
             The distance from the wall is \(\sqrt{4^2 - 3.8^2} = \sqrt{0.36} \approx 0.6\)m.                                                
             \end{solutionordottedlines}                                                                                                     
         \end{questions}                                                                                                                     
 \end{exercisebox}                                                                                                                           
|726192180|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \section*{Exercises}
        \begin{questions}
            \begin{multicols}{3}
                \Question[1]\,
                \drawTriangle{bottom left}{5.83}{3}{5}{$3$}{$5$}{$x$}{0.5}
                \begin{solutionordottedlines}[1in]
                \end{solutionordottedlines}
                \Question[1]\,
                \drawTriangle{top left}{8.944}{8}{4}{$4$}{$8$}{$y$}{0.5}
                \begin{solutionordottedlines}[1in]
                \end{solutionordottedlines}
                \Question[1]\,
                \drawTriangle{bottom right}{7}{2}{6.708}{$2$}{$x$}{$7$}{0.5}
                \begin{solutionordottedlines}[1in]
                \end{solutionordottedlines}
            \end{multicols}
        \Question[2] A door frame has height $1.8$m and width $1$m. Will a square piece of board $2.1$m wide fit through the opening?
            \begin{solutionordottedlines}[1.5in]
            \end{solutionordottedlines}
        \Question[3] A signwriter leans his ladder against a wall so that he can paint a sign. The wall is vertical and the ground in front of the wall is horizontal. The signwriter's ladder is $4$m long.\\If the signwriter wants the top of the ladder to be $3.8$m above the ground when leaning against the wall, how far, correct to $1$ decimal place should the foot of the ladder be placed from the wall?
            \begin{solutionordottedlines}[1.5in]
            \end{solutionordottedlines}
        \end{questions}
\end{exercisebox}
|726192184|kitty|/Applications/kitty.app|0|
 \begin{examplebox}                                                                                                                          
     \section*{Examples}                                                                                                                     
     Try to simplify the following:                                                                                                          
     \begin{multicols}{2}                                                                                                                    
         \begin{questions}                                                                                                                   
             \Question[1] $\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}$                                                                              
             \Question[1] $\sqrt{108}=\sqrt{36\times3}=6\sqrt{3}$                                                                            
             \Question[1] $\sqrt{45}=\sqrt{9\times5}=3\sqrt{5}$                                                                              
             \Question[1] $\sqrt{32}=\sqrt{16\times2}=4\sqrt{2}$                                                                             
         \end{questions}                                                                                                                     
     \end{multicols}                                                                                                                         
 \end{examplebox}                                                                                                                            
|726192215|kitty|/Applications/kitty.app|0|
\begin{examplebox}
    \section*{Examples}
    Try to simplify the following:
    \begin{multicols}{2}
        \begin{questions}
            \Question[1] $\sqrt{18}=\fillin[][1in]$
            \Question[1] $\sqrt{108}=\fillin[][1in]$
            \Question[1] $\sqrt{45}=\fillin[][1in]$
            \Question[1] $\sqrt{32}=\fillin[][1in]$
        \end{questions}
    \end{multicols}
\end{examplebox}
|726192218|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                                                         
     \section*{Exercises}                                                                                                                    
     Now try going backwards:                                                                                                                
     \begin{multicols}{2}                                                                                                                    
         \begin{questions}                                                                                                                   
             \Question[1] $3\sqrt{7}=\sqrt{9\times7}=\sqrt{63}$                                                                              
             \Question[1] $5\sqrt{3}=\sqrt{25\times3}=\sqrt{75}$                                                                             
         \end{questions}                                                                                                                     
     \end{multicols}                                                                                                                         
 \end{exercisebox}                                                                                                                           
|726192241|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                                                         
     \section*{Exercises}                                                                                                                    
         \begin{questions}                                                                                                                   
             \Question[12]                                                                                                                   
             \begin{parts}\begin{multicols}{2}                                                                                               
             \part \[4\sqrt{7} + 5\sqrt{7}=\]                                                                                                
                 \begin{solutionordottedlines}[2cm]                                                                                          
                 $9\sqrt{7}$                                                                                                                 
                 \end{solutionordottedlines}                                                                                                 
             \part \[6\sqrt{7}-2\sqrt{7}=\]                                                                                                  
                 \begin{solutionordottedlines}[2cm]                                                                                          
                 $4\sqrt{7}$                                                                                                                 
                 \end{solutionordottedlines}                                                                                                 
             \part \[2\sqrt{2}+5\sqrt{2}-3\sqrt{2}=\]                                                                                        
                 \begin{solutionordottedlines}[2cm]                                                                                          
                 $4\sqrt{2}$                                                                                                                 
                 \end{solutionordottedlines}                                                                                                 
             \part \[3\sqrt{5}+2\sqrt{7}-\sqrt{5}+4\sqrt{7}=\]                                                                               
                 \begin{solutionordottedlines}[2cm]                                                                                          
                 $2\sqrt{5}+6\sqrt{7}$                                                                                                       
                 \end{solutionordottedlines}                                                                                                 
             \part \[\sqrt{27}+2\sqrt{5}+\sqrt{20}-2\sqrt{3}=\]                                                                              
                 \begin{solutionordottedlines}[2cm]                                                                                          
                 $3\sqrt{3}+2\sqrt{5}+2\sqrt{5}-2\sqrt{3} = \sqrt{3}+4\sqrt{5}$                                                              
                 \end{solutionordottedlines}                                                                                                 
             \part \[\frac{\sqrt{3}}{2}+3\sqrt{3}=\]                                                                                         
                 \begin{solutionordottedlines}[2cm]                                                                                          
                 $\frac{1}{2}\sqrt{3}+\frac{6}{2}\sqrt{3} = \frac{7}{2}\sqrt{3}$                                                             
                 \end{solutionordottedlines}                                                                                                 
             \part \[6\sqrt{2} + 4\sqrt{2}=\]                                                                                                
                 \begin{solutionordottedlines}[2cm]                                                                                          
                 $10\sqrt{2}$                                                                                                                
                 \end{solutionordottedlines}                                                                                                 
             \part \[10\sqrt{7}-5\sqrt{7}=\]                                                                                                 
                 \begin{solutionordottedlines}[2cm]                                                                                          
                 $5\sqrt{7}$                                                                                                                 
                 \end{solutionordottedlines}                                                                                                 
             \part \[3\sqrt{11}-5\sqrt{11}=\]                                                                                                
                 \begin{solutionordottedlines}[2cm]                                                                                          
                 $-2\sqrt{11}$                                                                                                               
                 \end{solutionordottedlines}                                                                                                 
             \part \[2\sqrt{2}+3\sqrt{2}+\sqrt{2}=\]                                                                                         
                 \begin{solutionordottedlines}[2cm]                                                                                          
                 $6\sqrt{2}$                                                                                                                 
                 \end{solutionordottedlines}                                                                                                 
             \part \[10\sqrt{5}-\sqrt{5}-6\sqrt{5}=\]                                                                                        
                 \begin{solutionordottedlines}[2cm]                                                                                          
                 $3\sqrt{5}$                                                                                                                 
                 \end{solutionordottedlines}                                                                                                 
             \part \[\sqrt{7}-8\sqrt{7}+5\sqrt{7}=\]                                                                                         
                 \begin{solutionordottedlines}[2cm]                                                                                          
                 $-2\sqrt{7}$                                                                                                                
                 \end{solutionordottedlines}                                                                                                 
         \end{multicols}\end{parts}                                                                                                          
         \end{questions}                                                                                                                     
 \end{exercisebox}                                                                                                                           
|726192261|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \section*{Exercises}
        \begin{questions}
            \Question[12]
            \begin{parts}\begin{multicols}{2}
            \part \[4\sqrt{7} + 5\sqrt{7}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[6\sqrt{7}-2\sqrt{7}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[2\sqrt{2}+5\sqrt{2}-3\sqrt{2}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[3\sqrt{5}+2\sqrt{7}-\sqrt{5}+4\sqrt{7}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\sqrt{27}+2\sqrt{5}+\sqrt{20}-2\sqrt{3}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{\sqrt{3}}{2}+3\sqrt{3}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[6\sqrt{2} + 4\sqrt{2}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[10\sqrt{7}-5\sqrt{7}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[3\sqrt{11}-5\sqrt{11}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[2\sqrt{2}+3\sqrt{2}+\sqrt{2}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[10\sqrt{5}-\sqrt{5}-6\sqrt{5}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\sqrt{7}-8\sqrt{7}+5\sqrt{7}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \end{questions}
\end{exercisebox}
|726192265|kitty|/Applications/kitty.app|0|
 \begin{examplebox}                                                                                                                          
     \section*{Examples}                                                                                                                     
     \begin{multicols}{2}                                                                                                                    
         \begin{questions}                                                                                                                   
             \Question[2] $\sqrt{3}\times \sqrt{11}=$                                                                                        
                 \begin{solutionordottedlines}[1cm]                                                                                          
                 $\sqrt{33}$                                                                                                                 
                 \end{solutionordottedlines}                                                                                                 
             \question $\sqrt{15}\div\sqrt{3}=$                                                                                              
                 \begin{solutionordottedlines}[1cm]                                                                                          
                 $\sqrt{5}$                                                                                                                  
                 \end{solutionordottedlines}                                                                                                 
             \Question[2] $3\times 7\sqrt{5}=$                                                                                               
                 \begin{solutionordottedlines}[1cm]                                                                                          
                 $21\sqrt{5}$                                                                                                                
                 \end{solutionordottedlines}                                                                                                 
             \question $\sqrt{2}\times 4=$                                                                                                   
                 \begin{solutionordottedlines}[1cm]                                                                                          
                 $4\sqrt{2}$                                                                                                                 
                 \end{solutionordottedlines}                                                                                                 
             \Question[2] $\sqrt{6}\times\sqrt{6}=$                                                                                          
                 \begin{solutionordottedlines}[1cm]                                                                                          
                 $6$                                                                                                                         
                 \end{solutionordottedlines}                                                                                                 
             \question $(2\sqrt{6})^2=$                                                                                                      
                 \begin{solutionordottedlines}[1cm]                                                                                          
                 $24$                                                                                                                        
                 \end{solutionordottedlines}                                                                                                 
             \Question[2] $\sqrt{42}\div\sqrt{6}=$                                                                                           
                 \begin{solutionordottedlines}[1cm]                                                                                          
                 $\sqrt{7}$                                                                                                                  
                 \end{solutionordottedlines}                                                                                                 
             \question $\sqrt{35}\div\sqrt{10}=$                                                                                             
                 \begin{solutionordottedlines}[1cm]                                                                                          
                 $\sqrt{3.5}$ or $\sqrt{\frac{7}{2}}$                                                                                        
                 \end{solutionordottedlines}                                                                                                 
         \end{questions}                                                                                                                     
     \end{multicols}                                                                                                                         
 \end{examplebox}                                                                                                                            
|726192282|kitty|/Applications/kitty.app|0|
\begin{examplebox}
    \section*{Examples}
    \begin{multicols}{2}
        \begin{questions}
            \Question[2] $\sqrt{3}\times \sqrt{11}=$
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \question $\sqrt{15}\div\sqrt{3}=$
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \Question[2] $3\times 7\sqrt{5}=$
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \question $\sqrt{2}\times 4=$
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \Question[2] $\sqrt{6}\times\sqrt{6}=$
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \question $(2\sqrt{6})^2=$
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \Question[2] $\sqrt{42}\div\sqrt{6}=$
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
            \question $\sqrt{35}\div\sqrt{10}=$
                \begin{solutionordottedlines}[1cm]
                \end{solutionordottedlines}
        \end{questions}
    \end{multicols}
\end{examplebox}
|726192287|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                                                         
     \section*{Exercises}                                                                                                                    
     The distributive law is exactly the same as last week:                                                                                  
     \begin{equation}                                                                                                                        
         (\source{a}+\source{b})(\target{c}+\target{d})= ac + ad + bc + bd\mbox{\drawarrows}                                                 
     \end{equation}                                                                                                                          
                                                                                                                                             
     Expand and simplify the following:                                                                                                      
     \begin{questions}                                                                                                                       
         \Question[1] $2\sqrt{3}(4+3\sqrt3)=$                                                                                                
             \begin{solutionordottedlines}[2cm]                                                                                              
             $8\sqrt{3}+18$                                                                                                                  
             \end{solutionordottedlines}                                                                                                     
         \Question[1] $(3\sqrt{7}+1)(5\sqrt{7}-4)=$                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                              
             $15\sqrt{49} - 12\sqrt{7} + 5\sqrt{7} - 4 = 105 + 5\sqrt{7} - 12\sqrt{7} - 4 = 101 - 7\sqrt{7}$                                 
             \end{solutionordottedlines}                                                                                                     
         \Question[1] $(5\sqrt{2}-3)(2\sqrt{2}-4)=$                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                              
             $10\sqrt{4} - 20\sqrt{2} - 6\sqrt{2} + 12 = 20 - 26\sqrt{2} + 12 = 32 - 26\sqrt{2}$                                             
             \end{solutionordottedlines}                                                                                                     
         \Question[1] $(3\sqrt{2}-4\sqrt{3})(5\sqrt{3}-\sqrt{2})=$                                                                           
             \begin{solutionordottedlines}[2cm]                                                                                              
             $15\sqrt{6} - 3\sqrt{4} - 20\sqrt{9} + 4\sqrt{6} = 19\sqrt{6} - 6 - 60 = -66 + 19\sqrt{6}$                                      
             \end{solutionordottedlines}                                                                                                     
         \Question[1] $(1-\sqrt{2})(3+2\sqrt{2})=$                                                                                           
             \begin{solutionordottedlines}[2cm]                                                                                              
             $3 + 2\sqrt{4} - 3\sqrt{2} - 2\sqrt{4} = 3 + 4 - 3\sqrt{2} - 4 = -3\sqrt{2} + 3$                                                
             \end{solutionordottedlines}                                                                                                     
     \end{questions}                                                                                                                         
                                                                                                                                             
 \end{exercisebox}                                                                                                                           
|726192293|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \section*{Exercises}
    The distributive law is exactly the same as last week:
    \begin{equation}
        (\source{a}+\source{b})(\target{c}+\target{d})= ac + ad + bc + bd\mbox{\drawarrows} 
    \end{equation}

    Expand and simplify the following:
    \begin{questions}
        \Question[1] $2\sqrt{3}(4+3\sqrt3)=$
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Question[1] $(3\sqrt{7}+1)(5\sqrt{7}-4)=$
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Question[1] $(5\sqrt{2}-3)(2\sqrt{2}-4)=$
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Question[1] $(3\sqrt{2}-4\sqrt{3})(5\sqrt{3}-\sqrt{2})=$
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Question[1] $(1-\sqrt{2})(3+2\sqrt{2})=$
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{questions}

\end{exercisebox}
|726192298|kitty|/Applications/kitty.app|0|
\subsection*{Section 2}
% The Pythagorean Theorem
\begin{questions}
    \Question[3]
    \begin{parts}
    \begin{multicols}{3}
    \part\, \drawTriangle{top left}{11}{7}{8.49}{$y$}{$7$}{$11$}{0.3}
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
    \part\, \drawTriangle{top right}{18.44}{4}{18}{$4$}{$18$}{$a$}{0.3}
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
    \part\, \drawTriangle{bottom left}{10}{7.14}{7}{$7$}{$b$}{$10$}{0.3}
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
    \end{multicols}
    \end{parts}
    \Question[3] Determine whether or not the following side-lengths form a right-angled triangle:
    \begin{parts}\begin{multicols}{3}
    \part $4.5,7,7.5$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
    \part $6,10,12$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
    \part $20,21,29$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
    \end{multicols}
    \end{parts}
    \Question[2] As part of a design, an artist draws a circle passing through the four corners (vertices) of a square.
    \begin{parts}
        \part If the square has side lengths of $4$cm, what is the radius, to the nearest millimetre, of the circle?
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part If the circle has a radius of $3$cm, what are the side lengths, to the nearest millimetre, of the square?
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
    \end{parts}
    \Question[2] A parent is asked to make some scarves for the local Scout troop. Two scarves can be made from one square piece of material by cutting on the diagonal. If this diagonal side length is to be $100$cm long, what must be the side length of the square piece of material to the nearest mm?
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
    \Question[2] A girl planned to swim straight across a river of width $25$m. After she had swum across the river, the girl found she had been swept $4$m downstream. How far did she actually swim? Calculate your answer, in metres, correct to 1 decimal place.
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
\end{questions}

\subsection*{Section 3}
% Introduction to Surds
\begin{questions}
    \Question[3] Use Pythagoras' Theorem to find the value of $x$. Give your answer as a \emph{surd} which has been simplified.
    \begin{parts}
    \begin{multicols}{3}
        \part \drawTriangle{bottom left}{4.24}{3}{3}{$3$}{$x$}{$3$}{0.5}
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part \drawTriangle{bottom right}{2.82}{2.4}{1.73}{$\sqrt{3}$}{$\sqrt{5}$}{$x$}{0.7}
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part \drawTriangle{bottom right}{14}{12.65}{6}{$x$}{$6$}{$14$}{0.3}
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
    \end{multicols}
    \end{parts}
\end{questions}

\subsection*{Section 4}
% Simplifying Surds
\begin{questions}
    \Question[18] Simplify all of the following:
        \begin{parts}
        \begin{multicols}{3}
        \part $(\sqrt{231})^2$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $(3\sqrt{5})^2$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $(2\sqrt{11})^2$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $\sqrt{3}\times\sqrt{5}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $\sqrt{5}\times\sqrt{6}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $\sqrt{6}\div\sqrt{2}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $\frac{\sqrt{77}}{\sqrt{11}}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $2\times3\sqrt{2}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $6\times5\sqrt{7}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $(\sqrt{2})^3$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $(\sqrt{11})^2+(\sqrt{2})^2$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $(\sqrt{5})^2+(\sqrt{11})^2$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $\sqrt{45}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $\sqrt{54}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $\sqrt{126}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $\sqrt{72}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $\sqrt{96}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $\sqrt{200}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \end{multicols}
        \end{parts}
        \Question[4] Express the following surds as the square root of a whole number
            \begin{parts}
            \begin{multicols}{2}
                \part $2\sqrt{3}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
                \part $2\sqrt{13}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
                \part $4\sqrt{5}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
                \part $12\sqrt{10}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \end{multicols}
            \end{parts}
        \Question[2] Evaluate:
            \begin{parts}
            \begin{multicols}{2}
                \part $(\sqrt{\frac{2}{3}})^2$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
                \part $\sqrt{\frac{16}{25}}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \end{multicols}
            \end{parts}
\end{questions}

\subsection*{Section 5}
% Addition & Subtraction of Surds
\begin{questions}
    \Question[16] Simplify:
        \begin{parts}
        \begin{multicols}{2}
            \part $6\sqrt{2} + 4\sqrt{2}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \part $10\sqrt{7}-5\sqrt{7}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \part $3\sqrt{11}-5\sqrt{11}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \part $2\sqrt{2}+3\sqrt{2}+\sqrt{2}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \part $10\sqrt{5}-\sqrt{5}-6\sqrt{5}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \part $\sqrt{3}-2\sqrt{2}+2\sqrt{3}+\sqrt{2}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \part $5\sqrt{14}+4\sqrt{6}+\sqrt{14}+3\sqrt{6}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \part $\sqrt{5}-3\sqrt{2}-4\sqrt{5}+7\sqrt{2}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \part $\sqrt{50} + 3\sqrt{2}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \part $\sqrt{48} + 2\sqrt{3}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \part $\sqrt{8}+4\sqrt{2}+2\sqrt{18}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \part $\sqrt{8}+\sqrt{2}+\sqrt{18}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \part $4\sqrt{5}-4\sqrt{20}-\sqrt{45}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \part \[\frac{\sqrt{7}}{2}-\frac{\sqrt{7}}{5}\]
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \part \[\frac{3\sqrt{11}}{7}+\frac{2\sqrt{11}}{21}\]
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
            \part $\sqrt{80}-\sqrt{45} = \sqrt{x}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \end{multicols}
        \end{parts}
\end{questions}
|726192354|kitty|/Applications/kitty.app|0|
 \subsection*{Section 2}                                                                                                                     
 % The Pythagorean Theorem                                                                                                                   
 \begin{questions}                                                                                                                           
     \Question[3]                                                                                                                            
     \begin{parts}                                                                                                                           
     \begin{multicols}{3}                                                                                                                    
     \part\, \drawTriangle{top left}{11}{7}{8.49}{$y$}{$7$}{$11$}{0.3}                                                                       
         \begin{solutionordottedlines}[1in]                                                                                                  
         $y = \sqrt{11^2 - 7^2} = \sqrt{121 - 49} = \sqrt{72} = 8.49$                                                                        
         \end{solutionordottedlines}                                                                                                         
     \part\, \drawTriangle{top right}{18.44}{4}{18}{$4$}{$18$}{$a$}{0.3}                                                                     
         \begin{solutionordottedlines}[1in]                                                                                                  
         $a = \sqrt{18^2 - 4^2} = \sqrt{324 - 16} = \sqrt{308} = 17.55$                                                                      
         \end{solutionordottedlines}                                                                                                         
     \part\, \drawTriangle{bottom left}{10}{7.14}{7}{$7$}{$b$}{$10$}{0.3}                                                                    
         \begin{solutionordottedlines}[1in]                                                                                                  
         $b = \sqrt{10^2 - 7^2} = \sqrt{100 - 49} = \sqrt{51} = 7.14$                                                                        
         \end{solutionordottedlines}                                                                                                         
     \end{multicols}                                                                                                                         
     \end{parts}                                                                                                                             
     \Question[3] Determine whether or not the following side-lengths form a right-angled triangle:                                          
     \begin{parts}\begin{multicols}{3}                                                                                                       
     \part $4.5,7,7.5$                                                                                                                       
         \begin{solutionordottedlines}[1in]                                                                                                  
         $7.5^2 \neq 4.5^2 + 7^2$, so it does not form a right-angled triangle.                                                              
         \end{solutionordottedlines}                                                                                                         
     \part $6,10,12$                                                                                                                         
         \begin{solutionordottedlines}[1in]                                                                                                  
         $12^2 = 6^2 + 10^2$, so it forms a right-angled triangle.                                                                           
         \end{solutionordottedlines}                                                                                                         
     \part $20,21,29$                                                                                                                        
         \begin{solutionordottedlines}[1in]                                                                                                  
         $29^2 \neq 20^2 + 21^2$, so it does not form a right-angled triangle.                                                               
         \end{solutionordottedlines}                                                                                                         
     \end{multicols}                                                                                                                         
     \end{parts}                                                                                                                             
     \Question[2] As part of a design, an artist draws a circle passing through the four corners (vertices) of a square.                     
     \begin{parts}                                                                                                                           
         \part If the square has side lengths of $4$cm, what is the radius, to the nearest millimetre, of the circle?                        
         \begin{solutionordottedlines}[1in]                                                                                                  
         Radius $r = \frac{\sqrt{4^2 + 4^2}}{2} = \frac{\sqrt{32}}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \approx 2.828$cm                     
         \end{solutionordottedlines}                                                                                                         
         \part If the circle has a radius of $3$cm, what are the side lengths, to the nearest millimetre, of the square?                     
         \begin{solutionordottedlines}[1in]                                                                                                  
         Side length $s = \sqrt{2} \times r = \sqrt{2} \times 3 \approx 4.242$cm                                                             
         \end{solutionordottedlines}                                                                                                         
     \end{parts}                                                                                                                             
     \Question[2] A parent is asked to make some scarves for the local Scout troop. Two scarves can be made from one square piece of materia 
 by cutting on the diagonal. If this diagonal side length is to be $100$cm long, what must be the side length of the square piece of materia 
 to the nearest mm?                                                                                                                          
         \begin{solutionordottedlines}[1in]                                                                                                  
         Side length $s = \frac{100}{\sqrt{2}} \approx 70.71$cm                                                                              
         \end{solutionordottedlines}                                                                                                         
     \Question[2] A girl planned to swim straight across a river of width $25$m. After she had swum across the river, the girl found she had 
 been swept $4$m downstream. How far did she actually swim? Calculate your answer, in metres, correct to 1 decimal place.                    
         \begin{solutionordottedlines}[1in]                                                                                                  
         Distance swum $d = \sqrt{25^2 + 4^2} \approx 25.3$m                                                                                 
         \end{solutionordottedlines}                                                                                                         
 \end{questions}                                                                                                                             
                                                                                                                                             
 \subsection*{Section 3}                                                                                                                     
 % Introduction to Surds                                                                                                                     
 \begin{questions}                                                                                                                           
     \Question[3] Use Pythagoras' Theorem to find the value of $x$. Give your answer as a \emph{surd} which has been simplified.             
     \begin{parts}                                                                                                                           
     \begin{multicols}{3}                                                                                                                    
         \part \drawTriangle{bottom left}{4.24}{3}{3}{$3$}{$x$}{$3$}{0.5}                                                                    
         \begin{solutionordottedlines}[1in]                                                                                                  
         $x = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$                                                                                      
         \end{solutionordottedlines}                                                                                                         
         \part \drawTriangle{bottom right}{2.82}{2.4}{1.73}{$\sqrt{3}$}{$\sqrt{5}$}{$x$}{0.7}                                                
         \begin{solutionordottedlines}[1in]                                                                                                  
         $x = \sqrt{\sqrt{5}^2 - \sqrt{3}^2} = \sqrt{5 - 3} = \sqrt{2}$                                                                      
         \end{solutionordottedlines}                                                                                                         
         \part \drawTriangle{bottom right}{14}{12.65}{6}{$x$}{$6$}{$14$}{0.3}                                                                
         \begin{solutionordottedlines}[1in]                                                                                                  
         $x = \sqrt{14^2 - 6^2} = \sqrt{196 - 36} = \sqrt{160} = 4\sqrt{10}$                                                                 
         \end{solutionordottedlines}                                                                                                         
     \end{multicols}                                                                                                                         
     \end{parts}                                                                                                                             
 \end{questions}                                                                                                                             
                                                                                                                                             
 \subsection*{Section 4}                                                                                                                     
 % Simplifying Surds                                                                                                                         
 \begin{questions}                                                                                                                           
     \Question[18] Simplify all of the following:                                                                                            
         \begin{parts}                                                                                                                       
         \begin{multicols}{3}                                                                                                                
         \part $(\sqrt{231})^2$                                                                                                              
         \begin{solutionordottedlines}[1in]                                                                                                  
         $231$                                                                                                                               
         \end{solutionordottedlines}                                                                                                         
         \part $(3\sqrt{5})^2$                                                                                                               
         \begin{solutionordottedlines}[1in]                                                                                                  
         $9 \times 5 = 45$                                                                                                                   
         \end{solutionordottedlines}                                                                                                         
         \part $(2\sqrt{11})^2$                                                                                                              
         \begin{solutionordottedlines}[1in]                                                                                                  
         $4 \times 11 = 44$                                                                                                                  
         \end{solutionordottedlines}                                                                                                         
         \part $\sqrt{3}\times\sqrt{5}$                                                                                                      
         \begin{solutionordottedlines}[1in]                                                                                                  
         $\sqrt{15}$                                                                                                                         
         \end{solutionordottedlines}                                                                                                         
         \part $\sqrt{5}\times\sqrt{6}$                                                                                                      
         \begin{solutionordottedlines}[1in]                                                                                                  
         $\sqrt{30}$                                                                                                                         
         \end{solutionordottedlines}                                                                                                         
         \part $\sqrt{6}\div\sqrt{2}$                                                                                                        
         \begin{solutionordottedlines}[1in]                                                                                                  
         $\sqrt{3}$                                                                                                                          
         \end{solutionordottedlines}                                                                                                         
         \part $\frac{\sqrt{77}}{\sqrt{11}}$                                                                                                 
         \begin{solutionordottedlines}[1in]                                                                                                  
         $\sqrt{7}$                                                                                                                          
         \end{solutionordottedlines}                                                                                                         
         \part $2\times3\sqrt{2}$                                                                                                            
         \begin{solutionordottedlines}[1in]                                                                                                  
         $6\sqrt{2}$                                                                                                                         
         \end{solutionordottedlines}                                                                                                         
         \part $6\times5\sqrt{7}$                                                                                                            
         \begin{solutionordottedlines}[1in]                                                                                                  
         $30\sqrt{7}$                                                                                                                        
         \end{solutionordottedlines}                                                                                                         
         \part $(\sqrt{2})^3$                                                                                                                
         \begin{solutionordottedlines}[1in]                                                                                                  
         $2\sqrt{2}$                                                                                                                         
         \end{solutionordottedlines}                                                                                                         
         \part $(\sqrt{11})^2+(\sqrt{2})^2$                                                                                                  
         \begin{solutionordottedlines}[1in]                                                                                                  
         $11 + 2 = 13$                                                                                                                       
         \end{solutionordottedlines}                                                                                                         
         \part $(\sqrt{5})^2+(\sqrt{11})^2$                                                                                                  
         \begin{solutionordottedlines}[1in]                                                                                                  
         $5 + 11 = 16$                                                                                                                       
         \end{solutionordottedlines}                                                                                                         
         \part $\sqrt{45}$                                                                                                                   
         \begin{solutionordottedlines}[1in]                                                                                                  
         $3\sqrt{5}$                                                                                                                         
         \end{solutionordottedlines}                                                                                                         
         \part $\sqrt{54}$                                                                                                                   
         \begin{solutionordottedlines}[1in]                                                                                                  
         $3\sqrt{6}$                                                                                                                         
         \end{solutionordottedlines}                                                                                                         
         \part $\sqrt{126}$                                                                                                                  
         \begin{solutionordottedlines}[1in]                                                                                                  
         $3\sqrt{14}$                                                                                                                        
         \end{solutionordottedlines}                                                                                                         
         \part $\sqrt{72}$                                                                                                                   
         \begin{solutionordottedlines}[1in]                                                                                                  
         $6\sqrt{2}$                                                                                                                         
         \end{solutionordottedlines}                                                                                                         
         \part $\sqrt{96}$                                                                                                                   
         \begin{solutionordottedlines}[1in]                                                                                                  
         $4\sqrt{6}$                                                                                                                         
         \end{solutionordottedlines}                                                                                                         
         \part $\sqrt{200}$                                                                                                                  
         \begin{solutionordottedlines}[1in]                                                                                                  
         $10\sqrt{2}$                                                                                                                        
         \end{solutionordottedlines}                                                                                                         
         \end{multicols}                                                                                                                     
         \end{parts}                                                                                                                         
         \Question[4] Express the following surds as the square root of a whole number                                                       
             \begin{parts}                                                                                                                   
             \begin{multicols}{2}                                                                                                            
                 \part $2\sqrt{3}$                                                                                                           
         \begin{solutionordottedlines}[1in]                                                                                                  
         $\sqrt{12}$                                                                                                                         
         \end{solutionordottedlines}                                                                                                         
                 \part $2\sqrt{13}$                                                                                                          
         \begin{solutionordottedlines}[1in]                                                                                                  
         $\sqrt{52}$                                                                                                                         
         \end{solutionordottedlines}                                                                                                         
                 \part $4\sqrt{5}$                                                                                                           
         \begin{solutionordottedlines}[1in]                                                                                                  
         $\sqrt{80}$                                                                                                                         
         \end{solutionordottedlines}                                                                                                         
                 \part $12\sqrt{10}$                                                                                                         
         \begin{solutionordottedlines}[1in]                                                                                                  
         $\sqrt{1440}$                                                                                                                       
         \end{solutionordottedlines}                                                                                                         
             \end{multicols}                                                                                                                 
             \end{parts}                                                                                                                     
         \Question[2] Evaluate:                                                                                                              
             \begin{parts}                                                                                                                   
             \begin{multicols}{2}                                                                                                            
                 \part $(\sqrt{\frac{2}{3}})^2$                                                                                              
         \begin{solutionordottedlines}[1in]                                                                                                  
         $\frac{2}{3}$                                                                                                                       
         \end{solutionordottedlines}                                                                                                         
                 \part $\sqrt{\frac{16}{25}}$                                                                                                
         \begin{solutionordottedlines}[1in]                                                                                                  
         $\frac{4}{5}$                                                                                                                       
         \end{solutionordottedlines}                                                                                                         
             \end{multicols}                                                                                                                 
             \end{parts}                                                                                                                     
 \end{questions}                                                                                                                             
                                                                                                                                             
 \subsection*{Section 5}                                                                                                                     
 % Addition & Subtraction of Surds                                                                                                           
 \begin{questions}                                                                                                                           
     \Question[16] Simplify:                                                                                                                 
         \begin{parts}                                                                                                                       
         \begin{multicols}{2}                                                                                                                
             \part $6\sqrt{2} + 4\sqrt{2}$                                                                                                   
         \begin{solutionordottedlines}[1in]                                                                                                  
         $10\sqrt{2}$                                                                                                                        
         \end{solutionordottedlines}                                                                                                         
             \part $10\sqrt{7}-5\sqrt{7}$                                                                                                    
         \begin{solutionordottedlines}[1in]                                                                                                  
         $5\sqrt{7}$                                                                                                                         
         \end{solutionordottedlines}                                                                                                         
             \part $3\sqrt{11}-5\sqrt{11}$                                                                                                   
         \begin{solutionordottedlines}[1in]                                                                                                  
         $-2\sqrt{11}$                                                                                                                       
         \end{solutionordottedlines}                                                                                                         
             \part $2\sqrt{2}+3\sqrt{2}+\sqrt{2}$                                                                                            
         \begin{solutionordottedlines}[1in]                                                                                                  
         $6\sqrt{2}$                                                                                                                         
         \end{solutionordottedlines}                                                                                                         
             \part $10\sqrt{5}-\sqrt{5}-6\sqrt{5}$                                                                                           
         \begin{solutionordottedlines}[1in]                                                                                                  
         $3\sqrt{5}$                                                                                                                         
         \end{solutionordottedlines}                                                                                                         
             \part $\sqrt{3}-2\sqrt{2}+2\sqrt{3}+\sqrt{2}$                                                                                   
         \begin{solutionordottedlines}[1in]                                                                                                  
         $3\sqrt{3}-\sqrt{2}$                                                                                                                
         \end{solutionordottedlines}                                                                                                         
             \part $5\sqrt{14}+4\sqrt{6}+\sqrt{14}+3\sqrt{6}$                                                                                
         \begin{solutionordottedlines}[1in]                                                                                                  
         $6\sqrt{14}+7\sqrt{6}$                                                                                                              
         \end{solutionordottedlines}                                                                                                         
             \part $\sqrt{5}-3\sqrt{2}-4\sqrt{5}+7\sqrt{2}$                                                                                  
         \begin{solutionordottedlines}[1in]                                                                                                  
         $-3\sqrt{5}+4\sqrt{2}$                                                                                                              
         \end{solutionordottedlines}                                                                                                         
             \part $\sqrt{50} + 3\sqrt{2}$                                                                                                   
         \begin{solutionordottedlines}[1in]                                                                                                  
         $5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}$                                                                                                 
         \end{solutionordottedlines}                                                                                                         
             \part $\sqrt{48} + 2\sqrt{3}$                                                                                                   
         \begin{solutionordottedlines}[1in]                                                                                                  
         $4\sqrt{3} + 2\sqrt{3} = 6\sqrt{3}$                                                                                                 
         \end{solutionordottedlines}                                                                                                         
             \part $\sqrt{8}+4\sqrt{2}+2\sqrt{18}$                                                                                           
         \begin{solutionordottedlines}[1in]                                                                                                  
         $2\sqrt{2}+4\sqrt{2}+6\sqrt{2} = 12\sqrt{2}$                                                                                        
         \end{solutionordottedlines}                                                                                                         
             \part $\sqrt{8}+\sqrt{2}+\sqrt{18}$                                                                                             
         \begin{solutionordottedlines}[1in]                                                                                                  
         $2\sqrt{2}+\sqrt{2}+3\sqrt{2} = 6\sqrt{2}$                                                                                          
         \end{solutionordottedlines}                                                                                                         
             \part $4\sqrt{5}-4\sqrt{20}-\sqrt{45}$                                                                                          
         \begin{solutionordottedlines}[1in]                                                                                                  
         $4\sqrt{5}-8\sqrt{5}-3\sqrt{5} = -7\sqrt{5}$                                                                                        
         \end{solutionordottedlines}                                                                                                         
             \part \[\frac{\sqrt{7}}{2}-\frac{\sqrt{7}}{5}\]                                                                                 
         \begin{solutionordottedlines}[1in]                                                                                                  
         $\frac{5\sqrt{7}-2\sqrt{7}}{10} = \frac{3\sqrt{7}}{10}$                                                                             
         \end{solutionordottedlines}                                                                                                         
             \part \[\frac{3\sqrt{11}}{7}+\frac{2\sqrt{11}}{21}\]                                                                            
         \begin{solutionordottedlines}[1in]                                                                                                  
         $\frac{9\sqrt{11}+2\sqrt{11}}{21} = \frac{11\sqrt{11}}{21}$                                                                         
         \end{solutionordottedlines}                                                                                                         
             \part $\sqrt{80}-\sqrt{45} = \sqrt{x}$                                                                                          
         \begin{solutionordottedlines}[1in]                                                                                                  
         $4\sqrt{5}-3\sqrt{5} = \sqrt{5} = \sqrt{x}$, so $x = 5$                                                                             
         \end{solutionordottedlines}                                                                                                         
         \end{multicols}                                                                                                                     
         \end{parts}                                                                                                                         
 \end{questions}                                                                                                                             
|726192373|kitty|/Applications/kitty.app|0|
 \subsection*{Section 6}                                                                                                                                             
 % Multiplication and Division of Surds                                                                                                                              
 \begin{questions}                                                                                                                                                   
     \Question[8] Simplify:                                                                                                                                          
         \begin{parts}                                                                                                                                               
         \begin{multicols}{2}                                                                                                                                        
         \part $4\sqrt{7}\times\sqrt{2}$                                                                                                                             
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $4\sqrt{7}\times\sqrt{2} = 4\sqrt{14}$                                                                                                                      
         \end{solutionordottedlines}                                                                                                                                 
         \part $6\sqrt{3}\times 4\sqrt{7}$                                                                                                                           
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $6\sqrt{3}\times 4\sqrt{7} = 24\sqrt{21}$                                                                                                                   
         \end{solutionordottedlines}                                                                                                                                 
         \part $12\sqrt{33}\div3\sqrt{3}$                                                                                                                            
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $12\sqrt{33}\div3\sqrt{3} = 4\sqrt{11}$                                                                                                                     
         \end{solutionordottedlines}                                                                                                                                 
         \part $36\sqrt{15}\div4\sqrt{3}$                                                                                                                            
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $36\sqrt{15}\div4\sqrt{3} = 9\sqrt{5}$                                                                                                                      
         \end{solutionordottedlines}                                                                                                                                 
         \part $7\sqrt{28}\div4\sqrt{7}$                                                                                                                             
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $7\sqrt{28}\div4\sqrt{7} = 7\sqrt{4} = 14$                                                                                                                  
         \end{solutionordottedlines}                                                                                                                                 
         \part $\sqrt{7}\times2\sqrt{7}$                                                                                                                             
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $\sqrt{7}\times2\sqrt{7} = 2\sqrt{49} = 14$                                                                                                                 
         \end{solutionordottedlines}                                                                                                                                 
         \part $7\sqrt{10}\times3\sqrt{2}$                                                                                                                           
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $7\sqrt{10}\times3\sqrt{2} = 21\sqrt{20} = 42\sqrt{5}$                                                                                                      
         \end{solutionordottedlines}                                                                                                                                 
         \part $\sqrt{2}(2\sqrt{2}-\sqrt{6})$                                                                                                                        
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $\sqrt{2}(2\sqrt{2}-\sqrt{6}) = 2\cdot2 - \sqrt{12} = 4 - 2\sqrt{3}$                                                                                        
         \end{solutionordottedlines}                                                                                                                                 
         \end{multicols}                                                                                                                                             
         \end{parts}                                                                                                                                                 
     \Question[12] Expand and simplify:                                                                                                                              
         \begin{parts}                                                                                                                                               
         \begin{multicols}{2}                                                                                                                                        
         \part $5\sqrt{5}(4\sqrt{2}-3)$                                                                                                                              
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $5\sqrt{5}(4\sqrt{2}-3) = 20\sqrt{10} - 15\sqrt{5}$                                                                                                         
         \end{solutionordottedlines}                                                                                                                                 
         \part $2\sqrt{3}(3\sqrt{3}-5)$                                                                                                                              
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $2\sqrt{3}(3\sqrt{3}-5) = 6\cdot3 - 10\sqrt{3} = 18 - 10\sqrt{3}$                                                                                           
         \end{solutionordottedlines}                                                                                                                                 
         \part $3\sqrt{7}(2-\sqrt{14})$                                                                                                                              
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $3\sqrt{7}(2-\sqrt{14}) = 6\sqrt{7} - 3\cdot7 = 6\sqrt{7} - 21$                                                                                             
         \end{solutionordottedlines}                                                                                                                                 
         \part $\sqrt{2}(2\sqrt{2}-\sqrt{6})$                                                                                                                        
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $\sqrt{2}(2\sqrt{2}-\sqrt{6}) = 4 - 2\sqrt{3}$                                                                                                              
         \end{solutionordottedlines}                                                                                                                                 
         \part $(4\sqrt{5}+1)(3\sqrt{5}+2)$                                                                                                                          
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $(4\sqrt{5}+1)(3\sqrt{5}+2) = 12\cdot5 + 8\sqrt{5} + 3\sqrt{5} + 2 = 60 + 11\sqrt{5}$                                                                       
         \end{solutionordottedlines}                                                                                                                                 
         \part $(3\sqrt{2}+2)(3\sqrt{2}-1)$                                                                                                                          
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $(3\sqrt{2}+2)(3\sqrt{2}-1) = 9\cdot2 + 6\sqrt{2} - 3\sqrt{2} - 2 = 18 + 3\sqrt{2}$                                                                         
         \end{solutionordottedlines}                                                                                                                                 
         \part $(2\sqrt{3}-4)(3\sqrt{3}+5)$                                                                                                                          
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $(2\sqrt{3}-4)(3\sqrt{3}+5) = 6\cdot3 + 10\sqrt{3} - 12\sqrt{3} - 20 = 18 - 2\sqrt{3} - 20 = -2 - 2\sqrt{3}$                                                
         \end{solutionordottedlines}                                                                                                                                 
         \part $(7\sqrt{2}+5)^2$                                                                                                                                     
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $(7\sqrt{2}+5)^2 = 49\cdot2 + 70\sqrt{2} + 25 = 98 + 70\sqrt{2} + 25 = 123 + 70\sqrt{2}$                                                                    
         \end{solutionordottedlines}                                                                                                                                 
         \part $(3\sqrt{5}+2)(\sqrt{2}+3)$                                                                                                                           
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $(3\sqrt{5}+2)(\sqrt{2}+3) = 3\sqrt{10} + 9\sqrt{5} + 2\sqrt{2} + 6$                                                                                        
         \end{solutionordottedlines}                                                                                                                                 
         \part $(4+2\sqrt{3})(2\sqrt{7}-5)$                                                                                                                          
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $(4+2\sqrt{3})(2\sqrt{7}-5) = 8\sqrt{7} - 20 + 4\sqrt{21} - 10\sqrt{3}$                                                                                     
         \end{solutionordottedlines}                                                                                                                                 
         \part $(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})$                                                                                                              
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5}) = 7 - 5 = 2$                                                                                                        
         \end{solutionordottedlines}                                                                                                                                 
         \part $(4\sqrt{7}-\sqrt{5})(2\sqrt{5}+\sqrt{7})$                                                                                                            
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $(4\sqrt{7}-\sqrt{5})(2\sqrt{5}+\sqrt{7}) = 8\sqrt{35} + 4\cdot7 - 2\cdot5 - \sqrt{35} = 28 - 10 + 7\sqrt{35}$                                              
         \end{solutionordottedlines}                                                                                                                                 
         \end{multicols}                                                                                                                                             
         \end{parts}                                                                                                                                                 
         \question \textbf{Challenge:}                                                                                                                               
         \Question[2] If $x = \sqrt{2} - 1$ and $y = \sqrt{3} + 1$, find:                                                                                            
         \begin{solutionordottedlines}[1in]                                                                                                                          
         $x + y = (\sqrt{2} - 1) + (\sqrt{3} + 1) = \sqrt{2} + \sqrt{3}$                                                                                             
         \end{solutionordottedlines}                                                                                                                                 
         \Question[2] \[\frac{6\sqrt{10}}{x+1}\]                                                                                                                     
         \begin{solutionordottedlines}[1.5in]                                                                                                                        
         \[\frac{6\sqrt{10}}{\sqrt{2}} = 6\sqrt{5}\]                                                                                                                 
         \end{solutionordottedlines}                                                                                                                                 
         \Question[2] \[x + \frac{1}{x}\]                                                                                                                            
         \begin{solutionordottedlines}[1.5in]                                                                                                                        
         \[\sqrt{2} - 1 + \frac{1}{\sqrt{2} - 1} = \sqrt{2} - 1 + \frac{\sqrt{2} + 1}{1} = 2\sqrt{2}\]                                                               
         \end{solutionordottedlines}                                                                                                                                 
 \end{questions}                                                                                                                                                     
                                                                                                                                                                     
|726192512|kitty|/Applications/kitty.app|0|
\subsection*{Section 6}
% Multiplication and Division of Surds
\begin{questions}
    \Question[8] Simplify:
        \begin{parts}
        \begin{multicols}{2}
        \part $4\sqrt{7}\times\sqrt{2}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $6\sqrt{3}\times 4\sqrt{7}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $12\sqrt{33}\div3\sqrt{3}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $36\sqrt{15}\div4\sqrt{3}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $7\sqrt{28}\div4\sqrt{7}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $\sqrt{7}\times2\sqrt{7}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $7\sqrt{10}\times3\sqrt{2}$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $\sqrt{2}(2\sqrt{2}-\sqrt{6})$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \end{multicols}
        \end{parts}
    \Question[12] Expand and simplify:
        \begin{parts}
        \begin{multicols}{2}
        \part $5\sqrt{5}(4\sqrt{2}-3)$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $2\sqrt{3}(3\sqrt{3}-5)$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $3\sqrt{7}(2-\sqrt{14})$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $\sqrt{2}(2\sqrt{2}-\sqrt{6})$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $(4\sqrt{5}+1)(3\sqrt{5}+2)$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $(3\sqrt{2}+2)(3\sqrt{2}-1)$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $(2\sqrt{3}-4)(3\sqrt{3}+5)$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $(7\sqrt{2}+5)^2$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $(3\sqrt{5}+2)(\sqrt{2}+3)$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $(4+2\sqrt{3})(2\sqrt{7}-5)$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \part $(4\sqrt{7}-\sqrt{5})(2\sqrt{5}+\sqrt{7})$
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \end{multicols}
        \end{parts}
        \question \textbf{Challenge:}
        \Question[2] If $x = \sqrt{2} - 1$ and $y = \sqrt{3} + 1$, find:
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \Question[2] \[\frac{6\sqrt{10}}{x+1}\]
        \begin{solutionordottedlines}[1.5in]
        \end{solutionordottedlines}
        \Question[2] \[x + \frac{1}{x}\]
        \begin{solutionordottedlines}[1.5in]
        \end{solutionordottedlines}
\end{questions}
            


|726192524|kitty|/Applications/kitty.app|0|
\printanswers
|726192835|kitty|/Applications/kitty.app|0|
         \begin{multicols}{2}                                                                                                   
         \begin{parts}                                                                                                          
             \part \((x+5)(\fillin[x+3])=x^2+8x+15\)                                                                            
             \part \((x+3)(\fillin[x-5])=x^2-2x-15\)                                                                            
             \part \((3x+4)(\fillin[\frac{1}{3}x-4])=3x^2+x-4\)                                                                 
             \part \((x+\fillin[3])(x+6)=x^2+9x+\fillin[18]\)                                                                   
             \part \((2x+3)(\fillin[$\frac{1}{2}x+5$])=2x^2+7x+\fillin[15]\)                                                      
             \part \((\fillin[4] x - 3)(\fillin[-3] x 5 \fillin[-1]) = 12x^2-x-6\)                                              
         \end{parts}                                                                                                            
         \end{multicols}                                                                                                        
|726193303|kitty|/Applications/kitty.app|0|
! Missing $ inserted.
<inserted text> 
$
l.282 ...    \part \((3x+4)(\fillin[\frac{1}{3}x-4])
                                                  =3x^2+x-4\)
? 
! Emergency stop.
<inserted text> 
$
l.282 ...    \part \((3x+4)(\fillin[\frac{1}{3}x-4])
                                                  =3x^2+x-4\)
End of file on the terminal!
|726193328|kitty|/Applications/kitty.app|0|
$|726193415|kitty|/Applications/kitty.app|0|
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|726263316|kitty|/Applications/kitty.app|0|
https://api.openai.com/v1/chat/completion|726263444|kitty|/Applications/kitty.app|0|
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|726263842|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                                                  
     \subsection{Exercises:}                                                                                                          
     \begin{doublespace}                                                                                                              
     \begin{questions}                                                                                                                
         \Question[4] Simplify:                                                                                                       
             \begin{multicols}{2}                                                                                                     
                 \begin{parts}                                                                                                        
                 \part \(\sqrt{108}=\fillin[6\sqrt{3}]\)                                                                              
                 \part \(\sqrt{90}=\fillin[3\sqrt{10}]\)                                                                              
                 \part \(\sqrt{52}=\fillin[2\sqrt{13}]\)                                                                              
                 \part \(\sqrt{98}=\fillin[7\sqrt{2}]\)                                                                               
                 \end{parts}                                                                                                          
             \end{multicols}                                                                                                          
         \Question[4] Add or Subtract:                                                                                                
             \begin{multicols}{2}                                                                                                     
                 \begin{parts}                                                                                                        
                     \part $3\sqrt{2}+2\sqrt{2}=\fillin[5\sqrt{2}]$                                                                   
                     \part $\sqrt{32}-\sqrt{18}=\fillin[2\sqrt{2}]$                                                                   
                     \part $\sqrt{45}+2\sqrt{5}-\sqrt{80}=\fillin[5\sqrt{5}]$                                                         
                     \part $\sqrt{28}+2\sqrt{63}-5\sqrt{7}=\fillin[9\sqrt{7}]$                                                        
                 \end{parts}                                                                                                          
             \end{multicols}                                                                                                          
         \Question[4] Multiply or Divide:                                                                                             
             \begin{multicols}{2}                                                                                                     
                 \begin{parts}                                                                                                        
                     \part $2\sqrt{3}\times 5\sqrt{6}=\fillin[30\sqrt{2}]$                                                            
                     \part $3\sqrt{5}\times 2\sqrt{10}=\fillin[30\sqrt{2}]$                                                           
                     \part $14\sqrt{40}\div 7\sqrt{5}=\fillin[4\sqrt{8}]$                                                             
                     \part $3\sqrt{2}\div \sqrt{2}=\fillin[3]$                                                                        
                 \end{parts}                                                                                                          
             \end{multicols}                                                                                                          
     \end{questions}                                                                                                                  
     \end{doublespace}                                                                                                                
 \end{exercisebox}                                                                                                                    
|726265041|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises:}
    \begin{doublespace}
    \begin{questions}
        \Question[4] Simplify:
            \begin{multicols}{2}
                \begin{parts}
                \part \(\sqrt{108}=\fillin[]\)
                \part \(\sqrt{90}=\fillin[]\)
                \part \(\sqrt{52}=\fillin[]\)
                \part \(\sqrt{98}=\fillin[]\)
                \end{parts}
            \end{multicols}
        \Question[4] Add or Subtract:
            \begin{multicols}{2}
                \begin{parts}
                    \part $3\sqrt{2}+2\sqrt{2}=\fillin[]$
                    \part $\sqrt{32}-\sqrt{18}=\fillin[]$
                    \part $\sqrt{45}+2\sqrt{5}-\sqrt{80}=\fillin[]$
                    \part $\sqrt{28}+2\sqrt{63}-5\sqrt{7}=\fillin[]$
                \end{parts}
            \end{multicols}
        \Question[4] Multiply or Divide:
            \begin{multicols}{2}
                \begin{parts}
                    \part $2\sqrt{3}\times 5\sqrt{6}=\fillin[]$
                    \part $3\sqrt{5}\times 2\sqrt{10}=\fillin[]$
                    \part $14\sqrt{40}\div 7\sqrt{5}=\fillin[]$
                    \part $3\sqrt{2}\div \sqrt{2}=\fillin[]$
                \end{parts}
            \end{multicols}
    \end{questions}
    \end{doublespace}
\end{exercisebox}
|726265047|kitty|/Applications/kitty.app|0|
 \begin{examplebox}                                                                                                                   
     \subsection{Examples:}                                                                                                           
     \begin{questions}                                                                                                                
         \Question[5]                                                                                                                 
         \begin{doublespace}                                                                                                          
         \begin{parts}                                                                                                                
         \part \[(\sqrt{5}+\sqrt{3})^{2}=\]                                                                                           
             \begin{solutionordottedlines}[2cm]                                                                                       
             $(\sqrt{5}+\sqrt{3})^{2} = (\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{3}) + (\sqrt{3})^2 = 5 + 2\sqrt{15} + 3 = 8 + 2\sqrt{15}$    
             \end{solutionordottedlines}                                                                                              
         \part \[(3 \sqrt{2}-2 \sqrt{3})^{2}=\]                                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                       
             $(3 \sqrt{2}-2 \sqrt{3})^{2} = (3\sqrt{2})^2 - 2(3\sqrt{2})(2\sqrt{3}) + (2\sqrt{3})^2 = 9\cdot2 - 2\cdot6\sqrt{6} +     
 4\cdot3 = 18 - 12\sqrt{6} + 12$                                                                                                      
             \end{solutionordottedlines}                                                                                              
         \part \[(2 \sqrt{3}+4 \sqrt{6})^{2}=\]                                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                       
             $(2 \sqrt{3}+4 \sqrt{6})^{2} = (2\sqrt{3})^2 + 2(2\sqrt{3})(4\sqrt{6}) + (4\sqrt{6})^2 = 4\cdot3 + 2\cdot8\sqrt{18} +    
 16\cdot6 = 12 + 16\sqrt{18} + 96$                                                                                                    
             \end{solutionordottedlines}                                                                                              
         \part \[(\sqrt{7}-5)(\sqrt{7}+5)=\]                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                       
             $(\sqrt{7}-5)(\sqrt{7}+5) = (\sqrt{7})^2 - (5)^2 = 7 - 25 = -18$                                                         
             \end{solutionordottedlines}                                                                                              
         \part \[(5 \sqrt{6}-2 \sqrt{5})(5 \sqrt{6}+2 \sqrt{5})=\]                                                                    
             \begin{solutionordottedlines}[2cm]                                                                                       
             $(5 \sqrt{6}-2 \sqrt{5})(5 \sqrt{6}+2 \sqrt{5}) = (5\sqrt{6})^2 - (2\sqrt{5})^2 = 25\cdot6 - 4\cdot5 = 150 - 20 = 130$   
             \end{solutionordottedlines}                                                                                              
         \end{parts}                                                                                                                  
         \end{doublespace}                                                                                                            
     \end{questions}                                                                                                                  
 \end{examplebox}                                                                                                                     
|726265068|kitty|/Applications/kitty.app|0|
\begin{examplebox}
    \subsection{Examples:}
    \begin{questions}
        \Question[5]
        \begin{doublespace}
        \begin{parts}
        \part \[(\sqrt{5}+\sqrt{3})^{2}=\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[(3 \sqrt{2}-2 \sqrt{3})^{2}=\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[(2 \sqrt{3}+4 \sqrt{6})^{2}=\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[(\sqrt{7}-5)(\sqrt{7}+5)=\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[(5 \sqrt{6}-2 \sqrt{5})(5 \sqrt{6}+2 \sqrt{5})=\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \end{doublespace}
    \end{questions}
\end{examplebox}
|726265072|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \Question[12]
        \begin{parts}\begin{multicols}{2}
            \part \((5+\sqrt{3})^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((\sqrt{2}+6)^{2}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \((5 \sqrt{6}+2 \sqrt{3})^{2}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \((\sqrt{21}+\sqrt{3})^{2}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \((4+2 \sqrt{5})^{2}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \((\sqrt{7}-2)^{2}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \((4-\sqrt{3})^{2}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \((2 \sqrt{35}+\sqrt{5})^{2}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \((\sqrt{70}-3 \sqrt{10})^{2}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \((3-\sqrt{5})(3+\sqrt{5})\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \((2 \sqrt{5}-1)^{2}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \((\sqrt{6}-1)(\sqrt{6}+1)\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            %\item \((2 \sqrt{3}-5 \sqrt{6})(2 \sqrt{3}+5 \sqrt{6})\)
            %    \mdots{2}
            %\item \((\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})\)
            %    \mdots{2}
        \end{multicols}\end{parts}
        \end{questions}
\end{exercisebox}
|726265176|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                                                  
     \subsection{Exercises:}                                                                                                          
     \begin{questions}                                                                                                                
         \Question[12]                                                                                                                
         \begin{parts}\begin{multicols}{2}                                                                                            
             \part \((5+\sqrt{3})^{2}\)                                                                                               
             \begin{solutionordottedlines}[2cm]                                                                                       
             $(5+\sqrt{3})^{2} = 5^2 + 2(5)(\sqrt{3}) + (\sqrt{3})^2 = 25 + 10\sqrt{3} + 3 = 28 + 10\sqrt{3}$                         
             \end{solutionordottedlines}                                                                                              
             \part \((\sqrt{2}+6)^{2}\)                                                                                               
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 $(\sqrt{2}+6)^{2} = (\sqrt{2})^2 + 2(\sqrt{2})(6) + 6^2 = 2 + 12\sqrt{2} + 36 = 38 + 12\sqrt{2}$                     
                 \end{solutionordottedlines}                                                                                          
             \part \((5 \sqrt{6}+2 \sqrt{3})^{2}\)                                                                                    
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 $(5 \sqrt{6}+2 \sqrt{3})^{2} = (5\sqrt{6})^2 + 2(5\sqrt{6})(2\sqrt{3}) + (2\sqrt{3})^2 = 25\cdot6 + 2\cdot10\sqrt{18 + 4\cdot3 = 150 + 20\sqrt{18} + 12$                                \end{solutionordottedlines}                                                                                          
             \part \((\sqrt{21}+\sqrt{3})^{2}\)                                                                                       
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 $(\sqrt{21}+\sqrt{3})^{2} = (\sqrt{21})^2 + 2(\sqrt{21})(\sqrt{3}) + (\sqrt{3})^2 = 21 + 2\sqrt{63} + 3 = 24 + 2\sqrt{63}$                                                              \end{solutionordottedlines}                                                                                          
             \part \((4+2 \sqrt{5})^{2}\)                                                                                             
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 $(4+2 \sqrt{5})^{2} = 4^2 + 2(4)(2\sqrt{5}) + (2\sqrt{5})^2 = 16 + 16\sqrt{5} + 4\cdot5 = 16 + 16\sqrt{5} + 20 = 36  16\sqrt{5}$                                                        \end{solutionordottedlines}                                                                                          
             \part \((\sqrt{7}-2)^{2}\)                                                                                               
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 $(\sqrt{7}-2)^{2} = (\sqrt{7})^2 - 2(\sqrt{7})(2) + 2^2 = 7 - 4\sqrt{7} + 4 = 11 - 4\sqrt{7}$                        
                 \end{solutionordottedlines}                                                                                          
             \part \((4-\sqrt{3})^{2}\)                                                                                               
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 $(4-\sqrt{3})^{2} = 4^2 - 2(4)(\sqrt{3}) + (\sqrt{3})^2 = 16 - 8\sqrt{3} + 3 = 19 - 8\sqrt{3}$                       
                 \end{solutionordottedlines}                                                                                          
             \part \((2 \sqrt{35}+\sqrt{5})^{2}\)                                                                                     
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 $(2 \sqrt{35}+\sqrt{5})^{2} = (2\sqrt{35})^2 + 2(2\sqrt{35})(\sqrt{5}) + (\sqrt{5})^2 = 4\cdot35 + 2\cdot2\sqrt{175} 5 = 140 + 4\sqrt{175} + 5$                                                                                                           
                 \end{solutionordottedlines}                                                                                          
             \part \((\sqrt{70}-3 \sqrt{10})^{2}\)                                                                                    
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 $(\sqrt{70}-3 \sqrt{10})^{2} = (\sqrt{70})^2 - 2(\sqrt{70})(3\sqrt{10}) + (3\sqrt{10})^2 = 70 - 2\cdot3\sqrt{700} + 9\cdot10 = 70 - 6\sqrt{700} + 90$                                                                                                    
                 \end{solutionordottedlines}                                                                                          
             \part \((3-\sqrt{5})(3+\sqrt{5})\)                                                                                       
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 $(3-\sqrt{5})(3+\sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4$                                                          
                 \end{solutionordottedlines}                                                                                          
             \part \((2 \sqrt{5}-1)^{2}\)                                                                                             
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 $(2 \sqrt{5}-1)^{2} = (2\sqrt{5})^2 - 2(2\sqrt{5})(1) + 1^2 = 4\cdot5 - 2\cdot2\sqrt{5} + 1 = 20 - 4\sqrt{5} + 1 = 2 - 4\sqrt{5}$                                                                                                                         
                 \end{solutionordottedlines}                                                                                          
             \part \((\sqrt{6}-1)(\sqrt{6}+1)\)                                                                                       
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 $(\sqrt{6}-1)(\sqrt{6}+1) = (\sqrt{6})^2 - 1^2 = 6 - 1 = 5$                                                          
                 \end{solutionordottedlines}                                                                                          
             %\item \((2 \sqrt{3}-5 \sqrt{6})(2 \sqrt{3}+5 \sqrt{6})\)                                                                
             %    \mdots{2}                                                                                                           
             %\item \((\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})\)                                                                        
             %    \mdots{2}                                                                                                           
         \end{multicols}\end{parts}                                                                                                   
         \end{questions}                                                                                                              
 \end{exercisebox}                                                                                                                    
|726265184|kitty|/Applications/kitty.app|0|
 \begin{examplebox}                                                                                                                   
     \subsection{Examples:}                                                                                                           
     \begin{questions}                                                                                                                
         \Question[2] Express the following with a \emph{rational} denominator                                                        
         \begin{doublespace}                                                                                                          
         \begin{parts}                                                                                                                
             \part \(\frac{4}{\sqrt{2}}=\)                                                                                            
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 \(\frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}\)                            
                 \end{solutionordottedlines}                                                                                          
             \part \(\frac{9}{4 \sqrt{3}}=\)                                                                                          
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 \(\frac{9}{4\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{9\sqrt{3}}{12} = \frac{3\sqrt{3}}{4}\)                
                 \end{solutionordottedlines}                                                                                          
             \part \[\frac{2}{\sqrt{3}}+\frac{\sqrt{3}}{2}=\]                                                                         
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 \(\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} + \frac{\sqrt{3}}{2} \times \frac{2}{2} = \frac{2\sqrt{3}}{3}  
 \frac{2\sqrt{3}}{4} = \frac{4\sqrt{3} + 3\sqrt{3}}{6} = \frac{7\sqrt{3}}{6}\)                                                        
                 \end{solutionordottedlines}                                                                                          
         \end{parts}                                                                                                                  
         \end{doublespace}                                                                                                            
     \end{questions}                                                                                                                  
 \end{examplebox}                                                                                                                     
|726265195|kitty|/Applications/kitty.app|0|
\begin{examplebox}
    \subsection{Examples:}
    \begin{questions}
        \Question[2] Express the following with a \emph{rational} denominator
        \begin{doublespace}
        \begin{parts}
            \part \(\frac{4}{\sqrt{2}}=\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \(\frac{9}{4 \sqrt{3}}=\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{2}{\sqrt{3}}+\frac{\sqrt{3}}{2}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
        \end{parts}
        \end{doublespace}
    \end{questions}
\end{examplebox}
|726265200|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                                                  
     \subsection{Exercises:}                                                                                                          
     \begin{questions}                                                                                                                
         \Question[6]                                                                                                                 
         \begin{parts}\begin{multicols}{2}                                                                                            
             \begin{doublespace}                                                                                                      
             \part \[\frac{\sqrt{3}}{\sqrt{7}}=\]                                                                                     
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 \(\frac{\sqrt{3}}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{21}}{7}\)                                 
                 \end{solutionordottedlines}                                                                                          
             \part \[\frac{2 \sqrt{7}}{\sqrt{3}}=\]                                                                                   
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 \(\frac{2 \sqrt{7}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{21}}{3}\)                              
                 \end{solutionordottedlines}                                                                                          
             \part \[\frac{\sqrt{2}}{3 \sqrt{10}}=\]                                                                                  
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 \(\frac{\sqrt{2}}{3\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} = \frac{\sqrt{20}}{30} = \frac{\sqrt{5}}{15}\)      
                 \end{solutionordottedlines}                                                                                          
             \part \[\frac{\sqrt{15}}{3 \sqrt{5}}=\]                                                                                  
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 \(\frac{\sqrt{15}}{3\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{75}}{15} = \frac{5}{3}\)                
                 \end{solutionordottedlines}                                                                                          
             \part \[\frac{2}{\sqrt{3}}+\frac{3}{2 \sqrt{3}}=\]                                                                       
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 \(\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} + \frac{3}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} =       
 \frac{2\sqrt{3}}{3} + \frac{3\sqrt{3}}{6} = \frac{4\sqrt{3} + 3\sqrt{3}}{6} = \frac{7\sqrt{3}}{6}\)                                  
                 \end{solutionordottedlines}                                                                                          
             \part \[\frac{5 \sqrt{2}}{3}-\frac{1}{\sqrt{3}}=\]                                                                       
                 \begin{solutionordottedlines}[2cm]                                                                                   
                 \(\frac{5 \sqrt{2}}{3} - \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{5 \sqrt{2}}{3} -                
 \frac{\sqrt{3}}{3} = \frac{5\sqrt{2} - \sqrt{3}}{3}\)                                                                                
                 \end{solutionordottedlines}                                                                                          
             \end{doublespace}                                                                                                        
         \end{multicols}\end{parts}                                                                                                   
         \Question[4] Find the value of \(x\). Express your answer with a rational denominator.                                       
             \begin{parts}                                                                                                            
                 \begin{multicols}{2}                                                                                                 
                     \part \includegraphics[width=0.4\linewidth]{img/4a}                                                              
                         \begin{solutionordottedlines}[4cm]                                                                           
                         % Since the image is not available, I cannot provide a solution.                                             
                         \end{solutionordottedlines}                                                                                  
                     \part \includegraphics[width=0.4\linewidth]{img/4b}                                                              
                         \begin{solutionordottedlines}[4cm]                                                                           
                         % Since the image is not available, I cannot provide a solution.                                             
                         \end{solutionordottedlines}                                                                                  
                 \end{multicols}                                                                                                      
             \end{parts}                                                                                                              
     \end{questions}                                                                                                                  
 \end{exercisebox}                                                                                                                    
|726265210|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \begin{doublespace}
            \part \[\frac{\sqrt{3}}{\sqrt{7}}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{2 \sqrt{7}}{\sqrt{3}}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{\sqrt{2}}{3 \sqrt{10}}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{\sqrt{15}}{3 \sqrt{5}}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{2}{\sqrt{3}}+\frac{3}{2 \sqrt{3}}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{5 \sqrt{2}}{3}-\frac{1}{\sqrt{3}}=\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \end{doublespace}
        \end{multicols}\end{parts}
        \Question[4] Find the value of \(x\). Express your answer with a rational denominator.
            \begin{parts}
                \begin{multicols}{2}
                    \part \includegraphics[width=0.4\linewidth]{img/4a}
                        \begin{solutionordottedlines}[4cm]
                        \end{solutionordottedlines}
                    \part \includegraphics[width=0.4\linewidth]{img/4b}
                        \begin{solutionordottedlines}[4cm]
                        \end{solutionordottedlines}
                \end{multicols}
            \end{parts}
    \end{questions}
\end{exercisebox}
|726265215|kitty|/Applications/kitty.app|0|
     To find the length of the edge $VC$, we can use the Pythagorean theorem on the triangle formed by the height of the pyramid, hal 
 the base, and the slant height (edge $VC$). The half of the base is $2$ cm (half of $4$ cm). So, $VC^2 = 5^2 + 2^2 = 25 + 4 = 29$,   
 thus $VC = \sqrt{29} \approx 5.39$ cm.                                                                                               
|726265254|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                                                  
     \subsection*{Exercises:}                                                                                                         
     \begin{questions}                                                                                                                
         \Question[6]                                                                                                                 
             \begin{center}                                                                                                           
             \includegraphics[width=0.4\textwidth]{img/5c}                                                                            
             \end{center}                                                                                                             
             The rectangular prism in the diagram has a length of \(12 \mathrm{~cm}\), a width of \(5 \mathrm{~cm}\) and a height of  
 \(6 \mathrm{~cm}\).                                                                                                                  
             \begin{parts}                                                                                                            
                 \part Consider triangle \(E F G\). Find:                                                                             
                     \begin{subparts}                                                                                                 
                         \subpart \(E F\)                                                                                             
                             \begin{solutionordottedlines}[2cm]                                                                       
                             $EF = 12$ cm (given as the length of the prism).                                                         
                             \end{solutionordottedlines}                                                                              
                         \subpart the size of \(E F G\)                                                                               
                             \begin{solutionordottedlines}[2cm]                                                                       
                             $\angle EFG$ is a right angle since $EF$ and $FG$ are edges of the rectangular prism meeting at a right  
 angle.                                                                                                                               
                             \end{solutionordottedlines}                                                                              
                     \end{subparts}                                                                                                   
                 \part Find \(E G\).                                                                                                  
                     \begin{solutionordottedlines}[2cm]                                                                               
                     To find $EG$, we use Pythagoras' theorem on triangle $EFG$: $EG^2 = EF^2 + FG^2 = 12^2 + 6^2 = 144 + 36 = 180$,  
 $EG = \sqrt{180} = 6\sqrt{5} \approx 13.42$ cm.                                                                                      
                     \end{solutionordottedlines}                                                                                      
                 \part Find \(A G\), correct to 1 decimal place.                                                                      
                     \begin{solutionordottedlines}[2cm]                                                                               
                     To find $AG$, we use Pythagoras' theorem on triangle $AFG$: $AG^2 = AF^2 + FG^2 = 5^2 + 6^2 = 25 + 36 = 61$, so  
 $AG = \sqrt{61} \approx 7.8$ cm.                                                                                                     
                     \end{solutionordottedlines}                                                                                      
             \end{parts}                                                                                                              
         \question[] Note: \(A G\) is called the space diagonal of the rectangular prism.                                             
         \Question[3] Find the length of the longest pencil that can fit inside a cylindrical pencil case of length \(15 \mathrm{~cm} 
 and radius \(2 \mathrm{~cm}\).                                                                                                       
             \begin{solutionordottedlines}[2in]                                                                                       
             The longest pencil that can fit inside the cylindrical pencil case would be the length of the diagonal of the cylinder.  
 This can be found using Pythagoras' theorem with the length and diameter of the cylinder: $pencil^2 = length^2 + diameter^2 = 15^2 + 
 (2 \times 2)^2 = 225 + 16 = 241$, so the pencil length is $\sqrt{241} \approx 15.52$ cm.                                             
             \end{solutionordottedlines}                                                                                              
     \end{questions}                                                                                                                  
 \end{exercisebox}                                                                                                                    
|726265278|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection*{Exercises:}
    \begin{questions}
        \Question[6] 
            \begin{center}
            \includegraphics[width=0.4\textwidth]{img/5c}
            \end{center}
            The rectangular prism in the diagram has a length of \(12 \mathrm{~cm}\), a width of \(5 \mathrm{~cm}\) and a height of \(6 \mathrm{~cm}\).
            \begin{parts}
                \part Consider triangle \(E F G\). Find:
                    \begin{subparts}
                        \subpart \(E F\)
                            \begin{solutionordottedlines}[2cm]
                            \end{solutionordottedlines}
                        \subpart the size of \(E F G\)
                            \begin{solutionordottedlines}[2cm]
                            \end{solutionordottedlines}
                    \end{subparts}
                \part Find \(E G\).
                    \begin{solutionordottedlines}[2cm]
                    \end{solutionordottedlines}
                \part Find \(A G\), correct to 1 decimal place.
                    \begin{solutionordottedlines}[2cm]
                    \end{solutionordottedlines}
            \end{parts}
        \question[] Note: \(A G\) is called the space diagonal of the rectangular prism.
        \Question[3] Find the length of the longest pencil that can fit inside a cylindrical pencil case of length \(15 \mathrm{~cm}\) and radius \(2 \mathrm{~cm}\).
            \begin{solutionordottedlines}[2in]
            \end{solutionordottedlines}
    \end{questions}
\end{exercisebox}
|726265281|kitty|/Applications/kitty.app|0|
 \begin{examplebox}                                                                                                                   
     \subsection{Examples:}                                                                                                           
     \begin{questions}                                                                                                                
         \Question[1] \[\frac{2 \sqrt{5}}{2 \sqrt{5}-2}=\]                                                                            
             \begin{solutionordottedlines}[3cm]                                                                                       
                 Multiply by the conjugate: $\frac{2 \sqrt{5}}{2 \sqrt{5}-2} \cdot \frac{2 \sqrt{5}+2}{2 \sqrt{5}+2} = \frac{4 \cdot  
 + 4 \sqrt{5}}{20 - 4} = \frac{20 + 4 \sqrt{5}}{16} = \frac{5}{4} + \sqrt{5}$                                                         
             \end{solutionordottedlines}                                                                                              
         \Question[1] \[\frac{\sqrt{3}+\sqrt{2}}{3 \sqrt{2}+2 \sqrt{3}}\]                                                             
             \begin{solutionordottedlines}[3cm]                                                                                       
                 Multiply by the conjugate: $\frac{\sqrt{3}+\sqrt{2}}{3 \sqrt{2}+2 \sqrt{3}} \cdot \frac{3 \sqrt{2}-2 \sqrt{3}}{3     
 \sqrt{2}-2 \sqrt{3}} = \frac{3 \cdot 2 - 2 \cdot 3}{6 - 4 \sqrt{6} + 4 \sqrt{6} - 12} = \frac{6 - 6}{-6} = 0$                        
             \end{solutionordottedlines}                                                                                              
     \end{questions}                                                                                                                  
 \end{examplebox}                                                                                                                     
|726265293|kitty|/Applications/kitty.app|0|
\begin{examplebox}
    \subsection{Examples:}
    \begin{questions}
        \Question[1] \[\frac{2 \sqrt{5}}{2 \sqrt{5}-2}=\]
            \begin{solutionordottedlines}[3cm]
            \end{solutionordottedlines}
        \Question[1] \[\frac{\sqrt{3}+\sqrt{2}}{3 \sqrt{2}+2 \sqrt{3}}\]
            \begin{solutionordottedlines}[3cm]
            \end{solutionordottedlines}
    \end{questions}
\end{examplebox}
|726265297|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                                                  
     \subsection{Exercises:}                                                                                                          
     \begin{questions}                                                                                                                
         \Question[6]                                                                                                                 
         \begin{parts}\begin{multicols}{2}                                                                                            
             \part \(\frac{3}{\sqrt{2}-1}\)                                                                                           
                 \begin{solutionordottedlines}[3cm]                                                                                   
                     Multiply by the conjugate: $\frac{3}{\sqrt{2}-1} \cdot \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{3(\sqrt{2}+1)}{2-1} 
 3\sqrt{2} + 3$                                                                                                                       
                 \end{solutionordottedlines}                                                                                          
             \part \(\frac{2}{3+\sqrt{5}}\)                                                                                           
                 \begin{solutionordottedlines}[3cm]                                                                                   
                     Multiply by the conjugate: $\frac{2}{3+\sqrt{5}} \cdot \frac{3-\sqrt{5}}{3-\sqrt{5}} = \frac{6 - 2\sqrt{5}}{9 -  
 = \frac{6 - 2\sqrt{5}}{4} = \frac{3}{2} - \frac{\sqrt{5}}{2}$                                                                        
                 \end{solutionordottedlines}                                                                                          
             \part \(\frac{4}{\sqrt{5}+\sqrt{2}}\)                                                                                    
                 \begin{solutionordottedlines}[3cm]                                                                                   
                     Multiply by the conjugate: $\frac{4}{\sqrt{5}+\sqrt{2}} \cdot \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} =      
 \frac{4\sqrt{5} - 4\sqrt{2}}{5 - 2} = \frac{4\sqrt{5} - 4\sqrt{2}}{3}$                                                               
                 \end{solutionordottedlines}                                                                                          
             \part \(\frac{1}{\sqrt{7}-\sqrt{5}}\)                                                                                    
                 \begin{solutionordottedlines}[3cm]                                                                                   
                     Multiply by the conjugate: $\frac{1}{\sqrt{7}-\sqrt{5}} \cdot \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}} =      
 \frac{\sqrt{7}+\sqrt{5}}{7 - 5} = \frac{\sqrt{7}+\sqrt{5}}{2}$                                                                       
                 \end{solutionordottedlines}                                                                                          
             \part \(\frac{1}{\sqrt{2}-1}\)                                                                                           
                 \begin{solutionordottedlines}[3cm]                                                                                   
                     Multiply by the conjugate: $\frac{1}{\sqrt{2}-1} \cdot \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{\sqrt{2}+1}{2-1} =  
 \sqrt{2} + 1$                                                                                                                        
                 \end{solutionordottedlines}                                                                                          
             \part \(\frac{1}{\sqrt{3}-\sqrt{2}}\)                                                                                    
                 \begin{solutionordottedlines}[3cm]                                                                                   
                     Multiply by the conjugate: $\frac{1}{\sqrt{3}-\sqrt{2}} \cdot \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} =      
 \frac{\sqrt{3}+\sqrt{2}}{3 - 2} = \sqrt{3} + \sqrt{2}$                                                                               
                 \end{solutionordottedlines}                                                                                          
         \end{multicols}\end{parts}                                                                                                   
     \end{questions}                                                                                                                  
 \end{exercisebox}                                                                                                                    
|726265302|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3}{\sqrt{2}-1}\)
                \begin{solutionordottedlines}[3cm]
                \end{solutionordottedlines}
            \part \(\frac{2}{3+\sqrt{5}}\)
                \begin{solutionordottedlines}[3cm]
                \end{solutionordottedlines}
            \part \(\frac{4}{\sqrt{5}+\sqrt{2}}\)
                \begin{solutionordottedlines}[3cm]
                \end{solutionordottedlines}
            \part \(\frac{1}{\sqrt{7}-\sqrt{5}}\)
                \begin{solutionordottedlines}[3cm]
                \end{solutionordottedlines}
            \part \(\frac{1}{\sqrt{2}-1}\)
                \begin{solutionordottedlines}[3cm]
                \end{solutionordottedlines}
            \part \(\frac{1}{\sqrt{3}-\sqrt{2}}\)
                \begin{solutionordottedlines}[3cm]
                \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \end{questions}
\end{exercisebox}
|726265305|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                                                  
     \subsection{Exercises}                                                                                                           
     \begin{questions}\begin{onehalfspacing}                                                                                          
         \Question[10]                                                                                                                
         \begin{parts}                                                                                                                
             \begin{multicols}{2}                                                                                                     
             \part \(0 . \dot{5}=\frac{5}{9}\)                                                                                        
             \part \(0 . \dot{7}=\frac{7}{9}\)                                                                                        
             \part \(0 . \dot{9}=\frac{9}{9}=1\)                                                                                      
             \part \(0.1 \dot{4}=\frac{14}{99}\)                                                                                      
             \part \(0.2 \dot{3}=\frac{23}{99}\)                                                                                      
             \part \(0.6 \dot{2}=\frac{62}{99}\)                                                                                      
             \part \(0 . \dot{1} \dot{3}=\frac{13}{99}\)                                                                              
             \part \(0 . \dot{0} 7=\frac{7}{90}\)                                                                                     
             \part \(0 . \dot{9} \dot{1}=\frac{91}{99}\)                                                                              
             \part \(0 . \dot{2} 4 \dot{1}=\frac{239}{990}\)                                                                          
             \end{multicols}                                                                                                          
             \begin{multicols}{4}                                                                                                     
             \part Which of these numbers are irrational?                                                                             
                 \begin{enumerate}                                                                                                    
                     \item[\CheckedBox] \(\sqrt{7}\)                                                                                  
                     \item[\Square] \(\sqrt{25}\)                                                                                     
                     \item[\Square] \(0 . \dot{6}\)                                                                                   
                     \item[\CheckedBox] \(\frac{\pi}{3}\)                                                                             
                 \end{enumerate}                                                                                                      
             \end{multicols}                                                                                                          
         \end{parts}                                                                                                                  
     \end{onehalfspacing}\end{questions}                                                                                              
 \end{exercisebox}                                                                                                                    
|726265320|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises}
    \begin{questions}\begin{onehalfspacing}
        \Question[10]
        \begin{parts}
            \begin{multicols}{2}
            \part \(0 . \dot{5}=\fillin[]\)
            \part \(0 . \dot{7}=\fillin[]\)
            \part \(0 . \dot{9}=\fillin[]\)
            \part \(0.1 \dot{4}=\fillin[]\)
            \part \(0.2 \dot{3}=\fillin[]\)
            \part \(0.6 \dot{2}=\fillin[]\)
            \part \(0 . \dot{1} \dot{3}=\fillin[]\)
            \part \(0 . \dot{0} 7=\fillin[]\)
            \part \(0 . \dot{9} \dot{1}=\fillin[]\)
            \part \(0 . \dot{2} 4 \dot{1}=\fillin[]\)
            \end{multicols}
            \begin{multicols}{4}
            \part Which of these numbers are irrational?
                \begin{enumerate}
                    \item[\Square] \(\sqrt{7}\)
                    \item[\Square] \(\sqrt{25}\)
                    \item[\Square] \(0 . \dot{6}\)
                    \item[\Square] \(\frac{\pi}{3}\)
                \end{enumerate}
            \end{multicols}
        \end{parts}
    \end{onehalfspacing}\end{questions}
\end{exercisebox}
|726265324|kitty|/Applications/kitty.app|0|
\begin{questions}
    \Question[6] Simplify the following:
    \begin{parts}
        \begin{multicols}{2}
            \part \((2 \sqrt{3}+\sqrt{2})^{2}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \((2 \sqrt{5}+4 \sqrt{3})^{2}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \((\sqrt{x y}+1)^{2}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \((4 \sqrt{2}-3 \sqrt{7})^{2}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \((\sqrt{x}-\sqrt{y})^{2}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \((\sqrt{11}-2 \sqrt{22})^{2}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
        \end{multicols}

        \part If \(x=\sqrt{2}-1\) and \(y=\sqrt{2}+1\), find:
            \begin{subparts}\begin{multicols}{2}
                \subpart \(x y\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
                \subpart \(x^{2} y\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
                \subpart \(y^{2} x\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
                \subpart \(\frac{1}{x}+\frac{1}{y}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \end{multicols}\end{subparts}

        \part \[(\sqrt{5}+\sqrt{3})^{2}-(\sqrt{5}-\sqrt{3})^{2}\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
        \part \[(\sqrt{3}+\sqrt{2})^{2}-(\sqrt{3}-\sqrt{2})^{2}\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
    \end{parts}
    \Question[2] Simplify \[(a \sqrt{b}+c \sqrt{d})(a \sqrt{b}-c \sqrt{d})\].
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
    \Question[9] For each of the figures shown, find:
        \begin{enumerate}[(i)]
            \item the value of \(x\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \item the area of the triangle
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \item  the perimeter of the triangle
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
        \end{enumerate}
        \begin{parts}\begin{multicols}{3}
            \part
                \begin{center}
                \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04(3)}
                \end{center}
                \begin{solutionordottedlines}[3in]
                \end{solutionordottedlines}
            \part
                \begin{center}
                \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04(1)}
                \end{center}
                \begin{solutionordottedlines}[3in]
                \end{solutionordottedlines}
            \part
                \begin{center}
                \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04(4)}
                \end{center}
                \begin{solutionordottedlines}[3in]
                \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \Question[4] For the 2 figures below, find:
        \begin{enumerate}[(i)]
            \item the perimeter
            \item the area
        \end{enumerate}
        \begin{parts}\begin{multicols}{2}
            \part
                \begin{center}
                \includegraphics[width=0.2\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04}
                \end{center}
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part
                \begin{center}
                \includegraphics[width=0.2\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04(2)}
                \end{center}
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \Question[8] Rationalise the following:
        \begin{parts}\begin{multicols}{2}
            \part \[\frac{14}{\sqrt{7}}\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{\sqrt{14}}{\sqrt{7}}\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{\sqrt{5}}{3 \sqrt{7}}\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{\sqrt{3}}{4 \sqrt{6}}\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{5}{\sqrt{3}}\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{1}{\sqrt{18}}\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{1}{\sqrt{2}}+\sqrt{2}\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{\sqrt{72}}{\sqrt{3}}+\frac{3}{\sqrt{2}}-\frac{2}{2 \sqrt{2}}\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}+\frac{7}{2 \sqrt{3}}\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \Question[6] If \(x=2 \sqrt{14}\) and \(y=4 \sqrt{2}\), find and rationalise the denominator.
        \begin{parts}\begin{multicols}{3}
                \part \(\frac{x}{y}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
                \part \(\frac{y}{x}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
                \part \(\frac{\sqrt{2} x}{\sqrt{3} y}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \Question[3] A bowl in the shape of a hemisphere of radius length \(5 \mathrm{~cm}\) is partially filled with water. The surface of the water is a circle of radius \(4 \mathrm{~cm}\) when the rim of the bowl is horizontal. Find the depth of the water.
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
|726265340|kitty|/Applications/kitty.app|0|
 \begin{questions}                                                                                                                    
     \Question[6] Simplify the following:                                                                                             
     \begin{parts}                                                                                                                    
         \begin{multicols}{2}                                                                                                         
             \part \((2 \sqrt{3}+\sqrt{2})^{2}\)                                                                                      
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $(2 \sqrt{3}+\sqrt{2})^{2} = 4 \cdot 3 + 4 \sqrt{3} \sqrt{2} + 2 = 12 + 4 \sqrt{6} + 2 = 14 + 4 \sqrt{6}$        
                 \end{solutionordottedlines}                                                                                          
             \part \((2 \sqrt{5}+4 \sqrt{3})^{2}\)                                                                                    
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $(2 \sqrt{5}+4 \sqrt{3})^{2} = 4 \cdot 5 + 16 \cdot 3 + 2 \cdot 8 \sqrt{15} = 20 + 48 + 16 \sqrt{15} = 68 + 16   
 \sqrt{15}$                                                                                                                           
                 \end{solutionordottedlines}                                                                                          
             \part \((\sqrt{x y}+1)^{2}\)                                                                                             
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $(\sqrt{x y}+1)^{2} = xy + 2\sqrt{xy} + 1$                                                                       
                 \end{solutionordottedlines}                                                                                          
             \part \((4 \sqrt{2}-3 \sqrt{7})^{2}\)                                                                                    
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $(4 \sqrt{2}-3 \sqrt{7})^{2} = 16 \cdot 2 + 9 \cdot 7 - 2 \cdot 12 \sqrt{14} = 32 + 63 - 24 \sqrt{14} = 95 - 24  
 \sqrt{14}$                                                                                                                           
                 \end{solutionordottedlines}                                                                                          
             \part \((\sqrt{x}-\sqrt{y})^{2}\)                                                                                        
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $(\sqrt{x}-\sqrt{y})^{2} = x - 2\sqrt{xy} + y$                                                                   
                 \end{solutionordottedlines}                                                                                          
             \part \((\sqrt{11}-2 \sqrt{22})^{2}\)                                                                                    
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $(\sqrt{11}-2 \sqrt{22})^{2} = 11 - 4 \cdot 22 + 4 \cdot 11 = 11 - 88 + 44 = -33 + 44 = 11$                      
                 \end{solutionordottedlines}                                                                                          
         \end{multicols}                                                                                                              
                                                                                                                                      
         \part If \(x=\sqrt{2}-1\) and \(y=\sqrt{2}+1\), find:                                                                        
             \begin{subparts}\begin{multicols}{2}                                                                                     
                 \subpart \(x y\)                                                                                                     
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $x y = (\sqrt{2}-1)(\sqrt{2}+1) = (\sqrt{2})^2 - (1)^2 = 2 - 1 = 1$                                              
                 \end{solutionordottedlines}                                                                                          
                 \subpart \(x^{2} y\)                                                                                                 
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $x^{2} y = (\sqrt{2}-1)^2(\sqrt{2}+1) = (2 - 2\sqrt{2} + 1)(\sqrt{2}+1) = (3 - 2\sqrt{2})(\sqrt{2}+1) = 3\sqrt{2 
 + 3 - 2\cdot 2 - 2\sqrt{2} = 3\sqrt{2} + 3 - 4 - 2\sqrt{2} = \sqrt{2} - 1$                                                           
                 \end{solutionordottedlines}                                                                                          
                 \subpart \(y^{2} x\)                                                                                                 
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $y^{2} x = (\sqrt{2}+1)^2(\sqrt{2}-1) = (2 + 2\sqrt{2} + 1)(\sqrt{2}-1) = (3 + 2\sqrt{2})(\sqrt{2}-1) = 3\sqrt{2 
 - 3 + 2\cdot 2 - 2\sqrt{2} = 3\sqrt{2} - 3 + 4 - 2\sqrt{2} = \sqrt{2} + 1$                                                           
                 \end{solutionordottedlines}                                                                                          
                 \subpart \(\frac{1}{x}+\frac{1}{y}\)                                                                                 
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $\frac{1}{x}+\frac{1}{y} = \frac{1}{\sqrt{2}-1} + \frac{1}{\sqrt{2}+1} = \frac{\sqrt{2}+1}{2-1} +                
 \frac{\sqrt{2}-1}{2-1} = \sqrt{2}+1 + \sqrt{2}-1 = 2\sqrt{2}$                                                                        
                 \end{solutionordottedlines}                                                                                          
             \end{multicols}\end{subparts}                                                                                            
                                                                                                                                      
         \part \[(\sqrt{5}+\sqrt{3})^{2}-(\sqrt{5}-\sqrt{3})^{2}\]                                                                    
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $(\sqrt{5}+\sqrt{3})^{2}-(\sqrt{5}-\sqrt{3})^{2} = (5 + 2\sqrt{15} + 3) - (5 - 2\sqrt{15} + 3) = 8 + 2\sqrt{15}  
 8 + 2\sqrt{15} = 4\sqrt{15}$                                                                                                         
                 \end{solutionordottedlines}                                                                                          
         \part \[(\sqrt{3}+\sqrt{2})^{2}-(\sqrt{3}-\sqrt{2})^{2}\]                                                                    
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $(\sqrt{3}+\sqrt{2})^{2}-(\sqrt{3}-\sqrt{2})^{2} = (3 + 2\sqrt{6} + 2) - (3 - 2\sqrt{6} + 2) = 5 + 2\sqrt{6} - 5 
 2\sqrt{6} = 4\sqrt{6}$                                                                                                               
                 \end{solutionordottedlines}                                                                                          
     \end{parts}                                                                                                                      
     \Question[2] Simplify \[(a \sqrt{b}+c \sqrt{d})(a \sqrt{b}-c \sqrt{d})\].                                                        
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $(a \sqrt{b}+c \sqrt{d})(a \sqrt{b}-c \sqrt{d}) = (a \sqrt{b})^2 - (c \sqrt{d})^2 = a^2 b - c^2 d$               
                 \end{solutionordottedlines}                                                                                          
     \Question[9] For each of the figures shown, find:                                                                                
         \begin{enumerate}[(i)]                                                                                                       
             \item the value of \(x\)                                                                                                 
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     % Since no figures are provided, no solution can be given.                                                       
                 \end{solutionordottedlines}                                                                                          
             \item the area of the triangle                                                                                           
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     % Since no figures are provided, no solution can be given.                                                       
                 \end{solutionordottedlines}                                                                                          
             \item  the perimeter of the triangle                                                                                     
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     % Since no figures are provided, no solution can be given.                                                       
                 \end{solutionordottedlines}                                                                                          
         \end{enumerate}                                                                                                              
         \begin{parts}\begin{multicols}{3}                                                                                            
             \part                                                                                                                    
                 \begin{center}                                                                                                       
                 \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04(3)}                                    
                 \end{center}                                                                                                         
                 \begin{solutionordottedlines}[3in]                                                                                   
                     % Since no figures are provided, no solution can be given.                                                       
                 \end{solutionordottedlines}                                                                                          
             \part                                                                                                                    
                 \begin{center}                                                                                                       
                 \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04(1)}                                    
                 \end{center}                                                                                                         
                 \begin{solutionordottedlines}[3in]                                                                                   
                     % Since no figures are provided, no solution can be given.                                                       
                 \end{solutionordottedlines}                                                                                          
             \part                                                                                                                    
                 \begin{center}                                                                                                       
                 \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04(4)}                                    
                 \end{center}                                                                                                         
                 \begin{solutionordottedlines}[3in]                                                                                   
                     % Since no figures are provided, no solution can be given.                                                       
                 \end{solutionordottedlines}                                                                                          
         \end{multicols}\end{parts}                                                                                                   
     \Question[4] For the 2 figures below, find:                                                                                      
         \begin{enumerate}[(i)]                                                                                                       
             \item the perimeter                                                                                                      
             \item the area                                                                                                           
         \end{enumerate}                                                                                                              
         \begin{parts}\begin{multicols}{2}                                                                                            
             \part                                                                                                                    
                 \begin{center}                                                                                                       
                 \includegraphics[width=0.2\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04}                                       
                 \end{center}                                                                                                         
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     % Since no figures are provided, no solution can be given.                                                       
                 \end{solutionordottedlines}                                                                                          
             \part                                                                                                                    
                 \begin{center}                                                                                                       
                 \includegraphics[width=0.2\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04(2)}                                    
                 \end{center}                                                                                                         
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     % Since no figures are provided, no solution can be given.                                                       
                 \end{solutionordottedlines}                                                                                          
         \end{multicols}\end{parts}                                                                                                   
     \Question[8] Rationalise the following:                                                                                          
         \begin{parts}\begin{multicols}{2}                                                                                            
             \part \[\frac{14}{\sqrt{7}}\]                                                                                            
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $\frac{14}{\sqrt{7}} = \frac{14}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{14\sqrt{7}}{7} = 2\sqrt{7}$   
                 \end{solutionordottedlines}                                                                                          
             \part \[\frac{\sqrt{14}}{\sqrt{7}}\]                                                                                     
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $\frac{\sqrt{14}}{\sqrt{7}} = \sqrt{\frac{14}{7}} = \sqrt{2}$                                                    
                 \end{solutionordottedlines}                                                                                          
             \part \[\frac{\sqrt{5}}{3 \sqrt{7}}\]                                                                                    
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $\frac{\sqrt{5}}{3 \sqrt{7}} = \frac{\sqrt{5}}{3 \sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{35}}{21 
                 \end{solutionordottedlines}                                                                                          
             \part \[\frac{\sqrt{3}}{4 \sqrt{6}}\]                                                                                    
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $\frac{\sqrt{3}}{4 \sqrt{6}} = \frac{\sqrt{3}}{4 \sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{\sqrt{18}}{24 
 = \frac{\sqrt{2}}{8}$                                                                                                                
                 \end{solutionordottedlines}                                                                                          
             \part \[\frac{5}{\sqrt{3}}\]                                                                                             
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $\frac{5}{\sqrt{3}} = \frac{5}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3}$                  
                 \end{solutionordottedlines}                                                                                          
             \part \[\frac{1}{\sqrt{18}}\]                                                                                            
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $\frac{1}{\sqrt{18}} = \frac{1}{\sqrt{18}} \cdot \frac{\sqrt{18}}{\sqrt{18}} = \frac{\sqrt{18}}{18} =            
 \frac{\sqrt{2}}{6}$                                                                                                                  
                 \end{solutionordottedlines}                                                                                          
             \part \[\frac{1}{\sqrt{2}}+\sqrt{2}\]                                                                                    
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $\frac{1}{\sqrt{2}}+\sqrt{2} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} + \sqrt{2} = \frac{\sqrt{2}}{2 
 + \sqrt{2} = \frac{3\sqrt{2}}{2}$                                                                                                    
                 \end{solutionordottedlines}                                                                                          
             \part \[\frac{\sqrt{72}}{\sqrt{3}}+\frac{3}{\sqrt{2}}-\frac{2}{2 \sqrt{2}}\]                                             
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $\frac{\sqrt{72}}{\sqrt{3}}+\frac{3}{\sqrt{2}}-\frac{2}{2 \sqrt{2}} = \sqrt{24} + \frac{3\sqrt{2}}{2} -          
 \frac{\sqrt{2}}{2} = 2\sqrt{6} + \frac{2\sqrt{2}}{2} = 2\sqrt{6} + \sqrt{2}$                                                         
                 \end{solutionordottedlines}                                                                                          
             \part \[\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}+\frac{7}{2 \sqrt{3}}\]                                                     
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}+\frac{7}{2 \sqrt{3}} = \frac{1+2+\frac{7}{2}}{\sqrt{3}} =                 
 \frac{\frac{2+4+7}{2}}{\sqrt{3}} = \frac{\frac{13}{2}}{\sqrt{3}} = \frac{13}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} =            
 \frac{13\sqrt{3}}{6}$                                                                                                                
                 \end{solutionordottedlines}                                                                                          
         \end{multicols}\end{parts}                                                                                                   
     \Question[6] If \(x=2 \sqrt{14}\) and \(y=4 \sqrt{2}\), find and rationalise the denominator.                                    
         \begin{parts}\begin{multicols}{3}                                                                                            
                 \part \(\frac{x}{y}\)                                                                                                
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $\frac{x}{y} = \frac{2 \sqrt{14}}{4 \sqrt{2}} = \frac{\sqrt{14}}{2 \sqrt{2}} = \frac{\sqrt{14}}{2 \sqrt{2}} \cdo 
 \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{28}}{4} = \frac{2\sqrt{7}}{4} = \frac{\sqrt{7}}{2}$                                          
                 \end{solutionordottedlines}                                                                                          
                 \part \(\frac{y}{x}\)                                                                                                
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $\frac{y}{x} = \frac{4 \sqrt{2}}{2 \sqrt{14}} = \frac{2 \sqrt{2}}{\sqrt{14}} = \frac{2 \sqrt{2}}{\sqrt{14}} \cdo 
 \frac{\sqrt{14}}{\sqrt{14}} = \frac{2\sqrt{28}}{14} = \frac{2\cdot 2\sqrt{7}}{14} = \frac{\sqrt{7}}{7}$                              
                 \end{solutionordottedlines}                                                                                          
                 \part \(\frac{\sqrt{2} x}{\sqrt{3} y}\)                                                                              
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $\frac{\sqrt{2} x}{\sqrt{3} y} = \frac{\sqrt{2} \cdot 2 \sqrt{14}}{\sqrt{3} \cdot 4 \sqrt{2}} =                  
 \frac{2\sqrt{28}}{4\sqrt{6}} = \frac{\sqrt{28}}{2\sqrt{6}} = \frac{\sqrt{28}}{2\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} =           
 \frac{\sqrt{168}}{12} = \frac{2\sqrt{42}}{12} = \frac{\sqrt{42}}{6}$                                                                 
                 \end{solutionordottedlines}                                                                                          
         \end{multicols}\end{parts}                                                                                                   
     \Question[3] A bowl in the shape of a hemisphere of radius length \(5 \mathrm{~cm}\) is partially filled with water. The surface 
 of the water is a circle of radius \(4 \mathrm{~cm}\) when the rim of the bowl is horizontal. Find the depth of the water.           
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     Using the Pythagorean theorem for the right triangle formed by the radius of the water's surface, the radius of  
 the bowl, and the depth of the water:                                                                                                
                     $5^2 = 4^2 + d^2$                                                                                                
                     $25 = 16 + d^2$                                                                                                  
                     $d^2 = 25 - 16$                                                                                                  
                     $d^2 = 9$                                                                                                        
                     $d = 3 \mathrm{~cm}$                                                                                             
                     The depth of the water is 3 cm.                                                                                  
                 \end{solutionordottedlines}                                                                                          
|726265393|kitty|/Applications/kitty.app|0|
     \Question[3] A bobbin for an industrial knitting machine is in the shape of a truncated cone. The diameter of the top is \(4 \mathrm{~cm}\), t 
 diameter of the base is \(6 \mathrm{~cm}\) and the length of the slant is \(10 \mathrm{~cm}\). Find the height of the bobbin.                      
         \begin{center}                                                                                                                             
         \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-09}                                                             
         \end{center}                                                                                                                               
                 \begin{solutionordottedlines}[2cm]                                                                                                 
                 Let the height of the bobbin be \(h\), the radius of the top be \(r_1 = 2 \mathrm{~cm}\), and the radius of the base be \(r_2 = 3  
 \mathrm{~cm}\). By the Pythagorean theorem, we have:                                                                                               
                 \[l^2 = h^2 + (r_2 - r_1)^2\]                                                                                                      
                 Substituting the given values:                                                                                                     
                 \[10^2 = h^2 + (3 - 2)^2\]                                                                                                         
                 \[100 = h^2 + 1\]                                                                                                                  
                 \[h^2 = 99\]                                                                                                                       
                 \[h = \sqrt{99}\]                                                                                                                  
                 \[h \approx 9.95 \mathrm{~cm}\]                                                                                                    
                 The height of the bobbin is approximately \(9.95 \mathrm{~cm}\).                                                                   
                 \end{solutionordottedlines}                                                                                                        
     \Question[4] Rationalise the following:                                                                                                        
         \begin{parts}\begin{multicols}{2}                                                                                                          
             \part \[\frac{1}{\sqrt{3}+2}\]                                                                                                         
                 \begin{solutionordottedlines}[2cm]                                                                                                 
                 Multiply the numerator and denominator by the conjugate of the denominator:                                                        
                 \[\frac{1}{\sqrt{3}+2} \cdot \frac{\sqrt{3}-2}{\sqrt{3}-2} = \frac{\sqrt{3}-2}{(\sqrt{3}+2)(\sqrt{3}-2)} = \frac{\sqrt{3}-2}{3-4}  
 2-\sqrt{3}\]                                                                                                                                       
                 \end{solutionordottedlines}                                                                                                        
             \part \[\frac{1}{\sqrt{3}+\sqrt{2}}\]                                                                                                  
                 \begin{solutionordottedlines}[2cm]                                                                                                 
                 Multiply the numerator and denominator by the conjugate of the denominator:                                                        
                 \[\frac{1}{\sqrt{3}+\sqrt{2}} \cdot \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{\sqrt{3}-\sqrt{2}}{3-2} =                  
 \sqrt{3}-\sqrt{2}\]                                                                                                                                
                 \end{solutionordottedlines}                                                                                                        
         \end{multicols}\end{parts}                                                                                                                 
     \Question[3] Find the integers \(p\) and \(q\) such that \[\frac{\sqrt{5}}{\sqrt{5}-2}=p+q \sqrt{5}\].                                         
                 \begin{solutionordottedlines}[2cm]                                                                                                 
                 Multiply the numerator and denominator by the conjugate of the denominator:                                                        
                 \[\frac{\sqrt{5}}{\sqrt{5}-2} \cdot \frac{\sqrt{5}+2}{\sqrt{5}+2} = \frac{5+2\sqrt{5}}{5-4} = \frac{5+2\sqrt{5}}{1} = 5+2\sqrt{5}\ 
                 Thus, \(p = 5\) and \(q = 2\).                                                                                                     
                 \end{solutionordottedlines}                                                                                                        
     \Question[9] Simplify the following:                                                                                                           
         \begin{parts}\begin{multicols}{2}                                                                                                          
             \part \(\frac{3}{\sqrt{5}-2}+\frac{2}{\sqrt{5}+2}\)                                                                                    
                 \begin{solutionordottedlines}[2cm]                                                                                                 
                 Multiply each term by the conjugate of its denominator:                                                                            
                 \[\frac{3}{\sqrt{5}-2} \cdot \frac{\sqrt{5}+2}{\sqrt{5}+2} + \frac{2}{\sqrt{5}+2} \cdot \frac{\sqrt{5}-2}{\sqrt{5}-2}\]            
                 \[= \frac{3(\sqrt{5}+2)}{5-4} + \frac{2(\sqrt{5}-2)}{5-4}\]                                                                        
                 \[= 3\sqrt{5} + 6 + 2\sqrt{5} - 4\]                                                                                                
                 \[= 5\sqrt{5} + 2\]                                                                                                                
                 \end{solutionordottedlines}                                                                                                        
             \part \(\frac{5}{(\sqrt{7}-\sqrt{2})^{2}}\)                                                                                            
                 \begin{solutionordottedlines}[2cm]                                                                                                 
                 Expand the denominator and simplify:                                                                                               
                 \[\frac{5}{7 - 2\sqrt{14} + 2}\]                                                                                                   
                 \[= \frac{5}{9 - 2\sqrt{14}}\]                                                                                                     
                 Multiply the numerator and denominator by the conjugate of the denominator:                                                        
                 \[\frac{5}{9 - 2\sqrt{14}} \cdot \frac{9 + 2\sqrt{14}}{9 + 2\sqrt{14}}\]                                                           
                 \[= \frac{45 + 10\sqrt{14}}{81 - 56}\]                                                                                             
                 \[= \frac{45 + 10\sqrt{14}}{25}\]                                                                                                  
                 \[= \frac{9}{5} + 2\sqrt{14}\]                                                                                                     
                 \end{solutionordottedlines}                                                                                                        
             \part \(0.0 \mathrm{i} \dot{6}\)                                                                                                       
                 \begin{solutionordottedlines}[2cm]                                                                                                 
                 This is a repeating decimal. Let \(x = 0.0\dot{6}\), then:                                                                         
                 \[10x = 0.\dot{6}\]                                                                                                                
                 \[10x - x = 0.\dot{6} - 0.0\dot{6}\]                                                                                               
                 \[9x = 0.6\]                                                                                                                       
                 \[x = \frac{0.6}{9}\]                                                                                                              
                 \[x = \frac{2}{30}\]                                                                                                               
                 \[x = \frac{1}{15}\]                                                                                                               
                 \end{solutionordottedlines}                                                                                                        
             \part \(0.3 \dot{2} \dot{4}\)                                                                                                          
                 \begin{solutionordottedlines}[2cm]                                                                                                 
                 This is a repeating decimal. Let \(x = 0.3\dot{2}\dot{4}\), then:                                                                  
                 \[1000x = 324.\dot{2}\dot{4}\]                                                                                                     
                 \[1000x - x = 324.\dot{2}\dot{4} - 0.3\dot{2}\dot{4}\]                                                                             
                 \[999x = 324\]                                                                                                                     
                 \[x = \frac{324}{999}\]                                                                                                            
                 \[x = \frac{108}{333}\]                                                                                                            
                 \[x = \frac{36}{111}\]                                                                                                             
                 \[x = \frac{12}{37}\]                                                                                                              
                 \end{solutionordottedlines}                                                                                                        
             \part \(0.51 \dot{2} \dot{6}\)                                                                                                         
                 \begin{solutionordottedlines}[2cm]                                                                                                 
                 This is a repeating decimal. Let \(x = 0.51\dot{2}\dot{6}\), then:                                                                 
                 \[1000x = 512.\dot{2}\dot{6}\]                                                                                                     
                 \[1000x - 10x = 512.\dot{2}\dot{6} - 5.1\dot{2}\dot{6}\]                                                                           
                 \[990x = 507.1\]                                                                                                                   
                 \[x = \frac{507.1}{990}\]                                                                                                          
                 \[x = \frac{5071}{9900}\]                                                                                                          
                 \[x = \frac{5071}{9900}\]                                                                                                          
                 Simplifying the fraction, we get:                                                                                                  
                 \[x = \frac{5071}{9900}\]                                                                                                          
                 \[x = \frac{563}{1100}\]                                                                                                           
                 \[x = \frac{563}{1100}\]                                                                                                           
                 \end{solutionordottedlines}                                                                                                        
             \part \(0.001 \dot{1}\)                                                                                                                
                 \begin{solutionordottedlines}[2cm]                                                                                                 
                 This is a repeating decimal. Let \(x = 0.001\dot{1}\), then:                                                                       
                 \[1000x = 1.\dot{1}\]                                                                                                              
                 \[1000x - x = 1.\dot{1} - 0.001\dot{1}\]                                                                                           
                 \[999x = 1.1\]                                                                                                                     
                 \[x = \frac{1.1}{999}\]                                                                                                            
                 \[x = \frac{11}{9990}\]                                                                                                            
                 \[x = \frac{1}{909}\]                                                                                                              
                 \end{solutionordottedlines}                                                                                                        
         \end{multicols}\end{parts}                                                                                                                 
     \Question[6] \textbf{Challenge:}                                                                                                               
         \begin{parts}                                                                                                                              
             \part Show that \[\frac{138}{19}=7+\frac{1}{3+\frac{1}{1+\frac{1}{4}}}\]                                                               
                 \begin{solutionordottedlines}[2cm]                                                                                                 
                 Start by evaluating the continued fraction from the innermost fraction outward:                                                    
                 \[1 + \frac{1}{4} = 1 + \frac{1}{4} = \frac{5}{4}\]                                                                                
                 \[3 + \frac{1}{\frac{5}{4}} = 3 + \frac{4}{5} = \frac{15}{5} + \frac{4}{5} = \frac{19}{5}\]                                        
                 \[7 + \frac{1}{\frac{19}{5}} = 7 + \frac{5}{19} = \frac{133}{19} + \frac{5}{19} = \frac{138}{19}\]                                 
                 Therefore, the continued fraction equals \(\frac{138}{19}\).                                                                       
                 \end{solutionordottedlines}                                                                                                        
             \part Express \[\frac{153}{11}\] as a continued fraction with all numerators 1 .                                                       
                 \begin{solutionordottedlines}[2cm]                                                                                                 
                 Perform the Euclidean algorithm to find the continued fraction:                                                                    
                 \[\frac{153}{11} = 13 + \frac{10}{11}\]                                                                                            
                 \[\frac{11}{10} = 1 + \frac{1}{10}\]                                                                                               
                 \[\frac{10}{1} = 10\]                                                                                                              
                 So, \[\frac{153}{11} = 13 + \frac{1}{1 + \frac{1}{10}}\]                                                                           
                 Therefore, the continued fraction is \(13 + \frac{1}{1 + \frac{1}{10}}\).                                                          
                 \end{solutionordottedlines}                                                                                                        
         \end{parts}                                                                                                                                
 \end{questions}                                                                                                                                    
                                                                                                                                                    
|726265470|kitty|/Applications/kitty.app|0|
    \Question[3] A bobbin for an industrial knitting machine is in the shape of a truncated cone. The diameter of the top is \(4 \mathrm{~cm}\), the diameter of the base is \(6 \mathrm{~cm}\) and the length of the slant is \(10 \mathrm{~cm}\). Find the height of the bobbin.
        \begin{center}
        \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-09}
        \end{center}
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
    \Question[4] Rationalise the following:
        \begin{parts}\begin{multicols}{2}
            \part \[\frac{1}{\sqrt{3}+2}\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \[\frac{1}{\sqrt{3}+\sqrt{2}}\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \Question[3] Find the integers \(p\) and \(q\) such that \[\frac{\sqrt{5}}{\sqrt{5}-2}=p+q \sqrt{5}\].
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
    \Question[9] Simplify the following:
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3}{\sqrt{5}-2}+\frac{2}{\sqrt{5}+2}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \(\frac{5}{(\sqrt{7}-\sqrt{2})^{2}}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \(0.0 \mathrm{i} \dot{6}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \(0.3 \dot{2} \dot{4}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \(0.51 \dot{2} \dot{6}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
            \part \(0.001 \dot{1}\)
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \Question[6] \textbf{Challenge:}
        \begin{parts}
            \part Show that \[\frac{138}{19}=7+\frac{1}{3+\frac{1}{1+\frac{1}{4}}}\]
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
The expression on the right is called a continued fraction.
            \part Express \[\frac{153}{11}\] as a continued fraction with all numerators 1 .
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
        \end{parts}
\end{questions}
|726265482|kitty|/Applications/kitty.app|0|
                     $(\sqrt{3}+\sqrt{2})^{2}-(\sqrt{3}-\sqrt{2})^{2} = (3 + 2\sqrt{6} + 2) - (3 - 2\sqrt{6} + 2) = 5 + 2\sqrt{6} - 5 2\sqrt{6} = 4\sqrt{6}$                                                                                   
|726265842|kitty|/Applications/kitty.app|0|
                 \begin{solutionordottedlines}[2cm]                                                                                   
                     $\frac{1}{x}+\frac{1}{y} = \frac{1}{\sqrt{2}-1} + \frac{1}{\sqrt{2}+1} = \frac{\sqrt{2}+1}{2-1} + \frac{\sqrt{2}-1}{2-1} = \sqrt{2}+1 + \sqrt{2}-1 = 2\sqrt{2}$                                                                        
                 \end{solutionordottedlines}                                                                                          
             \end{multicols}\end{subparts}                                                                                            
|726265945|kitty|/Applications/kitty.app|0|
Image: 413x513 (3.2 MB)|726548400|Brave Browser Beta|/Applications/Brave Browser Beta.app|1|2d199d3c04ceda906a33fb9c6a06c7d73170a225.tiff
\usepackage[top=15mm,bottom=20mm,right=20mm,left=20mm]{geometry}
\usepackage{pgffor}
\usepackage{setspace}
\usepackage{mathtools}
\usepackage{enumitem}
\usepackage{multicol}
\usepackage{cancel}
\usepackage{fontspec}
\usepackage{background}
\usepackage{hyperref}
\usepackage{xcolor}
\usepackage{wasysym}
\usepackage{rotating}

\usepackage{ifthen}   % Required for conditional statements
\usepackage{pgf}      % Required for random number generation
\graphicspath{ {./img/} }
|726635659|kitty|/Applications/kitty.app|0|
% that's all folks

\usetikzlibrary{decorations.text, shadows}
\newfontfamily\looney[]{That's Font Folks!}
\definecolor{darkblueOuter}{RGB}{1,11,23}
\definecolor{darkblueInner}{RGB}{1,18,37}
%

\setcounter{tocdepth}{1}
|726635666|kitty|/Applications/kitty.app|0|
F|726635671|kitty|/Applications/kitty.app|0|
\setlength\parindent{0pt}
|726635676|kitty|/Applications/kitty.app|0|
\marksnotpoints
\pointsinrightmargin
\boxedpoints
\setlength{\rightpointsmargin}{1cm}
|726635680|kitty|/Applications/kitty.app|0|
\marksnotpoints
\pointsinrightmargin
\boxedpoints
\setlength{\rightpointsmargin}{1cm}

|726635695|kitty|/Applications/kitty.app|0|
% some exam environment options
|726635695|kitty|/Applications/kitty.app|0|
\newcommand{\randomword}[1]{%
  \pgfmathsetmacro{\angle}{random(-45,45)}% Generate a random angle between -45 and 45
  \rotatebox{\angle}{#1}%
}

|726635697|kitty|/Applications/kitty.app|0|
\cfoot{\thepage}
|726635703|kitty|/Applications/kitty.app|0|
\section{Introduction}
    \setcounter{secmarks}{0}
    \input{1-intro}
    \setcounter{sec1marks}{\thesecmarks}
\section{Using linear equations to solve problems}
    \setcounter{secmarks}{0}
    \input{2-using-linear}
    \setcounter{sec2marks}{\thesecmarks}
\section{Literal Equations}
    \setcounter{secmarks}{0}
    \input{3-literal-eq}
    \setcounter{sec3marks}{\thesecmarks}
\section{Inequalities}
    \setcounter{secmarks}{0}
    \input{4-inequalities}
    \setcounter{sec4marks}{\thesecmarks}
\section{Solving linear inequalities}
    \setcounter{secmarks}{0}
    \input{5-solving-ineq}
    \setcounter{sec5marks}{\thesecmarks}
\newpage
\section{Homework}
    \setcounter{secmarks}{0}
    \input{0-homework}
    \setcounter{sec6marks}{\thesecmarks}


\foreach \i in {1,2,...,10} {
  \addtocounter{totalmarks}{\value{sec\i marks}}
}

\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    & 1
    & 2
    & 3
    & 4
    & 5
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    & $\dfrac{}{\arabic{sec1marks}}$ 
    & $\dfrac{}{\arabic{sec2marks}}$  
    & $\dfrac{}{\arabic{sec3marks}}$ 
    & $\dfrac{}{\arabic{sec4marks}}$ 
    & $\dfrac{}{\arabic{sec5marks}}$ 
    & $\dfrac{}{\arabic{sec6marks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
%\section{The End}
%\hspace{4.5cm}
%    \input{0-outro}

%playground

\end{document}
|726635734|kitty|/Applications/kitty.app|0|
\section{Introduction}
    \input{1-intro}
\section{}
    \input{}
\section{}
    \input{}
\section{}
    \input{}
\section{}
    \input{}
\section{}
    \input{}
\newpage
\section{Homework}
    \input{0-homework}

\begin{center}
\gradetable[h][questions]
\end{center}
%\section{The End}
%\hspace{4.5cm}
%    \input{0-outro}

%playground

\end{document}
|726635744|kitty|/Applications/kitty.app|0|
%\section{The End}
%\hspace{4.5cm}
%    \input{0-outro}

%playground
|726635752|kitty|/Applications/kitty.app|0|
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    & 1
    & 2
    & 3
    & 4
    & 5
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    & $\dfrac{}{\arabic{sec1marks}}$ 
    & $\dfrac{}{\arabic{sec2marks}}$  
    & $\dfrac{}{\arabic{sec3marks}}$ 
    & $\dfrac{}{\arabic{sec4marks}}$ 
    & $\dfrac{}{\arabic{sec5marks}}$ 
    & $\dfrac{}{\arabic{sec6marks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
|726635780|kitty|/Applications/kitty.app|0|
2-using-linear|726636043|kitty|/Applications/kitty.app|0|
3-literal-eq|726636045|kitty|/Applications/kitty.app|0|
4-inequalities|726636047|kitty|/Applications/kitty.app|0|
5-solving-ineq|726636049|kitty|/Applications/kitty.app|0|
\mygradingtable{5}
|726636441|kitty|/Applications/kitty.app|0|
\newcommand{\mygradingtable}[1]{
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||*{#1}{>{\centering}X|}>{\centering}X|>{\centering}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    \foreach \n in {1,...,#1}{
    & \n
    }
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    \foreach \n in {1,...,#1}{
    & $\dfrac{}{\csname sec\n marks\endcsname}$ 
    }
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
}

|726636444|kitty|/Applications/kitty.app|0|
\newcommand{\mygradingtable}[1]{
|726636447|kitty|/Applications/kitty.app|0|
}
|726636492|kitty|/Applications/kitty.app|0|
! Missing number, treated as zero.
<to be read again> 
##
l.75 \end{tabularx}
|726636562|kitty|/Applications/kitty.app|0|
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||*{#1}{>{\centering}X|}>{\centering}X|>{\centering}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    \foreach \n in {1,...,#1}{
    & \n
    }
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    \foreach \n in {1,...,#1}{
    & $\dfrac{}{\csname sec\n marks\endcsname}$ 
    }
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
|726636587|kitty|/Applications/kitty.app|0|
\newcommand{\gradingtable}[1]{
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||*{#1}{>{\centering}X|}>{\centering}X|>{\centering}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    \foreach \n in {1,...,#1}{
    & \n
    }
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    \foreach \n in {1,...,#1}{
    & $\dfrac{}{\csname sec\n marks\endcsname}$ 
    }
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
}
|726636593|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
! Missing \endgroup inserted.
<inserted text> 
\endgroup 
l.56 \gradingtable{5}
|726636617|kitty|/Applications/kitty.app|0|
\newcommand{\gradingtable}[1]{
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||*{#1}{>{\centering}X|}>{\centering}X|>{\centering}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    % Sections
    \forloop{section}{1}{\value{section} <= #1}{
        & \arabic{section}
    }
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    % Marks
    \forloop{section}{1}{\value{section} <= #1}{
        & $\dfrac{}{\csname sec\arabic{section}marks\endcsname}$ 
    }
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
}

\newcounter{section}
|726636737|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\newcommand{\gradingtable}[1]{
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||*{#1}{>{\centering}X|}>{\centering}X|>{\centering}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    \foreach \n in {1,...,#1}{
    & \n
    }
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    \foreach \n in {1,...,#1}{
    & $\dfrac{}{\csname sec\n marks\endcsname}$ 
    }
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
}

|726636741|kitty|/Applications/kitty.app|0|
\usepackage{forloop}
|726636747|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\newcounter{section}
|726636778|kitty|/Applications/kitty.app|0|
\usepackage{forloop}

\newcounter{secti}

\newcommand{\gradingtable}[1]{
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||*{#1}{>{\centering}X|}>{\centering}X|>{\centering}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    % Sections
    \forloop{secti}{1}{\value{secti} <= #1}{
        & \arabic{secti}
    }
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    % Marks
    \forloop{secti}{1}{\value{secti} <= #1}{
        & $\dfrac{}{\csname sec\arabic{secti}marks\endcsname}$ 
    }
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
}

|726636952|kitty|/Applications/kitty.app|0|
! Undefined control sequence.
\marksRow ... \endgroup \@@over \relax \sec1marks 
                                                  }}}$
l.56 \gradingtable{5}
                   
? 
! Emergency stop.
\marksRow ... \endgroup \@@over \relax \sec1marks 
                                                  }}}$
l.56 \gradingtable{5}
|726637154|kitty|/Applications/kitty.app|0|
\newcommand{\gradingtable}[1]{
    % Define macros to hold the contents for the table
    \def\sectionHeaderRow{}
    \def\marksRow{}
    % Create the section headers and the marks rows
    \foreach \n in {1,...,#1}{
        \xdef\sectionHeaderRow{\sectionHeaderRow & \n}
        \xdef\marksRow{\marksRow & $\dfrac{}{\csname sec\n marks\endcsname}$}
    }
    \begin{center}
    \begin{minipage}[m]{\linewidth}
    Marker's use only. \\ \large
    \begin{tabularx}{\linewidth}{||>{\bfseries}l||*{#1}{>{\centering}X|}>{\centering}X|>{\centering}X||>{\bfseries}c||}
    \hline \hline
    \rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
        \sectionHeaderRow % Insert the section headers
        & HW
        & Total
        \\ \hline
    \rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
        \marksRow % Insert the marks rows
        & $\dfrac{}{\arabic{hwmarks}}$
        & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
    \end{tabularx}    
    \end{minipage}
    \end{center}
}
|726637069|kitty|/Applications/kitty.app|0|
\marksRow ...dgroup \@@over \relax }}}{\sec1marks 
                                                  }$
l.56 \gradingtable{6}
                   
? 
! Emergency stop.
\marksRow ...dgroup \@@over \relax }}}{\sec1marks 
                                                  }$
l.56 \gradingtable{6}
                   
End of file on the terminal!
|726637162|kitty|/Applications/kitty.app|0|
! Undefined control sequence.
\marksRow ...dgroup \@@over \relax }}}{\sec1marks 
                                                  }$
l.56 \gradingtable{6}
                   
? 
! Emergency stop.
\marksRow ...dgroup \@@over \relax }}}{\sec1marks 
                                                  }$
l.56 \gradingtable{6}
                   
End of file on the terminal!
|726637165|kitty|/Applications/kitty.app|0|
\newcommand{\gradingtable}[1]{
    % Define macros to hold the contents for the table
    \def\sectionHeaderRow{}
    \def\marksRow{}
    % Create the section headers and the marks rows
    \foreach \n in {1,...,#1}{
        \expandafter\xdef\expandafter\sectionHeaderRow\expandafter{\sectionHeaderRow & \n}
        \expandafter\xdef\expandafter\marksRow\expandafter{\marksRow & $\expandafter\dfrac\expandafter{}\expandafter{\csname sec\n marks\endcsname}$}
    }
    \begin{center}
    \begin{minipage}[m]{\linewidth}
    Marker's use only. \\ \large
    \begin{tabularx}{\linewidth}{||>{\bfseries}l||*{#1}{>{\centering}X|}>{\centering}X|>{\centering}X||>{\bfseries}c||}
    \hline \hline
    \rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
        \sectionHeaderRow % Insert the section headers
        & HW
        & Total
        \\ \hline
    \rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
        \marksRow % Insert the marks rows
        & $\dfrac{}{\arabic{hwmarks}}$
        & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
    \end{tabularx}    
    \end{minipage}
    \end{center}
}
|726637241|kitty|/Applications/kitty.app|0|
! Missing = inserted for \ifnum.
<to be read again> 
\leq 
l.56 \gradingtable{2}
                   
? 
! Emergency stop.
<to be read again> 
\leq 
l.56 \gradingtable{2}
                   
End of file on the terminal!
|726637321|kitty|/Applications/kitty.app|0|
\newcommand{\gradingtable}[1]{
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||*{#1}{>{\centering}X|}>{\centering}X|>{\centering}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    \ifnum #1 > 0 1\fi
    \ifnum #1 > 1 & 2\fi
    \ifnum #1 > 2 & 3\fi
    % ... Continue this pattern up to the number before #1
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    \ifnum #1 > 0 & $\dfrac{}{\arabic{sec1marks}}$\fi
    \ifnum #1 > 1 & $\dfrac{}{\arabic{sec2marks}}$\fi
    \ifnum #1 > 2 & $\dfrac{}{\arabic{sec3marks}}$\fi
    % ... Continue this pattern up to the number before #1
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
}
|726637358|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\newcommand{\gradingtable}[1]{
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||*{10}{>{\centering}X|}>{\centering}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    % Sections
    \ifnum1\leq#1 1\fi
    \ifnum2\leq#1 & 2\fi
    \ifnum3\leq#1 & 3\fi
    % ... Continue this pattern up to 10
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    % Marks
    \ifnum1\leq#1 & $\dfrac{}{\arabic{sec1marks}}$\fi
    \ifnum2\leq#1 & $\dfrac{}{\arabic{sec2marks}}$\fi
    \ifnum3\leq#1 & $\dfrac{}{\arabic{sec3marks}}$\fi
    % ... Continue this pattern up to 10
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
}
|726637362|kitty|/Applications/kitty.app|0|
! Incomplete \ifnum; all text was ignored after line 56.
<inserted text> 
\fi 
l.56 \gradingtable{2}
                   
? 
! Emergency stop.
<inserted text> 
\fi 
l.56 \gradingtable{2}
                   
End of file on the terminal!
|726637406|kitty|/Applications/kitty.app|0|
\newcommand{\gradingtable}[1]{
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||*{#1}{>{\centering}X|}>{\centering}X|>{\centering}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    \ifnum #1 > 0 & 1\fi
    \ifnum #1 > 1 & 2\fi
    \ifnum #1 > 2 & 3\fi
    % ... Continue this pattern up to the number before #1
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    \ifnum #1 > 0 & $\dfrac{}{\arabic{sec1marks}}$\fi
    \ifnum #1 > 1 & $\dfrac{}{\arabic{sec2marks}}$\fi
    \ifnum #1 > 2 & $\dfrac{}{\arabic{sec3marks}}$\fi
    % ... Continue this pattern up to the number before #1
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
}
|726637454|kitty|/Applications/kitty.app|0|
Runaway argument?
\ifnum \mycount \leq 2 \unskip \relax \d@llarend \do@row@strut \hfil \ETC.
! Forbidden control sequence found while scanning use of \loop.
<inserted text> 
\par 
l.56 \gradingtable{2}
                   
? 
! Emergency stop.
<inserted text> 
\par 
l.56 \gradingtable{2}
                   
End of file on the terminal!

|726637482|kitty|/Applications/kitty.app|0|
\newcommand{\gradingtable}[1]{
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||*{#1}{>{\centering}X|}>{\centering\arraybackslash}X|>{\centering\arraybackslash}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    % Sections
    \newcount\mycount
    \mycount=1
    \loop
        \ifnum\mycount\leq#1
        & \the\mycount
        \advance\mycount by 1
    \repeat
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    % Marks
    \mycount=1
    \loop
        \ifnum\mycount\leq#1
        & $\dfrac{}{\csname sec\the\mycount marks\endcsname}$
        \advance\mycount by 1
    \repeat
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
}
|726637532|kitty|/Applications/kitty.app|0|
! Extra alignment tab has been changed to \cr.
<recently read> \endtemplate 
             
l.56 \gradingtable{2}
                   
? 
! Emergency stop.
<recently read> \endtemplate 
             
l.56 \gradingtable{2}
                   
End of file on the terminal!
|726637579|kitty|/Applications/kitty.app|0|
\newcommand{\createsection}[1]{
    & #1 & $\dfrac{}{\csname sec#1marks\endcsname}$
}

\newcommand{\gradingtable}[1]{
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||*{#1}{>{\centering}X|}>{\centering\arraybackslash}X|>{\centering\arraybackslash}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    % Section numbers
    \ifnum #1 >0 1\fi
    \ifnum #1 >1 & 2\fi
    \ifnum #1 >2 & 3\fi
    % ... Add similar lines for each section up to 10
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    % Section marks
    \ifnum #1 >0 & $\dfrac{}{\arabic{sec1marks}}$\fi
    \ifnum #1 >1 & $\dfrac{}{\arabic{sec2marks}}$\fi
    \ifnum #1 >2 & $\dfrac{}{\arabic{sec3marks}}$\fi
    % ... Add similar lines for each section's marks up to 10
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
}
|726637632|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\newcommand{\createsection}[1]{
    & #1 & $\dfrac{}{\csname sec#1marks\endcsname}$
}

\newcommand{\gradingtable}[1]{
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||*{#1}{>{\centering}X|}>{\centering\arraybackslash}X|>{\centering\arraybackslash}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    \ifnum #1 >0 \createsection{1}\fi
    \ifnum #1 >1 \createsection{2}\fi
    \ifnum #1 >2 \createsection{3}\fi
    % Add more \createsection commands up to 10
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}
    \ifnum #1 >0 \createsection{1}\fi
    \ifnum #1 >1 \createsection{2}\fi
    \ifnum #1 >2 \createsection{3}\fi
    % Add more \createsection commands up to 10
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
}
|726637639|kitty|/Applications/kitty.app|0|
! Incomplete \ifnum; all text was ignored after line 56.
<inserted text> 
\fi 
l.56 \gradingtable{2}
                   
? 
! Emergency stop.
<inserted text> 
\fi 
l.56 \gradingtable{2}
|726637754|kitty|/Applications/kitty.app|0|
\newcommand{\gradingtable}[1]{
    \begin{center}
    \begin{minipage}[m]{\linewidth}
    Marker's use only. \\ \large
    \begin{tabularx}{\linewidth}{||>{\bfseries}l||*{#1}{>{\centering\arraybackslash}X|}>{\centering\arraybackslash}X|>{\centering\arraybackslash}X||>{\bfseries}c||}
    \hline \hline
    \rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    % Section numbers
    \ifnum #1 > 0 1\fi
    \ifnum #1 > 1 & 2\fi
    \ifnum #1 > 2 & 3\fi
    % ... Add similar lines for each section up to 10
    & HW
    & Total
    \\ \hline
    \rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    % Section marks
    \ifnum #1 > 0 & $\dfrac{}{\arabic{sec1marks}}$\fi
    \ifnum #1 > 1 & $\dfrac{}{\arabic{sec2marks}}$\fi
    \ifnum #1 > 2 & $\dfrac{}{\arabic{sec3marks}}$\fi
    % ... Add similar lines for each section's marks up to 10
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
    \end{tabularx}    
    \end{minipage}
    \end{center}
}
|726637828|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\newcommand{\createsection}[1]{
    & #1 & $\dfrac{}{\csname sec#1marks\endcsname}$
}

\newcommand{\gradingtable}[1]{
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||*{#1}{>{\centering}X|}>{\centering\arraybackslash}X|>{\centering\arraybackslash}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    % Section numbers
    \ifnum#1>0 1\fi
    \ifnum#1>1 & 2\fi
    \ifnum#1>2 & 3\fi
    % ... Add similar lines for each section up to 10
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    % Section marks
    \ifnum#1>0 & $\dfrac{}{\arabic{sec1marks}}$\fi
    \ifnum#1>1 & $\dfrac{}{\arabic{sec2marks}}$\fi
    \ifnum#1>2 & $\dfrac{}{\arabic{sec3marks}}$\fi
    % ... Add similar lines for each section's marks up to 10
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
}
|726637833|kitty|/Applications/kitty.app|0|
    \foreach \n in {1,...,#1}{
    & \n
    }|726638125|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
    & 1
    & 2
    & 3
    & 4
    & 5
|726638165|kitty|/Applications/kitty.app|0|
    \foreach \n in {1,...,#1}{
    & $\dfrac{}{\csname sec\n marks\endcsname}$ 
    }|726638177|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
    & $\dfrac{}{\arabic{sec1marks}}$ 
    & $\dfrac{}{\arabic{sec2marks}}$  
    & $\dfrac{}{\arabic{sec3marks}}$ 
    & $\dfrac{}{\arabic{sec4marks}}$ 
    & $\dfrac{}{\arabic{sec5marks}}$ 
|726638243|kitty|/Applications/kitty.app|0|
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    \foreach \n in {1,...,#1}{
    & \n
    }
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    \foreach \n in {1,...,#1}{
        & $\dfrac{}{\arabic{sec\n marks}}$ 
    }
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
|726638250|kitty|/Applications/kitty.app|0|
    \begin{center}
    \begin{minipage}[m]{\linewidth}
    Marker's use only. \\ \large
    \begin{tabularx}{\linewidth}{||>{\bfseries}l||*{#1}{>{\centering\arraybackslash}X|}>{\centering\arraybackslash}X|>{\centering\arraybackslash}X||>{\bfseries}c||}
    \hline \hline
    \rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    % Section numbers
    \ifnum #1 > 0 & 1\fi
    \ifnum #1 > 1 & 2\fi
    \ifnum #1 > 2 & 3\fi
    % ... Add similar lines for each section up to 10
    & HW
    & Total
    \\ \hline
    \rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    % Section marks
    \ifnum #1 > 0 {& $\dfrac{}{\arabic{sec1marks}}$}\fi
    \ifnum #1 > 1 & $\dfrac{}{\arabic{sec2marks}}$\fi
    \ifnum #1 > 2 & $\dfrac{}{\arabic{sec3marks}}$\fi
    % ... Add similar lines for each section's marks up to 10
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
    \end{tabularx}    
    \end{minipage}
    \end{center}
|726638254|kitty|/Applications/kitty.app|0|
\mygradetable{5}
|726638482|kitty|/Applications/kitty.app|0|
\newcommand{\mygradetable}[1]{
\begin{center}
    \begin{minipage}[m]{\linewidth}
    Marker's use only. \\ \large
        \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
        \hline \hline
        \rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
            \foreach \n in {1,...,#1}{
            & \n
            }
            & HW
            & Total
            \\ \hline
        \rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
            \foreach \n in {1,...,#1}{
                & $\dfrac{}{\arabic{sec\n marks}}$ 
            }
            & $\dfrac{}{\arabic{hwmarks}}$
            & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
        \end{tabularx}
    \end{minipage}
\end{center}
}
|726638516|kitty|/Applications/kitty.app|0|
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
\begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    & 1
    & 2
    & 3
    & 4
    & 5
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    & $\dfrac{}{\arabic{sec1marks}}$ 
    & $\dfrac{}{\arabic{sec2marks}}$  
    & $\dfrac{}{\arabic{sec3marks}}$ 
    & $\dfrac{}{\arabic{sec4marks}}$ 
    & $\dfrac{}{\arabic{sec5marks}}$ 
    & $\dfrac{}{\arabic{hwmarks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
\end{minipage}
\end{center}
|726638689|kitty|/Applications/kitty.app|0|
    \ifnum#1 & 4
|726638875|kitty|/Applications/kitty.app|0|
    \ifnum#1 > 2 & 3\fi
|726638876|kitty|/Applications/kitty.app|0|
    \ifnum#1 > 7 & 3\fi
|726638883|kitty|/Applications/kitty.app|0|
    & 5
|726638898|kitty|/Applications/kitty.app|0|
    \ifnum#1 > 6 & 7\fi
    \ifnum#1 > 7 & 8\fi
    \ifnum#1 > 8 & 9\fi
    \ifnum#1 > 9 & 10\fi
|726638909|kitty|/Applications/kitty.app|0|
    \ifnum#1 > 4 & 5\fi
    \ifnum#1 > 5 & 6\fi
|726638916|kitty|/Applications/kitty.app|0|
    \ifnum#1 > 3 & 4\fi
    \ifnum#1 > 4 & 5\fi
    \ifnum#1 > 5 & 6\fi
    \ifnum#1 > 6 & 7\fi
    \ifnum#1 > 7 & 8\fi
    \ifnum#1 > 8 & 9\fi
    \ifnum#1 > 9 & 10\fi
|726638992|kitty|/Applications/kitty.app|0|
    \ifnum#1 > 4 & 5\fi
    \ifnum#1 > 5 & 6\fi
    \ifnum#1 > 6 & 7\fi
    \ifnum#1 > 7 & 8\fi
    \ifnum#1 > 8 & 9\fi
    \ifnum#1 > 9 & 10\fi
|726639001|kitty|/Applications/kitty.app|0|
    \ifnum#1 > 3 & 4\fi
|726639060|kitty|/Applications/kitty.app|0|
    & 4
|726639064|kitty|/Applications/kitty.app|0|
>{\centering}X|>{\centering}X||726639161|kitty|/Applications/kitty.app|0|
\begin{tabularx}{\linewidth}{||>{\bfseries}l||\addcols{#1}>{\bfseries}c||}|726639261|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\newcounter{colcount}
\newcommand{\addcols}[1]{
    \setcounter{colcount}{0}
    \whiledo{\value{colcount} < #1}{
        >{\centering\arraybackslash}X|
        \stepcounter{colcount}
    }
}|726639298|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
||726639365|kitty|/Applications/kitty.app|0|
! Package array Error:  Illegal pream-token (\maddcols): `c' used.
|726639540|kitty|/Applications/kitty.app|0|
! Package array Error:  Illegal pream-token (\addcols): `c' used.
|726639545|kitty|/Applications/kitty.app|0|
{\coldef}|726639597|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
{||>{\bfseries}l||\addcols{6}|>{\bfseries}c||}|726639622|kitty|/Applications/kitty.app|0|
\newcounter{colcount}
\newcommand{\buildcoldef}[1]{
    \setcounter{colcount}{1}
    \def\coldef{||>{\bfseries}l||} % Start with the first fixed column definition
    \whiledo{\value{colcount} < #1}{
        \edef\coldef{\coldef >{\centering\arraybackslash}X|} % Append new column definitions
        \stepcounter{colcount}
    }
    \edef\coldef{\coldef >{\centering\arraybackslash}X||>{\bfseries}c||} % Append the last fixed column definitions
}|726639631|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\buildcoldef{#1}|726639665|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\newcounter{colcount}
\newcommand{\addcols}[1]{
    \setcounter{colcount}{0}
    \whiledo{\value{colcount} < #1}{
        >{\centering}X|
        \stepcounter{colcount}
    }
}
|726639696|kitty|/Applications/kitty.app|0|
! Illegal parameter number in definition of \coldef.
|726639745|kitty|/Applications/kitty.app|0|
\newcounter{colcount}
\newcommand{\appendcol}[1]{
    \ifnum\value{colcount} < #1
        \edef\coldef{\coldef >{\centering\arraybackslash}X|} % Append column definition
        \stepcounter{colcount}
        \expandafter\appendcol\expandafter{#1} % Recursive call
    \fi
}

\newcommand{\buildcoldef}[1]{
    \setcounter{colcount}{1}
    \def\coldef{||>{\bfseries}l||} % Start with the first fixed column definition
    \appendcol{#1} % Recursively append column definitions
    \edef\coldef{\coldef >{\centering\arraybackslash}X||>{\bfseries}c||} % Append the last fixed column definitions
}|726639849|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
\newcounter{colcount}
\newcommand{\buildcoldef}[1]{
    \setcounter{colcount}{1}
    \def\coldef{||>{\bfseries}l||} % Start with the first fixed column definition
    \whiledo{\value{colcount} < #1}{
        \edef\coldef{\coldef >{\centering\arraybackslash}X|} % Append new column definitions
        \stepcounter{colcount}
    }
    \edef\coldef{\coldef >{\centering\arraybackslash}X||>{\bfseries}c||} % Append the last fixed column definitions
}
|726639852|kitty|/Applications/kitty.app|0|
\begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
|726639997|kitty|/Applications/kitty.app|0|
   \ifnum#1=1
        \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X||>{\bfseries}c||}
    \fi|726640113|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
        \ifnum#1=1
            \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X||>{\bfseries}c||}
        \fi
|726640143|kitty|/Applications/kitty.app|0|
        \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
|726640184|kitty|/Applications/kitty.app|0|
            \begin{tabularx}{\linewidth}{||>{\bfseries}l||\addcols{#1}|>{\bfseries}c||}
|726640186|kitty|/Applications/kitty.app|0|
            \ifnum#1 > 4 & 5\fi
|726640191|kitty|/Applications/kitty.app|0|
            if
|726640228|kitty|/Applications/kitty.app|0|
            \ifnum#1 > 0 & 1\fi
            \ifnum#1 > 1 & 2\fi
            \ifnum#1 > 2 & 3\fi
            \ifnum#1 > 3 & 4\fi
            \ifnum#1 > 4 & 5\fi
            \ifnum#1 > 5 & 6\fi
            \ifnum#1 > 6 & 7\fi
            \ifnum#1 > 7 & 8\fi
            \ifnum#1 > 8 & 9\fi
            \ifnum#1 > 9 & 10\fi
|726640236|kitty|/Applications/kitty.app|0|
$\dfrac{}{\arabic{sec1marks}}$|726640249|kitty|/Applications/kitty.app|0|
            \ifnum#1 > 0 & $\dfrac{}{\arabic{sec1marks}}$\fi
|726640255|kitty|/Applications/kitty.app|0|
            \ifnum#1 > 10 & $\dfrac{}{\arabic{sec1marks}}$\fi
|726640275|kitty|/Applications/kitty.app|0|
            \ifnum#1 > 1 & 2\fi
            \ifnum#1 > 2 & 3\fi
            \ifnum#1 > 3 & 4\fi
            \ifnum#1 > 4 & 5\fi
            \ifnum#1 > 5 & 6\fi
            \ifnum#1 > 6 & 7\fi
            \ifnum#1 > 7 & 8\fi
            \ifnum#1 > 8 & 9\fi
            \ifnum#1 > 9 & 10\fi
|726640279|kitty|/Applications/kitty.app|0|
            & $\dfrac{}{\arabic{sec1marks}}$ 
            & $\dfrac{}{\arabic{sec2marks}}$  
            & $\dfrac{}{\arabic{sec3marks}}$ 
            & $\dfrac{}{\arabic{sec4marks}}$ 
            & $\dfrac{}{\arabic{sec5marks}}$ 
|726640285|kitty|/Applications/kitty.app|0|
        \ifnum#1=4
            \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
        \fi
|726640377|kitty|/Applications/kitty.app|0|
\newcommand{\gradingtable}[1]{
    \begin{center}
        \begin{minipage}[m]{\linewidth}
        Marker's use only. \\ \large
            \ifnum#1=1
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \fi
            \ifnum#1=2
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \fi
            \ifnum#1=3
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \fi
            \ifnum#1=4
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \fi
            \ifnum#1=5
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \fi
            \ifnum#1=6
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \fi
            \ifnum#1=7
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \fi
            \ifnum#1=8
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \fi
            \ifnum#1=9
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \fi
            \ifnum#1=10
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \fi
            \hline \hline
            \rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
                \ifnum#1 > 0 & 1\fi
                \ifnum#1 > 1 & 2\fi
                \ifnum#1 > 2 & 3\fi
                \ifnum#1 > 3 & 4\fi
                \ifnum#1 > 4 & 5\fi
                \ifnum#1 > 5 & 6\fi
                \ifnum#1 > 6 & 7\fi
                \ifnum#1 > 7 & 8\fi
                \ifnum#1 > 8 & 9\fi
                \ifnum#1 > 9 & 10\fi
                & HW
                & Total
                \\ \hline
            \rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
                \ifnum#1 > 0 & $\dfrac{}{\arabic{sec1marks}}$\fi
                \ifnum#1 > 1 & $\dfrac{}{\arabic{sec2marks}}$\fi
                \ifnum#1 > 2 & $\dfrac{}{\arabic{sec3marks}}$\fi
                \ifnum#1 > 3 & $\dfrac{}{\arabic{sec4marks}}$\fi
                \ifnum#1 > 4 & $\dfrac{}{\arabic{sec5marks}}$\fi
                \ifnum#1 > 5 & $\dfrac{}{\arabic{sec6marks}}$\fi
                \ifnum#1 > 6 & $\dfrac{}{\arabic{sec7marks}}$\fi
                \ifnum#1 > 7 & $\dfrac{}{\arabic{sec8marks}}$\fi
                \ifnum#1 > 8 & $\dfrac{}{\arabic{sec9marks}}$\fi
                \ifnum#1 > 9 & $\dfrac{}{\arabic{sec10marks}}$\fi
                & $\dfrac{}{\arabic{hwmarks}}$
                & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
            \end{tabularx}    
        \end{minipage}
    \end{center}
}
|726640705|kitty|/Applications/kitty.app|0|
            \fi
|726640798|kitty|/Applications/kitty.app|0|
                \ifnum#1 > 0 & 1\fi
                \ifnum#1 > 1 & 2\fi
                \ifnum#1 > 2 & 3\fi
                \ifnum#1 > 3 & 4\fi
                \ifnum#1 > 4 & 5\fi
                \ifnum#1 > 5 & 6\fi
                \ifnum#1 > 6 & 7\fi
                \ifnum#1 > 7 & 8\fi
                \ifnum#1 > 8 & 9\fi
                \ifnum#1 > 9 & 10\fi
|726642071|kitty|/Applications/kitty.app|0|
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=2
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=3
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=4
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=5
|726642141|kitty|/Applications/kitty.app|0|
            \ifnum#1=5
|726642146|kitty|/Applications/kitty.app|0|
                \ifnum#1 > 5 & $\dfrac{}{\arabic{sec6marks}}$\fi
                \ifnum#1 > 6 & $\dfrac{}{\arabic{sec7marks}}$\fi
                \ifnum#1 > 7 & $\dfrac{}{\arabic{sec8marks}}$\fi
                \ifnum#1 > 8 & $\dfrac{}{\arabic{sec9marks}}$\fi
                \ifnum#1 > 9 & $\dfrac{}{\arabic{sec10marks}}$\fi
|726642201|kitty|/Applications/kitty.app|0|
\ifnum#1 > 0 
\ifnum#1 > 1 
\ifnum#1 > 2 
\ifnum#1 > 3 
\ifnum#1 > 4 
|726642289|kitty|/Applications/kitty.app|0|
\fi
\fi
\fi
\fi
\fi
|726642292|kitty|/Applications/kitty.app|0|
                \ifnum#1 > 5 & 6\fi
                \ifnum#1 > 6 & 7\fi
                \ifnum#1 > 7 & 8\fi
                \ifnum#1 > 8 & 9\fi
                \ifnum#1 > 9 & 10\fi
|726642294|kitty|/Applications/kitty.app|0|
\ifnum#1 > 0 
\ifnum#1 > 1 
\ifnum#1 > 2 
\ifnum#1 > 3 
\ifnum#1 > 4 
\ifnum#1 > 5 
|726642305|kitty|/Applications/kitty.app|0|
fi
fi
fi
fi
fi
fi
|726642309|kitty|/Applications/kitty.app|0|
\
\
\
\
\
\
|726642311|kitty|/Applications/kitty.app|0|
                & $\dfrac{}{\arabic{sec6marks}}$
                \ifnum#1 > 6 & $\dfrac{}{\arabic{sec7marks}}$\fi
                \ifnum#1 > 7 & $\dfrac{}{\arabic{sec8marks}}$\fi
                \ifnum#1 > 8 & $\dfrac{}{\arabic{sec9marks}}$\fi
                \ifnum#1 > 9 & $\dfrac{}{\arabic{sec10marks}}$\fi
|726642314|kitty|/Applications/kitty.app|0|
            \ifnum#1=1
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=2
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=3
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=4
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=5
|726642345|kitty|/Applications/kitty.app|0|
            \elifnum#1=6
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=7
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=8
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=9
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=10
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \fi
|726642349|kitty|/Applications/kitty.app|0|
\hline ->\noalign 
                  {\ifnum 0=`}\fi \hrule \@height \arrayrulewidth \futurelet...
l.56 \gradingtable{5}
                   
? 
! Emergency stop.
\hline ->\noalign 
                  {\ifnum 0=`}\fi \hrule \@height \arrayrulewidth \futurelet...
l.56 \gradingtable{5}
                   
End of file on the terminal!
|726642372|kitty|/Applications/kitty.app|0|
! Misplaced \noalign.
\hline ->\noalign 
                  {\ifnum 0=`}\fi \hrule \@height \arrayrulewidth \futurelet...
l.56 \gradingtable{5}
                   
? 
! Emergency stop.
\hline ->\noalign 
                  {\ifnum 0=`}\fi \hrule \@height \arrayrulewidth \futurelet...
l.56 \gradingtable{5}
                   
End of file on the terminal!
|726642375|kitty|/Applications/kitty.app|0|
gradingtable{5}|726642490|kitty|/Applications/kitty.app|0|
\newcommand{\gradingtable}[1]{
    \begin{center}
        \begin{minipage}[m]{\linewidth}
        Marker's use only. \\ \large
            \ifnum#1=1
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=2
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=3
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=4
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=5
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \elifnum#1=6
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \fi
            \ifnum#1=7
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \fi
            \ifnum#1=8
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \fi
            \ifnum#1=9
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \fi
            \ifnum#1=10
                \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
            \fi
            \hline \hline
            \rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
                \ifnum#1 > 0 & 1\fi
                \ifnum#1 > 1 & 2\fi
                \ifnum#1 > 2 & 3\fi
                \ifnum#1 > 3 & 4\fi
                \ifnum#1 > 4 & 5\fi
                \ifnum#1 > 5 & 6\fi
                \ifnum#1 > 6 & 7\fi
                \ifnum#1 > 7 & 8\fi
                \ifnum#1 > 8 & 9\fi
                \ifnum#1 > 9 & 10\fi
                & HW
                & Total
                \\ \hline
            \rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
                \ifnum#1 > 0 & $\dfrac{}{\arabic{sec1marks}}$\fi
                \ifnum#1 > 1 & $\dfrac{}{\arabic{sec2marks}}$\fi
                \ifnum#1 > 2 & $\dfrac{}{\arabic{sec3marks}}$\fi
                \ifnum#1 > 3 & $\dfrac{}{\arabic{sec4marks}}$\fi
                \ifnum#1 > 4 & $\dfrac{}{\arabic{sec5marks}}$\fi
                \ifnum#1 > 5 & $\dfrac{}{\arabic{sec6marks}}$\fi
                \ifnum#1 > 6 & $\dfrac{}{\arabic{sec7marks}}$\fi
                \ifnum#1 > 7 & $\dfrac{}{\arabic{sec8marks}}$\fi
                \ifnum#1 > 8 & $\dfrac{}{\arabic{sec9marks}}$\fi
                \ifnum#1 > 9 & $\dfrac{}{\arabic{sec10marks}}$\fi
                & $\dfrac{}{\arabic{hwmarks}}$
                & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
            \end{tabularx}    
        \end{minipage}
    \end{center}
}
|726642530|kitty|/Applications/kitty.app|0|
\newcounter{colcount}
\newcommand{\addcols}[1]{
    \setcounter{colcount}{0}
    \whiledo{\value{colcount} < #1}{
        >{\centering}X|
        \stepcounter{colcount}
    }
}


|726642533|kitty|/Applications/kitty.app|0|
\begin{center}
\begin{minipage}[m]{\linewidth}
Marker's use only. \\ \large
|726642665|kitty|/Applications/kitty.app|0|
\begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
\hline \hline
\rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
    & 1
    & 2
    & 3
    & 4
    & 5
    & 6
    & HW
    & Total
    \\ \hline
\rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
    & $\dfrac{}{\arabic{sec1marks}}$ 
    & $\dfrac{}{\arabic{sec2marks}}$  
    & $\dfrac{}{\arabic{sec3marks}}$ 
    & $\dfrac{}{\arabic{sec4marks}}$ 
    & $\dfrac{}{\arabic{sec5marks}}$ 
    & $\dfrac{}{\arabic{sec6marks}}$
    & $\dfrac{}{\arabic{sec7marks}}$
    & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
\end{tabularx}    
|726642677|kitty|/Applications/kitty.app|0|

\newcounter{totalmarks}
\newcounter{secmarks}
\newcounter{sec1marks}
\newcounter{sec2marks}
\newcounter{sec3marks}
\newcounter{sec4marks}
\newcounter{sec5marks}
\newcounter{sec6marks}
\newcounter{sec7marks}
\newcounter{sec8marks}
\newcounter{sec9marks}
\newcounter{sec10marks}
\newcounter{hwmarks}
|726642737|kitty|/Applications/kitty.app|0|
\end{minipage}
\end{center}
|726642760|kitty|/Applications/kitty.app|0|
\gradetable{6}
|726642843|kitty|/Applications/kitty.app|0|
\newcommand{\tableformat6}{
    \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
    \hline \hline
    \rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
        & 1
        & 2
        & 3
        & 4
        & 5
        & 6
        & HW
        & Total
        \\ \hline
    \rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
        & $\dfrac{}{\arabic{sec1marks}}$ 
        & $\dfrac{}{\arabic{sec2marks}}$  
        & $\dfrac{}{\arabic{sec3marks}}$ 
        & $\dfrac{}{\arabic{sec4marks}}$ 
        & $\dfrac{}{\arabic{sec5marks}}$ 
        & $\dfrac{}{\arabic{sec6marks}}$
        & $\dfrac{}{\arabic{hwmarks}}$
        & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
    \end{tabularx}    

}

\newcommand{\gradetable}[1]{
    \begin{center}
        \begin{minipage}[m]{\linewidth}
            Marker's use only. \\ \large
            \tableformat#1
        \end{minipage}
    \end{center}
}


|726642862|kitty|/Applications/kitty.app|0|
\usepackage{tabularx}
|726642927|kitty|/Applications/kitty.app|0|
\newcounter{totalmarks}
\newcounter{secmarks}
\newcounter{sec1marks}
\newcounter{sec2marks}
\newcounter{sec3marks}
\newcounter{sec4marks}
\newcounter{sec5marks}
\newcounter{sec6marks}
\newcounter{sec7marks}
\newcounter{sec8marks}
\newcounter{sec9marks}
\newcounter{sec10marks}
\newcounter{hwmarks}

\newcommand{\tableformat6}{
    \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
    \hline \hline
    \rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
        & 1
        & 2
        & 3
        & 4
        & 5
        & 6
        & HW
        & Total
        \\ \hline
    \rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
        & $\dfrac{}{\arabic{sec1marks}}$ 
        & $\dfrac{}{\arabic{sec2marks}}$  
        & $\dfrac{}{\arabic{sec3marks}}$ 
        & $\dfrac{}{\arabic{sec4marks}}$ 
        & $\dfrac{}{\arabic{sec5marks}}$ 
        & $\dfrac{}{\arabic{sec6marks}}$
        & $\dfrac{}{\arabic{hwmarks}}$
        & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
    \end{tabularx}    

}

\newcommand{\gradetable}[1]{
    \begin{center}
        \begin{minipage}[m]{\linewidth}
            Marker's use only. \\ \large
            \tableformat#1
        \end{minipage}
    \end{center}
}


|726642928|kitty|/Applications/kitty.app|0|
\newcounter{totalmarks}
\newcounter{secmarks}
\newcounter{sec1marks}
\newcounter{sec2marks}
\newcounter{sec3marks}
\newcounter{sec4marks}
\newcounter{sec5marks}
\newcounter{sec6marks}
\newcounter{sec7marks}
\newcounter{sec8marks}
\newcounter{sec9marks}
\newcounter{sec10marks}
\newcounter{hwmarks}

\newcommand{\tableformatsix}{
    \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
    \hline \hline
    \rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
        & 1
        & 2
        & 3
        & 4
        & 5
        & 6
        & HW
        & Total
        \\ \hline
    \rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
        & $\dfrac{}{\arabic{sec1marks}}$ 
        & $\dfrac{}{\arabic{sec2marks}}$  
        & $\dfrac{}{\arabic{sec3marks}}$ 
        & $\dfrac{}{\arabic{sec4marks}}$ 
        & $\dfrac{}{\arabic{sec5marks}}$ 
        & $\dfrac{}{\arabic{sec6marks}}$
        & $\dfrac{}{\arabic{hwmarks}}$
        & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
    \end{tabularx}    

}

\newcommand{\mygradetable}[1]{
    \begin{center}
        \begin{minipage}[m]{\linewidth}
            Marker's use only. \\ \large
            \tableformatsix
        \end{minipage}
    \end{center}
}


|726643045|kitty|/Applications/kitty.app|0|
how do I call the function tableformat6 in latex?|726643250|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
            \arabic{\tableformat#1}
|726643379|kitty|/Applications/kitty.app|0|
>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|726643446|kitty|/Applications/kitty.app|0|
        & 2
        & 3
        & 4
        & 5
        & 6
|726643449|kitty|/Applications/kitty.app|0|
        & $\dfrac{}{\arabic{sec2marks}}$  
        & $\dfrac{}{\arabic{sec3marks}}$ 
        & $\dfrac{}{\arabic{sec4marks}}$ 
        & $\dfrac{}{\arabic{sec5marks}}$ 
        & $\dfrac{}{\arabic{sec6marks}}$
|726643454|kitty|/Applications/kitty.app|0|
>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|726643470|kitty|/Applications/kitty.app|0|
        & 3
        & 4
        & 5
        & 6
|726643473|kitty|/Applications/kitty.app|0|
        & $\dfrac{}{\arabic{sec3marks}}$ 
        & $\dfrac{}{\arabic{sec4marks}}$ 
        & $\dfrac{}{\arabic{sec5marks}}$ 
        & $\dfrac{}{\arabic{sec6marks}}$
|726643478|kitty|/Applications/kitty.app|0|
>{\centering}X|>{\centering}X|>{\centering}X|726643491|kitty|/Applications/kitty.app|0|
        & 4
        & 5
        & 6
|726643495|kitty|/Applications/kitty.app|0|
        & $\dfrac{}{\arabic{sec4marks}}$ 
        & $\dfrac{}{\arabic{sec5marks}}$ 
        & $\dfrac{}{\arabic{sec6marks}}$
|726643499|kitty|/Applications/kitty.app|0|
>{\centering}X|>{\centering}X|726643512|kitty|/Applications/kitty.app|0|
        & 5
        & 6
|726643515|kitty|/Applications/kitty.app|0|
        & $\dfrac{}{\arabic{sec5marks}}$ 
        & $\dfrac{}{\arabic{sec6marks}}$
|726643518|kitty|/Applications/kitty.app|0|
>{\centering}X||726643578|kitty|/Applications/kitty.app|0|
        & 6
|726643659|kitty|/Applications/kitty.app|0|
        & $\dfrac{}{\arabic{sec6marks}}$
|726643673|kitty|/Applications/kitty.app|0|
\newcommand{\tableformatsix}{
    \begin{tabularx}{\linewidth}{||>{\bfseries}l||>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X||>{\bfseries}c||}
    \hline \hline
    \rule[-0.75em]{0em}{2em}\parbox[m]{2.75cm}{\MakeUppercase{Section}}
        & 1
        & 2
        & 3
        & 4
        & 5
        & 6
        & HW
        & Total
        \\ \hline
    \rule[-1.75em]{0em}{4em}\parbox[m]{0.15\linewidth}{\MakeUppercase{Marks}}  
        & $\dfrac{}{\arabic{sec1marks}}$ 
        & $\dfrac{}{\arabic{sec2marks}}$  
        & $\dfrac{}{\arabic{sec3marks}}$ 
        & $\dfrac{}{\arabic{sec4marks}}$ 
        & $\dfrac{}{\arabic{sec5marks}}$ 
        & $\dfrac{}{\arabic{sec6marks}}$
        & $\dfrac{}{\arabic{hwmarks}}$
        & $\dfrac{}{\thetotalmarks}$  \\ \hline \hline
    \end{tabularx}    
}
|726643688|kitty|/Applications/kitty.app|0|
            \elifnum#1 = 2 \tableformattwo\fi
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\fi|726643762|kitty|/Applications/kitty.app|0|
    \begin{center}
        \begin{minipage}[m]{\linewidth}
            Marker's use only. \\ \large
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        \end{minipage}
    \end{center}
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one|726643996|kitty|/Applications/kitty.app|0|
            \elifnum#1 = 2 \tableformattwo
            \elifnum#1 = 3 \tableformatthree
            \elifnum#1 = 4 \tableformatfour
            \elifnum#1 = 5 \tableformatfive
            \elifnum#1 = 6 \tableformatsix
            \elifnum#1 = 7 \tableformatseven
            \elifnum#1 = 8 \tableformateight
            \elifnum#1 = 9 \tableformatnine
            \elifnum#1 = 10 \tableformatten
|726644000|kitty|/Applications/kitty.app|0|
            \ifnum#1 = 6 \tableformatsix
            \fi
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six|726644034|kitty|/Applications/kitty.app|0|
el
el
el
el
el
el
el
el
el
|726644071|kitty|/Applications/kitty.app|0|
Square|727077304|kitty|/Applications/kitty.app|0|
Using linear equations to solve problems|727077494|kitty|/Applications/kitty.app|0|
u|727077504|kitty|/Applications/kitty.app|0|
Literal Equations|727077516|kitty|/Applications/kitty.app|0|
Inequalities|727077532|kitty|/Applications/kitty.app|0|
\section{Solving linear inequalities}
    \setcounter{secmarks}{0}
    \input{}
    \setcounter{sec5marks}{\thesecmarks}
|727077548|kitty|/Applications/kitty.app|0|


\documentclass[10pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage[version=4]{mhchem}
\usepackage{stmaryrd}
\usepackage{graphicx}
\usepackage[export]{adjustbox}
\graphicspath{ {./images/} }
|727077868|kitty|/Applications/kitty.app|0|

\title{Formulas }

\author{}
\date{}
|727077873|kitty|/Applications/kitty.app|0|

\begin{document}
\maketitle
|727077886|kitty|/Applications/kitty.app|0|



The area, \(A\) square units, of a circle with radius \(r\) units is given by \(A=\pi r^{2}\).

The volume, \(V\) cubic units, of a cylinder with radius \(r\) units and height \(h\) units is given by \(V=\pi r^{2} h\).

Einstein discovered the remarkable formula \(E=m c^{2}\), where:
|727078043|kitty|/Applications/kitty.app|0|

\[
\begin{aligned}
E & =\text { energy } \\
m & =\text { mass }
|727078044|kitty|/Applications/kitty.app|0|
\end{aligned}
\]

and \(c=\) the speed of light.

These are examples of formulas.
\includegraphics[max width=\textwidth, center]{2023_12_09_5c1d855de2ec0db78c30g-01}
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common|727078068|kitty|/Applications/kitty.app|0|
factor|727078071|kitty|/Applications/kitty.app|0|
difference|727078076|kitty|/Applications/kitty.app|0|
of two squares|727078079|kitty|/Applications/kitty.app|0|
monic|727078086|kitty|/Applications/kitty.app|0|
non-monic|727078091|kitty|/Applications/kitty.app|0|
quadratic|727078094|kitty|/Applications/kitty.app|0|
perfect square|727078112|kitty|/Applications/kitty.app|0|
\randomword{factorisation}%
\randomword{expansion}%
|727078117|kitty|/Applications/kitty.app|0|
\hspace{\fill}
|727078171|kitty|/Applications/kitty.app|0|
A formula relates different quantities.|727078214|kitty|/Applications/kitty.app|0|
\section*{Substitution into formulas}
 For instance, the formula \(A=\pi r^{2}\) relates the radius, \(r\) units, of a circle to the area, \(A\) square units, of the circle. That is, it describes how the area of a circle depends on its radius.

\section*{Formulas with a subject}
In many formulas that you see there is a single pronumeral on the left-hand side of the equal sign. This pronumeral is called the subject of the formula.

For instance, in the formula \(E=m c^{2}, E\) is the subject.

\section*{Substitution}
If the values of all the pronumerals except the subject of a formula are known, then we can find the value of the subject by substitution.

#|727078256|kitty|/Applications/kitty.app|0|
what is a formula?
answer.
A formula relates different quantities.

what is the most famous formula that you know?
answer.

can you derive this formula?
answer.

what is the subject of your formula?
answer.
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        \begin{solutionordottedlines}[2cm]
        \end{solutionordottedlines}
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answer.
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    \{nd}
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reasonably|727079487|kitty|/Applications/kitty.app|0|
\section*{Example 1}
Find the value of the subject when the pronumerals in the formula have the values indicated.
a \(F=m a\), where \(a=10, m=3.5\)
b \(m=\frac{a+b}{2}\), where \(a=12, b=26\)

\section*{Solution}
a \(F=m a\)
b \(m=\frac{a+b}{2}\)
\(=10 \times 3.5\)
\(=\frac{12+26}{2}\)
\(=35\)
\(=19\)

\section*{Example 2}
The formula for the circumference \(C\) of a circle of radius \(r\) is \(C=2 \pi r\).

Find the value of \(C\) when \(r=20\) :

a in terms of \(\pi\) (that is, exactly) \(\quad\) b correct to 2 decimal places

\section*{Solution}
a \(C=2 \pi r\)

\(=40 \pi\) b \(C=40 \pi\)

\(=125.663 \ldots\)

\(\approx 125.66\) (using a calculator)

(correct to 2 decimal places)

\section*{Example 3}
a The area of a triangle \(A \mathrm{~cm}^{2}\) is given by \(A=\frac{1}{2} b h\), where \(b \mathrm{~cm}\) is the base length and \(h \mathrm{~cm}\) is the height. Calculate the area of a triangle with base length \(16 \mathrm{~cm}\) and height \(11 \mathrm{~cm}\).

b The simple interest payable when \(\$ P\) is invested at a rate of \(r \%\) per year for \(t\) years is given by \(I=\frac{P r t}{100}\). Calculate the simple interest payable when \(\$ 1000\) is invested at \(3.5 \%\) per year for 6 years.

\section*{Solution}
\[
\text { a } \begin{aligned}
A & =\frac{1}{2} b h \\
& =\frac{1}{2} \times 16 \times 11 \\
& =88
\end{aligned}
\]

The area of the triangle is \(88 \mathrm{~cm}^{2}\).

\[
\text { b } \begin{aligned}
I & =\frac{P r t}{100} \\
& =\frac{1000 \times 3.5 \times 6}{100} \\
& =210
\end{aligned}
\]

The interest payable is \(\$ 210\).

\section*{Substitution into a formula}
When the pronumeral whose value is to be found is not the subject of the formula, it is necessary to solve an equation to find the value of this pronumeral. This is shown in the following examples.

\section*{Example 4}
For a car travelling in a straight line with initial velocity \(u \mathrm{~m} / \mathrm{s}\) and acceleration \(a \mathrm{~m} / \mathrm{s}^{2}\), the formula for the velocity \(v \mathrm{~m} / \mathrm{s}\) at time \(t\) seconds is \(v=u+a t\).
a Find \(u\) if \(a=2, v=15\) and \(t=7\).
b Find \(a\) if \(v=10, u=6\) and \(t=3\).

\section*{Solution}
a \(v=u+a t\)

When \(a=2, v=15\) and \(t=7\).

\[
\begin{aligned}
15 & =u+2 \times 7 \\
15 & =u+14 \\
u & =1
\end{aligned}
\]

The initial velocity is \(1 \mathrm{~m} / \mathrm{s}\). b \(v=u+a t\)

When \(v=10, u=6\) and \(t=3\).

\[
\begin{aligned}
10 & =6+3 a \\
4 & =3 a \\
a & =\frac{4}{3}
\end{aligned}
\]

The acceleration is \(\frac{4}{3} \mathrm{~m} / \mathrm{s}^{2}\).

\section*{Example 5}
The thin lens formula states that

\[
\frac{1}{f}=\frac{1}{u}+\frac{1}{v}
\]

where \(u\) is the distance from the object to the lens, \(v\) is the distance of the image from the lens and \(f\) is the focal length of the lens.
a Find \(f\) if \(u=2\) and \(v=5\).
b Find \(u\) if \(f=2\) and \(v=6\).

\section*{Solution}
a \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)

When \(u=2\) and \(v=5\),

\(\frac{1}{f}=\frac{1}{2}+\frac{1}{5}\)

\[
=\frac{7}{10}
\]

Taking reciprocals of both sides

of the equation gives \(f=\frac{10}{7}\). b \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)

If \(f=2\) and \(v=6\),

\(\frac{1}{2}=\frac{1}{u}+\frac{1}{6}\)

\(\frac{1}{u}=\frac{1}{2}-\frac{1}{6}\)

\(=\frac{1}{3}\)

Taking reciprocals of both sides of the equation gives \(u=3\).

\section*{Example 6}
The area of a circle \(A \mathrm{~cm}^{2}\) is given by \(A=\pi r^{2}\), where \(r \mathrm{~cm}\) is the radius of the circle.

If \(A=20\), find \(r\) :
a exactly
b correct to 2 decimal places

\section*{Solution}
a \(A=\pi r^{2}\)

When \(A=20\),

\(20=\pi r^{2}\)

\(\frac{20}{\pi}=r^{2}\)

(Divide both sides of equation by \(\pi\).)

\(r=\sqrt{\frac{20}{\pi}} \quad(r\) is positive. \()\)

b \(r \approx 2.52\)

(correct to 2 decimal places)
|727079546|kitty|/Applications/kitty.app|0|

\section*{Solution}
a \(F=m a\)
b \(m=\frac{a+b}{2}\)
\(=10 \times 3.5\)
\(=\frac{12+26}{2}\)
\(=35\)
\(=19\)
|727079741|kitty|/Applications/kitty.app|0|
a \(C=2 \pi r\)

\(=40 \pi\) b \(C=40 \pi\)
|727079817|kitty|/Applications/kitty.app|0|
\(=40 \pi\)
|727079826|kitty|/Applications/kitty.app|0|
                b \(C=40 \pi\)
|727079828|kitty|/Applications/kitty.app|0|
\(=125.663 \ldots\)
|727079830|kitty|/Applications/kitty.app|0|
\(\approx 125.66\) (using a calculator)
|727079849|kitty|/Applications/kitty.app|0|
(correct to 2 decimal places)
|727079858|kitty|/Applications/kitty.app|0|
\[
\text { a } \begin{aligned}
A & =\frac{1}{2} b h \\
& =\frac{1}{2} \times 16 \times 11 \\
& =88
\end{aligned}
\]

The area of the triangle is \(88 \mathrm{~cm}^{2}\).
|727079942|kitty|/Applications/kitty.app|0|
\[
\text { b } \begin{aligned}
I & =\frac{P r t}{100} \\
& =\frac{1000 \times 3.5 \times 6}{100} \\
& =210
\end{aligned}
\]

The interest payable is \(\$ 210\).
|727079969|kitty|/Applications/kitty.app|0|
\section*{Solution}


\section*{Substitution into a formula}
When the pronumeral whose value is to be found is not the subject of the formula, it is necessary to solve an equation to find the value of this pronumeral. This is shown in the following examples.
|727079978|kitty|/Applications/kitty.app|0|
(using a calculator, |727081081|kitty|/Applications/kitty.app|0|
)|727081089|kitty|/Applications/kitty.app|0|
                    \begin{solutionordottedlines}[2cm]
                    \end{solutionordottedlines}
|727081121|kitty|/Applications/kitty.app|0|
a \(v=u+a t\)

When \(a=2, v=15\) and \(t=7\).

\[
\begin{aligned}
15 & =u+2 \times 7 \\
15 & =u+14 \\
u & =1
\end{aligned}
\]

The initial velocity is \(1 \mathrm{~m} / \mathrm{s}\). b \(v=u+a t\)
|727081131|kitty|/Applications/kitty.app|0|
 b \(v=u+a t\)|727081138|kitty|/Applications/kitty.app|0|
When \(v=10, u=6\) and \(t=3\).

\[
\begin{aligned}
10 & =6+3 a \\
4 & =3 a \\
a & =\frac{4}{3}
\end{aligned}
\]

The acceleration is \(\frac{4}{3} \mathrm{~m} / \mathrm{s}^{2}\).

|727081143|kitty|/Applications/kitty.app|0|

\section*{Solution}

\section*{Example 5}
|727081192|kitty|/Applications/kitty.app|0|
                \part The thin lens formula states that
                    \[
                    \frac{1}{f}=\frac{1}{u}+\frac{1}{v}
                    \]
                where \(u\) is the distance from the object to the lens, \(v\) is the distance of the image from the lens and \(f\) is the focal length of the lens.
|727081225|kitty|/Applications/kitty.app|0|
\(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)

When \(u=2\) and \(v=5\),

\(\frac{1}{f}=\frac{1}{2}+\frac{1}{5}\)

\[
=\frac{7}{10}
\]

Taking reciprocals of both sides

of the equation gives \(f=\frac{10}{7}\). b \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
|727081257|kitty|/Applications/kitty.app|0|
                b \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
|727081269|kitty|/Applications/kitty.app|0|
If \(f=2\) and \(v=6\),

\(\frac{1}{2}=\frac{1}{u}+\frac{1}{6}\)

\(\frac{1}{u}=\frac{1}{2}-\frac{1}{6}\)

\(=\frac{1}{3}\)

Taking reciprocals of both sides of the equation gives \(u=3\).
|727081278|kitty|/Applications/kitty.app|0|
end|727081312|kitty|/Applications/kitty.app|0|
If \(A=20\), find \(r\) :
|727081346|kitty|/Applications/kitty.app|0|
a \(A=\pi r^{2}\)

When \(A=20\),

\(20=\pi r^{2}\)

\(\frac{20}{\pi}=r^{2}\)

(Divide both sides of equation by \(\pi\).)

\(r=\sqrt{\frac{20}{\pi}} \quad(r\) is positive. \()\)
|727081373|kitty|/Applications/kitty.app|0|
b \(r \approx 2.52\)

(correct to 2 decimal places)
|727081382|kitty|/Applications/kitty.app|0|
\section*{Solution}


\end{examplebox}
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\randomword{Einstein}%
\hspace{\fill}
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W hat is a formula?|727081605|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
 \newcommand{\Question}[2][]{%
   \ifx&#1&%
     \question #2%
   \else
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 }
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\newcommand{\Question}[2][]{
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  \question[#1] #2
}
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\hrulefill
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        \begin{parts}
            \part 
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 
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        \end{parts}
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            \part 
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
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        \begin{parts}\begin{multicols}{2}
            \part 
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            \end{solutionordottedlines}
            \part 
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            \end{solutionordottedlines}
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            \end{solutionordottedlines}
            \part 
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
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        \Question[2]
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        \Question[3]
        \begin{parts}
            \part
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            \end{solutionordottedlines}
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            \end{solutionordottedlines}
        \end{parts}
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            \part
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            \part
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
|727086310|kitty|/Applications/kitty.app|0|
Example 1 1 For each part, find the value of the subject when the other pronumerals have the value indicated.

a \(A=\ell w\), where \(\ell=5, w=8\)

b \(s=\frac{d}{t}\), where \(d=120, t=6\)

c \(A=\frac{1}{2} x y\), where \(x=10, y=7\)
|727086556|kitty|/Applications/kitty.app|0|
a \(A=\ell w\), where \(\ell=5, w=8\)

b \(s=\frac{d}{t}\), where \(d=120, t=6\)

c \(A=\frac{1}{2} x y\), where \(x=10, y=7\)
|727086567|kitty|/Applications/kitty.app|0|

b \(s=\frac{d}{t}\), where \(d=120, t=6\)

c \(A=\frac{1}{2} x y\), where \(x=10, y=7\)
|727086572|kitty|/Applications/kitty.app|0|

c \(A=\frac{1}{2} x y\), where \(x=10, y=7\)
|727086577|kitty|/Applications/kitty.app|0|
a \(A=\ell w\), where \(\ell=5, w=8\)

b \(s=\frac{d}{t}\), where \(d=120, t=6\)

c \(A=\frac{1}{2} x y\), where \(x=10, y=7\)

|727086610|kitty|/Applications/kitty.app|0|
Example 1 1 For each part, find the value of the subject when the other pronumerals have the value indicated.

d \(A=\frac{1}{2}(a+b) h\), where \(a=4, b=6, h=10\)

e \(t=a+(n-1) d\), where \(a=30, n=8, d=4\)

f \(E=\frac{1}{2} m v^{2}\), where \(m=8, v=4\)
|727086613|kitty|/Applications/kitty.app|0|
\section{Introduction}
    \setcounter{secmarks}{0}
    \input{1-intro}
    \setcounter{sec1marks}{\thesecmarks}
\section{Substitution into formulas}
    \setcounter{secmarks}{0}
    \input{2-subs}
    \setcounter{sec2marks}{\thesecmarks}
\section{Changing the subject of a formula}
    \setcounter{secmarks}{0}
    \input{3-subj}
    \setcounter{sec3marks}{\thesecmarks}
\section{Constructing Formulas}
    \setcounter{secmarks}{0}
    \input{4-constr}
    \setcounter{sec4marks}{\thesecmarks}
\newpage
\section{Homework}
    \setcounter{secmarks}{0}
    \input{0-homework}
    \setcounter{hwmarks}{\thesecmarks}
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\section{Introduction}
    \setcounter{secmarks}{0}
    \input{1-intro}
    \setcounter{sec1marks}{\thesecmarks}
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    \setcounter{secmarks}{0}
    \input{2-subs}
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    \setcounter{secmarks}{0}
    \input{3-subj}
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    \setcounter{secmarks}{0}
    \input{4-constr}
    \setcounter{sec4marks}{\thesecmarks}
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\newpage
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\section{Homework}
    \setcounter{secmarks}{0}
    \input{0-homework}
    \setcounter{hwmarks}{\thesecmarks}
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\section{Changing the subject of a formula}
\section{Constructing Formulas}
|727086638|kitty|/Applications/kitty.app|0|
2 For each part, find the value of the subject when the other pronumerals have the value indicated. Calculate \(\mathbf{a}-\mathbf{c}\) correct to 3 decimal places and \(\mathbf{d}\) correct to 2 .

a \(x=\sqrt{a b}\), where \(a=40, b=50\)

b \(V=\pi r^{2} h\), where \(r=12, h=20\)

c \(T=2 \pi \sqrt{\frac{\ell}{g}}\), where \(\ell=88.2, g=9.8\)

d \(A=P(1+R)^{n}\), where \(P=10000, R=0.065, n=10\)
|727086715|kitty|/Applications/kitty.app|0|
        \Question[3]
        \begin{parts}
            \part 
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
|727086726|kitty|/Applications/kitty.app|0|
3 For the formula \(v=u+a t\), find:
a \(v\) if \(u=6, a=3\) and \(t=5\)
b \(u\) if \(v=40, a=5\) and \(t=2\)
c \(a\) if \(v=60, u=0\) and \(t=5\)
d \(t\) if \(v=100, u=20\) and \(a=6\)
|727086752|kitty|/Applications/kitty.app|0|
a \(v\) if \(u=6, a=3\) and \(t=5\)
b \(u\) if \(v=40, a=5\) and \(t=2\)
c \(a\) if \(v=60, u=0\) and \(t=5\)
d \(t\) if \(v=100, u=20\) and \(a=6\)
|727086757|kitty|/Applications/kitty.app|0|
b \(u\) if \(v=40, a=5\) and \(t=2\)
c \(a\) if \(v=60, u=0\) and \(t=5\)
d \(t\) if \(v=100, u=20\) and \(a=6\)
|727086762|kitty|/Applications/kitty.app|0|
c \(a\) if \(v=60, u=0\) and \(t=5\)
d \(t\) if \(v=100, u=20\) and \(a=6\)
|727086765|kitty|/Applications/kitty.app|0|
d \(t\) if \(v=100, u=20\) and \(a=6\)
|727086769|kitty|/Applications/kitty.app|0|

4 a For the formula \(S=2(a-b)\), find \(a\) if \(S=60\) and \(b=10\).

b For the formula \(I=\frac{180 n-360}{n}\), find \(n\) if \(I=120\).

c For the formula \(a=\frac{m+n}{2}\), find \(m\) if \(a=20\) and \(n=6\).

d For the formula \(A=\frac{P R T}{100}\), find \(P\) if \(A=1600, R=4\) and \(T=10\).
|727086787|kitty|/Applications/kitty.app|0|

b For the formula \(I=\frac{180 n-360}{n}\), find \(n\) if \(I=120\).

c For the formula \(a=\frac{m+n}{2}\), find \(m\) if \(a=20\) and \(n=6\).

d For the formula \(A=\frac{P R T}{100}\), find \(P\) if \(A=1600, R=4\) and \(T=10\).
|727086808|kitty|/Applications/kitty.app|0|

c For the formula \(a=\frac{m+n}{2}\), find \(m\) if \(a=20\) and \(n=6\).

d For the formula \(A=\frac{P R T}{100}\), find \(P\) if \(A=1600, R=4\) and \(T=10\).
|727086813|kitty|/Applications/kitty.app|0|

d For the formula \(A=\frac{P R T}{100}\), find \(P\) if \(A=1600, R=4\) and \(T=10\).
|727086817|kitty|/Applications/kitty.app|0|

e For the formula \(S=2(\ell w+\ell h+h w)\), find \(h\) if \(S=592, \ell=10\) and \(w=8\).

f For the formula \(s=u t+\frac{1}{2} a t^{2}\), find \(a\) if \(s=1000, u=20\) and \(t=5\).

g For the formula \(t=a+(n-1) d\), find \(n\) if \(t=58, d=3\) and \(a=7\).
|727086857|kitty|/Applications/kitty.app|0|
5 Given \(v^{2}=u^{2}+2 a x\) and \(v>0\), find the value of \(v\) (correct to 1 decimal place) when:
a \(u=0, a=5\) and \(x=10\)
b \(u=2, a=9.8\) and \(x=22\)
|727086905|kitty|/Applications/kitty.app|0|
7 For the formula \(s=u t+\frac{1}{2} a t^{2}\), find the value of:
a \(u\), when \(s=10, t=20\) and \(a=2\)
b \(a\), when \(s=20, u=5\) and \(t=2\)
|727086937|kitty|/Applications/kitty.app|0|
a \(u\), when \(s=10, t=20\) and \(a=2\)
b \(a\), when \(s=20, u=5\) and \(t=2\)
|727086955|kitty|/Applications/kitty.app|0|
b \(a\), when \(s=20, u=5\) and \(t=2\)
|727086958|kitty|/Applications/kitty.app|0|

8 Given \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\), find the value of:
a \(u\) when \(f=2\) and \(v=4\)
b \(u\) when \(f=3\) and \(v=4\)
|727086968|kitty|/Applications/kitty.app|0|

9 Given that \(P=\frac{M+m}{M-m}\), find the value of \(P\) when:
a \(M=8\) and \(m=4\)
b \(M=26\) and \(m=17\)
|727086984|kitty|/Applications/kitty.app|0|
a \(M=8\) and \(m=4\)
b \(M=26\) and \(m=17\)
|727086990|kitty|/Applications/kitty.app|0|
b \(M=26\) and \(m=17\)
|727086994|kitty|/Applications/kitty.app|0|
c \(M=19.2\) and \(m=5.9\)
d \(M=\frac{3}{4}\) and \(m=\frac{2}{5}\)
|727087002|kitty|/Applications/kitty.app|0|
Example 6 10 The area \(A \mathrm{~cm}^{2}\) of a square with side length \(x \mathrm{~cm}\) is given by \(A=x^{2}\). If \(A=20\), find:

a the value of \(x\)

b the value of \(x\) correct to 2 decimal places.
|727087006|kitty|/Applications/kitty.app|0|

a the value of \(x\)

b the value of \(x\) correct to 2 decimal places.
|727087014|kitty|/Applications/kitty.app|0|

b the value of \(x\) correct to 2 decimal places.
|727087018|kitty|/Applications/kitty.app|0|
11 For a rectangle of length \(\ell \mathrm{cm}\) and width \(w \mathrm{~cm}\), the perimeter \(P \mathrm{~cm}\) is given by \(P=2(\ell+w)\).

Use this formula to calculate the length of a rectangle which has width \(15 \mathrm{~cm}\) and perimeter \(57 \mathrm{~cm}\).
|727087049|kitty|/Applications/kitty.app|0|
12 The formula for finding the number of degrees Fahrenheit \((F)\) for a temperature given as a number of degrees Celsius \((C)\) is \(F=\frac{9}{5} C+32\).

Fahrenheit temperatures are still used in the USA, but in Australia we commonly use Celsius.

Calculate the Fahrenheit temperatures which people in the USA would recognise for:

a the freezing point of water, \(0^{\circ} \mathrm{C}\)

b the boiling point of water, \(100^{\circ} \mathrm{C}\)

c a nice summer temperature of \(25^{\circ} \mathrm{C}\)

Now calculate the Celsius temperatures which people in Australia would recognise for:
d \(50^{\circ} \mathrm{F}\)
e \(104^{\circ} \mathrm{F}\)
|727087103|kitty|/Applications/kitty.app|0|
p
|727087109|kitty|/Applications/kitty.app|0|
13 The area \(A \mathrm{~cm}^{2}\) of a triangle with side lengths \(a \mathrm{~cm}, b \mathrm{~cm}\) and \(c \mathrm{~cm}\) is given by Heron's formula

\(A^{2}=s(s-a)(s-b)(s-c)\),

where \(s=\frac{a+b+c}{2}=\) half the perimeter.

\begin{center}
\includegraphics[max width=\textwidth]{2023_12_09_5c1d855de2ec0db78c30g-06}
\end{center}

|727087123|kitty|/Applications/kitty.app|0|
Find the exact areas of the triangles whose side lengths are given below.
a \(6 \mathrm{~cm}, 8 \mathrm{~cm}\) and \(10 \mathrm{~cm}\)
b \(5 \mathrm{~cm}, 12 \mathrm{~cm}\) and \(13 \mathrm{~cm}\)
c \(8 \mathrm{~cm}, 10 \mathrm{~cm}\) and \(14 \mathrm{~cm}\)
d \(13 \mathrm{~cm}, 14 \mathrm{~cm}\) and \(15 \mathrm{~cm}\)
|727087128|kitty|/Applications/kitty.app|0|
        a \(6 \mathrm{~cm}, 8 \mathrm{~cm}\) and \(10 \mathrm{~cm}\)
        b \(5 \mathrm{~cm}, 12 \mathrm{~cm}\) and \(13 \mathrm{~cm}\)
        c \(8 \mathrm{~cm}, 10 \mathrm{~cm}\) and \(14 \mathrm{~cm}\)
        d \(13 \mathrm{~cm}, 14 \mathrm{~cm}\) and \(15 \mathrm{~cm}\)
|727087133|kitty|/Applications/kitty.app|0|
        b \(5 \mathrm{~cm}, 12 \mathrm{~cm}\) and \(13 \mathrm{~cm}\)
        c \(8 \mathrm{~cm}, 10 \mathrm{~cm}\) and \(14 \mathrm{~cm}\)
        d \(13 \mathrm{~cm}, 14 \mathrm{~cm}\) and \(15 \mathrm{~cm}\)
|727087137|kitty|/Applications/kitty.app|0|
        c \(8 \mathrm{~cm}, 10 \mathrm{~cm}\) and \(14 \mathrm{~cm}\)
        d \(13 \mathrm{~cm}, 14 \mathrm{~cm}\) and \(15 \mathrm{~cm}\)
|727087141|kitty|/Applications/kitty.app|0|
        d \(13 \mathrm{~cm}, 14 \mathrm{~cm}\) and \(15 \mathrm{~cm}\)
|727087146|kitty|/Applications/kitty.app|0|
14 Sam throws a stone down to the ground from the top of a cliff \(s\) metres high, with an initial speed of \(u \mathrm{~m} / \mathrm{s}\). It accelerates at \(a \mathrm{~m} / \mathrm{s}^{2}\). The stone hits the ground with a speed of \(v \mathrm{~m} / \mathrm{s}\) given by the formula \(v^{2}=u^{2}+2 a s\). Find the speed at which the stone hits the ground, correct to 2 decimal places, if:
a \(u=0, a=9.8\) and \(s=50\)
b \(u=5, a=9.8\) and \(s=35\)
|727087162|kitty|/Applications/kitty.app|0|
15 The distance \(d\) metres Jim's car takes to stop once the brakes are applied is given by the formula \(d=0.2 v+0.005 v^{2}\), where \(v \mathrm{~km} / \mathrm{h}\) is the speed of the car when the brakes are applied.

Find the distance the car takes to stop if the brakes are applied when it is travelling at each of the speeds given below. Calculate your answers correct to 3 decimal places where appropriate.
a \(60 \mathrm{~km} / \mathrm{h}\)
b \(65 \mathrm{~km} / \mathrm{h}\)
c \(70 \mathrm{~km} / \mathrm{h}\)
d \(80 \mathrm{~km} / \mathrm{h}\)
e \(100 \mathrm{~km} / \mathrm{h}\)
f \(120 \mathrm{~km} / \mathrm{h}\)
|727087175|kitty|/Applications/kitty.app|0|
        \begin{theorembox}
            \subsection*{Heron's Formula}
            \[A^2 = s(s-a)(s-b)(s-c)\]
            where $s = \frac{a+b+c}{2} = $ half the perimeter.
            This is helpful for finding the area of non-right-angled triangles for which you do not even have an angle for.
            Try to determine the area of this shape yourself without the formula if you dare.
            \begin{tikzpicture}
            \end{tikzpicture}
        \end{theorembox}
|727087309|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \begin{questions}
        \Question[3] For each part, find the value of the subject when the other pronumerals have the value indicated.
        \begin{parts}
            \part \(A=\ell w\), where \(\ell=5, w=8\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(s=\frac{d}{t}\), where \(d=120, t=6\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(A=\frac{1}{2} x y\), where \(x=10, y=7\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[4] For the formula \(v=u+a t\), find:
        \begin{parts}\begin{multicols}{2}
            \part \(v\) if \(u=6, a=3\) and \(t=5\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(u\) if \(v=40, a=5\) and \(t=2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(a\) if \(v=60, u=0\) and \(t=5\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(t\) if \(v=100, u=20\) and \(a=6\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}
            \part For the formula \(S=2(a-b)\), find \(a\) if \(S=60\) and \(b=10\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part For the formula \(I=\frac{180 n-360}{n}\), find \(n\) if \(I=120\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part For the formula \(a=\frac{m+n}{2}\), find \(m\) if \(a=20\) and \(n=6\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part For the formula \(A=\frac{P R T}{100}\), find \(P\) if \(A=1600, R=4\) and \(T=10\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] For the formula \(s=u t+\frac{1}{2} a t^{2}\), find the value of:
        \begin{parts}
            \part \(u\), when \(s=10, t=20\) and \(a=2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(a\), when \(s=20, u=5\) and \(t=2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] Given that \(P=\frac{M+m}{M-m}\), find the value of \(P\) when:
        \begin{parts}
            \part \(M=8\) and \(m=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(M=26\) and \(m=17\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] The area \(A \mathrm{~cm}^{2}\) of a square with side length \(x \mathrm{~cm}\) is given by \(A=x^{2}\). If \(A=20\), find:
        \begin{parts}
            \part the value of \(x\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part the value of \(x\) correct to 2 decimal places.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2.5] For a rectangle of length \(\ell \mathrm{cm}\) and width \(w \mathrm{~cm}\), the perimeter \(P \mathrm{~cm}\) is given by \(P=2(\ell+w)\). \\Use this formula to calculate the length of a rectangle which has width \(15 \mathrm{~cm}\) and perimeter \(57 \mathrm{~cm}\).
        \Question[6] The area $A\,\text{cm}^2$ of a triangle with side lengths, $a$ cm, $b$ cm and $c$ cm is given by \textit{Heron's formula}:
        \begin{theorembox}
            \subsection*{Heron's Formula}
            \[A^2 = s(s-a)(s-b)(s-c)\]
            where $s = \frac{a+b+c}{2} = $ half the perimeter.
            This is helpful for finding the area of non-right-angled triangles for which you do not even have an angle for.
            Try to determine the area of this shape yourself without the formula if you dare.
            \begin{tikzpicture}
            \end{tikzpicture}
        \end{theorembox}
        Find the exact areas of the triangles whose side lengths are given below.
        \begin{parts}
            \part \(6 \mathrm{~cm}, 8 \mathrm{~cm}\) and \(10 \mathrm{~cm}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(5 \mathrm{~cm}, 12 \mathrm{~cm}\) and \(13 \mathrm{~cm}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(8 \mathrm{~cm}, 10 \mathrm{~cm}\) and \(14 \mathrm{~cm}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(13 \mathrm{~cm}, 14 \mathrm{~cm}\) and \(15 \mathrm{~cm}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{exercisebox}
|727087417|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \begin{questions}
        \Question[3] For each part, find the value of the subject when the other pronumerals have the value indicated.
        \begin{parts}
            \part \(A=\ell w\), where \(\ell=5, w=8\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(s=\frac{d}{t}\), where \(d=120, t=6\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(A=\frac{1}{2} x y\), where \(x=10, y=7\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[4] For the formula \(v=u+a t\), find:
        \begin{parts}\begin{multicols}{2}
            \part \(v\) if \(u=6, a=3\) and \(t=5\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(u\) if \(v=40, a=5\) and \(t=2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(a\) if \(v=60, u=0\) and \(t=5\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(t\) if \(v=100, u=20\) and \(a=6\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}
            \part For the formula \(S=2(a-b)\), find \(a\) if \(S=60\) and \(b=10\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part For the formula \(I=\frac{180 n-360}{n}\), find \(n\) if \(I=120\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part For the formula \(a=\frac{m+n}{2}\), find \(m\) if \(a=20\) and \(n=6\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part For the formula \(A=\frac{P R T}{100}\), find \(P\) if \(A=1600, R=4\) and \(T=10\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] For the formula \(s=u t+\frac{1}{2} a t^{2}\), find the value of:
        \begin{parts}
            \part \(u\), when \(s=10, t=20\) and \(a=2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(a\), when \(s=20, u=5\) and \(t=2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] Given that \(P=\frac{M+m}{M-m}\), find the value of \(P\) when:
        \begin{parts}
            \part \(M=8\) and \(m=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(M=26\) and \(m=17\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] The area \(A \mathrm{~cm}^{2}\) of a square with side length \(x \mathrm{~cm}\) is given by \(A=x^{2}\). If \(A=20\), find:
        \begin{parts}
            \part the value of \(x\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part the value of \(x\) correct to 2 decimal places.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2.5] For a rectangle of length \(\ell \mathrm{cm}\) and width \(w \mathrm{~cm}\), the perimeter \(P \mathrm{~cm}\) is given by \(P=2(\ell+w)\). \\Use this formula to calculate the length of a rectangle which has width \(15 \mathrm{~cm}\) and perimeter \(57 \mathrm{~cm}\).
        \Question[6] The area $A\,\text{cm}^2$ of a triangle with side lengths, $a$ cm, $b$ cm and $c$ cm is given by \textit{Heron's formula}:
        \begin{theorembox}
            \subsection*{Heron's Formula}
            \[A^2 = s(s-a)(s-b)(s-c)\]
            where $s = \frac{a+b+c}{2} = $ half the perimeter.
            This is helpful for finding the area of non-right-angled triangles for which you do not even have an angle for.
            Try to determine the area of this shape yourself without the formula if you dare.
            \begin{tikzpicture}
            \end{tikzpicture}
        \end{theorembox}
        Find the exact areas of the triangles whose side lengths are given below.
        \begin{parts}
            \part \(6 \mathrm{~cm}, 8 \mathrm{~cm}\) and \(10 \mathrm{~cm}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(5 \mathrm{~cm}, 12 \mathrm{~cm}\) and \(13 \mathrm{~cm}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(8 \mathrm{~cm}, 10 \mathrm{~cm}\) and \(14 \mathrm{~cm}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(13 \mathrm{~cm}, 14 \mathrm{~cm}\) and \(15 \mathrm{~cm}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{exercisebox}


|727087460|kitty|/Applications/kitty.app|0|
Package mdframed Info: mdframed works in twoside mode on input line 154.
! Missing \endcsname inserted.
<to be read again> 
\relax 
l.183         \begin
                  {parts}
? 
! Emergency stop.
<to be read again> 
\relax 
l.183         \begin
                  {parts}
End of file on the terminal!
|727087525|kitty|/Applications/kitty.app|0|
        \Question[4]
        \begin{parts}
            \part For the formula \(S=2(a-b)\), find \(a\) if \(S=60\) and \(b=10\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part For the formula \(I=\frac{180 n-360}{n}\), find \(n\) if \(I=120\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part For the formula \(a=\frac{m+n}{2}\), find \(m\) if \(a=20\) and \(n=6\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part For the formula \(A=\frac{P R T}{100}\), find \(P\) if \(A=1600, R=4\) and \(T=10\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
|727087771|kitty|/Applications/kitty.app|0|
\newcommand{\Question}[2][]{%
    \ifx&#1&%
     \question #2%
    \else
     \addtocounter{secmarks}{#1}%
     \question[#1] #2%
    \fi%
}
|727087755|kitty|/Applications/kitty.app|0|
\randomword{equation}%
\hspace{\fill}
|727092428|kitty|/Applications/kitty.app|0|
The manager of a bed-and-breakfast guest house finds that the weekly profit \(\$ P\) is given by the formula

\[
P=40 G-600
\]

where \(G\) is the number of guests who stay during the week. Make \(G\) the subject of the formula and use the result to find the number of guests needed to make a profit of \(\$ 800\).
|727092760|kitty|/Applications/kitty.app|0|
\[
\begin{aligned}
P & =40 G-600 \\
P+600 & =40 G \\
G & =\frac{P+600}{40} \\
\text { When } P & =800, \\
G & =\frac{800+600}{40} \\
& =\frac{1400}{40}=35
\end{aligned}
\]

Thirty-five guests are required to make a profit of \(\$ 800\).
|727092773|kitty|/Applications/kitty.app|0|
Given the formula \(v^{2}=u^{2}+2 a s\) :

a rearrange the formula to make \(s\) the subject

b find the value of \(s\) when \(u=4, v=10\) and \(a=2\)

c find the value of \(s\) when \(u=4, v=12\) and \(a=3\)
|727092794|kitty|/Applications/kitty.app|0|

a rearrange the formula to make \(s\) the subject

b find the value of \(s\) when \(u=4, v=10\) and \(a=2\)

c find the value of \(s\) when \(u=4, v=12\) and \(a=3\)
|727092799|kitty|/Applications/kitty.app|0|

b find the value of \(s\) when \(u=4, v=10\) and \(a=2\)

c find the value of \(s\) when \(u=4, v=12\) and \(a=3\)
|727092805|kitty|/Applications/kitty.app|0|

c find the value of \(s\) when \(u=4, v=12\) and \(a=3\)
|727092809|kitty|/Applications/kitty.app|0|
a \(\quad v^{2}=u^{2}+2 a s\)

\[
\begin{array}{rlrl}
v^{2}-u^{2} & =2 a s & & \text { (Subtract } u^{2} \text { from both sides of the formula.) } \\
s & =\frac{v^{2}-u^{2}}{2 a} & \text { (Divide both sides of the equation by } 2 a .)
\end{array}
\]

b When \(u=4, v=10\) and \(a=2\).

\[
\begin{aligned}
s & =\frac{10^{2}-4^{2}}{2 \times 2} \\
& =\frac{100-16}{4} \\
& =\frac{84}{4} \\
& =21
\end{aligned}
\]

c When \(u=4, v=12\) and \(a=3\).

\[
\begin{aligned}
s & =\frac{12^{2}-4^{2}}{2 \times 3} \\
& =\frac{144-16}{6} \\
& =\frac{128}{6} \\
& =21 \frac{1}{3}
\end{aligned}
\]
|727092833|kitty|/Applications/kitty.app|0|

b When \(u=4, v=10\) and \(a=2\).

\[
\begin{aligned}
s & =\frac{10^{2}-4^{2}}{2 \times 2} \\
& =\frac{100-16}{4} \\
& =\frac{84}{4} \\
& =21
\end{aligned}
\]

c When \(u=4, v=12\) and \(a=3\).

\[
\begin{aligned}
s & =\frac{12^{2}-4^{2}}{2 \times 3} \\
& =\frac{144-16}{6} \\
& =\frac{128}{6} \\
& =21 \frac{1}{3}
\end{aligned}
\]
|727092839|kitty|/Applications/kitty.app|0|

c When \(u=4, v=12\) and \(a=3\).

\[
\begin{aligned}
s & =\frac{12^{2}-4^{2}}{2 \times 3} \\
& =\frac{144-16}{6} \\
& =\frac{128}{6} \\
& =21 \frac{1}{3}
\end{aligned}
\]
|727092846|kitty|/Applications/kitty.app|0|
Rearrange each of these formulas to make the pronumeral in brackets the subject.
a \(E=\frac{p^{2}}{2 m}\)
\((m)\)
b \(T=2 \pi \sqrt{\frac{\ell}{g}}\)
c \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
(u)
d \(P=\sqrt{h+c}-a\)

\((h)\)
|727092973|kitty|/Applications/kitty.app|0|
a \(E=\frac{p^{2}}{2 m}\)
\((m)\)
b \(T=2 \pi \sqrt{\frac{\ell}{g}}\)
c \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
(u)
d \(P=\sqrt{h+c}-a\)

\((h)\)
|727092983|kitty|/Applications/kitty.app|0|
b \(T=2 \pi \sqrt{\frac{\ell}{g}}\)
c \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
(u)
d \(P=\sqrt{h+c}-a\)

\((h)\)
|727093004|kitty|/Applications/kitty.app|0|
c \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
(u)
d \(P=\sqrt{h+c}-a\)

\((h)\)
|727093011|kitty|/Applications/kitty.app|0|
d \(P=\sqrt{h+c}-a\)

\((h)\)
|727093078|kitty|/Applications/kitty.app|0|
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(T=2 \pi \sqrt{\frac{\ell}{g}}\) \hfill\(\ell\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\) \hfill\((u)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
|727093108|kitty|/Applications/kitty.app|0|
a \(\quad E=\frac{p^{2}}{2 m}\)

\[
\begin{aligned}
m E & =\frac{p^{2}}{2} & & \text { (Multiply both sides of the equation by } m .) \\
m & =\frac{p^{2}}{2 E} & & \text { (Divide both sides by } E .)
\end{aligned}
\]
|727093113|kitty|/Applications/kitty.app|0|
\[
\begin{array}{rlr}
T & =2 \pi \sqrt{\frac{\ell}{g}} & \\
\frac{T}{2 \pi} & \left.=\sqrt{\frac{\ell}{g}} \quad \text { (Divide both sides of the equation by } 2 \pi .\right) \\
\frac{\ell}{g} & =\left(\frac{T}{2 \pi}\right)^{2} & \text { (Square both sides of the equation.) } \\
& =\frac{T^{2}}{4 \pi^{2}} & \\
\ell & \left.=\frac{T^{2}}{4 \pi^{2}} \times g \quad \text { (Multiply both sides of the equation by } g .\right)
\end{array}
\]

That is, \(\ell=\frac{T^{2} g}{4 \pi^{2}}\)
|727093129|kitty|/Applications/kitty.app|0|
c \(\quad \frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
\(\frac{1}{f}-\frac{1}{v}=\frac{1}{u}\)
(Subtract \(\frac{1}{v}\) from both sides.)
\(\frac{v-f}{f v}=\frac{1}{u}\)
(common denominator on LHS of equation)

\[
u=\frac{f v}{v-f}
\]

(Take reciprocals of both sides.)

Note: \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\) does not imply \(f=u+v\).
|727093145|kitty|/Applications/kitty.app|0|
\[
P=\sqrt{h+c}-a
\]

\[
\begin{array}{rlrl}
P+a & =\sqrt{h+c} \quad & \text { (Add } a \text { to both sides of the equation. }) \\
(P+a)^{2} & =h+c \quad \quad \text { (Square both sides. }) \\
h & =(P+a)^{2}-c
\end{array}
\]

The previous example shows some of the techniques that can be used to rearrange a formula.
|727093155|kitty|/Applications/kitty.app|0|
Make \(q\) the subject of the formula \(\frac{3 p}{4}-\frac{5}{q}=\frac{p^{2}}{3 q}\).
|727093187|kitty|/Applications/kitty.app|0|
\[
\begin{gathered}
\frac{3 p}{4}-\frac{5}{q}=\frac{p^{2}}{3 q} \\
12 q\left(\frac{3 p}{4}-\frac{5}{q}\right)=12 q \times \frac{p^{2}}{3 q}
\end{gathered}
\]

\(3 p \times 3 q-5 \times 12=p^{2} \times 4 \quad\) (Multiply both sides by the lowest common denominator, \(12 q\).)

\[
\begin{aligned}
9 p q-60 & =4 p^{2} \\
9 p q & =4 p^{2}+60 \\
q & =\frac{4 p^{2}+60}{9 p}
\end{aligned}
\]
|727093234|kitty|/Applications/kitty.app|0|
\end{}
|727093383|kitty|/Applications/kitty.app|0|
            \subpart
            \subpart
|727093431|kitty|/Applications/kitty.app|0|
    \begin{parts}
        \Part[1]
        \Part[1]
        \Part[2]
        \begin{subparts}
            \subpart
            \subpart
            \subpart
            \subpart
        \end{subparts}
    \end{parts}
|727094001|kitty|/Applications/kitty.app|0|
        \Part[1]
|727094004|kitty|/Applications/kitty.app|0|
1 The profit \(\$ P\) made each day by a store owner who sells \(C D\) s is given by the formula \(P=5 n-150\), where \(n\) is the number of CDs sold.

a What profit is made if the store owner sells 60 CDs?

b Make \(n\) the subject of the formula.

c How many CDs were sold if the store made:
i a profit of \(\$ 275\) ?
ii a profit of \(\$ 400\) ?
iii a loss of \(\$ 100\) ?
iv no profit?
|727094264|kitty|/Applications/kitty.app|0|

a What profit is made if the store owner sells 60 CDs?

b Make \(n\) the subject of the formula.

c How many CDs were sold if the store made:
i a profit of \(\$ 275\) ?
ii a profit of \(\$ 400\) ?
iii a loss of \(\$ 100\) ?
iv no profit?
|727094271|kitty|/Applications/kitty.app|0|

b Make \(n\) the subject of the formula.

c How many CDs were sold if the store made:
i a profit of \(\$ 275\) ?
ii a profit of \(\$ 400\) ?
iii a loss of \(\$ 100\) ?
iv no profit?
|727094277|kitty|/Applications/kitty.app|0|

c How many CDs were sold if the store made:
i a profit of \(\$ 275\) ?
ii a profit of \(\$ 400\) ?
iii a loss of \(\$ 100\) ?
iv no profit?
|727094284|kitty|/Applications/kitty.app|0|
i a profit of \(\$ 275\) ?
ii a profit of \(\$ 400\) ?
iii a loss of \(\$ 100\) ?
iv no profit?
|727094289|kitty|/Applications/kitty.app|0|
ii a profit of \(\$ 400\) ?
iii a loss of \(\$ 100\) ?
iv no profit?

i|727094297|kitty|/Applications/kitty.app|0|
iii a loss of \(\$ 100\) ?
iv no profit?

i
i|727094301|kitty|/Applications/kitty.app|0|
iv no profit?
|727094302|kitty|/Applications/kitty.app|0|
Example 82 The cost \(\$ C\) of hiring a reception room for a function is given by the formula \(C=12 n+250\), where \(n\) is the number of people attending the function.

a Rearrange the formula to make \(n\) the subject.

b How many people attended the function if the cost of hiring the reception room was:
i \(\$ 730\) ?
ii \(\$ 1090\) ?
iii \(\$ 1210\) ?
iv \(\$ 1690\) ?
|727094310|kitty|/Applications/kitty.app|0|
E|727094313|kitty|/Applications/kitty.app|0|
8|727094314|kitty|/Applications/kitty.app|0|

a Rearrange the formula to make \(n\) the subject.

b How many people attended the function if the cost of hiring the reception room was:
i \(\$ 730\) ?
ii \(\$ 1090\) ?
iii \(\$ 1210\) ?
iv \(\$ 1690\) ?
|727094318|kitty|/Applications/kitty.app|0|

b How many people attended the function if the cost of hiring the reception room was:
i \(\$ 730\) ?
ii \(\$ 1090\) ?
iii \(\$ 1210\) ?
iv \(\$ 1690\) ?
|727094322|kitty|/Applications/kitty.app|0|
i \(\$ 730\) ?
ii \(\$ 1090\) ?
iii \(\$ 1210\) ?
iv \(\$ 1690\) ?
|727094326|kitty|/Applications/kitty.app|0|
ii \(\$ 1090\) ?
iii \(\$ 1210\) ?
iv \(\$ 1690\) ?

i|727094334|kitty|/Applications/kitty.app|0|
iii \(\$ 1210\) ?
iv \(\$ 1690\) ?

i
i|727094339|kitty|/Applications/kitty.app|0|
iv \(\$ 1690\) ?
|727094340|kitty|/Applications/kitty.app|0|
v|727094344|kitty|/Applications/kitty.app|0|
4 Given the formula \(t=a+(n-1) d\) :

a rearrange the formula to make \(a\) the subject

b find the value of \(a\) when:
i \(t=11, n=4\) and \(d=3\)
ii \(t=8, n=5\) and \(d=-3\)

c rearrange the formula to make \(d\) the subject

d find the value of \(d\) when:
i \(t=48, a=3\) and \(n=16\)
ii \(t=120, a=-30\) and \(n=101\)

e rearrange the formula to make \(n\) the subject and find the value of \(n\) when \(t=150, a=5\) and \(d=5\)
|727094360|kitty|/Applications/kitty.app|0|

a rearrange the formula to make \(a\) the subject

b find the value of \(a\) when:
i \(t=11, n=4\) and \(d=3\)
ii \(t=8, n=5\) and \(d=-3\)

c rearrange the formula to make \(d\) the subject

d find the value of \(d\) when:
i \(t=48, a=3\) and \(n=16\)
ii \(t=120, a=-30\) and \(n=101\)

e rearrange the formula to make \(n\) the subject and find the value of \(n\) when \(t=150, a=5\) and \(d=5\)
|727094393|kitty|/Applications/kitty.app|0|

b find the value of \(a\) when:
i \(t=11, n=4\) and \(d=3\)
ii \(t=8, n=5\) and \(d=-3\)

c rearrange the formula to make \(d\) the subject

d find the value of \(d\) when:
i \(t=48, a=3\) and \(n=16\)
ii \(t=120, a=-30\) and \(n=101\)

e rearrange the formula to make \(n\) the subject and find the value of \(n\) when \(t=150, a=5\) and \(d=5\)
|727094400|kitty|/Applications/kitty.app|0|
i \(t=11, n=4\) and \(d=3\)
ii \(t=8, n=5\) and \(d=-3\)

c rearrange the formula to make \(d\) the subject

d find the value of \(d\) when:
i \(t=48, a=3\) and \(n=16\)
ii \(t=120, a=-30\) and \(n=101\)

e rearrange the formula to make \(n\) the subject and find the value of \(n\) when \(t=150, a=5\) and \(d=5\)
|727094407|kitty|/Applications/kitty.app|0|
ii \(t=8, n=5\) and \(d=-3\)

c rearrange the formula to make \(d\) the subject

d find the value of \(d\) when:
i \(t=48, a=3\) and \(n=16\)
ii \(t=120, a=-30\) and \(n=101\)

e rearrange the formula to make \(n\) the subject and find the value of \(n\) when \(t=150, a=5\) and \(d=5\)

i|727094414|kitty|/Applications/kitty.app|0|

c rearrange the formula to make \(d\) the subject

d find the value of \(d\) when:
i \(t=48, a=3\) and \(n=16\)
ii \(t=120, a=-30\) and \(n=101\)

e rearrange the formula to make \(n\) the subject and find the value of \(n\) when \(t=150, a=5\) and \(d=5\)
|727094417|kitty|/Applications/kitty.app|0|

d find the value of \(d\) when:
i \(t=48, a=3\) and \(n=16\)
ii \(t=120, a=-30\) and \(n=101\)

e rearrange the formula to make \(n\) the subject and find the value of \(n\) when \(t=150, a=5\) and \(d=5\)
|727094422|kitty|/Applications/kitty.app|0|
i \(t=48, a=3\) and \(n=16\)
ii \(t=120, a=-30\) and \(n=101\)

e rearrange the formula to make \(n\) the subject and find the value of \(n\) when \(t=150, a=5\) and \(d=5\)
|727094426|kitty|/Applications/kitty.app|0|
ii \(t=120, a=-30\) and \(n=101\)

e rearrange the formula to make \(n\) the subject and find the value of \(n\) when \(t=150, a=5\) and \(d=5\)

i|727094433|kitty|/Applications/kitty.app|0|

e rearrange the formula to make \(n\) the subject and find the value of \(n\) when \(t=150, a=5\) and \(d=5\)
|727094434|kitty|/Applications/kitty.app|0|

3 Given the formula \(v=u+a t\) :

a rearrange the formula to make \(u\) the subject

b find the value of \(u\) when:

i \(v=20, a=2\) and \(t=5 \quad\) ii \(v=40, a=-6\) and \(t=4\)

c rearrange the formula to make \(a\) the subject

d find the value of \(a\) when:

i \(v=20, u=15\) and \(t=2 \quad\) ii \(v=-26.8, u=-14.4\) and \(t=2\)

iii \(v=\frac{1}{2}, u=\frac{2}{3}\) and \(t=\frac{5}{6}\)

e rearrange the formula to make \(t\) the subject and find \(t\) when \(v=6, u=7\) and \(a=-3\).
|727094540|kitty|/Applications/kitty.app|0|
5 Rearrange each of these formulas to make the pronumeral in brackets the subject.
a \(y=m x+c\) (c)
b \(y=m x+c\)
d \(C=2 \pi r\)
f \(s=u t+\frac{1}{2} a t^{2}\)
h \(V=\frac{1}{3} \pi r^{2} h\)
j \(S=\frac{n}{2}(a+\ell)\)
c \(A=\frac{1}{2} b h\)
e \(P=A+2 \ell h\)
i \(s=\frac{n}{2}(a+\ell)\)
l \(E=m g h+\frac{1}{2} m v^{2}\)
\(\mathbf{k} V=\pi r^{2}+\pi r s\)
|727094580|kitty|/Applications/kitty.app|0|
a \(y=m x+c\) (c)
|727094584|kitty|/Applications/kitty.app|0|
c \(A=\frac{1}{2} b h\)
e \(P=A+2 \ell h\)
i \(s=\frac{n}{2}(a+\ell)\)
|727094596|kitty|/Applications/kitty.app|0|
\(\mathbf{k} V=\pi r^{2}+\pi r s\)
|727094631|kitty|/Applications/kitty.app|0|
b \(y=m x+c\)
|727094639|kitty|/Applications/kitty.app|0|
c \(A=\frac{1}{2} b h\)
|727094644|kitty|/Applications/kitty.app|0|
d \(C=2 \pi r\)
f \(s=u t+\frac{1}{2} a t^{2}\)
h \(V=\frac{1}{3} \pi r^{2} h\)
j \(S=\frac{n}{2}(a+\ell)\)
|727094653|kitty|/Applications/kitty.app|0|
l \(E=m g h+\frac{1}{2} m v^{2}\)
|727094682|kitty|/Applications/kitty.app|0|

\[
2
\]
|727094702|kitty|/Applications/kitty.app|0|
a \(y=m x+c\) (c)
c \(A=\frac{1}{2} b h\) for x
e \(P=A+2 \ell h\) for l
i \(s=\frac{n}{2}(a+\ell)\) for a
k \(V=\pi r^{2}+\pi r s\) for s
|727094704|kitty|/Applications/kitty.app|0|
c \(A=\frac{1}{2} b h\) for x
e \(P=A+2 \ell h\) for l
i \(s=\frac{n}{2}(a+\ell)\) for a
k \(V=\pi r^{2}+\pi r s\) for s
|727094743|kitty|/Applications/kitty.app|0|
e \(P=A+2 \ell h\) for l
i \(s=\frac{n}{2}(a+\ell)\) for a
k \(V=\pi r^{2}+\pi r s\) for s
|727094747|kitty|/Applications/kitty.app|0|
i \(s=\frac{n}{2}(a+\ell)\) for a
k \(V=\pi r^{2}+\pi r s\) for s
|727094750|kitty|/Applications/kitty.app|0|
k \(V=\pi r^{2}+\pi r s\) for s
|727094753|kitty|/Applications/kitty.app|0|
6 The formula for the sum \(S\) of the interior angles in a convex \(n\)-sided polygon is \(S=180(n-2)\). Rearrange the formula to make \(n\) the subject and use this to find the number of sides in the polygon if the sum of the interior angles is:
a \(1080^{\circ}\)
b \(1800^{\circ}\)
c \(3240^{\circ}\)
|727094873|kitty|/Applications/kitty.app|0|
a \(1080^{\circ}\)
b \(1800^{\circ}\)
c \(3240^{\circ}\)
|727094881|kitty|/Applications/kitty.app|0|
b \(1800^{\circ}\)
c \(3240^{\circ}\)
|727094884|kitty|/Applications/kitty.app|0|
c \(3240^{\circ}\)
|727094886|kitty|/Applications/kitty.app|0|
7 The kinetic energy \(E\) joules of a moving object is given by \(E=\frac{1}{2} m v^{2}\), where \(m \mathrm{~kg}\) is the mass of the object and \(v \mathrm{~m} / \mathrm{s}\) is its speed.

Rearrange the formula to make \(m\) the subject and use this to find the mass of the object when its energy and speed are, respectively:
a 400 joules, \(10 \mathrm{~m} / \mathrm{s}\)
b 28 joules, \(4 \mathrm{~m} / \mathrm{s}\)
|727094904|kitty|/Applications/kitty.app|0|
8 When an object is shot up into the air with a speed of \(u\) metres per second, its height above the ground \(h\) metres and time of flight \(t\) seconds are related (ignoring air resistance) by \(h=u t-4.9 t^{2}\).

Find the speed at which an object was fired if it reached a height of 27.5 metres after 5 seconds.
|727094921|kitty|/Applications/kitty.app|0|
b \(K=3 \ell^{2}+4 m\)
|727094935|kitty|/Applications/kitty.app|0|
b \(K=3 \ell^{2}+4 m\)
d \(d=8 \sqrt{\frac{h}{5}}\)
|727094937|kitty|/Applications/kitty.app|0|
b \(K=3 \ell^{2}+4 m\)
d \(d=8 \sqrt{\frac{h}{5}}\)
\[
\text { f } D=\frac{m}{v}
\]
|727094939|kitty|/Applications/kitty.app|0|
b \(K=3 \ell^{2}+4 m\)
d \(d=8 \sqrt{\frac{h}{5}}\)
\[
\text { f } D=\frac{m}{v}
\]
\(\mathbf{h} T=\frac{a}{4 r^{2}}\)
|727094944|kitty|/Applications/kitty.app|0|

9 Rearrange each of these formulas to make the pronumeral in brackets the subject. (All pronumerals represent positive numbers.)
a \(c=a^{2}+b^{2}\)
c \(x=\sqrt{a b}\)
|727094967|kitty|/Applications/kitty.app|0|
a \(c=a^{2}+b^{2}\)
c \(x=\sqrt{a b}\)
|727094977|kitty|/Applications/kitty.app|0|
c \(x=\sqrt{a b}\)
|727094982|kitty|/Applications/kitty.app|0|
10 Temperatures can be measured in either degrees Fahrenheit or degrees Celsius. To convert from one scale to the other, the following formula is used: \(F=\frac{9}{5} C+32\).

a Rearrange the formula to make \(C\) the subject.

b On a particular day in Melbourne, the temperature was \(28^{\circ} \mathrm{C}\). What is this temperature measured in Fahrenheit?

c In Boston, USA, the minimum overnight temperature was \(4^{\circ} \mathrm{F}\). What is this temperature measured in Celsius?

d What number represents the same temperature in \({ }^{\circ} \mathrm{C}\) and \({ }^{\circ} \mathrm{F}\) ?

e An approximate conversion formula, used frequently when converting oven temperatures, is \(F=2 C+30\). Use this to convert these temperatures:
i an oven temperature of \(180^{\circ} \mathrm{C}\)
ii an oven temperature of \(530^{\circ} \mathrm{F}\)
|727095589|kitty|/Applications/kitty.app|0|
\[
|727095650|kitty|/Applications/kitty.app|0|
\]
|727095651|kitty|/Applications/kitty.app|0|
now |727095784|kitty|/Applications/kitty.app|0|
        \part
|727095886|kitty|/Applications/kitty.app|0|
qusetinos|727095899|kitty|/Applications/kitty.app|0|
qusetions|727095901|kitty|/Applications/kitty.app|0|
com.logitech.Logi-Options
/Applications/Logi Options.app/Contents/MacOS/Logi Options
|727096023|Karabiner-EventViewer|/Applications/Karabiner-EventViewer.app|0|
        \beg
|727096363|kitty|/Applications/kitty.app|0|
        \Question[6]
        \begin{parts}
            \part
            \part
            \part
            \part
            \part
            \part
        \end{parts}
|727096377|kitty|/Applications/kitty.app|0|
Find a formula for \(n\), the number of cents in \(x\) dollars and \(y\) cents.

\section*{Solution}
In \(\$ x\) there are \(100 x\) cents.

In \(\$ x\) and \(y\) cents there are \((100 x+y)\) cents.

The formula is \(n=100 x+y\).
|727096406|kitty|/Applications/kitty.app|0|
In \(\$ x\) there are \(100 x\) cents.

In \(\$ x\) and \(y\) cents there are \((100 x+y)\) cents.

The formula is \(n=100 x+y\).
|727096415|kitty|/Applications/kitty.app|0|
Here is an isosceles triangle with equal base angles marked.

Find a formula for \(\beta\) in terms of \(\alpha\).

\begin{center}
\includegraphics[max width=\textwidth]{2023_12_09_5c1d855de2ec0db78c30g-14}
\end{center}

\section*{Solution}
\[
\begin{aligned}
\alpha+2 \beta & =180 \quad \text { (angle sum of triangle) } \\
2 \beta & =180-\alpha \\
\beta & =\frac{180-\alpha}{2}
\end{aligned}
\]
|727096428|kitty|/Applications/kitty.app|0|

\begin{center}
\includegraphics[max width=\textwidth]{2023_12_09_5c1d855de2ec0db78c30g-14}
\end{center}
|727096436|kitty|/Applications/kitty.app|0|
\[
\begin{aligned}
\alpha+2 \beta & =180 \quad \text { (angle sum of triangle) } \\
2 \beta & =180-\alpha \\
\beta & =\frac{180-\alpha}{2}
\end{aligned}
\]
|727096446|kitty|/Applications/kitty.app|0|
1 Construct a formula for:

a \(D\) in terms of \(n\), where \(D\) is the number of degrees in \(n\) right angles

b \(c\) in terms of \(D\), where \(c\) is the number of cents in \(\$ D\)

c \(m\) in terms of \(h\), where \(m\) is the number of minutes in \(h\) hours

d \(d\) in terms of \(m\), where \(d\) is the number of days in \(m\) weeks
|727096463|kitty|/Applications/kitty.app|0|

a \(D\) in terms of \(n\), where \(D\) is the number of degrees in \(n\) right angles

b \(c\) in terms of \(D\), where \(c\) is the number of cents in \(\$ D\)

c \(m\) in terms of \(h\), where \(m\) is the number of minutes in \(h\) hours

d \(d\) in terms of \(m\), where \(d\) is the number of days in \(m\) weeks
|727096469|kitty|/Applications/kitty.app|0|

b \(c\) in terms of \(D\), where \(c\) is the number of cents in \(\$ D\)

c \(m\) in terms of \(h\), where \(m\) is the number of minutes in \(h\) hours

d \(d\) in terms of \(m\), where \(d\) is the number of days in \(m\) weeks
|727096474|kitty|/Applications/kitty.app|0|

c \(m\) in terms of \(h\), where \(m\) is the number of minutes in \(h\) hours

d \(d\) in terms of \(m\), where \(d\) is the number of days in \(m\) weeks
|727096478|kitty|/Applications/kitty.app|0|

d \(d\) in terms of \(m\), where \(d\) is the number of days in \(m\) weeks
|727096481|kitty|/Applications/kitty.app|0|
2 Construct a formula for:

a the number of centimetres \(n\) in \(p\) metres

b the number of millilitres \(s\) in \(t\) litres

c the number of centimetres \(q\) in \(5 p\) metres

d the number of grams \(x\) in \(\frac{y}{2}\) kilograms
|727096497|kitty|/Applications/kitty.app|0|

a the number of centimetres \(n\) in \(p\) metres

b the number of millilitres \(s\) in \(t\) litres

c the number of centimetres \(q\) in \(5 p\) metres

d the number of grams \(x\) in \(\frac{y}{2}\) kilograms
|727096504|kitty|/Applications/kitty.app|0|

b the number of millilitres \(s\) in \(t\) litres

c the number of centimetres \(q\) in \(5 p\) metres

d the number of grams \(x\) in \(\frac{y}{2}\) kilograms
|727096508|kitty|/Applications/kitty.app|0|

c the number of centimetres \(q\) in \(5 p\) metres

d the number of grams \(x\) in \(\frac{y}{2}\) kilograms
|727096511|kitty|/Applications/kitty.app|0|

d the number of grams \(x\) in \(\frac{y}{2}\) kilograms
|727096515|kitty|/Applications/kitty.app|0|
3 Find a formula for:

a the number of cents \(z\) in \(x\) dollars and \(y\) cents

b the number of minutes \(x\) in \(y\) minutes and \(z\) seconds

c the number of hours \(x\) in \(y\) minutes and \(z\) seconds

d the cost \(\$ m\) of 1 book if 20 books cost \(\$ c\)

e the cost \(\$ n\) of 1 suit if 5 suits cost \(\$ m\)

f the cost \(\$ m\) of 1 tyre if \(x\) tyres cost \(\$ y\)

g the cost \(\$ p\) of \(n\) suits if 4 suits cost \(\$ k\)

\(\mathbf{h}\) the cost \(\$ q\) of \(x\) cars if 8 cars cost \(\$ b\)
|727096527|kitty|/Applications/kitty.app|0|
Example 13 Find a formula relating \(x\) and \(y\) for each of these statements, making \(y\) the subject.

a \(y\) is three less than \(x\).

b \(y\) is four more than the square of \(x\).

c \(y\) is eight times the square root of one-fifth of \(x\).

d \(x\) and \(y\) are supplementary angles.

e A car travelled \(80 \mathrm{~km}\) in \(x\) hours at an average speed of \(y \mathrm{~km} / \mathrm{h}\).

f A car used \(x\) litres of petrol on a trip of \(80 \mathrm{~km}\) and the fuel consumption was \(y\) litres \(/ 100 \mathrm{~km}\).

5 Find a formula relating the given pronumerals for each of these statements.

a The number of square \(\mathrm{cm} x\) in \(y\) square metres

b The selling price \(\$ S\) of an article with an original price of \(\$ m\) when a discount of \(20 \%\) is given

c The length \(c \mathrm{~cm}\) of the hypotenuse and the lengths \(a \mathrm{~cm}\) and \(b \mathrm{~cm}\) of the other two sides in a right-angled triangle

d The area \(A \mathrm{~cm}^{2}\) of a sector of a circle with a radius of length \(r \mathrm{~cm}\) and angle \(\theta\) at the centre of the circle

e The distance \(d \mathrm{~km}\) travelled by a car in \(t\) hours at an average speed of \(75 \mathrm{~km} / \mathrm{h}\)

f The number of hectares \(h\) in a rectangular paddock of length \(400 \mathrm{~m}\) and width \(w \mathrm{~m}\)
|727096545|kitty|/Applications/kitty.app|0|
5 Find a formula relating the given pronumerals for each of these statements.

a The number of square \(\mathrm{cm} x\) in \(y\) square metres

b The selling price \(\$ S\) of an article with an original price of \(\$ m\) when a discount of \(20 \%\) is given

c The length \(c \mathrm{~cm}\) of the hypotenuse and the lengths \(a \mathrm{~cm}\) and \(b \mathrm{~cm}\) of the other two sides in a right-angled triangle

d The area \(A \mathrm{~cm}^{2}\) of a sector of a circle with a radius of length \(r \mathrm{~cm}\) and angle \(\theta\) at the centre of the circle

e The distance \(d \mathrm{~km}\) travelled by a car in \(t\) hours at an average speed of \(75 \mathrm{~km} / \mathrm{h}\)

f The number of hectares \(h\) in a rectangular paddock of length \(400 \mathrm{~m}\) and width \(w \mathrm{~m}\)
|727096562|kitty|/Applications/kitty.app|0|

a The number of square \(\mathrm{cm} x\) in \(y\) square metres

b The selling price \(\$ S\) of an article with an original price of \(\$ m\) when a discount of \(20 \%\) is given

c The length \(c \mathrm{~cm}\) of the hypotenuse and the lengths \(a \mathrm{~cm}\) and \(b \mathrm{~cm}\) of the other two sides in a right-angled triangle

d The area \(A \mathrm{~cm}^{2}\) of a sector of a circle with a radius of length \(r \mathrm{~cm}\) and angle \(\theta\) at the centre of the circle

e The distance \(d \mathrm{~km}\) travelled by a car in \(t\) hours at an average speed of \(75 \mathrm{~km} / \mathrm{h}\)

f The number of hectares \(h\) in a rectangular paddock of length \(400 \mathrm{~m}\) and width \(w \mathrm{~m}\)
|727096568|kitty|/Applications/kitty.app|0|

b The selling price \(\$ S\) of an article with an original price of \(\$ m\) when a discount of \(20 \%\) is given

c The length \(c \mathrm{~cm}\) of the hypotenuse and the lengths \(a \mathrm{~cm}\) and \(b \mathrm{~cm}\) of the other two sides in a right-angled triangle

d The area \(A \mathrm{~cm}^{2}\) of a sector of a circle with a radius of length \(r \mathrm{~cm}\) and angle \(\theta\) at the centre of the circle

e The distance \(d \mathrm{~km}\) travelled by a car in \(t\) hours at an average speed of \(75 \mathrm{~km} / \mathrm{h}\)

f The number of hectares \(h\) in a rectangular paddock of length \(400 \mathrm{~m}\) and width \(w \mathrm{~m}\)
|727096573|kitty|/Applications/kitty.app|0|

c The length \(c \mathrm{~cm}\) of the hypotenuse and the lengths \(a \mathrm{~cm}\) and \(b \mathrm{~cm}\) of the other two sides in a right-angled triangle

d The area \(A \mathrm{~cm}^{2}\) of a sector of a circle with a radius of length \(r \mathrm{~cm}\) and angle \(\theta\) at the centre of the circle

e The distance \(d \mathrm{~km}\) travelled by a car in \(t\) hours at an average speed of \(75 \mathrm{~km} / \mathrm{h}\)

f The number of hectares \(h\) in a rectangular paddock of length \(400 \mathrm{~m}\) and width \(w \mathrm{~m}\)
|727096577|kitty|/Applications/kitty.app|0|

d The area \(A \mathrm{~cm}^{2}\) of a sector of a circle with a radius of length \(r \mathrm{~cm}\) and angle \(\theta\) at the centre of the circle

e The distance \(d \mathrm{~km}\) travelled by a car in \(t\) hours at an average speed of \(75 \mathrm{~km} / \mathrm{h}\)

f The number of hectares \(h\) in a rectangular paddock of length \(400 \mathrm{~m}\) and width \(w \mathrm{~m}\)
|727096581|kitty|/Applications/kitty.app|0|

e The distance \(d \mathrm{~km}\) travelled by a car in \(t\) hours at an average speed of \(75 \mathrm{~km} / \mathrm{h}\)

f The number of hectares \(h\) in a rectangular paddock of length \(400 \mathrm{~m}\) and width \(w \mathrm{~m}\)
|727096585|kitty|/Applications/kitty.app|0|

f The number of hectares \(h\) in a rectangular paddock of length \(400 \mathrm{~m}\) and width \(w \mathrm{~m}\)
|727096590|kitty|/Applications/kitty.app|0|
a \(y\) is three less than \(x\).

b \(y\) is four more than the square of \(x\).

c \(y\) is eight times the square root of one-fifth of \(x\).

d \(x\) and \(y\) are supplementary angles.

e A car travelled \(80 \mathrm{~km}\) in \(x\) hours at an average speed of \(y \mathrm{~km} / \mathrm{h}\).

f A car used \(x\) litres of petrol on a trip of \(80 \mathrm{~km}\) and the fuel consumption was \(y\) litres \(/ 100 \mathrm{~km}\).
|727096600|kitty|/Applications/kitty.app|0|

b \(y\) is four more than the square of \(x\).

c \(y\) is eight times the square root of one-fifth of \(x\).

d \(x\) and \(y\) are supplementary angles.

e A car travelled \(80 \mathrm{~km}\) in \(x\) hours at an average speed of \(y \mathrm{~km} / \mathrm{h}\).

f A car used \(x\) litres of petrol on a trip of \(80 \mathrm{~km}\) and the fuel consumption was \(y\) litres \(/ 100 \mathrm{~km}\).
            \part
|727096605|kitty|/Applications/kitty.app|0|

b \(y\) is four more than the square of \(x\).

c \(y\) is eight times the square root of one-fifth of \(x\).

d \(x\) and \(y\) are supplementary angles.

e A car travelled \(80 \mathrm{~km}\) in \(x\) hours at an average speed of \(y \mathrm{~km} / \mathrm{h}\).

f A car used \(x\) litres of petrol on a trip of \(80 \mathrm{~km}\) and the fuel consumption was \(y\) litres \(/ 100 \mathrm{~km}\).
|727096608|kitty|/Applications/kitty.app|0|

c \(y\) is eight times the square root of one-fifth of \(x\).

d \(x\) and \(y\) are supplementary angles.

e A car travelled \(80 \mathrm{~km}\) in \(x\) hours at an average speed of \(y \mathrm{~km} / \mathrm{h}\).

f A car used \(x\) litres of petrol on a trip of \(80 \mathrm{~km}\) and the fuel consumption was \(y\) litres \(/ 100 \mathrm{~km}\).
|727096612|kitty|/Applications/kitty.app|0|

d \(x\) and \(y\) are supplementary angles.

e A car travelled \(80 \mathrm{~km}\) in \(x\) hours at an average speed of \(y \mathrm{~km} / \mathrm{h}\).

f A car used \(x\) litres of petrol on a trip of \(80 \mathrm{~km}\) and the fuel consumption was \(y\) litres \(/ 100 \mathrm{~km}\).
|727096616|kitty|/Applications/kitty.app|0|

e A car travelled \(80 \mathrm{~km}\) in \(x\) hours at an average speed of \(y \mathrm{~km} / \mathrm{h}\).

f A car used \(x\) litres of petrol on a trip of \(80 \mathrm{~km}\) and the fuel consumption was \(y\) litres \(/ 100 \mathrm{~km}\).
|727096620|kitty|/Applications/kitty.app|0|

f A car used \(x\) litres of petrol on a trip of \(80 \mathrm{~km}\) and the fuel consumption was \(y\) litres \(/ 100 \mathrm{~km}\).
|727096633|kitty|/Applications/kitty.app|0|
6 In each part, find a formula from the information given.

a A hire car firm charges \(\$ 20\) per day plus 40 cents per \(\mathrm{km}\). What is the total cost \(\$ C\) for a day in which \(x \mathrm{~km}\) was travelled?

b If there are 50 litres of petrol in the tank of a car and petrol is used at the rate of 4 litres per day, what is the number of litres \(y\) that remains after \(x\) days?
c Cooking instructions for a forequarter of lamb are as follows: preheat oven to \(220^{\circ} \mathrm{C}\) and cook for \(45 \mathrm{~min}\) per kg plus an additional \(20 \mathrm{~min}\). What is the formula relating the cooking time \(T\) minutes and weight \(w \mathrm{~kg}\) ?

d In a sequence of numbers the first number is 2 , the second number is 4 , the third is 8 , the fourth is 16, etc. Assuming the doubling pattern continues, what is the formula you would use to calculate \(t\), the \(n\)th number?

e A piece of wire of length \(x \mathrm{~cm}\) is bent into a circle of area \(A \mathrm{~cm}^{2}\). What is the formula relating \(A\) and \(x\) ?
|727096700|kitty|/Applications/kitty.app|0|
7 Gareth the gardener has a large rectangular vegetable patch and he wishes to put in a path around it using concrete pavers that measure \(50 \mathrm{~cm} \times 50 \mathrm{~cm}\). The path is to be 1 paver wide. Let \(n\) be the number of pavers required. If the vegetable patch measures \(x\) metres by \(y\) metres, find a formula for \(n\) in terms of \(x\) and \(y\).
|727096717|kitty|/Applications/kitty.app|0|
\section*{Review exercise}
1 If \(s=\frac{n}{2}(2 a+(n-1) d)\) :

a find the value of \(s\) when \(n=10, a=6\) and \(d=3\)

b find the value of \(a\) when \(s=350, n=20\) and \(d=4\)

c find the value of \(d\) when \(s=460, n=10\) and \(a=10\)

2 The formula for the geometric mean \(m\) of two positive numbers \(a\) and \(b\) is \(m=\sqrt{a b}\).

a Find \(m\) if \(a=16\) and \(b=25\).

b Find \(a\) if \(m=7\) and \(b=16\).

3 If \(x=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\) :

a find \(x\) if \(b=4, a=1\) and \(c=-24\)

b find \(c\) if \(a=1, x=6\) and \(b=2\)

4 A pillar is in the shape of a cylinder with a hemispherical top. If \(r\) metres is the radius of the base and \(h\) metres is the total height, the volume \(V\) cubic metres is given by the formula \(V=\frac{1}{3} \pi r^{2}(3 h-r)\).

a Rearrange the formula to make \(h\) the subject.

b Find the height of the pillar, correct to the nearest centimetre, if the radius of the pillar is \(0.5 \mathrm{~m}\) and the volume is \(10 \mathrm{~m}^{3}\).

5 Rearrange each of these formulas to make the pronumeral in brackets the subject. (All of the pronumerals represent positive numbers.)
a \(A=\ell \times w\)
b \(C=2 \pi r\)
c \(V=\pi r^{2} h\)
d \(A=2 h(\ell+b)\)
e \(A=4 \pi r^{2}\)
f \(w=10 \sqrt{\frac{x}{a}}\)
g \(w=\sqrt{\frac{3 V}{\pi h}}\)
\(\mathbf{h} \frac{1}{x}+\frac{1}{y}=\frac{2}{z}\)

6 If a stone is dropped off a cliff, the number of metres it has fallen after a certain number of seconds is found by multiplying the square of the number of seconds by 4.9.

a Find the formula for the distance \(d\) metres fallen by the stone in \(t\) seconds.

b Find the distance fallen in 1.5 seconds.

7 If \(t=\sqrt{\frac{M}{M-m}}\) :

a express the formula with \(m\) as the subject

b express the formula with \(M\) as the subject

c find the value of \(M\) if \(m=3\) and \(t=\sqrt{2}\).

8 The total surface area \(S \mathrm{~cm}^{2}\) of a cylinder is given in terms of its radius \(r \mathrm{~cm}\) and height \(h \mathrm{~cm}\) by the formula \(S=2 \pi r(r+h)\).

a Express this formula with \(h\) as the subject.

b What is the height of such a cylinder if the radius is \(7 \mathrm{~cm}\) and the total surface area is \(500 \mathrm{~cm}^{2}\) ? Calculate your answer in centimetres, correct to 2 decimal places.

9 The sum \(S\) of the squares of the first \(n\) whole numbers is given by the formula \(S=\frac{n(n+1)(2 n+1)}{6}\). Find the sum of the squares of:

a the first 20 whole numbers

b all the numbers from 5 to 21 inclusive

10 a For the formula \(T=2 \pi \sqrt{\frac{W}{g F}}\), make \(F\) the subject.

b For the formula \(P=\frac{\pi r x}{180}+2 r\), make \(x\) the subject.

c For the formula \(D=\sqrt{\frac{f+x}{f-x}}\), make \(x\) the subject.

11 Cans in a supermarket are displayed in a triangular stack with one can at the top, two cans in the second row from the top, three cans in the third row from the top, and so on. What is the number of cans in the display if the number of rows is:
a 4 ?
b 5 ?
c \(n\) ?
d 35 ?

12 Rearrange each of these formulas to make the pronumeral in brackets the subject. Assume all pronumerals take non-negative values.
a \(M=\sqrt{\frac{a b}{t}+a^{2}}\)
b \(V=\frac{1}{3} a^{2} \sqrt{\ell^{2}-\frac{a^{2}}{2}}\)
c \(E=\frac{w^{2} a}{\left(w^{2}+m\right) b^{3}}\)
d \(T=2 \pi \sqrt{\frac{\ell+r}{g}}\)
e \(x=a \sqrt{\frac{k}{p h+y}}\)
f \(v^{2}=u^{2}+2 a s\)

13 The volume \(V \mathrm{~cm}^{3}\) of metal in a tube is given by the formula \(V=\pi \ell\left(r^{2}-(r-t)^{2}\right)\) where \(\ell \mathrm{cm}\) is the length, \(r \mathrm{~cm}\) is the radius of the outside surface and \(t \mathrm{~cm}\) the thickness of the material.
a Find \(V\) if \(\ell=50, r=5\) and \(t=0.25\).
b Make \(r\) the subject of the formula.

14 For the formula \(I=\frac{180 n-360}{n}\) :

a find \(I\) when \(n=6\)

b make \(n\) the subject of the formula and find \(n\) when \(I=108\)

15 Find the formula connecting \(x\) and \(y\) for each of these statements making \(y\) the subject.

a \(y\) is four more than twice the square of \(x\).

b \(x\) and \(y\) are complementary angles.

c A car travelled \(x \mathrm{~km}\) in \(y\) hours at a speed of \(100 \mathrm{~km} / \mathrm{h}\).

d A car travelled \(100 \mathrm{~km}\) in \(y\) hours at a speed of \(x \mathrm{~km} / \mathrm{h}\).

\section*{Challenge exercise}
1 Rearrange each of these formulas to make the pronumeral in brackets the subject. (All of the pronumerals represent positive numbers.)
a \(P=\frac{M+m}{M-m}\)
c \(\frac{1}{S}=\frac{1}{R}+\frac{1}{T}\)
e \(T=t-\frac{h}{\ell}\)
b \(F=\frac{M P}{M+\ell}\)
d \(E=\frac{R}{i}+P\)
f \(I=\frac{M}{4}\left(R^{2}+\frac{L^{2}}{3}\right)\)
g \(L=\ell \sqrt{1+\frac{v^{2}}{c^{2}}}\)
i \(v=c\left(\frac{1}{r}-\frac{1}{s}\right)\)
k \(P=p \sqrt{1+\frac{1}{\ell}}\)
\(\mathbf{m}(a+b)^{2}+c^{2}=(a-d)^{2}\)
h \(T=2 \pi\left(\frac{1+e}{1-e}\right)\)
j \(P=2 \pi \sqrt{\frac{h+k}{g}}\)
l \(A=\pi\left(R^{2}-r^{2}\right)\)
n \((M-m)^{2}+p^{2}=M^{2}\)

2 Sophie is playing with a box of connecting rods which can be joined together to produce a rectangular grid. The grid shown opposite is of size \(4 \times 2\) (that is, length \(=4\), width \(=2\) ) and uses 22 rods.

\begin{center}
\includegraphics[max width=\textwidth]{2023_12_09_5c1d855de2ec0db78c30g-19(1)}
\end{center}

a How many rods are needed to make these grids?
i \(1 \times 2\)
ii \(3 \times 2\)
iii \(10 \times 2\)
iv \(n \times 2\)

b How many rods are needed to make these grids?
i \(1 \times 3\)
ii \(2 \times 3\)
iii \(10 \times 3\)
iv \(n \times 3\)

c How many rods are needed to make a grid of size \(n \times m\) ?

d In the \(4 \times 2\) grid shown above, how many rods have been placed:
i vertically?
ii horizontally?

e In an \(n \times m\) grid, how many rods have been placed:
i vertically?
ii horizontally?
iii in total?

f Does your answer to part \(\mathbf{e}\) iii agree with your answer to part \(\mathbf{c}\) ?

g Sophie makes a \(4 \times 7\) grid. How many rods did she use?

h If she used 97 rods to make a grid of width 6 , what was the length of the grid?

i If she has 100 rods, can she make a grid that uses all 100 rods? If so, what size grid can she make?

j If the rods can be joined to make a three-dimensional grid, find the formula for the number of rods required to make a grid as shown of size \(m \times n \times p\) (that is, length \(=m\),

\includegraphics[max width=\textwidth, center]{2023_12_09_5c1d855de2ec0db78c30g-19(2)}
width \(=n\), height \(=p\) ). The diagram shows a rectangular prism measuring 3 rods by 2 rods by 1 rod.

3 A builder wishes to place a circular cap of a given height above an existing window. To do this he needs to know the location of the centre of the circle (the cap is not necessarily a semicircle) and the radius of the circle. \(O\) is the centre of the required circle, the radius of the required circle is \(r \mathrm{~cm}\), the width of the window is \(2 d \mathrm{~cm}\) and the height of the circular cap is \(h \mathrm{~cm}\).

\includegraphics[max width=\textwidth, center]{2023_12_09_5c1d855de2ec0db78c30g-19}
a Express each of these in terms of \(r, d\) and \(h\).
i \(A B\)
ii \(O A\)

b Show that \(r=\frac{h^{2}+d^{2}}{2 h}\).

c If the window is \(120 \mathrm{~cm}\) wide and the cap is \(40 \mathrm{~cm}\) high, find:

i the radius of the circle

ii how far below the top of the window the centre of the circle must be placed

d If the builder used a circle of radius \(50 \mathrm{~cm}\) and this produced a cap of height \(20 \mathrm{~cm}\), what was the width of the window?

4 A group of \(n\) people attend a club meeting. Before the meeting begins, they all shake hands with each other. Write a formula to find \(H\), the number of handshakes exchanged.

5 A cyclic quadrilateral has all its vertices on a circle. Its area \(A\) is given by Brahmagupta's formula

\[
A^{2}=(s-a)(s-b)(s-c)(s-d)
\]

where \(a, b, c\) and \(d\) are the side lengths of the quadrilateral and

\(s=\frac{a+b+c+d}{2}\) is the 'semi-perimeter'. Find the exact area of a cyclic quadrilateral with side lengths:
a \(4,5,6,7\)
b \(7,4,4,3\)
c \(8,9,10,13\)
d \(39,52,25,60\)
e \(51,40,68,75\)

6 a The population of a town decreases by \(2 \%\) each year. The population was initially \(P\), and is \(Q\) after \(n\) years. What is the formula relating \(Q, P\) and \(n\) ?

b The population of a town decreases by \(5 \%\) each year. The percentage decrease over a period of \(n\) years is \(a \%\). What is the formula relating \(a\) and \(n\) ?


\end{document}
|727096737|kitty|/Applications/kitty.app|0|
b \(C=2 \pi r\)
|727096758|kitty|/Applications/kitty.app|0|
d \(A=2 h(\ell+b)\)
e \(A=4 \pi r^{2}\)
f \(w=10 \sqrt{\frac{x}{a}}\)
g \(w=\sqrt{\frac{3 V}{\pi h}}\)
|727096763|kitty|/Applications/kitty.app|0|
a For the formula \(T=2 \pi \sqrt{\frac{W}{g F}}\), make \(F\) the subject.|727096777|kitty|/Applications/kitty.app|0|
b For the formula \(P=\frac{\pi r x}{180}+2 r\), make \(x\) the subject.
|727096778|kitty|/Applications/kitty.app|0|
a 4 ?
b 5 ?
|727096785|kitty|/Applications/kitty.app|0|
a \(M=\sqrt{\frac{a b}{t}+a^{2}}\)
b \(V=\frac{1}{3} a^{2} \sqrt{\ell^{2}-\frac{a^{2}}{2}}\)
c \(E=\frac{w^{2} a}{\left(w^{2}+m\right) b^{3}}\)
d \(T=2 \pi \sqrt{\frac{\ell+r}{g}}\)
e \(x=a \sqrt{\frac{k}{p h+y}}\)
|727096794|kitty|/Applications/kitty.app|0|
13 The volume \(V \mathrm{~cm}^{3}\) of metal in a tube is given by the formula \(V=\pi \ell\left(r^{2}-(r-t)^{2}\right)\) where \(\ell \mathrm{cm}\) is the length, \(r \mathrm{~cm}\) is the radius of the outside surface and \(t \mathrm{~cm}\) the thickness of the material.
a Find \(V\) if \(\ell=50, r=5\) and \(t=0.25\).
b Make \(r\) the subject of the formula.

|727096823|kitty|/Applications/kitty.app|0|
1 Rearrange each of these formulas to make the pronumeral in brackets the subject. (All of the pronumerals represent positive numbers.)
a \(P=\frac{M+m}{M-m}\)
c \(\frac{1}{S}=\frac{1}{R}+\frac{1}{T}\)
e \(T=t-\frac{h}{\ell}\)
b \(F=\frac{M P}{M+\ell}\)
d \(E=\frac{R}{i}+P\)
f \(I=\frac{M}{4}\left(R^{2}+\frac{L^{2}}{3}\right)\)
g \(L=\ell \sqrt{1+\frac{v^{2}}{c^{2}}}\)
i \(v=c\left(\frac{1}{r}-\frac{1}{s}\right)\)
k \(P=p \sqrt{1+\frac{1}{\ell}}\)
\(\mathbf{m}(a+b)^{2}+c^{2}=(a-d)^{2}\)
h \(T=2 \pi\left(\frac{1+e}{1-e}\right)\)
j \(P=2 \pi \sqrt{\frac{h+k}{g}}\)
l \(A=\pi\left(R^{2}-r^{2}\right)\)
n \((M-m)^{2}+p^{2}=M^{2}\)

2 Sophie is playing with a box of connecting rods which can be joined together to produce a rectangular grid. The grid shown opposite is of size \(4 \times 2\) (that is, length \(=4\), width \(=2\) ) and uses 22 rods.

\begin{center}
\includegraphics[max width=\textwidth]{2023_12_09_5c1d855de2ec0db78c30g-19(1)}
\end{center}

a How many rods are needed to make these grids?
i \(1 \times 2\)
ii \(3 \times 2\)
iii \(10 \times 2\)
iv \(n \times 2\)

b How many rods are needed to make these grids?
i \(1 \times 3\)
ii \(2 \times 3\)
iii \(10 \times 3\)
iv \(n \times 3\)

c How many rods are needed to make a grid of size \(n \times m\) ?

d In the \(4 \times 2\) grid shown above, how many rods have been placed:
i vertically?
ii horizontally?

e In an \(n \times m\) grid, how many rods have been placed:
i vertically?
ii horizontally?
iii in total?

f Does your answer to part \(\mathbf{e}\) iii agree with your answer to part \(\mathbf{c}\) ?

g Sophie makes a \(4 \times 7\) grid. How many rods did she use?

h If she used 97 rods to make a grid of width 6 , what was the length of the grid?

i If she has 100 rods, can she make a grid that uses all 100 rods? If so, what size grid can she make?

j If the rods can be joined to make a three-dimensional grid, find the formula for the number of rods required to make a grid as shown of size \(m \times n \times p\) (that is, length \(=m\),

\includegraphics[max width=\textwidth, center]{2023_12_09_5c1d855de2ec0db78c30g-19(2)}
width \(=n\), height \(=p\) ). The diagram shows a rectangular prism measuring 3 rods by 2 rods by 1 rod.

|727096841|kitty|/Applications/kitty.app|0|
5 A cyclic quadrilateral has all its vertices on a circle. Its area \(A\) is given by Brahmagupta's formula

\[
A^{2}=(s-a)(s-b)(s-c)(s-d)
\]

where \(a, b, c\) and \(d\) are the side lengths of the quadrilateral and

\(s=\frac{a+b+c+d}{2}\) is the 'semi-perimeter'. Find the exact area of a cyclic quadrilateral with side lengths:
a \(4,5,6,7\)
b \(7,4,4,3\)
c \(8,9,10,13\)
d \(39,52,25,60\)
e \(51,40,68,75\)
|727096874|kitty|/Applications/kitty.app|0|

6 a The population of a town decreases by \(2 \%\) each year. The population was initially \(P\), and is \(Q\) after \(n\) years. What is the formula relating \(Q, P\) and \(n\) ?

b The population of a town decreases by \(5 \%\) each year. The percentage decrease over a period of \(n\) years is \(a \%\). What is the formula relating \(a\) and \(n\) ?


\end{document}
|727096898|kitty|/Applications/kitty.app|0|
,|727096922|kitty|/Applications/kitty.app|0|
2023_12_09_5c1d855de2ec0db78c30g-19|727096931|kitty|/Applications/kitty.app|0|
    \Question[]
|727098672|kitty|/Applications/kitty.app|0|
iii \(v=\frac{1}{2}, u=\frac{2}{3}\) and \(t=\frac{5}{6}\)

i
i|727099062|kitty|/Applications/kitty.app|0|
\(K=3 \ell^{2}+4 m\hfill{}(\ell)\)
\(d=8 \sqrt{\frac{h}{5}}\hfill{}(h)\)
\(D=\frac{m}{v}\hfill{}(v)\)
\(T=\frac{a}{4 r^{2}}\hfill{}(r)\)
|727099205|kitty|/Applications/kitty.app|0|
    \setcounter{question}{\ContinuedQuestion}
|727099374|kitty|/Applications/kitty.app|0|
s|727099435|kitty|/Applications/kitty.app|0|
    \Question[10] \textbf{Challenge:} A cyclic quadrilateral has all its vertices on a circle. Its area \(A\) is given by Brahmagupta's formula \[ A^{2}=(s-a)(s-b)(s-c)(s-d) \] where \(a, b, c\) and \(d\) are the side lengths of the quadrilateral and \(s=\frac{a+b+c+d}{2}\) is the 'semi-perimeter'. Find the exact area of a cyclic quadrilateral with side lengths:
    \begin{parts}
        \part \(4,5,6,7\)
        \part \(7,4,4,3\)
        \part \(8,9,10,13\)
        \part \(39,52,25,60\)
        \part \(51,40,68,75\)
    \end{parts}
|727099535|kitty|/Applications/kitty.app|0|
\textbf{Challenge:} |727099566|kitty|/Applications/kitty.app|0|
5|727099600|kitty|/Applications/kitty.app|0|
\begin{}
|727145726|kitty|/Applications/kitty.app|0|
use 'lervag/vimtex'|727146899|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
local M = {}

function M.getCompletionItems(prefix)
  -- define your total completion items
  local items = vim.api.nvim_call_function('vimtex#complete#omnifunc',{0, prefix})
  return items
end

M.complete_item = {
  item = M.getCompletionItems
}

return M|727148673|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
lua require'completion'.addCompletionSource('vimtex', require'vimtex'.complete_item)|727148684|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
[quote="zeertzjq, post:2, topic:2013, full:true"]
I use vimtex for compiling and TexLab LSP for completion, linting, and formatting. (TexLab supports linting using chktex and formatting using latexindent.) There are some other plugins like texmagic.nvim for compiling, but I am too lazy to switch from vimtex. vimtex also has some other useful features like more concealing items and better syntax highlighting. vim-surround is also really helpful when inserting an environment.
[/quote]

|727148859|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Image: 1366x1024 (16.0 MB)|727154507|||1|7e15dc987cc58ff7b4c61fce3fa07e0785ff8242.tiff
local M = {}

function M.getCompletionItems(prefix)
  -- define your total completion items
  local items = vim.api.nvim_call_function('vimtex#complete#omnifunc',{0, prefix})
  return items
end

M.complete_item = {
  item = M.getCompletionItems
}

return M

|727191972|kitty|/Applications/kitty.app|0|
function ()
  
end
|727193029|kitty|/Applications/kitty.app|0|
Introduction|727193157|kitty|/Applications/kitty.app|0|
Substitution into formulas|727193954|kitty|/Applications/kitty.app|0|
2-subs|727193964|kitty|/Applications/kitty.app|0|
Changing the subject of a formula|727193976|kitty|/Applications/kitty.app|0|
Constructing Formulas|727193990|kitty|/Applications/kitty.app|0|
4-constr|727194003|kitty|/Applications/kitty.app|0|
\section{Scientific N}
    \setcounter{secmarks}{0}
    \input{4-fra-ind}
    \setcounter{sec4marks}{\thesecmarks}
|727194104|kitty|/Applications/kitty.app|0|
\section{Fractional Indices}
    \setcounter{secmarks}{0}
    \input{4-fra-ind}
    \setcounter{sec4marks}{\thesecmarks}
|727194106|kitty|/Applications/kitty.app|0|
x|727194138|kitty|/Applications/kitty.app|0|
-|727194138|kitty|/Applications/kitty.app|0|
\docu
|727194281|kitty|/Applications/kitty.app|0|
require'lspconfig'.pyright.setup{}|727194363|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
pyright|727194390|kitty|/Applications/kitty.app|0|
\document
|727194413|kitty|/Applications/kitty.app|0|
let g:vimtex_compiler_latexmk_engines = {'pdf': ['lualatex', 'latexmk']}
|727195117|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
%! TEX program = lualatex|727195127|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
%! TEX program = lualatex
|727195149|kitty|/Applications/kitty.app|0|
let g:vimtex_view_method = 'zathura'
|727195263|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
zathura|727195289|kitty|/Applications/kitty.app|0|
let g:vimtex_view_method = 'custom'
let g:vimtex_view_general_viewer = 'brave'
let g:vimtex_view_general_options = '-a'|727195429|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
let g:vimtex_view_method = 'custom'
let g:vimtex_view_general_viewer = 'brave'
let g:vimtex_view_general_options = '-a'
|727195607|kitty|/Applications/kitty.app|0|
vim.cmd "let g:vimtex_view_method = 'custom'"
|727196104|kitty|/Applications/kitty.app|0|
th
|727196169|kitty|/Applications/kitty.app|0|
t
|727196175|kitty|/Applications/kitty.app|0|
"|727196613|kitty|/Applications/kitty.app|0|
0|727198384|kitty|/Applications/kitty.app|0|
formula|727198410|kitty|/Applications/kitty.app|0|
subject|727198415|kitty|/Applications/kitty.app|0|
pronumeral|727198420|kitty|/Applications/kitty.app|0|
substitution|727198424|kitty|/Applications/kitty.app|0|
expression|727198427|kitty|/Applications/kitty.app|0|
construction|727198437|kitty|/Applications/kitty.app|0|
equation|727198445|kitty|/Applications/kitty.app|0|
rearrange|727198456|kitty|/Applications/kitty.app|0|
a formula?|727198496|kitty|/Applications/kitty.app|0|
        A formula is an expression that which relates different quantities.
|727198507|kitty|/Applications/kitty.app|0|
What is the most famous formula that you know?|727198522|kitty|/Applications/kitty.app|0|
        \[E = mc^2\]
|727198544|kitty|/Applications/kitty.app|0|
    \Question[4] How do you interpret this?
    \begin{solutionordottedlines}[2cm]
        $2^4 = 2\times 2\times 2\times 2 = 16$
    \end{solutionordottedlines}
|727198562|kitty|/Applications/kitty.app|0|
How do you interpret this?|727198567|kitty|/Applications/kitty.app|0|
solutionordottedlines|727198597|kitty|/Applications/kitty.app|0|
    \begin{parts}
        \part Can you derive this formula?
        \begin{solutionorbox}[1in]
        \end{solutionorbox}
        \part What is the subject of your formula?
        \begin{solutionordottedlines}[1cm]
        \end{solutionordottedlines}
        \part What do all of the variables stand for?
        \begin{solutionordottedlines}[3cm]
        \end{solutionordottedlines}
    \end{parts}
|727198602|kitty|/Applications/kitty.app|0|
    \subsubsection*{1}
|727198795|kitty|/Applications/kitty.app|0|
thebibliography|727198820|kitty|/Applications/kitty.app|0|
            \part 
|727198867|kitty|/Applications/kitty.app|0|
        \Questions[]
|727198965|kitty|/Applications/kitty.app|0|
\begin{boxdef}
    \textbf{Base:}
    \begin{solutionordottedlines}[1cm]
    \end{solutionorbox}
\end{boxdef}
|727199666|kitty|/Applications/kitty.app|0|
Base:|727199671|kitty|/Applications/kitty.app|0|
  \item The number 2 in \(2^{4}\) is called the base.
  \item The number 4 in \(2^{4}\) is called the index or exponent.
|727199680|kitty|/Applications/kitty.app|0|
  \item The number 4 in \(2^{4}\) is called the index or exponent.
|727199690|kitty|/Applications/kitty.app|0|
\section*{Example 1}
Express as a power or as a product of powers.
a \(5 \times 5 \times 5\)
b \(3 \times 3 \times 7 \times 7 \times 7 \times 7\)

Solution

a \(5 \times 5 \times 5=5^{3} \quad\) b \(3 \times 3 \times 7 \times 7 \times 7 \times 7=3^{2} \times 7^{4}\)

\section*{Example 2}
Express each number as a power of a prime.
a 81
b 128

\section*{Solution}
a \(\quad 81=3 \times 3 \times 3 \times 3\)
b \(128=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\)
\(=3^{4}\)
\(=2^{7}\)
|727199715|kitty|/Applications/kitty.app|0|
 \section*{Example 1}|727199742|kitty|/Applications/kitty.app|0|
Express as a power or as a product of powers.|727199745|kitty|/Applications/kitty.app|0|

Solution

|727199768|kitty|/Applications/kitty.app|0|
\(5 \times 5 \times 5=5^{3} \quad\) 
|727199779|kitty|/Applications/kitty.app|0|

\section*{Example 2}
Express each number as a power of a prime.
a 81
b 128

\section*{Solution}
a \(\quad 81=3 \times 3 \times 3 \times 3\)
b \(128=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\)
\(=3^{4}\)
\(=2^{7}\)
|727199833|kitty|/Applications/kitty.app|0|
\section*{Example 2}|727199838|kitty|/Applications/kitty.app|0|
a 81
b 128

\section*{Solution}
a \(\quad 81=3 \times 3 \times 3 \times 3\)
b \(128=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\)
\(=3^{4}\)
\(=2^{7}\)
|727199846|kitty|/Applications/kitty.app|0|
b 128

|727199855|kitty|/Applications/kitty.app|0|
a \(\quad 81=3 \times 3 \times 3 \times 3\)
|727199868|kitty|/Applications/kitty.app|0|
b \(128=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\)
\(=3^{4}\)
\(=2^{7}\)
|727199878|kitty|/Applications/kitty.app|0|
solutionorbox|727199898|kitty|/Applications/kitty.app|0|
\section*{Index laws}
\section*{Index law 1}
To multiply powers of the same base, add the indices.

\[
a^{m} a^{n}=a^{m+n}
\]

\section*{Index law 2}
To divide powers of the same base, subtract the indices.

\[
\frac{a^{m}}{a^{n}}=a^{m-n} \quad \text { where } m>n \text { and } a \neq 0
\]

\section*{Index law 3}
To raise a power to a power, multiply the indices.

\[
\left(a^{m}\right)^{n}=a^{m n}
\]

\section*{Index law 4}
A power of a product is the product of the powers.

\((a b)^{m}=a^{m} b^{m}\)

\section*{Index law 5}
A power of a quotient is the quotient of the powers.

\[
\left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}} \quad \text { where } b \neq 0
\]
|727199932|kitty|/Applications/kitty.app|0|
\section*{Index laws}
\section*{Index law 1}
|727199936|kitty|/Applications/kitty.app|0|
\section*{Index law 2}
|727199943|kitty|/Applications/kitty.app|0|
To divide powers of the same base, subtract the indices.

\[
\frac{a^{m}}{a^{n}}=a^{m-n} \quad \text { where } m>n \text { and } a \neq 0
\]

\section*{Index law 3}
To raise a power to a power, multiply the indices.

\[
\left(a^{m}\right)^{n}=a^{m n}
\]

\section*{Index law 4}
A power of a product is the product of the powers.

\((a b)^{m}=a^{m} b^{m}\)

\section*{Index law 5}
A power of a quotient is the quotient of the powers.

\[
\left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}} \quad \text { where } b \neq 0
\]
|727199947|kitty|/Applications/kitty.app|0|

\section*{Index law 3}
To raise a power to a power, multiply the indices.

\[
\left(a^{m}\right)^{n}=a^{m n}
\]

\section*{Index law 4}
A power of a product is the product of the powers.

\((a b)^{m}=a^{m} b^{m}\)

\section*{Index law 5}
A power of a quotient is the quotient of the powers.

\[
\left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}} \quad \text { where } b \neq 0
\]
|727199956|kitty|/Applications/kitty.app|0|
        \section*{Index law 3}
|727199961|kitty|/Applications/kitty.app|0|

\section*{Index law 4}
A power of a product is the product of the powers.

\((a b)^{m}=a^{m} b^{m}\)

\section*{Index law 5}
A power of a quotient is the quotient of the powers.

\[
\left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}} \quad \text { where } b \neq 0
\]
|727199968|kitty|/Applications/kitty.app|0|
        \section*{Index law 4}
|727199973|kitty|/Applications/kitty.app|0|
\section*{Index law 5}
A power of a quotient is the quotient of the powers.

\[
\left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}} \quad \text { where } b \neq 0
\]
|727199978|kitty|/Applications/kitty.app|0|
\section*{Index law 5}
|727199980|kitty|/Applications/kitty.app|0|
Simplify, expressing the answer in index form.
a \(3^{2} \times 3^{4}\)
b \(a^{3} \times a^{5}\)
c \(3 x^{2} \times x^{3}\)
d \(2 a^{2} b^{3} \times 5 a b^{2}\)
|727200001|kitty|/Applications/kitty.app|0|
a \(3^{2} \times 3^{4}\)
b \(a^{3} \times a^{5}\)
c \(3 x^{2} \times x^{3}\)
d \(2 a^{2} b^{3} \times 5 a b^{2}\)
|727200007|kitty|/Applications/kitty.app|0|
b \(a^{3} \times a^{5}\)
c \(3 x^{2} \times x^{3}\)
d \(2 a^{2} b^{3} \times 5 a b^{2}\)
|727200010|kitty|/Applications/kitty.app|0|
c \(3 x^{2} \times x^{3}\)
d \(2 a^{2} b^{3} \times 5 a b^{2}\)
|727200013|kitty|/Applications/kitty.app|0|
d \(2 a^{2} b^{3} \times 5 a b^{2}\)
|727200015|kitty|/Applications/kitty.app|0|
a \(3^{2} \times 3^{4}=3^{6}\)
b \(a^{3} \times a^{5}=a^{8}\)
c \(3 x^{2} \times x^{3}=3 x^{5}\)
d \(2 a^{2} b^{3} \times 5 a b^{2}=10 \times a^{2+1} \times b^{3+2}\)

\[
=10 a^{3} b^{5}
\]
|727200029|kitty|/Applications/kitty.app|0|
a \(\frac{3^{5}}{3^{2}}\)
b \(\frac{9^{5}}{9^{4}}\)
c \(10^{6} \div 10^{4}\)
d \(a^{7} \div a^{4}\)
e \(\frac{3 y^{4}}{y}\)
f \(\frac{6 x^{5}}{2 x^{3}}\)
|727200039|kitty|/Applications/kitty.app|0|
b \(\frac{9^{5}}{9^{4}}\)
c \(10^{6} \div 10^{4}\)
d \(a^{7} \div a^{4}\)
e \(\frac{3 y^{4}}{y}\)
f \(\frac{6 x^{5}}{2 x^{3}}\)
|727200043|kitty|/Applications/kitty.app|0|
c \(10^{6} \div 10^{4}\)
d \(a^{7} \div a^{4}\)
e \(\frac{3 y^{4}}{y}\)
f \(\frac{6 x^{5}}{2 x^{3}}\)
|727200045|kitty|/Applications/kitty.app|0|
d \(a^{7} \div a^{4}\)
e \(\frac{3 y^{4}}{y}\)
f \(\frac{6 x^{5}}{2 x^{3}}\)
|727200047|kitty|/Applications/kitty.app|0|
e \(\frac{3 y^{4}}{y}\)
f \(\frac{6 x^{5}}{2 x^{3}}\)
|727200048|kitty|/Applications/kitty.app|0|
f \(\frac{6 x^{5}}{2 x^{3}}\)
|727200050|kitty|/Applications/kitty.app|0|
a \(\frac{3^{5}}{3^{2}}=3^{5-2}\)
b \(\frac{9^{5}}{9^{4}}=9^{1}\)
c \(10^{6} \div 10^{4}=10^{6-4}\)
\(=3^{3}\)
\(=9\)
\(=10^{2}\)
\(=27\)
\(=100\)
d \(\begin{aligned} a^{7} \div a^{4} & =a^{7-4} \\ & =a^{3}\end{aligned}\)
e \(\frac{3 y^{4}}{y}=3 \times \frac{y^{4}}{y}\)
f \(\frac{6 x^{5}}{2 x^{3}}=\frac{6}{2} \times \frac{x^{5}}{x^{3}}\)
\(=3 \times y^{4-1}\)
\(=3 \times x^{5-3}\)
\(=3 y^{3}\)
\(=3 x^{2}\)
|727200069|kitty|/Applications/kitty.app|0|
                \begin{solutionordottedlines}[2cm]
                \end{solutionordottedlines}
|727200078|kitty|/Applications/kitty.app|0|
f \(\frac{6 x^{5}}{2 x^{3}}=\frac{6}{2} \times \frac{x^{5}}{x^{3}}\)
\(=3 \times y^{4-1}\)
\(=3 \times x^{5-3}\)
\(=3 y^{3}\)
\(=3 x^{2}\)
|727200091|kitty|/Applications/kitty.app|0|
e \(\frac{3 y^{4}}{y}=3 \times \frac{y^{4}}{y}\)
|727200100|kitty|/Applications/kitty.app|0|
d \(\begin{aligned} a^{7} \div a^{4} & =a^{7-4} \\ & =a^{3}\end{aligned}\)
|727200110|kitty|/Applications/kitty.app|0|
c \(10^{6} \div 10^{4}=10^{6-4}\)
\(=3^{3}\)
\(=9\)
\(=10^{2}\)
\(=27\)
\(=100\)
|727200122|kitty|/Applications/kitty.app|0|
b \(\frac{9^{5}}{9^{4}}=9^{1}\)
|727200136|kitty|/Applications/kitty.app|0|
a \(\frac{3^{5}}{3^{2}}=3^{5-2}\)
|727200145|kitty|/Applications/kitty.app|0|
a \(3^{2} \times 3^{4}=3^{6}\)
|727200160|kitty|/Applications/kitty.app|0|
b \(a^{3} \times a^{5}=a^{8}\)
|727200170|kitty|/Applications/kitty.app|0|
c \(3 x^{2} \times x^{3}=3 x^{5}\)
|727200173|kitty|/Applications/kitty.app|0|
d \(2 a^{2} b^{3} \times 5 a b^{2}=10 \times a^{2+1} \times b^{3+2}\)

\[
=10 a^{3} b^{5}
\]
|727200178|kitty|/Applications/kitty.app|0|
a \(\frac{3 x^{3} y^{2}}{4 x y} \times \frac{6 x^{2} y^{3}}{x^{3} y^{2}}\)
b \(\frac{8 a^{2} b^{3}}{3 a^{3} b} \div \frac{4 a b^{2}}{9 a^{3} b^{5}}\)
|727200204|kitty|/Applications/kitty.app|0|
b \(\frac{8 a^{2} b^{3}}{3 a^{3} b} \div \frac{4 a b^{2}}{9 a^{3} b^{5}}\)
|727200218|kitty|/Applications/kitty.app|0|
\(=\frac{9 x y^{2}}{2}\)
|727200247|kitty|/Applications/kitty.app|0|
a \(\frac{3 x^{3} y^{2}}{4 x y} \times \frac{6 x^{2} y^{3}}{x^{3} y^{2}}=\frac{18 x^{5} y^{5}}{4 x^{4} y^{3}}\)
\(=\frac{9 x y^{2}}{2}\)
|727200252|kitty|/Applications/kitty.app|0|
            \begi
|727200271|kitty|/Applications/kitty.app|0|
b \(\frac{8 a^{2} b^{3}}{3 a^{3} b} \div \frac{4 a b^{2}}{9 a^{3} b^{5}}=\frac{8 a^{2} b^{3}}{3 a^{3} b} \times \frac{9 a^{3} b^{5}}{4 a b^{2}}\)

\[
=\frac{72 \times a^{5} \times b^{8}}{12 \times a^{4} \times b^{3}}
\]

\[
=6 a b^{5}
\]
|727200275|kitty|/Applications/kitty.app|0|
a \(\left(\frac{2}{3}\right)^{2}\)
b \(\left(\frac{m}{n}\right)^{5}\)
c \(\left(\frac{x^{3}}{y^{2}}\right)^{2} \times\left(\frac{y}{x}\right)^{4}\)
d \(\left(\frac{2 x^{2}}{3}\right)^{2} \div \frac{4 x^{3}}{9}\)
|727200298|kitty|/Applications/kitty.app|0|
b \(\left(\frac{m}{n}\right)^{5}\)
c \(\left(\frac{x^{3}}{y^{2}}\right)^{2} \times\left(\frac{y}{x}\right)^{4}\)
d \(\left(\frac{2 x^{2}}{3}\right)^{2} \div \frac{4 x^{3}}{9}\)
|727200301|kitty|/Applications/kitty.app|0|
c \(\left(\frac{x^{3}}{y^{2}}\right)^{2} \times\left(\frac{y}{x}\right)^{4}\)
d \(\left(\frac{2 x^{2}}{3}\right)^{2} \div \frac{4 x^{3}}{9}\)
|727200302|kitty|/Applications/kitty.app|0|
d \(\left(\frac{2 x^{2}}{3}\right)^{2} \div \frac{4 x^{3}}{9}\)
|727200304|kitty|/Applications/kitty.app|0|
a \(\begin{aligned}\left(\frac{2}{3}\right)^{3} & =\frac{2^{3}}{3^{3}} \\ & =\frac{8}{27}\end{aligned}\)

\(\mathbf{b}\left(\frac{m}{n}\right)^{5}=\frac{m^{5}}{n^{5}}\)

c \(\left(\frac{x^{3}}{y^{2}}\right)^{2} \times\left(\frac{y}{x}\right)^{4}=\frac{x^{6}}{y^{4}} \times \frac{y^{4}}{x^{4}}\)

d \(\left(\frac{2 x^{2}}{3}\right)^{2} \div \frac{4 x^{3}}{9}=\frac{4 x^{4}}{9} \times \frac{9}{4 x^{3}}\)

\(=x^{2}\)

\(=x\)

There are different possible interpretations of the word 'simplify'. There may be more than one acceptable simplified form.
|727200365|kitty|/Applications/kitty.app|0|
\(\mathbf{b}\left(\frac{m}{n}\right)^{5}=\frac{m^{5}}{n^{5}}\)

c \(\left(\frac{x^{3}}{y^{2}}\right)^{2} \times\left(\frac{y}{x}\right)^{4}=\frac{x^{6}}{y^{4}} \times \frac{y^{4}}{x^{4}}\)

d \(\left(\frac{2 x^{2}}{3}\right)^{2} \div \frac{4 x^{3}}{9}=\frac{4 x^{4}}{9} \times \frac{9}{4 x^{3}}\)

\(=x^{2}\)

\(=x\)

There are different possible interpretations of the word 'simplify'. There may be more than one acceptable simplified form.
|727200371|kitty|/Applications/kitty.app|0|
c \(\left(\frac{x^{3}}{y^{2}}\right)^{2} \times\left(\frac{y}{x}\right)^{4}=\frac{x^{6}}{y^{4}} \times \frac{y^{4}}{x^{4}}\)

d \(\left(\frac{2 x^{2}}{3}\right)^{2} \div \frac{4 x^{3}}{9}=\frac{4 x^{4}}{9} \times \frac{9}{4 x^{3}}\)

\(=x^{2}\)

\(=x\)
|727200376|kitty|/Applications/kitty.app|0|
c \(\left(\frac{x^{3}}{y^{2}}\right)^{2} \times\left(\frac{y}{x}\right)^{4}=\frac{x^{6}}{y^{4}} \times \frac{y^{4}}{x^{4}}\)

d \(\left(\frac{2 x^{2}}{3}\right)^{2} \div \frac{4 x^{3}}{9}=\frac{4 x^{4}}{9} \times \frac{9}{4 x^{3}}\)

\(=x^{2}\)

\(=x\)

There are different possible interpretations of the word 'simplify'. There may be more than one acceptable simplified form.
|727200379|kitty|/Applications/kitty.app|0|
d \(\left(\frac{2 x^{2}}{3}\right)^{2} \div \frac{4 x^{3}}{9}=\frac{4 x^{4}}{9} \times \frac{9}{4 x^{3}}\)

\(=x^{2}\)

\(=x\)

There are different possible interpretations of the word 'simplify'. There may be more than one acceptable simplified form.
|727200384|kitty|/Applications/kitty.app|0|
There are different possible interpretations of the word 'simplify'. There may be more than one acceptable simplified form.
|727200387|kitty|/Applications/kitty.app|0|
\mathbf{b}|727200433|kitty|/Applications/kitty.app|0|
a \(\left(5 a^{3}\right)^{0}\)
b \(\frac{6 x^{2} y}{x y^{2}} \times \frac{y^{3} x}{2 y^{2} x^{2}}\)
|727200473|kitty|/Applications/kitty.app|0|
a \(\left(5 a^{3}\right)^{0}=1\)

\[
\text { b } \begin{aligned}
\frac{6 x^{2} y}{x y^{2}} \times \frac{y^{3} x}{2 y^{2} x^{2}} & =\frac{6}{2} \times \frac{x^{3}}{x^{3}} \times \frac{y^{4}}{y^{4}} \\
& =3 x^{0} y^{0} \\
& =3 \times 1 \times 1 \\
& =3
\end{aligned}
\]
|727200483|kitty|/Applications/kitty.app|0|
a \(\left(m n^{2}\right)^{0}\)
b \(\left(a^{4} b^{2}\right)^{3}\)
c \(\left(2 a^{4}\right)^{3}\)
d \(2\left(x^{2} y\right)^{0} \times\left(x^{2} y^{3}\right)^{3}\)

\section*{Solution}
a \(\left(m n^{2}\right)^{0}=1\)
b \(\left(a^{4} b^{2}\right)^{3}=a^{12} b^{6}\)
c \(\left(2 a^{4}\right)^{3}=2^{3} \times a^{12}\)
d \(2\left(x^{2} y\right)^{0} \times\left(x^{2} y^{3}\right)^{3}=2 \times 1 \times x^{6} y^{9}\)
\(=8 a^{12}\)
\(=2 x^{6} y^{9}\)
|727200497|kitty|/Applications/kitty.app|0|
b \(\left(a^{4} b^{2}\right)^{3}\)
c \(\left(2 a^{4}\right)^{3}\)
d \(2\left(x^{2} y\right)^{0} \times\left(x^{2} y^{3}\right)^{3}\)
|727200507|kitty|/Applications/kitty.app|0|
c \(\left(2 a^{4}\right)^{3}\)
d \(2\left(x^{2} y\right)^{0} \times\left(x^{2} y^{3}\right)^{3}\)
|727200509|kitty|/Applications/kitty.app|0|
d \(2\left(x^{2} y\right)^{0} \times\left(x^{2} y^{3}\right)^{3}\)
|727200511|kitty|/Applications/kitty.app|0|
b \(\left(a^{4} b^{2}\right)^{3}=a^{12} b^{6}\)
c \(\left(2 a^{4}\right)^{3}=2^{3} \times a^{12}\)
d \(2\left(x^{2} y\right)^{0} \times\left(x^{2} y^{3}\right)^{3}=2 \times 1 \times x^{6} y^{9}\)
|727200514|kitty|/Applications/kitty.app|0|
b \(\left(a^{4} b^{2}\right)^{3}=a^{12} b^{6}\)
c \(\left(2 a^{4}\right)^{3}=2^{3} \times a^{12}\)
d \(2\left(x^{2} y\right)^{0} \times\left(x^{2} y^{3}\right)^{3}=2 \times 1 \times x^{6} y^{9}\)
\(=8 a^{12}\)
\(=2 x^{6} y^{9}\)
|727200516|kitty|/Applications/kitty.app|0|
a \(\left(m n^{2}\right)^{0}=1\)
|727200556|kitty|/Applications/kitty.app|0|
c \(\left(2 a^{4}\right)^{3}=2^{3} \times a^{12}\)
d \(2\left(x^{2} y\right)^{0} \times\left(x^{2} y^{3}\right)^{3}=2 \times 1 \times x^{6} y^{9}\)
\(=8 a^{12}\)
\(=2 x^{6} y^{9}\)
|727200578|kitty|/Applications/kitty.app|0|
d \(2\left(x^{2} y\right)^{0} \times\left(x^{2} y^{3}\right)^{3}=2 \times 1 \times x^{6} y^{9}\)
\(=8 a^{12}\)
\(=2 x^{6} y^{9}\)
|727200582|kitty|/Applications/kitty.app|0|
a \(\left(5 a^{3}\right)^{0}=1\)
|727200643|kitty|/Applications/kitty.app|0|
\newlength{\lawboxheight}
\newenvironment{boxlaw}{
   %\par\noindent                        % No indentation
   \begin{flushleft}
   %\refstepcounter{theorem}             % Counter for this environment
   \begin{lrbox}{\lawbox}           % Name of box to contain text
   \begin{minipage}{0.95\linewidth-0.5cm}
}
{
  \end{minipage}
  \end{lrbox}%
  \setlength{\lawboxheight}{\totalheightof{\usebox{\lawbox}}+20.5pt}
  \rput[tl](0,0){   % Choose colours of framebox
    \psframebox[opacity=0.8,fillstyle=solid,fillcolor=yellow!30,framearc=0,framesep=10pt,linecolor=red,linestyle=none]{%
      \usebox{\lawbox}
    }
  }
  \rput[l](1cm,0){      % Patch white line
    \psline[linewidth=2.1\pslinewidth,linecolor=white](-0.25em,0)(\lawtextlength,0)%
  }%
  \rput[l](1cm,0.5ex){  % Place text
    % \textcolor{red}{\parbox[m]{0.66cm}{\includegraphics[width=0.66cm]{./include/law_red.eps}} \textbf{\lawboxtext}}%
    \color{red} \textbf{\lawboxtext}
  }
  \psline[linewidth=3pt,linecolor=red](0.22,0)(0.22,-\lawboxheight)
  \par
  \setlength{\margedef}{\ht\lawbox+\dp\lawbox+\marge}
  \vspace{\margedef}
  \end{flushleft}
}
|727201447|kitty|/Applications/kitty.app|0|
\newsavebox{\lawbox}
|727201532|kitty|/Applications/kitty.app|0|
\newcommand{\lawboxtext}{\raisebox{-0.5ex}{\Large \faGavel} \textsf{Laws/Results}}
|727201560|kitty|/Applications/kitty.app|0|
\newlength{\lawtextlength}
\setlength{\lawtextlength}{\widthof{\textbf{\lawboxtext}+0.5em}}
|727201566|kitty|/Applications/kitty.app|0|
\begin{lawbox}
|727201587|kitty|/Applications/kitty.app|0|
theorembox|727201603|kitty|/Applications/kitty.app|0|
Note, we will
|727201637|kitty|/Applications/kitty.app|0|
lawbox|727201920|kitty|/Applications/kitty.app|0|
Multiplication|727202062|kitty|/Applications/kitty.app|0|
\randomword{base}%
\hspace{\fill}
|727202097|kitty|/Applications/kitty.app|0|
base|727202100|kitty|/Applications/kitty.app|0|
 7 Negative Exponent Law: [ a^{-n} = \frac{1}{a^n} ] A negative exponent indicates the
   reciprocal of the base raised to the corresponding positive exponent.
|727202227|kitty|/Applications/kitty.app|0|
        \Question[3]
        \begin{parts}
            \part
            \part
            \part
        \end{parts}
|727202311|kitty|/Applications/kitty.app|0|
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part
            \part
            \part
            \part
            \part
            \part
        \end{multicols}\end{parts}
|727202382|kitty|/Applications/kitty.app|0|
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part
            \part
            \part
            \part
            \part
            \part
        \end{multicols}\end{parts}
|727202443|kitty|/Applications/kitty.app|0|
        \Question[2]
        \begin{parts}\begin{multicols}{2}
            \part
            \part
        \end{multicols}\end{parts}
|727202454|kitty|/Applications/kitty.app|0|
a \(6^{4}\)
b \(7^{3}\)
c \(8^{2}\)
|727202589|kitty|/Applications/kitty.app|0|
b \(7^{3}\)
c \(8^{2}\)
|727202593|kitty|/Applications/kitty.app|0|
a 8
b 27
c 64
|727202602|kitty|/Applications/kitty.app|0|
b 27
c 64
|727202605|kitty|/Applications/kitty.app|0|
c 64
|727202606|kitty|/Applications/kitty.app|0|
a \(3^{4}\)
b \(2^{7}\)
c \(5^{5}\)
|727202612|kitty|/Applications/kitty.app|0|
b \(2^{7}\)
c \(5^{5}\)
|727202615|kitty|/Applications/kitty.app|0|
c \(5^{5}\)
|727202617|kitty|/Applications/kitty.app|0|
4 Express as a product of powers of prime numbers.
a 18
b 24
c 144
|727202651|kitty|/Applications/kitty.app|0|
3 Evaluate:
|727202672|kitty|/Applications/kitty.app|0|
2 Express as a power of a prime number.
|727202674|kitty|/Applications/kitty.app|0|
1 State the base and index of:
|727202676|kitty|/Applications/kitty.app|0|
4 Express as a product of powers of prime numbers.
|727202682|kitty|/Applications/kitty.app|0|
\section{Introduction}
    \setcounter{secmarks}{0}
    \input{1-intro}
    \setcounter{sec1marks}{\thesecmarks}
\section{The Index Laws}
    \setcounter{secmarks}{0}
    \input{2-ind-laws}
    \setcounter{sec2marks}{\thesecmarks}
\section{Negative Indices}
    \setcounter{secmarks}{0}
    \input{3-neg-ind}
    \setcounter{sec3marks}{\thesecmarks}
|727202697|kitty|/Applications/kitty.app|0|
\subsection{Introduction}
    \setcounter{secmarks}{0}
    \input{1-intro}
    \setcounter{sec1marks}{\thesecmarks}
|727202703|kitty|/Applications/kitty.app|0|
    \setcounter{secmarks}{0}
|727202716|kitty|/Applications/kitty.app|0|
    \input{2-ind-laws}
|727202716|kitty|/Applications/kitty.app|0|
    \setcounter{sec2marks}{\thesecmarks}
|727202717|kitty|/Applications/kitty.app|0|
    \input{3-neg-ind}
|727202718|kitty|/Applications/kitty.app|0|
    \setcounter{sec3marks}{\thesecmarks}
|727202718|kitty|/Applications/kitty.app|0|
1 State the base and index of:
d \(10^{4}\)
e 5
f \(6^{0}\)

2 Express as a power of a prime number.
d 243
e 125
f 81

3 Evaluate:
d \(7^{4}\)
e \(2^{3} \times 3^{5}\)
f \(6^{4} \times 3^{2}\)

4 Express as a product of powers of prime numbers.
d 90
e 700
f 84
|727202729|kitty|/Applications/kitty.app|0|
3|727202761|kitty|/Applications/kitty.app|0|
a 18
b 24
c 144
|727202769|kitty|/Applications/kitty.app|0|
b 24
c 144
|727202771|kitty|/Applications/kitty.app|0|
c 144
|727202772|kitty|/Applications/kitty.app|0|
            \part
            \part
            \part
|727202805|kitty|/Applications/kitty.app|0|
b \(2^{7} \times 2^{3}\)
|727202812|kitty|/Applications/kitty.app|0|
b \(2^{7} \times 2^{3}\)
f \(3^{3} \times 3^{4} \times 3^{5}\)
|727202815|kitty|/Applications/kitty.app|0|
b \(2^{7} \times 2^{3}\)
f \(3^{3} \times 3^{4} \times 3^{5}\)
j \(3 x^{2} \times 4 x^{3}\)
|727202820|kitty|/Applications/kitty.app|0|
f \(3^{3} \times 3^{4} \times 3^{5}\)
j \(3 x^{2} \times 4 x^{3}\)
|727202824|kitty|/Applications/kitty.app|0|
j \(3 x^{2} \times 4 x^{3}\)
|727202825|kitty|/Applications/kitty.app|0|
5 Simplify, leaving the answer as a power or a product of powers.
a \(2^{2} \times 2^{3}\)
c \(3^{4} \times 3^{5}\)
d \(3^{4} \times 3^{7}\)
e \(3^{2} \times 3 \times 3^{4}\)
g \(a^{3} \times a^{8}\)
h \(b^{7} \times b^{12}\)
i \(3 a^{2} \times a^{3}\)
\(\mathbf{k} 2 y \times 3 y^{4}\)
l \(4 b^{2} \times 3 b^{4}\)
|727202846|kitty|/Applications/kitty.app|0|
a \(2^{2} \times 2^{3}\)
|727202852|kitty|/Applications/kitty.app|0|
e \(3^{2} \times 3 \times 3^{4}\)
|727202857|kitty|/Applications/kitty.app|0|
i \(3 a^{2} \times a^{3}\)
|727202877|kitty|/Applications/kitty.app|0|
c \(x^{2} y \times x^{3} y\)
|727202893|kitty|/Applications/kitty.app|0|
c \(x^{2} y \times x^{3} y\)
f \(5 a^{4} b \times 2 a b^{3}\)
|727202894|kitty|/Applications/kitty.app|0|
6 Simplify:
c \(x^{2} y \times x^{3} y\)
f \(5 a^{4} b \times 2 a b^{3}\)
|727202897|kitty|/Applications/kitty.app|0|
a \(a^{2} b^{3} \times b^{2}\)
b \(a^{3} b \times a^{2} b^{3}\)
d \(2 x y^{2} \times 3 x^{2} y\)
e \(4 a^{3} b^{2} \times a^{2} b^{4}\)
|727202909|kitty|/Applications/kitty.app|0|
b \(a^{3} b \times a^{2} b^{3}\)
d \(2 x y^{2} \times 3 x^{2} y\)
e \(4 a^{3} b^{2} \times a^{2} b^{4}\)
|727202924|kitty|/Applications/kitty.app|0|
d \(2 x y^{2} \times 3 x^{2} y\)
e \(4 a^{3} b^{2} \times a^{2} b^{4}\)
|727202925|kitty|/Applications/kitty.app|0|
e \(4 a^{3} b^{2} \times a^{2} b^{4}\)
|727202926|kitty|/Applications/kitty.app|0|
a \(\frac{3^{7}}{3^{2}}\)
b \(\frac{2^{6}}{2^{2}}\)
|727202951|kitty|/Applications/kitty.app|0|
a \(\frac{3^{7}}{3^{2}}\)
b \(\frac{2^{6}}{2^{2}}\)
e \(10^{7} \div 10^{2}\)
f \(\frac{10^{12}}{10^{4}}\)
|727202955|kitty|/Applications/kitty.app|0|
a \(\frac{3^{7}}{3^{2}}\)
b \(\frac{2^{6}}{2^{2}}\)
e \(10^{7} \div 10^{2}\)
f \(\frac{10^{12}}{10^{4}}\)
i \(\frac{2 x^{3}}{x^{2}}\)
j \(\frac{6 x^{5}}{2 x^{2}}\)
|727202959|kitty|/Applications/kitty.app|0|
b \(\frac{2^{6}}{2^{2}}\)
e \(10^{7} \div 10^{2}\)
f \(\frac{10^{12}}{10^{4}}\)
i \(\frac{2 x^{3}}{x^{2}}\)
j \(\frac{6 x^{5}}{2 x^{2}}\)
|727202963|kitty|/Applications/kitty.app|0|
e \(10^{7} \div 10^{2}\)
f \(\frac{10^{12}}{10^{4}}\)
i \(\frac{2 x^{3}}{x^{2}}\)
j \(\frac{6 x^{5}}{2 x^{2}}\)
|727202965|kitty|/Applications/kitty.app|0|
f \(\frac{10^{12}}{10^{4}}\)
i \(\frac{2 x^{3}}{x^{2}}\)
j \(\frac{6 x^{5}}{2 x^{2}}\)
|727202967|kitty|/Applications/kitty.app|0|
i \(\frac{2 x^{3}}{x^{2}}\)
j \(\frac{6 x^{5}}{2 x^{2}}\)
|727202969|kitty|/Applications/kitty.app|0|
j \(\frac{6 x^{5}}{2 x^{2}}\)
|727202970|kitty|/Applications/kitty.app|0|
Example 5 7 Simplify, leaving the answer as a power or a product of powers.
c \(5^{4} \div 5\)
d \(7^{5} \div 7^{3}\)
g \(\frac{a^{4}}{a}\)
h \(\frac{a^{5}}{a^{3}}\)
k \(\frac{10 y^{12}}{5 y^{3}}\)
l \(\frac{27 p^{4}}{9 p}\)
|727202994|kitty|/Applications/kitty.app|0|
a \(\frac{a^{3} b^{2}}{a^{2}}\)
b \(\frac{x^{3} y^{2}}{x y}\)
|727203009|kitty|/Applications/kitty.app|0|
a \(\frac{a^{3} b^{2}}{a^{2}}\)
b \(\frac{x^{3} y^{2}}{x y}\)
e \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
f \(\frac{15 x y^{3}}{3 y^{2}}\)
|727203015|kitty|/Applications/kitty.app|0|
b \(\frac{x^{3} y^{2}}{x y}\)
e \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
f \(\frac{15 x y^{3}}{3 y^{2}}\)
|727203019|kitty|/Applications/kitty.app|0|
e \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
f \(\frac{15 x y^{3}}{3 y^{2}}\)
|727203021|kitty|/Applications/kitty.app|0|
f \(\frac{15 x y^{3}}{3 y^{2}}\)
|727203022|kitty|/Applications/kitty.app|0|
8 Simplify:
c \(\frac{a^{5} b^{3}}{a^{4} b}\)
d \(\frac{x^{4} y^{7}}{x^{3} y^{2}}\)
g \(\frac{16 a^{4} b^{3}}{12 a^{2} b^{2}}\)
h \(\frac{27 x^{2} y^{3}}{18 x y^{2}}\)
|727203031|kitty|/Applications/kitty.app|0|
a \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
b \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
|727203041|kitty|/Applications/kitty.app|0|
a \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
b \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
e \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
f \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
|727203044|kitty|/Applications/kitty.app|0|
b \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
e \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
f \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
|727203047|kitty|/Applications/kitty.app|0|
e \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
f \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
|727203048|kitty|/Applications/kitty.app|0|
f \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
|727203050|kitty|/Applications/kitty.app|0|
Example 6 9 Simplify:
c \(\frac{2 a b^{2}}{3 a^{2} b^{4}} \times \frac{6 a^{4} b^{5}}{a b}\)
d \(\frac{12 x^{4} y^{3}}{3 x^{2} y} \times \frac{x^{2} y^{4}}{x^{3} y^{5}}\)
g \(\frac{14 a^{4} b^{3}}{3 a b^{2}} \div \frac{7 a^{5} b^{4}}{6 a^{3} b^{5}}\)
h \(\frac{12 x^{2} y}{x^{3} y^{4}} \div \frac{6 x y^{2}}{x^{6} y^{7}}\)
|727203060|kitty|/Applications/kitty.app|0|
a \(a^{4} \times \ldots=a^{10}\)
|727203066|kitty|/Applications/kitty.app|0|
a \(a^{4} \times \ldots=a^{10}\)
d \(9 d^{5} \times \ldots=27 d^{6}\)
|727203070|kitty|/Applications/kitty.app|0|
a \(a^{4} \times \ldots=a^{10}\)
d \(9 d^{5} \times \ldots=27 d^{6}\)
g \(15 d^{7} \div \ldots=3 d^{2}\)
|727203073|kitty|/Applications/kitty.app|0|
a \(a^{4} \times \ldots=a^{10}\)
d \(9 d^{5} \times \ldots=27 d^{6}\)
g \(15 d^{7} \div \ldots=3 d^{2}\)
j \(8 a b^{4} \times \ldots=24 a^{2} b^{6}\)
|727203075|kitty|/Applications/kitty.app|0|
a \(a^{4} \times \ldots=a^{10}\)
d \(9 d^{5} \times \ldots=27 d^{6}\)
g \(15 d^{7} \div \ldots=3 d^{2}\)
j \(8 a b^{4} \times \ldots=24 a^{2} b^{6}\)
\(\mathbf{m} \ell^{6} m^{7} \div \ldots=\ell^{2} m^{5}\)
|727203080|kitty|/Applications/kitty.app|0|
a \(a^{4} \times \ldots=a^{10}\)
d \(9 d^{5} \times \ldots=27 d^{6}\)
g \(15 d^{7} \div \ldots=3 d^{2}\)
j \(8 a b^{4} \times \ldots=24 a^{2} b^{6}\)
\(\mathbf{m} \ell^{6} m^{7} \div \ldots=\ell^{2} m^{5}\)
b \(b^{7} \times \ldots=b^{16}\)
|727203083|kitty|/Applications/kitty.app|0|
d \(9 d^{5} \times \ldots=27 d^{6}\)
g \(15 d^{7} \div \ldots=3 d^{2}\)
j \(8 a b^{4} \times \ldots=24 a^{2} b^{6}\)
\(\mathbf{m} \ell^{6} m^{7} \div \ldots=\ell^{2} m^{5}\)
b \(b^{7} \times \ldots=b^{16}\)
|727203100|kitty|/Applications/kitty.app|0|
g \(15 d^{7} \div \ldots=3 d^{2}\)
j \(8 a b^{4} \times \ldots=24 a^{2} b^{6}\)
\(\mathbf{m} \ell^{6} m^{7} \div \ldots=\ell^{2} m^{5}\)
b \(b^{7} \times \ldots=b^{16}\)
|727203103|kitty|/Applications/kitty.app|0|
j \(8 a b^{4} \times \ldots=24 a^{2} b^{6}\)
\(\mathbf{m} \ell^{6} m^{7} \div \ldots=\ell^{2} m^{5}\)
b \(b^{7} \times \ldots=b^{16}\)
|727203104|kitty|/Applications/kitty.app|0|
\(\mathbf{m} \ell^{6} m^{7} \div \ldots=\ell^{2} m^{5}\)
b \(b^{7} \times \ldots=b^{16}\)
|727203106|kitty|/Applications/kitty.app|0|
b \(b^{7} \times \ldots=b^{16}\)
|727203107|kitty|/Applications/kitty.app|0|
10 Copy and complete.
c \(4 a^{3} \times \ldots=12 a^{7}\)
e \(a^{8} \div \ldots=a^{4}\)
f \(x^{10} \div \ldots=x^{6}\)
h \(9 d^{6} \div \ldots=3 d\)
i \(m^{4} n^{5} \times \ldots=m^{10} n^{7}\)
k \(a^{7} b^{4} \div \ldots=a^{2} b\)
l \(14 x^{5} y^{2} \times \ldots=42 x^{10} y^{5}\)
n \(9 m^{7} n^{4} \div \ldots=3 m^{2}\)
o \(18 p^{2} q^{6} \div \ldots=3 p q\)
|727203143|kitty|/Applications/kitty.app|0|


11 Simplify each expression. Check your answer for part a by substituting \(x=2\) into both the original expression and the simplified expression. Repeat for \(x=3\) in part \(\mathbf{b}\) and \(x=-2\) in part \(\mathbf{c}\).
a \(\frac{6 x^{2}}{x}\)
\(\mathbf{b} \frac{6 x^{2}}{3 x}\)
c \(\frac{6 x^{2}}{6 x}\)
d \(\frac{8 x^{4}}{4 x}\)
e \(\frac{3 x^{5}}{x^{3}}\)
f \(\frac{10 x^{4}}{5 x^{3}}\)
g \(\frac{10 x^{4}}{2 x^{4}}\)
\(\mathbf{h} \frac{12 x^{4}}{6 x^{2}}\)
|727203153|kitty|/Applications/kitty.app|0|
Example 8 12 Simplify:
a \(a^{0}\)
b \(2 x^{0}\)
|727203157|kitty|/Applications/kitty.app|0|
 Example 8 12 Simplify:|727203161|kitty|/Applications/kitty.app|0|
b \(2 x^{0}\)
|727203163|kitty|/Applications/kitty.app|0|
c \(x y^{0}\)
d \(7 x^{0} y^{0}\)
|727203172|kitty|/Applications/kitty.app|0|
Example 8 12 Simplify:
c \(x y^{0}\)
d \(7 x^{0} y^{0}\)
|727203175|kitty|/Applications/kitty.app|0|
a \(a^{0}\)
b \(2 x^{0}\)
|727203182|kitty|/Applications/kitty.app|0|
a \(3 a^{0}\)
b \(6 a^{0}\)
|727203207|kitty|/Applications/kitty.app|0|
a \(3 a^{0}\)
b \(6 a^{0}\)
e \(4 a^{0}+3 b^{0}\)
f \(6 a^{0}+7 m^{0}\)
|727203211|kitty|/Applications/kitty.app|0|
a \(3 a^{0}\)
b \(6 a^{0}\)
e \(4 a^{0}+3 b^{0}\)
f \(6 a^{0}+7 m^{0}\)
i \((4 b)^{0}+2 b^{0}\)
j \((3 b)^{0}-5 d^{0}\)
|727203215|kitty|/Applications/kitty.app|0|
b \(6 a^{0}\)
e \(4 a^{0}+3 b^{0}\)
f \(6 a^{0}+7 m^{0}\)
i \((4 b)^{0}+2 b^{0}\)
j \((3 b)^{0}-5 d^{0}\)
|727203220|kitty|/Applications/kitty.app|0|
e \(4 a^{0}+3 b^{0}\)
f \(6 a^{0}+7 m^{0}\)
i \((4 b)^{0}+2 b^{0}\)
j \((3 b)^{0}-5 d^{0}\)
|727203222|kitty|/Applications/kitty.app|0|
f \(6 a^{0}+7 m^{0}\)
i \((4 b)^{0}+2 b^{0}\)
j \((3 b)^{0}-5 d^{0}\)
|727203224|kitty|/Applications/kitty.app|0|
i \((4 b)^{0}+2 b^{0}\)
j \((3 b)^{0}-5 d^{0}\)
|727203226|kitty|/Applications/kitty.app|0|
j \((3 b)^{0}-5 d^{0}\)
|727203227|kitty|/Applications/kitty.app|0|
i|727203238|kitty|/Applications/kitty.app|0|
j|727203240|kitty|/Applications/kitty.app|0|
Example 9 13 Simplify:
c \((4 a)^{0}\)
d \((3 b)^{0}\)
g \((2 a+1)^{0}\)
h \((4 a+3 b)^{0}\)
k \(\left(5 m^{0}+7 b\right)^{0}\)
l \(\left(6 m-2 c^{0}\right)^{0}\)
|727203247|kitty|/Applications/kitty.app|0|
14 Simplify, leaving the answer as a power.
a \(\left(2^{3}\right)^{4}\)
b \(\left(3^{2}\right)^{3}\)
|727203257|kitty|/Applications/kitty.app|0|
14 Simplify, leaving the answer as a power.
|727203283|kitty|/Applications/kitty.app|0|
4|727203264|kitty|/Applications/kitty.app|0|
b \(\left(3^{2}\right)^{3}\)
|727203267|kitty|/Applications/kitty.app|0|
c \(\left(a^{2}\right)^{5}\)
d \(\left(y^{5}\right)^{6}\)
|727203274|kitty|/Applications/kitty.app|0|
a \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
b \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
|727203289|kitty|/Applications/kitty.app|0|
a \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
b \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
e \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
f \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
|727203291|kitty|/Applications/kitty.app|0|
b \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
e \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
f \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
|727203294|kitty|/Applications/kitty.app|0|
e \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
f \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
|727203296|kitty|/Applications/kitty.app|0|
f \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
|727203297|kitty|/Applications/kitty.app|0|
d \(\frac{\left(y^{3}\right)^{4}}{\left(y^{4}\right)^{2}}\)
|727203326|kitty|/Applications/kitty.app|0|
d \(\frac{\left(y^{3}\right)^{4}}{\left(y^{4}\right)^{2}}\)
i \(\frac{3\left(x^{3} y\right)^{2}}{\left(x^{2} y\right)^{2}} \div \frac{12 x^{4} y^{2}}{\left(2 x^{3} y\right)^{2}}\)
|727203329|kitty|/Applications/kitty.app|0|
c \(\left(m^{5}\right)^{\cdots}=m^{10}\)
|727203352|kitty|/Applications/kitty.app|0|
f \((\ldots)^{5}=p^{25}\)
|727203353|kitty|/Applications/kitty.app|0|
i \((\ldots)^{\cdots}=m^{20}\)
|727203354|kitty|/Applications/kitty.app|0|
16 Copy and complete (using index law 3).
a \(\left(a^{6}\right)^{\cdots}=a^{24}\)
b \(\left(b^{3}\right)^{\cdots}=b^{21}\)
d \(\left(m^{6}\right) \cdots=m^{30}\)
e \((\ldots)^{6}=p^{36}\)
g \((\ldots)^{4}=a^{8}\)
h \((\ldots)^{3}=m^{15}\)
|727203355|kitty|/Applications/kitty.app|0|
17 a Is it true that \(\left(a^{2}\right)^{6}=\left(a^{6}\right)^{2}\) ? b Is it true that \(\left(b^{4}\right)^{7}=\left(b^{7}\right)^{4}\) ?

c Generalise your result.
|727203366|kitty|/Applications/kitty.app|0|
a \((3 a)^{2}\)
|727203374|kitty|/Applications/kitty.app|0|
a \((3 a)^{2}\)
b \((2 x)^{3}\)
|727203376|kitty|/Applications/kitty.app|0|
a \((3 a)^{2}\)
b \((2 x)^{3}\)
e \(\left(\frac{a}{5}\right)^{2}\)
\(\mathbf{f}\left(\frac{2}{x}\right)^{3}\)
|727203380|kitty|/Applications/kitty.app|0|
b \((2 x)^{3}\)
e \(\left(\frac{a}{5}\right)^{2}\)
\(\mathbf{f}\left(\frac{2}{x}\right)^{3}\)
|727203390|kitty|/Applications/kitty.app|0|
e \(\left(\frac{a}{5}\right)^{2}\)
\(\mathbf{f}\left(\frac{2}{x}\right)^{3}\)
|727203392|kitty|/Applications/kitty.app|0|
\(\mathbf{f}\left(\frac{2}{x}\right)^{3}\)
|727203393|kitty|/Applications/kitty.app|0|
Example 78 Simplify by expanding the brackets.
c \(\left(x y^{3}\right)^{2}\)
d \(\left(a^{2} b\right)^{4}\)
g \(\left(\frac{a}{b}\right)^{5}\)
h \(\left(\frac{x^{2}}{y}\right)^{3}\)
|727203419|kitty|/Applications/kitty.app|0|
a \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
b \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
|727203428|kitty|/Applications/kitty.app|0|
a \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
b \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
d \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
e \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
|727203434|kitty|/Applications/kitty.app|0|
b \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
d \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
e \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
|727203438|kitty|/Applications/kitty.app|0|
d \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
e \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
|727203439|kitty|/Applications/kitty.app|0|
e \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
|727203440|kitty|/Applications/kitty.app|0|


19 Simplify:
c \(\left(m^{2} n^{3}\right)^{2} \times(m n)^{3}\)
f \(\left(2 x y^{2}\right)^{0} \times\left(3 x^{2} y\right)^{3}\)
|727203475|kitty|/Applications/kitty.app|0|
\(\mathbf{a}\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
b \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
|727203479|kitty|/Applications/kitty.app|0|
b \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
|727203491|kitty|/Applications/kitty.app|0|
a \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
b \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
|727203499|kitty|/Applications/kitty.app|0|
b \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
|727203504|kitty|/Applications/kitty.app|0|
20 Simplify:
c \(\left(\frac{x^{3}}{y^{2}}\right)^{2} \div\left(\frac{x}{y^{2}}\right)^{3}\)
d \(\left(\frac{2 x^{4}}{y}\right)^{5} \div\left(\frac{4 x^{3}}{y^{3}}\right)^{2}\)

21 Simplify:
c \(\frac{\left(2 x^{2} y^{3}\right)^{3} \times\left(5 x y^{2}\right)^{2}}{\left(10 x^{2} y\right)^{2} \times(x y)^{3}}\)
d \(\frac{(6 a b)^{3} \times 2 a^{7} b^{4}}{(2 a b)^{4} \times\left(3 a^{2} b\right)^{2}}\)
|727203512|kitty|/Applications/kitty.app|0|
22 Copy and complete.
a \((\ldots)^{4}=a^{8} b^{12}\)
b \((\ldots)^{6}=m^{30} n^{24}\)
c \(\left(p^{3} q\right) \ldots=p^{9} q^{3}\)
d \(\left(x^{4} y^{7}\right) \ldots=1\)
e \((\ldots)^{4}=16 a^{8}\)
f \((\ldots)^{3}=27 q^{9}\)
g \((\ldots)^{2}=49 m^{6}\)
h \((\ldots)^{3}=64 \ell^{9} m^{3}\)
i \((\ldots)^{2}=25 m^{10} n^{6}\)
|727203527|kitty|/Applications/kitty.app|0|
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^{7}}{3^{2}}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
|727203624|kitty|/Applications/kitty.app|0|
        \Question[4] Simplify:
        \begin{parts}\begin{multicols}{2}
            \part \(a^{2} b^{3} \times b^{2}\)
            \part \(a^{3} b \times a^{2} b^{3}\)
            \part \(2 x y^{2} \times 3 x^{2} y\)
            \part \(4 a^{3} b^{2} \times a^{2} b^{4}\)
        \end{multicols}\end{parts}
|727203651|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection*{Exercises:}
    \begin{questions}
        \Question[3] State the base and index of:
        \begin{parts}
            \part \(6^{4}\)
            \part \(7^{3}\)
            \part \(8^{2}\)
        \end{parts}
        \Question[3] Express as a power of a prime number.
        \begin{parts}
            \part 8
            \part 27
            \part 64
        \end{parts}
        \Question[3] Evaluate:
        \begin{parts}
            \part \(3^{4}\)
            \part \(2^{7}\)
            \part \(5^{5}\)
        \end{parts}
        \Question[3] Express as a product of powers of prime numbers.
        \begin{parts}
            \part 18
            \part 24
            \part 144
        \end{parts}
        \Question[3] Simplify:
        \begin{parts}\begin{multicols}{3}
            \part \(2^{7} \times 2^{3}\)
            \part \(3^{3} \times 3^{4} \times 3^{5}\)
            \part \(3 x^{2} \times 4 x^{3}\)
        \end{multicols}\end{parts}
        \Question[4] Simplify:
        \begin{parts}\begin{multicols}{2}
            \part \(a^{2} b^{3} \times b^{2}\)
            \part \(a^{3} b \times a^{2} b^{3}\)
            \part \(2 x y^{2} \times 3 x^{2} y\)
            \part \(4 a^{3} b^{2} \times a^{2} b^{4}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^{7}}{3^{2}}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
            \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
            \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
            \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(a^{4} \times \fillin[]=a^{10}\)
            \part \(9 d^{5} \times \fillin[]=27 d^{6}\)
            \part \(15 d^{7} \div \fillin[]=3 d^{2}\)
            \part \(8 a b^{4} \times \fillin[]=24 a^{2} b^{6}\)
            \part \(\ell^{6} m^{7} \div \fillin[]=\ell^{2} m^{5}\)
            \part \(b^{7} \times \fillin[]=b^{16}\)
        \end{multicols}\end{parts}
        \Question[2]
        \begin{parts}\begin{multicols}{2}
            \part \(a^{0}\)
            \part \(2 x^{0}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(3 a^{0}\)
            \part \(6 a^{0}\)
            \part \(4 a^{0}+3 b^{0}\)
            \part \(6 a^{0}+7 m^{0}\)
            \part \((4 b)^{0}+2 b^{0}\)
            \part \((3 b)^{0}-5 d^{0}\)
        \end{multicols}\end{parts}
        \Question[2] Simplify, leaving the answer as a power.
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2^{3}\right)^{4}\)
            \part \(\left(3^{2}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
            \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
            \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
            \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \((3 a)^{2}\)
            \part \((2 x)^{3}\)
            \part \(\left(\frac{a}{5}\right)^{2}\)
            \part \(\left(\frac{2}{x}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
            \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
            \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
            \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
            \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
            \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
            \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
        \end{multicols}\end{parts}
    \end{questions}
\end{exercisebox}

|727203752|kitty|/Applications/kitty.app|0|
! Missing \endcsname inserted.
<to be read again> 
\relax 
l.273         \begin
                  {parts}\begin{multicols}{2}
? 
! Emergency stop.
<to be read again> 
\relax 
l.273         \begin
                  {parts}\begin{multicols}{2}
End of file on the terminal!
|727203944|kitty|/Applications/kitty.app|0|
        \Question[4] Simplify:
        \begin{parts}\begin{multicols}{2}
            \part \(a^{2} b^{3} \times b^{2}\)
            \part \(a^{3} b \times a^{2} b^{3}\)
            \part \(2 x y^{2} \times 3 x^{2} y\)
            \part \(4 a^{3} b^{2} \times a^{2} b^{4}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^{7}}{3^{2}}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
            \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
            \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
            \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(a^{4} \times \fillin[]=a^{10}\)
            \part \(9 d^{5} \times \fillin[]=27 d^{6}\)
            \part \(15 d^{7} \div \fillin[]=3 d^{2}\)
            \part \(8 a b^{4} \times \fillin[]=24 a^{2} b^{6}\)
            \part \(\ell^{6} m^{7} \div \fillin[]=\ell^{2} m^{5}\)
            \part \(b^{7} \times \fillin[]=b^{16}\)
        \end{multicols}\end{parts}
        \Question[2]
        \begin{parts}\begin{multicols}{2}
            \part \(a^{0}\)
            \part \(2 x^{0}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(3 a^{0}\)
            \part \(6 a^{0}\)
            \part \(4 a^{0}+3 b^{0}\)
            \part \(6 a^{0}+7 m^{0}\)
            \part \((4 b)^{0}+2 b^{0}\)
            \part \((3 b)^{0}-5 d^{0}\)
        \end{multicols}\end{parts}
        \Question[2] Simplify, leaving the answer as a power.
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2^{3}\right)^{4}\)
            \part \(\left(3^{2}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
            \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
            \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
            \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \((3 a)^{2}\)
            \part \((2 x)^{3}\)
            \part \(\left(\frac{a}{5}\right)^{2}\)
            \part \(\left(\frac{2}{x}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
            \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
            \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
            \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
        \end{multicols}\end{parts}
|727204064|kitty|/Applications/kitty.app|0|
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(a^{4} \times \fillin[]=a^{10}\)
            \part \(9 d^{5} \times \fillin[]=27 d^{6}\)
            \part \(15 d^{7} \div \fillin[]=3 d^{2}\)
            \part \(8 a b^{4} \times \fillin[]=24 a^{2} b^{6}\)
            \part \(\ell^{6} m^{7} \div \fillin[]=\ell^{2} m^{5}\)
            \part \(b^{7} \times \fillin[]=b^{16}\)
        \end{multicols}\end{parts}
        \Question[2]
        \begin{parts}\begin{multicols}{2}
            \part \(a^{0}\)
            \part \(2 x^{0}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(3 a^{0}\)
            \part \(6 a^{0}\)
            \part \(4 a^{0}+3 b^{0}\)
            \part \(6 a^{0}+7 m^{0}\)
            \part \((4 b)^{0}+2 b^{0}\)
            \part \((3 b)^{0}-5 d^{0}\)
        \end{multicols}\end{parts}
        \Question[2] Simplify, leaving the answer as a power.
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2^{3}\right)^{4}\)
            \part \(\left(3^{2}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
            \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
            \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
            \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \((3 a)^{2}\)
            \part \((2 x)^{3}\)
            \part \(\left(\frac{a}{5}\right)^{2}\)
            \part \(\left(\frac{2}{x}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
            \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
            \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
            \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
            \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
            \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
            \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
        \end{multicols}\end{parts}
|727204133|kitty|/Applications/kitty.app|0|
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^{7}}{3^{2}}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
            \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
            \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
            \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(a^{4} \times \fillin[]=a^{10}\)
            \part \(9 d^{5} \times \fillin[]=27 d^{6}\)
            \part \(15 d^{7} \div \fillin[]=3 d^{2}\)
            \part \(8 a b^{4} \times \fillin[]=24 a^{2} b^{6}\)
            \part \(\ell^{6} m^{7} \div \fillin[]=\ell^{2} m^{5}\)
            \part \(b^{7} \times \fillin[]=b^{16}\)
        \end{multicols}\end{parts}
        \Question[2]
        \begin{parts}\begin{multicols}{2}
            \part \(a^{0}\)
            \part \(2 x^{0}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(3 a^{0}\)
            \part \(6 a^{0}\)
            \part \(4 a^{0}+3 b^{0}\)
            \part \(6 a^{0}+7 m^{0}\)
            \part \((4 b)^{0}+2 b^{0}\)
            \part \((3 b)^{0}-5 d^{0}\)
        \end{multicols}\end{parts}
        \Question[2] Simplify, leaving the answer as a power.
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2^{3}\right)^{4}\)
            \part \(\left(3^{2}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
            \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
            \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
            \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \((3 a)^{2}\)
            \part \((2 x)^{3}\)
            \part \(\left(\frac{a}{5}\right)^{2}\)
            \part \(\left(\frac{2}{x}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
            \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
            \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
            \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
            \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
            \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
            \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
        \end{multicols}\end{parts}
|727204160|kitty|/Applications/kitty.app|0|
        \Question[4] Simplify:
        \begin{parts}\begin{multicols}{2}
            \part \(a^{2} b^{3} \times b^{2}\)
            \part \(a^{3} b \times a^{2} b^{3}\)
            \part \(2 x y^{2} \times 3 x^{2} y\)
            \part \(4 a^{3} b^{2} \times a^{2} b^{4}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^{7}}{3^{2}}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
|727204217|kitty|/Applications/kitty.app|0|
        \Question[4] Simplify:
        \begin{parts}\begin{multicols}{2}
            \part \(a^{2} b^{3} \times b^{2}\)
            \part \(a^{3} b \times a^{2} b^{3}\)
            \part \(2 x y^{2} \times 3 x^{2} y\)
            \part \(4 a^{3} b^{2} \times a^{2} b^{4}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^{7}}{3^{2}}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
        \end{multicols}\end{parts}
|727204226|kitty|/Applications/kitty.app|0|
        \Question[4] Simplify:
        \begin{parts}\begin{multicols}{2}
            \part \(a^{2} b^{3} \times b^{2}\)
            \part \(a^{3} b \times a^{2} b^{3}\)
            \part \(2 x y^{2} \times 3 x^{2} y\)
            \part \(4 a^{3} b^{2} \times a^{2} b^{4}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^{7}}{3^{2}}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
            \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
            \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
            \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
        \end{multicols}\end{parts}
|727204238|kitty|/Applications/kitty.app|0|
            \part \[\frac{3^7}{3^2}\]
|727204462|kitty|/Applications/kitty.app|0|
            \part \[\frac{3^{7}}{3^{2}}\]
|727204466|kitty|/Applications/kitty.app|0|
    \end{questions}
    \begin{questions}
|727204588|kitty|/Applications/kitty.app|0|
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part $\frac{3^{7}}{3^{2}}$
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
|727204605|kitty|/Applications/kitty.app|0|
            \part $\frac{3^{7}}{3^{2}}$
|727204766|kitty|/Applications/kitty.app|0|
\section*{Review exercise}
1 Evaluate:
a \(4^{3}\)
b \(2^{6}\)
c \(8^{2}\)
d \(10^{6}\)

2 Express as a product of powers of prime numbers.
a \(120^{2}\)
b \(900^{3}\)
c \(315^{4}\)
d \(490^{5}\)

3 Simplify and evaluate where possible.
a \(a^{6} \times a^{7}\)
b \(b^{4} \times b^{9}\)
c \(3 a^{4} \times 5 a^{5}\)
d \(2 x^{3} \times 5 x^{6}\)
e \(a^{7} \div a^{4}\)
f \(m^{12} \times m^{6}\)
g \(\frac{12 b^{7}}{6 b^{2}}\)
h \(\frac{18 p^{10}}{9 p}\)
i \(\left(a^{4}\right)^{3}\)
j \(\left(b^{6}\right)^{5}\)
k \(\left(2 a^{7}\right)^{3}\)
l \(\left(3 m^{2}\right)^{4}\)
\(\mathbf{m} a^{0}\)
n \(3 b^{0}\)
o \(5 m^{0}\)
p \((3 q)^{0}\)

4 Simplify and evaluate where possible.
a \(4 a^{2} b^{3} \times 5 a b^{4}\)
b \(2 m^{4} n^{3} \times 5 m^{6} n^{7}\)
c \(\frac{20 a^{4} b^{2}}{5 a^{2} b}\)
d \(\frac{24 m^{9} n^{4}}{18 m^{6} n^{2}}\)
e \(\left(3 a^{3} b\right)^{4}\)
f \(\left(5 a^{2} b\right)^{2} \times 4 a^{4} b^{3}\)
g \(\frac{5 a^{6} b^{7}}{4 a^{3} b^{2}} \times \frac{12 a^{10} b^{9}}{a^{6} b^{7}}\)
h \(\frac{8 m^{4} n^{2}}{7 m^{3} n} \div \frac{3 m^{3} n^{5}}{14 m^{9} n^{16}}\)
i \(\frac{\left(2 x^{2} y\right)^{3}}{5 x^{6} y^{2}} \times\left(\frac{x^{3}}{2 y^{2}}\right)^{3}\)
j \(\frac{(4 a b)^{3} \times 5 a^{2} b}{\left(10 a b^{2}\right)^{2}}\)
\(\mathbf{k} \frac{\left(3 x^{2} y\right)^{3} \times 2\left(x y^{2}\right)^{3}}{(3 x y)^{4}}\)
l \(\frac{\left(4 a^{2} b^{3}\right)^{2}}{3 a^{6} b^{4}} \div \frac{a b^{5}}{\left(3 a^{3} b^{2}\right)^{3}}\)

5 Evaluate:
a \(6^{-2}\)
b \(8^{-3}\)
c \(2^{-7}\)
d \(4^{-3}\)
e \(\left(\frac{4}{5}\right)^{-2}\)
f \(\left(\frac{2}{3}\right)^{-4}\)
\(\mathbf{g}\left(2^{\frac{1}{2}}\right)^{-3}\)
h \(\left(16^{\frac{1}{4}}\right)^{-3}\)

6 Simplify, expressing the answer with positive indices.
a \(\frac{4 m^{2} n^{5} p^{-6}}{16 m^{-2} n^{5} p^{3}}\)
b \((4 y)^{-3}\)
c \(\left(2^{2} y^{3}\right)^{-5}\)
d \(\left(5^{-2} x^{3}\right)^{-5}\)
e \(\left(3^{-3} a^{2} b^{-1}\right)^{-4}\)
\(\mathbf{f}\left(\frac{a^{3}}{b^{2}}\right)^{-2}\)
g \(\left(\frac{5 g^{2}}{h^{-3}}\right)^{-2}\)
\(\mathbf{h}\left(\frac{m^{-3}}{(2 n)^{-4}}\right)^{-2}\)

7 Simplify, expressing the answer with positive indices.
a \(4 a^{2} \times 5 a^{-3}\)
b \(8 m^{2} n^{-3} \times 5 m^{-4} n^{6}\)
c \(14 a^{-4} \div 7 a^{-5}\)
d \(\frac{18 m^{2} n^{-3}}{9 m^{4} n^{-1}}\)
e \(\left(4 m^{2} n^{-2}\right)^{-3}\)
f \(\frac{\left(5 m^{2} n\right)^{-2}}{10 m^{4} n^{3}}\)
g \(\left(3 a^{4} b^{-3}\right)^{-3} \div 6 a^{2} b^{-7}\)
h \(\frac{2 m^{3} n^{4}}{(5 m)^{2}} \times \frac{10 m}{3 n^{-4}}\)
i \(\frac{\left(5 a^{4} b^{-3}\right)^{2}}{a^{-2} b} \div \frac{5\left(a^{-1} b\right)^{-3}}{a b^{-4}}\)

8 Evaluate:
a \(49^{\frac{1}{2}}\)
b \(8^{\frac{1}{3}}\)
c \(125^{\frac{2}{3}}\)
d \(64^{\frac{2}{3}}\)
e \(9^{\frac{3}{2}}\)
f \(16^{-\frac{3}{2}}\)
g \(\left(\frac{1}{8}\right)^{-\frac{2}{3}}\)
h \(\left(\frac{64}{27}\right)^{-\frac{2}{3}}\)

9 Simplify, expressing the answer with positive indices.
a \(a^{\frac{1}{2}} \times a^{\frac{1}{3}}\)
b \(3 b^{\frac{2}{3}} \times 4 b\)
c \(p^{\frac{2}{3}} \div p^{\frac{1}{2}}\)
d \(m^{-\frac{1}{2}} \div m^{-2}\)
e \(\left(4 x^{\frac{1}{3}}\right)^{4}\)
f \(\left(2 x^{-\frac{1}{3}}\right)^{-2}\)
g \(\left(9 a^{\frac{1}{3}}\right)^{\frac{1}{2}}\)
h \(\left(8 p^{-2} q^{3}\right)^{\frac{1}{2}}\)

10 Write in scientific notation.
a 47000
b 164000000
c 0.0047
d 0.0035
e 840
f 0.840

11 Write in decimal form.
a \(6.8 \times 10^{4}\)
b \(7.5 \times 10^{3}\)
c \(2.6 \times 10^{-3}\)
d \(9.4 \times 10^{-2}\)
e \(6.7 \times 10^{0}\)
f \(3.2 \times 10^{-4}\)

12 Simplify, writing each answer in scientific notation.
a \(\left(3.1 \times 10^{4}\right) \times\left(2 \times 10^{-2}\right)\)
b \(\left(5.2 \times 10^{-7}\right) \div\left(2 \times 10^{3}\right)\)
c \(\left(3 \times 10^{4}\right)^{3}\)
d \(\left(5 \times 10^{-4}\right)^{2}\)
e \(\left(4 \times 10^{2}\right)^{2} \times\left(5 \times 10^{-6}\right)\)
f \(\frac{\left(3 \times 10^{4}\right)^{3}}{9 \times 10^{-2}}\)

13 Write in scientific notation correct to the number of significant figures indicated in the brackets.
a 18.62
(2)
b 18.62
(3)
c 18.62
d 0.004276
(2)
e 5973.4
(2)
f 0.473952

\section*{Challenge exercise}
1 The following table gives information regarding the population, approximate rate of population growth, and land area of several countries in 1996. Use the information to answer the questions below.

\begin{center}
\begin{tabular}{|l|c|c|c|}
\hline
Country & \begin{tabular}{c}
Population \\
(in millions) \\
\end{tabular} & \begin{tabular}{c}
Rate of population \\
growth p.a. \\
\end{tabular} & \begin{tabular}{c}
Area of country \\
(in \(\mathbf{k m}^{2}\) ) \\
\end{tabular} \\
\hline
Australia & 18.2 & \(1.4 \%\) & \(7.6 \times 10^{6}\) \\
\hline
China & 1200 & \(1.1 \%\) & \(9.6 \times 10^{6}\) \\
\hline
Indonesia & 201 & \(1.6 \%\) & \(1.9 \times 10^{6}\) \\
\hline
Singapore & 2.9 & \(1.1 \%\) & 633 \\
\hline
New Zealand & 3.4 & \(0.6 \%\) & \(2.7 \times 10^{5}\) \\
\hline
United States & 261 & \(1.0 \%\) & \(9.2 \times 10^{6}\) \\
\hline
\end{tabular}
\end{center}

a How many times larger is Australia than Singapore, correct to the nearest whole number?

b How many times larger is China than New Zealand, correct to 1 decimal place?

c Express Indonesia's population as a percentage of the United States' population. (Calculate your answer correct to the nearest per cent.)

d The population density of a country is defined to be the average number of people for each square kilometre of land. Calculate the population density for each country and find:

i how many times larger the population density of China is compared to Australia

ii the country with the lowest population density

iii the country with the highest population density

e If each country maintains its present rate of population growth, what will the population of each country be (assume compounding growth):
i 5 years from 1996 ?
ii 10 years from 1996 ?

iii 100 years from 1996 ?

f Find the population density of each country 100 years from 1996, assuming each country maintains its present growth rate.

2 a Consider the equilateral triangle shown opposite.

If each side of the triangle is of length \(3 \mathrm{~cm}\),

what is the perimeter of the triangle?

\begin{center}
\includegraphics[max width=\textwidth]{2023_12_09_892d2a0f5cd5401d8048g-28}
\end{center}

b Suppose that the following procedure is performed on the triangle above: 'On the outside of each side of the triangle, draw a triangle with side lengths one-third of the side lengths of the original triangle'.


\end{document}
|727204876|kitty|/Applications/kitty.app|0|
a \(4^{3}\)
|727204896|kitty|/Applications/kitty.app|0|
c \(8^{2}\)
|727204897|kitty|/Applications/kitty.app|0|
b \(900^{3}\)
c \(315^{4}\)
d \(490^{5}\)
|727204901|kitty|/Applications/kitty.app|0|
b \(b^{4} \times b^{9}\)
|727204906|kitty|/Applications/kitty.app|0|
c \(3 a^{4} \times 5 a^{5}\)
|727204906|kitty|/Applications/kitty.app|0|
f \(m^{12} \times m^{6}\)
|727204907|kitty|/Applications/kitty.app|0|
l \(\left(3 m^{2}\right)^{4}\)
|727204912|kitty|/Applications/kitty.app|0|
\(\mathbf{m} a^{0}\)
|727204913|kitty|/Applications/kitty.app|0|
o \(5 m^{0}\)
|727204914|kitty|/Applications/kitty.app|0|
p \((3 q)^{0}\)
|727204915|kitty|/Applications/kitty.app|0|
b \(2 m^{4} n^{3} \times 5 m^{6} n^{7}\)
|727204921|kitty|/Applications/kitty.app|0|
g \(\frac{5 a^{6} b^{7}}{4 a^{3} b^{2}} \times \frac{12 a^{10} b^{9}}{a^{6} b^{7}}\)
|727204923|kitty|/Applications/kitty.app|0|
j \(\frac{(4 a b)^{3} \times 5 a^{2} b}{\left(10 a b^{2}\right)^{2}}\)
|727204926|kitty|/Applications/kitty.app|0|
l \(\frac{\left(4 a^{2} b^{3}\right)^{2}}{3 a^{6} b^{4}} \div \frac{a b^{5}}{\left(3 a^{3} b^{2}\right)^{3}}\)
|727204929|kitty|/Applications/kitty.app|0|
\(\mathbf{k} \frac{\left(3 x^{2} y\right)^{3} \times 2\left(x y^{2}\right)^{3}}{(3 x y)^{4}}\)
|727204930|kitty|/Applications/kitty.app|0|
\(\mathbf{g}\left(2^{\frac{1}{2}}\right)^{-3}\)
|727204939|kitty|/Applications/kitty.app|0|
d \(4^{-3}\)
|727204940|kitty|/Applications/kitty.app|0|
b \((4 y)^{-3}\)
|727204945|kitty|/Applications/kitty.app|0|
g \(\left(\frac{5 g^{2}}{h^{-3}}\right)^{-2}\)
|727204947|kitty|/Applications/kitty.app|0|
\(\mathbf{h}\left(\frac{m^{-3}}{(2 n)^{-4}}\right)^{-2}\)
|727204948|kitty|/Applications/kitty.app|0|
\(\mathbf{f}\left(\frac{a^{3}}{b^{2}}\right)^{-2}\)
|727204949|kitty|/Applications/kitty.app|0|
a \(4 a^{2} \times 5 a^{-3}\)
|727204955|kitty|/Applications/kitty.app|0|
a \(4 a^{2} \times 5 a^{-3}\)
c \(14 a^{-4} \div 7 a^{-5}\)
|727204957|kitty|/Applications/kitty.app|0|
a \(4 a^{2} \times 5 a^{-3}\)
c \(14 a^{-4} \div 7 a^{-5}\)
h \(\frac{2 m^{3} n^{4}}{(5 m)^{2}} \times \frac{10 m}{3 n^{-4}}\)
|727204960|kitty|/Applications/kitty.app|0|
b \(8 m^{2} n^{-3} \times 5 m^{-4} n^{6}\)
d \(\frac{18 m^{2} n^{-3}}{9 m^{4} n^{-1}}\)
e \(\left(4 m^{2} n^{-2}\right)^{-3}\)
f \(\frac{\left(5 m^{2} n\right)^{-2}}{10 m^{4} n^{3}}\)
g \(\left(3 a^{4} b^{-3}\right)^{-3} \div 6 a^{2} b^{-7}\)
i \(\frac{\left(5 a^{4} b^{-3}\right)^{2}}{a^{-2} b} \div \frac{5\left(a^{-1} b\right)^{-3}}{a b^{-4}}\)
|727204964|kitty|/Applications/kitty.app|0|
a \(49^{\frac{1}{2}}\)
|727204969|kitty|/Applications/kitty.app|0|
a \(49^{\frac{1}{2}}\)
c \(125^{\frac{2}{3}}\)
|727204970|kitty|/Applications/kitty.app|0|
a \(49^{\frac{1}{2}}\)
c \(125^{\frac{2}{3}}\)
g \(\left(\frac{1}{8}\right)^{-\frac{2}{3}}\)
|727204972|kitty|/Applications/kitty.app|0|
b \(8^{\frac{1}{3}}\)
d \(64^{\frac{2}{3}}\)
e \(9^{\frac{3}{2}}\)
f \(16^{-\frac{3}{2}}\)
h \(\left(\frac{64}{27}\right)^{-\frac{2}{3}}\)
|727204975|kitty|/Applications/kitty.app|0|
a \(a^{\frac{1}{2}} \times a^{\frac{1}{3}}\)
|727204982|kitty|/Applications/kitty.app|0|
d \(m^{-\frac{1}{2}} \div m^{-2}\)
|727204984|kitty|/Applications/kitty.app|0|
e \(\left(4 x^{\frac{1}{3}}\right)^{4}\)
|727204984|kitty|/Applications/kitty.app|0|
g \(\left(9 a^{\frac{1}{3}}\right)^{\frac{1}{2}}\)
|727204985|kitty|/Applications/kitty.app|0|
a 47000
|727204991|kitty|/Applications/kitty.app|0|
e 840
|727204992|kitty|/Applications/kitty.app|0|
f 0.840
|727204992|kitty|/Applications/kitty.app|0|
b \(7.5 \times 10^{3}\)
|727204996|kitty|/Applications/kitty.app|0|
e \(6.7 \times 10^{0}\)
|727204997|kitty|/Applications/kitty.app|0|
c \(2.6 \times 10^{-3}\)
|727204997|kitty|/Applications/kitty.app|0|
b \(\left(5.2 \times 10^{-7}\right) \div\left(2 \times 10^{3}\right)\)
c \(\left(3 \times 10^{4}\right)^{3}\)
d \(\left(5 \times 10^{-4}\right)^{2}\)
e \(\left(4 \times 10^{2}\right)^{2} \times\left(5 \times 10^{-6}\right)\)
|727205002|kitty|/Applications/kitty.app|0|
\section*{Challenge exercise}
1 The following table gives information regarding the population, approximate rate of population growth, and land area of several countries in 1996. Use the information to answer the questions below.

\begin{center}
\begin{tabular}{|l|c|c|c|}
\hline
Country & \begin{tabular}{c}
Population \\
(in millions) \\
\end{tabular} & \begin{tabular}{c}
Rate of population \\
growth p.a. \\
\end{tabular} & \begin{tabular}{c}
Area of country \\
(in \(\mathbf{k m}^{2}\) ) \\
\end{tabular} \\
\hline
Australia & 18.2 & \(1.4 \%\) & \(7.6 \times 10^{6}\) \\
\hline
China & 1200 & \(1.1 \%\) & \(9.6 \times 10^{6}\) \\
\hline
Indonesia & 201 & \(1.6 \%\) & \(1.9 \times 10^{6}\) \\
\hline
Singapore & 2.9 & \(1.1 \%\) & 633 \\
\hline
New Zealand & 3.4 & \(0.6 \%\) & \(2.7 \times 10^{5}\) \\
\hline
United States & 261 & \(1.0 \%\) & \(9.2 \times 10^{6}\) \\
\hline
\end{tabular}
\end{center}

a How many times larger is Australia than Singapore, correct to the nearest whole number?

b How many times larger is China than New Zealand, correct to 1 decimal place?

c Express Indonesia's population as a percentage of the United States' population. (Calculate your answer correct to the nearest per cent.)

d The population density of a country is defined to be the average number of people for each square kilometre of land. Calculate the population density for each country and find:

i how many times larger the population density of China is compared to Australia

ii the country with the lowest population density

iii the country with the highest population density

e If each country maintains its present rate of population growth, what will the population of each country be (assume compounding growth):
i 5 years from 1996 ?
ii 10 years from 1996 ?

iii 100 years from 1996 ?

f Find the population density of each country 100 years from 1996, assuming each country maintains its present growth rate.

2 a Consider the equilateral triangle shown opposite.

If each side of the triangle is of length \(3 \mathrm{~cm}\),

what is the perimeter of the triangle?

\begin{center}
\includegraphics[max width=\textwidth]{2023_12_09_892d2a0f5cd5401d8048g-28}
\end{center}

b Suppose that the following procedure is performed on the triangle above: 'On the outside of each side of the triangle, draw a triangle with side lengths one-third of the side lengths of the original triangle'.


\end{document}
|727205008|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection*{Exercises:}
    \begin{questions}
        \Question[3] State the base and index of:
        \begin{parts}
            \part \(6^{4}\)
            \part \(7^{3}\)
            \part \(8^{2}\)
        \end{parts}
        \Question[3] Express as a power of a prime number.
        \begin{parts}
            \part 8
            \part 27
            \part 64
        \end{parts}
        \Question[3] Evaluate:
        \begin{parts}
            \part \(3^{4}\)
            \part \(2^{7}\)
            \part \(5^{5}\)
        \end{parts}
        \Question[3] Express as a product of powers of prime numbers.
        \begin{parts}
            \part 18
            \part 24
            \part 144
        \end{parts}
        \Question[3] Simplify:
        \begin{parts}\begin{multicols}{3}
            \part \(2^{7} \times 2^{3}\)
            \part \(3^{3} \times 3^{4} \times 3^{5}\)
            \part \(3 x^{2} \times 4 x^{3}\)
        \end{multicols}\end{parts}
        \Question[4] Simplify:
        \begin{parts}\begin{multicols}{2}
            \part \(a^{2} b^{3} \times b^{2}\)
            \part \(a^{3} b \times a^{2} b^{3}\)
            \part \(2 x y^{2} \times 3 x^{2} y\)
            \part \(4 a^{3} b^{2} \times a^{2} b^{4}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part $\frac{3^{7}}{3^{2}}$
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
            \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
            \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
            \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(a^{4} \times \fillin[]=a^{10}\)
            \part \(9 d^{5} \times \fillin[]=27 d^{6}\)
            \part \(15 d^{7} \div \fillin[]=3 d^{2}\)
            \part \(8 a b^{4} \times \fillin[]=24 a^{2} b^{6}\)
            \part \(\ell^{6} m^{7} \div \fillin[]=\ell^{2} m^{5}\)
            \part \(b^{7} \times \fillin[]=b^{16}\)
        \end{multicols}\end{parts}
        \Question[2]
        \begin{parts}\begin{multicols}{2}
            \part \(a^{0}\)
            \part \(2 x^{0}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(3 a^{0}\)
            \part \(6 a^{0}\)
            \part \(4 a^{0}+3 b^{0}\)
            \part \(6 a^{0}+7 m^{0}\)
            \part \((4 b)^{0}+2 b^{0}\)
            \part \((3 b)^{0}-5 d^{0}\)
        \end{multicols}\end{parts}
        \Question[2] Simplify, leaving the answer as a power.
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2^{3}\right)^{4}\)
            \part \(\left(3^{2}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
            \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
            \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
            \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \((3 a)^{2}\)
            \part \((2 x)^{3}\)
            \part \(\left(\frac{a}{5}\right)^{2}\)
            \part \(\left(\frac{2}{x}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
            \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
            \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
            \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
            \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
            \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
            \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
        \end{multicols}\end{parts}
    \end{questions}
\end{exercisebox}

|727236458|kitty|/Applications/kitty.app|0|
            \part \(\frac{3^{7}}{3^{2}}\)
|727236719|kitty|/Applications/kitty.app|0|
        \Question[6]
|727236763|kitty|/Applications/kitty.app|0|
Package hyperref Warning: Token not allowed in a PDF string (Unicode):
(hyperref)                removing `math shift' on input line 274.
|727236844|kitty|/Applications/kitty.app|0|
        \Question[4] Simplify:
        \begin{parts}\begin{multicols}{2}
            \part \(a^{2} b^{3} \times b^{2}\)
            \part \(a^{3} b \times a^{2} b^{3}\)
            \part \(2 x y^{2} \times 3 x^{2} y\)
            \part \(4 a^{3} b^{2} \times a^{2} b^{4}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^7}{3^2}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
|727236906|kitty|/Applications/kitty.app|0|
        \Question[4] Simplify:
        \begin{parts}\begin{multicols}{2}
            \part \(a^{2} b^{3} \times b^{2}\)
            \part \(a^{3} b \times a^{2} b^{3}\)
            \part \(2 x y^{2} \times 3 x^{2} y\)
            \part \(4 a^{3} b^{2} \times a^{2} b^{4}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^7}{3^2}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
            \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
            \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
            \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(a^{4} \times \fillin[]=a^{10}\)
            \part \(9 d^{5} \times \fillin[]=27 d^{6}\)
            \part \(15 d^{7} \div \fillin[]=3 d^{2}\)
            \part \(8 a b^{4} \times \fillin[]=24 a^{2} b^{6}\)
            \part \(\ell^{6} m^{7} \div \fillin[]=\ell^{2} m^{5}\)
            \part \(b^{7} \times \fillin[]=b^{16}\)
        \end{multicols}\end{parts}
        \Question[2]
        \begin{parts}\begin{multicols}{2}
            \part \(a^{0}\)
            \part \(2 x^{0}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(3 a^{0}\)
            \part \(6 a^{0}\)
            \part \(4 a^{0}+3 b^{0}\)
            \part \(6 a^{0}+7 m^{0}\)
            \part \((4 b)^{0}+2 b^{0}\)
            \part \((3 b)^{0}-5 d^{0}\)
        \end{multicols}\end{parts}
        \Question[2] Simplify, leaving the answer as a power.
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2^{3}\right)^{4}\)
            \part \(\left(3^{2}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
            \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
            \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
            \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \((3 a)^{2}\)
            \part \((2 x)^{3}\)
            \part \(\left(\frac{a}{5}\right)^{2}\)
            \part \(\left(\frac{2}{x}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
            \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
            \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
            \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
            \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
            \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
            \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
        \end{multicols}\end{parts}
|727236940|kitty|/Applications/kitty.app|0|
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
            \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
            \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
            \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(a^{4} \times \fillin[]=a^{10}\)
            \part \(9 d^{5} \times \fillin[]=27 d^{6}\)
            \part \(15 d^{7} \div \fillin[]=3 d^{2}\)
            \part \(8 a b^{4} \times \fillin[]=24 a^{2} b^{6}\)
            \part \(\ell^{6} m^{7} \div \fillin[]=\ell^{2} m^{5}\)
            \part \(b^{7} \times \fillin[]=b^{16}\)
        \end{multicols}\end{parts}
        \Question[2]
        \begin{parts}\begin{multicols}{2}
            \part \(a^{0}\)
            \part \(2 x^{0}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(3 a^{0}\)
            \part \(6 a^{0}\)
            \part \(4 a^{0}+3 b^{0}\)
            \part \(6 a^{0}+7 m^{0}\)
            \part \((4 b)^{0}+2 b^{0}\)
            \part \((3 b)^{0}-5 d^{0}\)
        \end{multicols}\end{parts}
        \Question[2] Simplify, leaving the answer as a power.
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2^{3}\right)^{4}\)
            \part \(\left(3^{2}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
            \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
            \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
            \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \((3 a)^{2}\)
            \part \((2 x)^{3}\)
            \part \(\left(\frac{a}{5}\right)^{2}\)
            \part \(\left(\frac{2}{x}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
            \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
            \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
            \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
            \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
            \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
            \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
        \end{multicols}\end{parts}
|727236968|kitty|/Applications/kitty.app|0|
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^7}{3^2}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
|727236992|kitty|/Applications/kitty.app|0|
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^7}{3^2}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
        \end{multicols}\end{parts}
|727237009|kitty|/Applications/kitty.app|0|
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
            \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
            \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
            \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
        \end{multicols}\end{parts}
|727237012|kitty|/Applications/kitty.app|0|
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(a^{4} \times \fillin[]=a^{10}\)
            \part \(9 d^{5} \times \fillin[]=27 d^{6}\)
            \part \(15 d^{7} \div \fillin[]=3 d^{2}\)
            \part \(8 a b^{4} \times \fillin[]=24 a^{2} b^{6}\)
            \part \(\ell^{6} m^{7} \div \fillin[]=\ell^{2} m^{5}\)
            \part \(b^{7} \times \fillin[]=b^{16}\)
        \end{multicols}\end{parts}
|727237015|kitty|/Applications/kitty.app|0|
        \Question[2]
        \begin{parts}\begin{multicols}{2}
            \part \(a^{0}\)
            \part \(2 x^{0}\)
        \end{multicols}\end{parts}
|727237018|kitty|/Applications/kitty.app|0|
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(3 a^{0}\)
            \part \(6 a^{0}\)
            \part \(4 a^{0}+3 b^{0}\)
            \part \(6 a^{0}+7 m^{0}\)
            \part \((4 b)^{0}+2 b^{0}\)
            \part \((3 b)^{0}-5 d^{0}\)
        \end{multicols}\end{parts}
|727237020|kitty|/Applications/kitty.app|0|
        \Question[2] Simplify, leaving the answer as a power.
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2^{3}\right)^{4}\)
            \part \(\left(3^{2}\right)^{3}\)
        \end{multicols}\end{parts}
|727237022|kitty|/Applications/kitty.app|0|
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
            \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
            \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
            \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
        \end{multicols}\end{parts}
|727237024|kitty|/Applications/kitty.app|0|
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \((3 a)^{2}\)
            \part \((2 x)^{3}\)
            \part \(\left(\frac{a}{5}\right)^{2}\)
            \part \(\left(\frac{2}{x}\right)^{3}\)
        \end{multicols}\end{parts}
|727237027|kitty|/Applications/kitty.app|0|
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
            \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
            \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
            \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
        \end{multicols}\end{parts}
|727237030|kitty|/Applications/kitty.app|0|
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^7}{3^2}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
            \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
            \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
            \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(a^{4} \times \fillin[]=a^{10}\)
            \part \(9 d^{5} \times \fillin[]=27 d^{6}\)
            \part \(15 d^{7} \div \fillin[]=3 d^{2}\)
            \part \(8 a b^{4} \times \fillin[]=24 a^{2} b^{6}\)
            \part \(\ell^{6} m^{7} \div \fillin[]=\ell^{2} m^{5}\)
            \part \(b^{7} \times \fillin[]=b^{16}\)
        \end{multicols}\end{parts}
        \Question[2]
        \begin{parts}\begin{multicols}{2}
            \part \(a^{0}\)
            \part \(2 x^{0}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(3 a^{0}\)
            \part \(6 a^{0}\)
            \part \(4 a^{0}+3 b^{0}\)
            \part \(6 a^{0}+7 m^{0}\)
            \part \((4 b)^{0}+2 b^{0}\)
            \part \((3 b)^{0}-5 d^{0}\)
        \end{multicols}\end{parts}
        \Question[2] Simplify, leaving the answer as a power.
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2^{3}\right)^{4}\)
            \part \(\left(3^{2}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
            \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
            \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
            \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \((3 a)^{2}\)
            \part \((2 x)^{3}\)
            \part \(\left(\frac{a}{5}\right)^{2}\)
            \part \(\left(\frac{2}{x}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
            \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
            \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
            \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
            \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
            \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
            \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
        \end{multicols}\end{parts}
|727237106|kitty|/Applications/kitty.app|0|
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^7}{3^2}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
            \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
            \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
            \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(a^{4} \times \fillin[]=a^{10}\)
            \part \(9 d^{5} \times \fillin[]=27 d^{6}\)
            \part \(15 d^{7} \div \fillin[]=3 d^{2}\)
            \part \(8 a b^{4} \times \fillin[]=24 a^{2} b^{6}\)
            \part \(\ell^{6} m^{7} \div \fillin[]=\ell^{2} m^{5}\)
            \part \(b^{7} \times \fillin[]=b^{16}\)
        \end{multicols}\end{parts}
        \Question[2]
        \begin{parts}\begin{multicols}{2}
            \part \(a^{0}\)
            \part \(2 x^{0}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(3 a^{0}\)
            \part \(6 a^{0}\)
            \part \(4 a^{0}+3 b^{0}\)
            \part \(6 a^{0}+7 m^{0}\)
            \part \((4 b)^{0}+2 b^{0}\)
            \part \((3 b)^{0}-5 d^{0}\)
        \end{multicols}\end{parts}
        \Question[2] Simplify, leaving the answer as a power.
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2^{3}\right)^{4}\)
            \part \(\left(3^{2}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
            \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
            \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
            \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \((3 a)^{2}\)
            \part \((2 x)^{3}\)
            \part \(\left(\frac{a}{5}\right)^{2}\)
            \part \(\left(\frac{2}{x}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
            \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
            \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
            \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
            \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
            \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
            \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
        \end{multicols}\end{parts}
|727237116|kitty|/Applications/kitty.app|0|
        \begin{parts}\begin{multicols}{3}
            \part \(2^{7} \times 2^{3}\)
            \part \(3^{3} \times 3^{4} \times 3^{5}\)
            \part \(3 x^{2} \times 4 x^{3}\)
        \end{multicols}\end{parts}
        \Question[4] Simplify:
        \begin{parts}\begin{multicols}{2}
            \part \(a^{2} b^{3} \times b^{2}\)
            \part \(a^{3} b \times a^{2} b^{3}\)
            \part \(2 x y^{2} \times 3 x^{2} y\)
            \part \(4 a^{3} b^{2} \times a^{2} b^{4}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^7}{3^2}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
            \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
            \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
            \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(a^{4} \times \fillin[]=a^{10}\)
            \part \(9 d^{5} \times \fillin[]=27 d^{6}\)
            \part \(15 d^{7} \div \fillin[]=3 d^{2}\)
            \part \(8 a b^{4} \times \fillin[]=24 a^{2} b^{6}\)
            \part \(\ell^{6} m^{7} \div \fillin[]=\ell^{2} m^{5}\)
            \part \(b^{7} \times \fillin[]=b^{16}\)
        \end{multicols}\end{parts}
        \Question[2]
        \begin{parts}\begin{multicols}{2}
            \part \(a^{0}\)
            \part \(2 x^{0}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(3 a^{0}\)
            \part \(6 a^{0}\)
            \part \(4 a^{0}+3 b^{0}\)
            \part \(6 a^{0}+7 m^{0}\)
            \part \((4 b)^{0}+2 b^{0}\)
            \part \((3 b)^{0}-5 d^{0}\)
        \end{multicols}\end{parts}
        \Question[2] Simplify, leaving the answer as a power.
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2^{3}\right)^{4}\)
            \part \(\left(3^{2}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
            \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
            \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
            \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \((3 a)^{2}\)
            \part \((2 x)^{3}\)
            \part \(\left(\frac{a}{5}\right)^{2}\)
            \part \(\left(\frac{2}{x}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
            \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
            \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
            \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
            \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
            \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
            \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
        \end{multicols}\end{parts}
|727237172|kitty|/Applications/kitty.app|0|

\begin{exercisebox}
    \subsection*{Exercises:}
    \begin{questions}
        \Question[3] State the base and index of:
        \begin{parts}
            \part \(6^{4}\)
            \part \(7^{3}\)
            \part \(8^{2}\)
        \end{parts}
        \Question[3] Express as a power of a prime number.
        \begin{parts}
            \part 8
            \part 27
            \part 64
        \end{parts}
        \Question[3] Evaluate:
        \begin{parts}
            \part \(3^{4}\)
            \part \(2^{7}\)
            \part \(5^{5}\)
        \end{parts}
        \Question[3] Express as a product of powers of prime numbers.
        \begin{parts}
            \part 18
            \part 24
            \part 144
        \end{parts}
        \Question[3] Simplify:
        \begin{parts}\begin{multicols}{3}
            \part \(2^{7} \times 2^{3}\)
            \part \(3^{3} \times 3^{4} \times 3^{5}\)
            \part \(3 x^{2} \times 4 x^{3}\)
        \end{multicols}\end{parts}
        \Question[4] Simplify:
        \begin{parts}\begin{multicols}{2}
            \part \(a^{2} b^{3} \times b^{2}\)
            \part \(a^{3} b \times a^{2} b^{3}\)
            \part \(2 x y^{2} \times 3 x^{2} y\)
            \part \(4 a^{3} b^{2} \times a^{2} b^{4}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^7}{3^2}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
            \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
            \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
            \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(a^{4} \times \fillin[]=a^{10}\)
            \part \(9 d^{5} \times \fillin[]=27 d^{6}\)
            \part \(15 d^{7} \div \fillin[]=3 d^{2}\)
            \part \(8 a b^{4} \times \fillin[]=24 a^{2} b^{6}\)
            \part \(\ell^{6} m^{7} \div \fillin[]=\ell^{2} m^{5}\)
            \part \(b^{7} \times \fillin[]=b^{16}\)
        \end{multicols}\end{parts}
        \Question[2]
        \begin{parts}\begin{multicols}{2}
            \part \(a^{0}\)
            \part \(2 x^{0}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(3 a^{0}\)
            \part \(6 a^{0}\)
            \part \(4 a^{0}+3 b^{0}\)
            \part \(6 a^{0}+7 m^{0}\)
            \part \((4 b)^{0}+2 b^{0}\)
            \part \((3 b)^{0}-5 d^{0}\)
        \end{multicols}\end{parts}
        \Question[2] Simplify, leaving the answer as a power.
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2^{3}\right)^{4}\)
            \part \(\left(3^{2}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
            \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
            \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
            \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \((3 a)^{2}\)
            \part \((2 x)^{3}\)
            \part \(\left(\frac{a}{5}\right)^{2}\)
            \part \(\left(\frac{2}{x}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
            \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
            \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
            \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
            \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
            \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
            \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
        \end{multicols}\end{parts}
    \end{questions}
\end{exercisebox}

|727237240|kitty|/Applications/kitty.app|0|
  and  |727237322|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection*{Exercises:}
    \begin{questions}
        \Question[3] State the base and index of:
        \begin{parts}
            \part \(6^{4}\)
            \part \(7^{3}\)
            \part \(8^{2}\)
        \end{parts}
        \Question[3] Express as a power of a prime number.
        \begin{parts}
            \part 8
            \part 27
            \part 64
        \end{parts}
        \Question[3] Evaluate:
        \begin{parts}
            \part \(3^{4}\)
            \part \(2^{7}\)
            \part \(5^{5}\)
        \end{parts}
        \Question[3] Express as a product of powers of prime numbers.
        \begin{parts}
            \part 18
            \part 24
            \part 144
        \end{parts}
        \Question[3] Simplify:
        \begin{parts}\begin{multicols}{3}
            \part \(2^{7} \times 2^{3}\)
            \part \(3^{3} \times 3^{4} \times 3^{5}\)
            \part \(3 x^{2} \times 4 x^{3}\)
        \end{multicols}\end{parts}
        \Question[4] Simplify:
        \begin{parts}\begin{multicols}{2}
            \part \(a^{2} b^{3} \times b^{2}\)
            \part \(a^{3} b \times a^{2} b^{3}\)
            \part \(2 x y^{2} \times 3 x^{2} y\)
            \part \(4 a^{3} b^{2} \times a^{2} b^{4}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^7}{3^2}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
            \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
            \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
            \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(a^{4} \times \fillin[]=a^{10}\)
            \part \(9 d^{5} \times \fillin[]=27 d^{6}\)
            \part \(15 d^{7} \div \fillin[]=3 d^{2}\)
            \part \(8 a b^{4} \times \fillin[]=24 a^{2} b^{6}\)
            \part \(\ell^{6} m^{7} \div \fillin[]=\ell^{2} m^{5}\)
            \part \(b^{7} \times \fillin[]=b^{16}\)
        \end{multicols}\end{parts}
        \Question[2]
        \begin{parts}\begin{multicols}{2}
            \part \(a^{0}\)
            \part \(2 x^{0}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(3 a^{0}\)
            \part \(6 a^{0}\)
            \part \(4 a^{0}+3 b^{0}\)
            \part \(6 a^{0}+7 m^{0}\)
            \part \((4 b)^{0}+2 b^{0}\)
            \part \((3 b)^{0}-5 d^{0}\)
        \end{multicols}\end{parts}
        \Question[2] Simplify, leaving the answer as a power.
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2^{3}\right)^{4}\)
            \part \(\left(3^{2}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
            \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
            \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
            \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \((3 a)^{2}\)
            \part \((2 x)^{3}\)
            \part \(\left(\frac{a}{5}\right)^{2}\)
            \part \(\left(\frac{2}{x}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
            \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
            \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
            \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
            \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
            \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
            \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
        \end{multicols}\end{parts}
    \end{questions}
\end{exercisebox}



|727237369|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}
 \subsection*{Exercises:}
 \begin{questions}
 \Question[3] State the base and index of:
 \begin{parts}
 \part \(6^{4}\)
 \part \(7^{3}\)
 \part \(8^{2}\)
 \end{parts}
 \Question[3] Express as a power of a prime number.
 \begin{parts}
 \part 8
 \part 27
 \part 64
 \end{parts}
 \Question[3] Evaluate:
 \begin{parts}
 \part \(3^{4}\)
 \part \(2^{7}\)
 \part \(5^{5}\)
 \end{parts}
 \Question[3] Express as a product of powers of prime numbers.
 \begin{parts}
 \part 18
 \part 24
 \part 144
 \end{parts}
 \Question[3] Simplify:
 \begin{parts}
 \begin{multicols}{3}
 \part \(2^{7} \times 2^{3}\)
 \part \(3^{3} \times 3^{4} \times 3^{5}\)
 \part \(3 x^{2} \times 4 x^{3}\)
 \end{multicols}
 \end{parts}
 \Question[4] Simplify:
 \begin{parts}
 \begin{multicols}{2}
 \part \(a^{2} b^{3} \times b^{2}\)
 \part \(a^{3} b \times a^{2} b^{3}\)
 \part \(2 x y^{2} \times 3 x^{2} y\)
 \part \(4 a^{3} b^{2} \times a^{2} b^{4}\)
 \end{multicols}
 \end{parts}
 \Question[6]
 \begin{parts}
 \begin{multicols}{2}
 \part \(\frac{3^7}{3^2}\)
 \part \(\frac{2^{6}}{2^{2}}\)
 \part \(10^{7} \div 10^{2}\)
 \part \(\frac{10^{12}}{10^{4}}\)
 \part \(\frac{2 x^{3}}{x^{2}}\)
 \part \(\frac{6 x^{5}}{2 x^{2}}\)
 \end{multicols}
 \end{parts}
 \Question[4]
 \begin{parts}
 \begin{multicols}{2}
 \part \(\frac{a^{3} b^{2}}{a^{2}}\)
 \part \(\frac{x^{3} y^{2}}{x y}\)
 \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
 \part \(\frac{15 x y^{3}}{3 y^{2}}\)
 \end{multicols}
 \end{parts}
 \Question[4]
 \begin{parts}
 \begin{multicols}{2}
 \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
 \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
 \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
 \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
 \end{multicols}
 \end{parts}
 \Question[6]
 \begin{parts}
 \begin{multicols}{3}
 \part \(a^{4} \times \fbox{a^{10}}\)
 \part \(9 d^{5} \times \fbox{27 d^{6}}\)
 \part \(15 d^{7} \div \fbox{3 d^{2}}\)
 \part \(8 a b^{4} \times \fbox{24 a^{2} b^{6}}\)
 \part \(\fbox{6} m^{7} \div \fbox{6 m^{5}}\)
 \part \(b^{7} \times \fbox{b^{16}}\)
 \end{multicols}
 \end{parts}
 \Question[2]
 \begin{parts}
 \begin{multicols}{2}
 \part \(a^{0}\)
 \part \(2 x^{0}\)
 \end{multicols}
 \end{parts}
 \Question[6]
 \begin{parts}
 \begin{multicols}{3}
 \part \(3 a^{0}\)
 \part \(6 a^{0}\)
 \part \(4 a^{0}+3 b^{0}\)
 \part \(6 a^{0}+7 m^{0}\)
 \part \((4 b)^{0}+2 b^{0}\)
 \part \((3 b)^{0}-5 d^{0}\)
 \end{multicols}
 \end{parts}
 \Question[2] Simplify, leaving the answer as a power.
 \begin{parts}
 \begin{multicols}{2}
 \part \(\left(2^{3}\right)^{4}\)
 \part \(\left(3^{2}\right)^{3}\)
 \end{multicols}
 \end{parts}
 \Question[4]
 \begin{parts}
 \begin{multicols}{2}
 \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
 \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
 \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
 \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
 \end{multicols}
 \end{parts}
 \Question[4]
 \begin{parts}
 \begin{multicols}{2}
 \part \((3 a)^{2}\)
 \part \((2 x)^{3}\)
 \part \(\left(\frac{a}{5}\right)^{2}\)
 \part \(\left(\frac{2}{x}\right)^{3}\)
 \end{multicols}
 \end{parts}
 \Question[4]
 \begin{parts}
 \begin{multicols}{2}
 \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
 \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
 \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
 \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
 \end{multicols}
 \end{parts}
 \Question[4]
 \begin{parts}
 \begin{multicols}{2}
 \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
 \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
 \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
 \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
 \end{multicols}
 \end{parts}
 \end{questions}
 \end{exercisebox}
|727237513|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection*{Exercises:}
    \begin{questions}
        \Question[3] State the base and index of:
        \begin{parts}
            \part \(6^{4}\)
            \part \(7^{3}\)
            \part \(8^{2}\)
        \end{parts}
        \Question[3] Express as a power of a prime number.
        \begin{parts}
            \part 8
            \part 27
            \part 64
        \end{parts}
        \Question[3] Evaluate:
        \begin{parts}
            \part \(3^{4}\)
            \part \(2^{7}\)
            \part \(5^{5}\)
        \end{parts}
        \Question[3] Express as a product of powers of prime numbers.
        \begin{parts}
            \part 18
            \part 24
            \part 144
        \end{parts}
        \Question[3] Simplify:
        \begin{parts}\begin{multicols}{3}
            \part \(2^{7} \times 2^{3}\)
            \part \(3^{3} \times 3^{4} \times 3^{5}\)
            \part \(3 x^{2} \times 4 x^{3}\)
        \end{multicols}\end{parts}
        \Question[4] Simplify:
        \begin{parts}\begin{multicols}{2}
            \part \(a^{2} b^{3} \times b^{2}\)
            \part \(a^{3} b \times a^{2} b^{3}\)
            \part \(2 x y^{2} \times 3 x^{2} y\)
            \part \(4 a^{3} b^{2} \times a^{2} b^{4}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^7}{3^2}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
            \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
            \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
            \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(a^{4} \times \fillin[]=a^{10}\)
\part \(9 d^{5} \times \fillin[]=27 d^{6}\)
            \part \(15 d^{7} \div \fillin[]=3 d^{2}\)
            \part \(8 a b^{4} \times \fillin[]=24 a^{2} b^{6}\)
            \part \(\ell^{6} m^{7} \div \fillin[]=\ell^{2} m^{5}\)
            \part \(b^{7} \times \fillin[]=b^{16}\)
        \end{multicols}\end{parts}
        \Question[2]
        \begin{parts}\begin{multicols}{2}
            \part \(a^{0}\)
            \part \(2 x^{0}\)
        \end{multicols}\end{parts}
        \Question[6]
        \begin{parts}\begin{multicols}{3}
            \part \(3 a^{0}\)
            \part \(6 a^{0}\)
            \part \(4 a^{0}+3 b^{0}\)
            \part \(6 a^{0}+7 m^{0}\)
            \part \((4 b)^{0}+2 b^{0}\)
            \part \((3 b)^{0}-5 d^{0}\)
        \end{multicols}\end{parts}
        \Question[2] Simplify, leaving the answer as a power.
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2^{3}\right)^{4}\)
            \part \(\left(3^{2}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
            \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
            \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
            \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \((3 a)^{2}\)
            \part \((2 x)^{3}\)
            \part \(\left(\frac{a}{5}\right)^{2}\)
            \part \(\left(\frac{2}{x}\right)^{3}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
            \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
            \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
            \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
            \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
            \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
            \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
        \end{multicols}\end{parts}
    \end{questions}
\end{exercisebox}



|727237535|kitty|/Applications/kitty.app|0|
        \Question[6]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^7}{3^2}\)
            \part \(\frac{2^{6}}{2^{2}}\)
            \part \(10^{7} \div 10^{2}\)
            \part \(\frac{10^{12}}{10^{4}}\)
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
        \end{multicols}\end{parts}
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
            \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
            \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
            \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
        \end{multicols}\end{parts}
|727237953|kitty|/Applications/kitty.app|0|
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
            \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
            \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
            \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
        \end{multicols}\end{parts}
|727237998|kitty|/Applications/kitty.app|0|
*|727238126|kitty|/Applications/kitty.app|0|
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
            \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
            \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
            \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
|727238138|kitty|/Applications/kitty.app|0|
        \begin{parts}\begin{multicols}{2}
|727238363|kitty|/Applications/kitty.app|0|
$\,$
|727238550|kitty|/Applications/kitty.app|0|
multline|727238788|kitty|/Applications/kitty.app|0|
1
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
|727239048|kitty|/Applications/kitty.app|0|
 \textbf{Proof:}

 Consider the product of \( a^n \) and \( a^{-n} \):
 \[ a^n \cdot a^{-n} = a^{n + (-n)} = a^0 \]

 Since anything raised to the power of 0 is 1, we have:
 \[ a^0 = 1 \]

 Therefore:
 \[ a^n \cdot a^{-n} = 1 \]

 Dividing both sides by \( a^n \), we get:
 \[ \frac{a^n \cdot a^{-n}}{a^n} = \frac{1}{a^n} \]

 Simplifying the left side, we have:
 \[ a^{-n} = \frac{1}{a^n} \]

 Hence proved.
|727242508|kitty|/Applications/kitty.app|0|
        \begin{parts}
            \part
            \part
            \part
            \part
        \end{parts}
|727242721|kitty|/Applications/kitty.app|0|
        \begin{parts}
            \part
            \part
            \part
        \end{parts}
|727242743|kitty|/Applications/kitty.app|0|
        \begin{parts}
            \part
            \part
            \part
            \part
            \part
            \part
            \part
            \part
        \end{parts}
|727242782|kitty|/Applications/kitty.app|0|
            \part
            \part
            \part
            \part
|727242786|kitty|/Applications/kitty.app|0|
        \Question[3]
|727242837|kitty|/Applications/kitty.app|0|
        \Question[4]
|727242854|kitty|/Applications/kitty.app|0|
            \part
|727242901|kitty|/Applications/kitty.app|0|
        \begin{parts}\begin{multicols}{3}
            \part
            \part
            \part
        \end{multicols}\end{parts}
|727242906|kitty|/Applications/kitty.app|0|
        \begin{parts}\begin{multicols}{2}
            \part
            \part
            \part
            \part
        \end{multicols}\end{parts}
|727242912|kitty|/Applications/kitty.app|0|
        \begin{parts}\begin{multicols}{2}
            \part
            \part
            \part
            \part
            \part
            \part
        \end{multicols}\end{parts}
|727242933|kitty|/Applications/kitty.app|0|
            \part
            \part
|727242935|kitty|/Applications/kitty.app|0|
a \(6^{-2}\)
b \(4^{-3}\)
c \(2^{-7}\)
d \(10^{-3}\)

\section*{Solution}
a \(6^{-2}=\frac{1}{6^{2}}\)
b \(4^{-3}=\frac{1}{4^{3}}\)
c \(2^{-7}=\frac{1}{2^{7}}\)
d \(10^{-3}=\frac{1}{10^{3}}\)
\(=\frac{1}{36}\)
\(=\frac{1}{64}\)
\(=\frac{1}{128}\)
\(=\frac{1}{1000}\)
|727243283|kitty|/Applications/kitty.app|0|
a \(6^{-2}=\frac{1}{6^{2}}\)
|727243293|kitty|/Applications/kitty.app|0|
\(=\frac{1}{36}\)
|727243298|kitty|/Applications/kitty.app|0|
b \(4^{-3}=\frac{1}{4^{3}}\)
|727243305|kitty|/Applications/kitty.app|0|
\(=\frac{1}{64}\)
|727243311|kitty|/Applications/kitty.app|0|
c \(2^{-7}=\frac{1}{2^{7}}\)
|727243315|kitty|/Applications/kitty.app|0|
\(=\frac{1}{128}\)
|727243317|kitty|/Applications/kitty.app|0|
a \(6^{-2}=\frac{1}{6^{2}}\)
\(=\frac{1}{36}\)
|727243328|kitty|/Applications/kitty.app|0|

\section*{Solution}
|727243339|kitty|/Applications/kitty.app|0|
b \(4^{-3}\)
b \(4^{-3}=\frac{1}{4^{3}}\)
\(=\frac{1}{64}\)
c \(2^{-7}\)
c \(2^{-7}=\frac{1}{2^{7}}\)
\(=\frac{1}{128}\)
d \(10^{-3}\)
d \(10^{-3}=\frac{1}{10^{3}}\)
\(=\frac{1}{1000}\)
|727243340|kitty|/Applications/kitty.app|0|
c \(2^{-7}\)
c \(2^{-7}=\frac{1}{2^{7}}\)
\(=\frac{1}{128}\)
d \(10^{-3}\)
d \(10^{-3}=\frac{1}{10^{3}}\)
\(=\frac{1}{1000}\)
|727243344|kitty|/Applications/kitty.app|0|
d \(10^{-3}\)
d \(10^{-3}=\frac{1}{10^{3}}\)
\(=\frac{1}{1000}\)
|727243347|kitty|/Applications/kitty.app|0|
d \(10^{-3}=\frac{1}{10^{3}}\)
\(=\frac{1}{1000}\)
|727243352|kitty|/Applications/kitty.app|0|
c \(2^{-7}=\frac{1}{2^{7}}\)
\(=\frac{1}{128}\)
|727243363|kitty|/Applications/kitty.app|0|
b \(4^{-3}=\frac{1}{4^{3}}\)
\(=\frac{1}{64}\)
|727243377|kitty|/Applications/kitty.app|0|
a \(\left(\frac{1}{3}\right)^{-1}\)
b \(\left(\frac{2}{7}\right)^{-2}\)
c \(\left(4 \frac{1}{4}\right)^{-2}\)

\section*{Solution}
a \(\left(\frac{1}{3}\right)^{-1}=\frac{3}{1}\)
b \(\left(\frac{2}{7}\right)^{-2}=\left(\frac{7}{2}\right)^{2}\)
c \(\left(4 \frac{1}{4}\right)^{-2}=\left(\frac{17}{4}\right)^{-2}\)
\(=3\)
\(=\frac{49}{4}\)
\(=\left(\frac{4}{17}\right)^{2}\)
\(=\frac{16}{289}\)

Note that in general \(\left(\frac{a}{b}\right)^{-1}=\frac{b}{a}\).
|727243404|kitty|/Applications/kitty.app|0|
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part \(6^{-2}\)
            \begin{solutionordottedlines}[2cm]
                \(6^{-2}=\frac{1}{6^{2}}\)
                \(=\frac{1}{36}\)
            \end{solutionordottedlines}
            \part \(4^{-3}\)
            \begin{solutionordottedlines}[2cm]
                \(4^{-3}=\frac{1}{4^{3}}\)
                \(=\frac{1}{64}\)
            \end{solutionordottedlines}
            \part \(2^{-7}\)
            \begin{solutionordottedlines}[2cm]
                \(2^{-7}=\frac{1}{2^{7}}\)
                \(=\frac{1}{128}\)
            \end{solutionordottedlines}
            \part \(10^{-3}\)
            \begin{solutionordottedlines}[2cm]
                \(10^{-3}=\frac{1}{10^{3}}\)
                \(=\frac{1}{1000}\)
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
|727243435|kitty|/Applications/kitty.app|0|
        \Question[4] Evaluate:
        \begin{parts}
            \part
            \part
            \part
            \part
        \end{parts}
|727243446|kitty|/Applications/kitty.app|0|
        \Question[4]
        \begin{parts}\begin{multicols}{2}
            \part
            \part
            \part
            \part
        \end{multicols}\end{parts}
|727243450|kitty|/Applications/kitty.app|0|
Note that in general \(\left(\frac{a}{b}\right)^{-1}=\frac{b}{a}\).
|727243456|kitty|/Applications/kitty.app|0|
a \(\left(\frac{1}{3}\right)^{-1}\)
b \(\left(\frac{2}{7}\right)^{-2}\)
c \(\left(4 \frac{1}{4}\right)^{-2}\)

\section*{Solution}
a \(\left(\frac{1}{3}\right)^{-1}=\frac{3}{1}\)
b \(\left(\frac{2}{7}\right)^{-2}=\left(\frac{7}{2}\right)^{2}\)
c \(\left(4 \frac{1}{4}\right)^{-2}=\left(\frac{17}{4}\right)^{-2}\)
\(=3\)
\(=\frac{49}{4}\)
\(=\left(\frac{4}{17}\right)^{2}\)
\(=\frac{16}{289}\)
|727243557|kitty|/Applications/kitty.app|0|
\(=3\)
|727243563|kitty|/Applications/kitty.app|0|
\(=\frac{49}{4}\)
|727243565|kitty|/Applications/kitty.app|0|
b \(\left(\frac{2}{7}\right)^{-2}\)
c \(\left(4 \frac{1}{4}\right)^{-2}\)
|727243585|kitty|/Applications/kitty.app|0|
c \(\left(4 \frac{1}{4}\right)^{-2}\)
|727243587|kitty|/Applications/kitty.app|0|

a \(\left(\frac{1}{3}\right)^{-1}=\frac{3}{1}\)
\(=3\)
|727243598|kitty|/Applications/kitty.app|0|
b \(\left(\frac{2}{7}\right)^{-2}=\left(\frac{7}{2}\right)^{2}\)
\(=\frac{49}{4}\)
|727243607|kitty|/Applications/kitty.app|0|
c \(\left(4 \frac{1}{4}\right)^{-2}=\left(\frac{17}{4}\right)^{-2}\)
\(=\left(\frac{4}{17}\right)^{2}\)
\(=\frac{16}{289}\)
|727243617|kitty|/Applications/kitty.app|0|
a \(3^{4} \times 3^{-2}\)
b \(5^{7} \times 5^{-8}\)
c \(13^{-8} \times 13^{15} \times 13^{-7}\)
d \(\left(\frac{2}{3}\right)^{-6} \times\left(\frac{2}{3}\right)^{4}\)
|727243652|kitty|/Applications/kitty.app|0|
b \(5^{7} \times 5^{-8}\)
c \(13^{-8} \times 13^{15} \times 13^{-7}\)
d \(\left(\frac{2}{3}\right)^{-6} \times\left(\frac{2}{3}\right)^{4}\)
|727243655|kitty|/Applications/kitty.app|0|
c \(13^{-8} \times 13^{15} \times 13^{-7}\)
d \(\left(\frac{2}{3}\right)^{-6} \times\left(\frac{2}{3}\right)^{4}\)
|727243657|kitty|/Applications/kitty.app|0|
d \(\left(\frac{2}{3}\right)^{-6} \times\left(\frac{2}{3}\right)^{4}\)
|727243658|kitty|/Applications/kitty.app|0|
a \(\frac{2^{4}}{2^{5}}\)
b \(\frac{3^{4}}{3^{7}}\)
c \(\frac{5}{5^{3}}\)
d \(\frac{3^{4}}{3^{6}}\)
|727243664|kitty|/Applications/kitty.app|0|
b \(\frac{3^{4}}{3^{7}}\)
c \(\frac{5}{5^{3}}\)
d \(\frac{3^{4}}{3^{6}}\)
|727243668|kitty|/Applications/kitty.app|0|
c \(\frac{5}{5^{3}}\)
d \(\frac{3^{4}}{3^{6}}\)
|727243670|kitty|/Applications/kitty.app|0|
d \(\frac{3^{4}}{3^{6}}\)
|727243671|kitty|/Applications/kitty.app|0|
b \(5^{7} \times 5^{-8}=5^{-1}\)

\[
=9
\]
|727243680|kitty|/Applications/kitty.app|0|
c \(\begin{aligned} 13^{-8} \times 13^{15} \times 13^{-7} & =13^{0} \\ & =1\end{aligned}\)

\[
=1
\]

\[
=\frac{1}{5}
\]
|727243685|kitty|/Applications/kitty.app|0|
\(=\frac{1}{2}\)
|727244506|kitty|/Applications/kitty.app|0|
\(=\frac{1}{3^{3}}\)
|727244513|kitty|/Applications/kitty.app|0|
\(=\frac{1}{3^{3}}\)
\(=\frac{1}{27}\)
|727244514|kitty|/Applications/kitty.app|0|
\(=\frac{1}{5^{2}}\)
|727244522|kitty|/Applications/kitty.app|0|
\(=\frac{1}{5^{2}}\)
\(=\frac{1}{25}\)
|727244524|kitty|/Applications/kitty.app|0|
a \(\frac{2^{4}}{2^{5}}=2^{-1}\)
\(=\frac{1}{2}\)
|727244550|kitty|/Applications/kitty.app|0|
b \(\frac{3^{4}}{3^{7}}=3^{-3}\)
\(=\frac{1}{3^{3}}\)
\(=\frac{1}{27}\)
|727244556|kitty|/Applications/kitty.app|0|
c \(\frac{5}{5^{3}}=5^{-2}\)
\(=\frac{1}{5^{2}}\)
\(=\frac{1}{25}\)
|727244565|kitty|/Applications/kitty.app|0|
d \(\frac{3^{4}}{3^{6}}=3^{-2}\)
\(=\frac{1}{3^{2}}\)
\(=\frac{1}{9}\)
|727244570|kitty|/Applications/kitty.app|0|
d \(\left(\frac{2}{3}\right)^{-6} \times\left(\frac{2}{3}\right)^{4}=\left(\frac{2}{3}\right)^{-2}\)

\[
=\left(\frac{3}{2}\right)^{2}
\]

\[
=\frac{9}{4}
\]
|727244698|kitty|/Applications/kitty.app|0|
a \(3^{4} \times 3^{-2}=3^{2}\)
|727244734|kitty|/Applications/kitty.app|0|
a \(a^{2} b^{-3} \times a^{-4} b^{5}\)
b \(\frac{x^{2} y^{3}}{x^{3} y^{2}}\)
c \(\left(2 a^{-2} b^{3}\right)^{-2}\)
d \(\left(\frac{3 m^{2}}{n}\right)^{-4}\)
|727244757|kitty|/Applications/kitty.app|0|
b \(\frac{x^{2} y^{3}}{x^{3} y^{2}}\)
c \(\left(2 a^{-2} b^{3}\right)^{-2}\)
d \(\left(\frac{3 m^{2}}{n}\right)^{-4}\)
|727244761|kitty|/Applications/kitty.app|0|
c \(\left(2 a^{-2} b^{3}\right)^{-2}\)
d \(\left(\frac{3 m^{2}}{n}\right)^{-4}\)
|727244762|kitty|/Applications/kitty.app|0|
d \(\left(\frac{3 m^{2}}{n}\right)^{-4}\)
|727244763|kitty|/Applications/kitty.app|0|
a \(a^{2} b^{-3} \times a^{-4} b^{5}=a^{2-4} \times b^{-3+5}\)
|727244767|kitty|/Applications/kitty.app|0|
\(=\frac{1}{a^{2}} \times b^{2}\)
|727244784|kitty|/Applications/kitty.app|0|
\(=\frac{1}{a^{2}} \times b^{2}\)
\(=\frac{b^{2}}{a^{2}}\)
|727244786|kitty|/Applications/kitty.app|0|
\(=\frac{1}{2^{2}} \times a^{4} \times \frac{1}{b^{6}}\)
\(=\frac{a^{4}}{4 b^{6}}\)
|727244812|kitty|/Applications/kitty.app|0|
d \(\left(\frac{3 m^{2}}{n}\right)^{-4}=\left(\frac{n}{3 m^{2}}\right)^{4}\)
\(=\frac{n^{4}}{81 m^{8}}\)
|727244847|kitty|/Applications/kitty.app|0|
c \(\left(2 a^{-2} b^{3}\right)^{-2}=2^{-2} \times a^{4} \times b^{-6}\)
\(=\frac{1}{2^{2}} \times a^{4} \times \frac{1}{b^{6}}\)
\(=\frac{a^{4}}{4 b^{6}}\)
|727244852|kitty|/Applications/kitty.app|0|
b \(\frac{x^{2} y^{3}}{x^{3} y^{2}}=x^{-1} y^{1}\)
\(=\frac{1}{x} \times y\)
\(=\frac{y}{x}\)
|727244858|kitty|/Applications/kitty.app|0|
inverse|727245098|kitty|/Applications/kitty.app|0|
a \(2^{-1}\)
|727245131|kitty|/Applications/kitty.app|0|
a \(2^{-1}\)
b \(5^{-1}\)
|727245133|kitty|/Applications/kitty.app|0|
l \(2^{-7}\)
|727245144|kitty|/Applications/kitty.app|0|
f \(10^{-2}\)
|727245146|kitty|/Applications/kitty.app|0|
1 Express with a positive index and then evaluate.
|727245148|kitty|/Applications/kitty.app|0|
https://www.fairtrading.nsw.gov.au/housing-and-property/renting/rental-bonds-online|727245853|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
https://www.fairtrading.nsw.gov.au/housing-and-property/renting/rental-bonds-online/for-tenants|727246533|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
Image: 585x726 (6.5 MB)|727246625|Brave Browser Beta|/Applications/Brave Browser Beta.app|1|135a4fc68c1cad4cff20c33f1594890e6fc53524.tiff
a \(\frac{1}{8}\)
b \(\frac{1}{9}\)
|727246821|kitty|/Applications/kitty.app|0|
2 Write each fraction as a power of a prime with a negative index.
|727246835|kitty|/Applications/kitty.app|0|
\(\mathbf{k} \frac{1}{31}\)
l \(\frac{1}{32}\)
|727246845|kitty|/Applications/kitty.app|0|
a \(a^{-3}\)
b \(x^{-7}\)
|727246857|kitty|/Applications/kitty.app|0|
3 Express with positive indices, evaluating where possible.
a \(a^{-3}\)
b \(x^{-7}\)
g \(\frac{3}{a^{-4}}\)
h \(\frac{5}{x^{-5}}\)
|727246863|kitty|/Applications/kitty.app|0|
\(\mathbf{k} \frac{13^{-2}}{n^{-6}}\)
l \(\frac{y^{-4}}{x^{-3}}\)
|727246871|kitty|/Applications/kitty.app|0|
Example
|727246878|kitty|/Applications/kitty.app|0|
\(\mathbf{a}\left(\frac{1}{4}\right)^{-1}\)
\(\mathbf{b}\left(\frac{2}{5}\right)^{-2}\)
|727246889|kitty|/Applications/kitty.app|0|
a \(\frac{2^{3}}{2^{6}}\)
b \(\frac{4^{2}}{4^{4}}\)
|727246926|kitty|/Applications/kitty.app|0|
\(\mathbf{j} \frac{2^{7}}{2^{13}}\)
\(\mathbf{k} \frac{10^{2}}{10^{6}}\)
|727246946|kitty|/Applications/kitty.app|0|
6 Express with negative index.
a \(\frac{3}{x}\)
b \(\frac{5}{x^{2}}\)
c \(\frac{8}{x^{4}}\)
|727246955|kitty|/Applications/kitty.app|0|
7 Evaluate.
\(\mathbf{a}\left(\frac{1}{2}\right)^{-1}\)
\(\mathbf{b}\left(\frac{2}{3}\right)^{-1}\)
c \(\left(\frac{1}{2}\right)^{-2}\)
|727246966|kitty|/Applications/kitty.app|0|
a \(x^{-6} y^{4} \times x^{2} y^{-2}\)
|727246978|kitty|/Applications/kitty.app|0|
a \(x^{-6} y^{4} \times x^{2} y^{-2}\)
d \(2 a^{-1} b^{5} \times 7 a b^{-3}\)
|727246979|kitty|/Applications/kitty.app|0|
a \(x^{-6} y^{4} \times x^{2} y^{-2}\)
d \(2 a^{-1} b^{5} \times 7 a b^{-3}\)
g \(\frac{8 a^{-4}}{2 a^{6}}\)
|727246982|kitty|/Applications/kitty.app|0|
Example 148 Simplify, expressing the answer with positive indices.
a \(x^{-6} y^{4} \times x^{2} y^{-2}\)
d \(2 a^{-1} b^{5} \times 7 a b^{-3}\)
g \(\frac{8 a^{-4}}{2 a^{6}}\)
l \(\frac{36 h^{-9}}{9 h^{-4}}\)
|727246988|kitty|/Applications/kitty.app|0|
\(\mathbf{k} \frac{56 t^{-7}}{8 t^{-2}}\)
|727247000|kitty|/Applications/kitty.app|0|
\(\mathbf{m} \frac{144 x^{7} y^{5}}{12 x^{-3} y^{4}}\)
|727247000|kitty|/Applications/kitty.app|0|
\(\mathbf{p} \frac{9 m^{3} n^{4} p^{-5}}{21 m^{-3} n^{4} p^{2}}\)
|727247001|kitty|/Applications/kitty.app|0|
a \(6^{4} \times \ldots=6^{2}\)
|727247004|kitty|/Applications/kitty.app|0|
a \(6^{4} \times \ldots=6^{2}\)
d \(m^{5} \times \ldots=m^{-6}\)
|727247006|kitty|/Applications/kitty.app|0|
a \(6^{4} \times \ldots=6^{2}\)
d \(m^{5} \times \ldots=m^{-6}\)
g \(d^{-7} \div \ldots=d^{15}\)
|727247008|kitty|/Applications/kitty.app|0|
a \(6^{4} \times \ldots=6^{2}\)
d \(m^{5} \times \ldots=m^{-6}\)
g \(d^{-7} \div \ldots=d^{15}\)
j \(\left(a^{5}\right) \cdots=a^{-15}\)
\(\mathbf{m}(\ldots)^{-2}=\frac{m^{6}}{25}\)
|727247015|kitty|/Applications/kitty.app|0|
i \(\left(m^{-2}\right) \cdots=m^{10}\)
|727247046|kitty|/Applications/kitty.app|0|
\(\mathbf{k}(\ldots)^{-4}=\frac{m^{8}}{16}\)
|727247047|kitty|/Applications/kitty.app|0|
\(\mathbf{q}(\ldots)^{\cdots}=a^{6} b^{-4}\)
|727247049|kitty|/Applications/kitty.app|0|
\(\mathbf{r}(\ldots) \cdots=\frac{m^{6}}{n^{9}}\)
|727247053|kitty|/Applications/kitty.app|0|
Write two possible alternatives for part \(\mathbf{q}\) and part \(\mathbf{r}\).
|727247054|kitty|/Applications/kitty.app|0|
a \(\left(3 a^{2} b^{-2}\right)^{3} \times\left(2 a^{4}\right)^{-2}\)
|727247059|kitty|/Applications/kitty.app|0|
a \(\left(3 a^{2} b^{-2}\right)^{3} \times\left(2 a^{4}\right)^{-2}\)
d \(\left(6 a^{5} b^{-4}\right)^{-3} \times 2\left(a^{3} b^{-3}\right)^{2}\)
|727247061|kitty|/Applications/kitty.app|0|
a \(\left(3 a^{2} b^{-2}\right)^{3} \times\left(2 a^{4}\right)^{-2}\)
d \(\left(6 a^{5} b^{-4}\right)^{-3} \times 2\left(a^{3} b^{-3}\right)^{2}\)
g \(\frac{\left(2 a^{4} b^{-2}\right)^{3}}{c^{2}} \times \frac{\left(2^{2} a^{-3} b^{2}\right)^{-1}}{c}\)
|727247064|kitty|/Applications/kitty.app|0|
10 Simplify, expressing the answers with positive indices. Evaluate powers where possible.
a \(\left(3 a^{2} b^{-2}\right)^{3} \times\left(2 a^{4}\right)^{-2}\)
d \(\left(6 a^{5} b^{-4}\right)^{-3} \times 2\left(a^{3} b^{-3}\right)^{2}\)
g \(\frac{\left(2 a^{4} b^{-2}\right)^{3}}{c^{2}} \times \frac{\left(2^{2} a^{-3} b^{2}\right)^{-1}}{c}\)
\(\mathbf{j} \frac{\left(2 a^{4}\right)^{2}}{b^{7}} \div \frac{\left(a^{2}\right)^{-3}}{2 b}\)
|727247072|kitty|/Applications/kitty.app|0|
1 Express with a positive index and then evaluate.
c \(3^{-2}\)
d \(6^{-2}\)
e \(9^{-2}\)
i \(5^{-3}\)
j \(3^{-4}\)
k \(10^{-5}\)

2 Write each fraction as a power of a prime with a negative index.
c \(\frac{1}{27}\)
d \(\frac{1}{49}\)
e \(\frac{1}{121}\)
f \(\frac{1}{125}\)
i \(\frac{1}{169}\)
j \(\frac{1}{81}\)

3 Express with positive indices, evaluating where possible.
c \(3 a^{-4}\)
d \(5 x^{-7}\)
e \(4 a^{-5}\)
f \(\frac{1}{x^{-3}}\)
i \(3^{-2} a^{-2}\)
j \(4^{-2} x^{-2}\)

Simplify where possible and then evaluate.
c \(\left(3 \frac{1}{3}\right)^{-2}\)
\(\mathbf{d}\left(\frac{2}{3}\right)^{-3}\)
g \(7^{3} \times 7^{-5}\)
h \(4^{3} \times 4^{-5}\)

Example 135 Write as a single power and then evaluate.
c \(\frac{3^{8}}{3^{9}}\)
d \(\frac{6^{5}}{6^{8}}\)
e \(\frac{7^{1}}{7^{3}}\)
f \(\frac{5^{7}}{5^{10}}\)
i \(\frac{3^{5}}{3^{9}}\)
l \(\frac{12^{12}}{12^{14}}\)

6 Express with negative index.
d \(\frac{3}{2 x^{4}}\)
e \(\frac{4}{3 x^{7}}\)
f \(\frac{2}{3 x^{5}}\)

7 Evaluate.
d \(\left(\frac{4}{5}\right)^{-2}\)
e \(\left(2 \frac{1}{4}\right)^{-2}\)
f \(\left(1 \frac{1}{5}\right)^{-3}\)

Example 148 Simplify, expressing the answer with positive indices.
b \(a^{-3} b^{-5} \times a^{5} b^{-3}\)
c \(3 x^{-2} y^{5} \times 5 x^{-7} y^{-2}\)
e \(7 a^{3} m^{-4} \times 8 a^{-5} m^{-3}\)
f \(3 r^{2} s^{3} \times 4 r^{-3} s^{-5}\)
h \(\frac{16 a^{-4}}{8 a^{5}}\)
i \(\frac{18 a^{-4}}{4 a^{5}}\)
j \(\frac{27 m^{-3}}{9 m^{-2}}\)
n \(\frac{72 a^{4} b^{-3}}{36 a b^{-2}}\)
o \(\frac{7 a^{2} b^{-3} c^{-4}}{21 a^{5} b^{-7} c^{-9}}\)

9 Copy and complete.
b \(9^{5} \times \ldots=9^{4}\)
c \(b^{9} \times \ldots=b^{7}\)
e \(a^{11} \div \ldots=a^{14}\)
f \(b^{7} \div \ldots=b^{15}\)
h \(e^{-7} \div \ldots=e^{-5}\)
i \(\left(m^{-2}\right) \cdots=m^{10}\)
l \((\ldots)^{-3}=\frac{1}{27 a^{9}}\)
n \((\ldots)^{-3}=\frac{a^{6}}{b^{9}}\)
o \((\ldots)^{-6}=\frac{m^{12} n^{18}}{p^{6}}\)

10 Simplify, expressing the answers with positive indices. Evaluate powers where possible.
b \(\left(5 x^{4} y^{6}\right)^{-3} \times\left(5^{2} x y^{-1}\right)^{3}\)
c \(\left(5 m^{2} n^{-3}\right)^{-2} \times 2\left(m^{-2} n^{3}\right)^{2}\)
e \(\frac{\left(x^{2}\right)^{2}}{y} \times \frac{\left(y^{2}\right)^{-3}}{x^{3}}\)
f \(\frac{\left(2 x^{3}\right)^{-2}}{y^{4}} \times \frac{\left(2 x^{7}\right)^{2}}{3 y^{5}}\)
h \(\frac{\left(m^{2} n^{3}\right)^{2}}{p^{-3}} \times\left(m n p^{-2}\right)^{-3}\)
i \(\frac{\left(a^{2}\right)^{3}}{b^{3}} \div\left(\frac{a}{b^{2}}\right)^{-2}\)
\(\mathbf{k} \frac{\left(4 c^{4} d^{-3}\right)^{2}}{9} \div \frac{3 c^{-2}}{d}\)
l \(\frac{\left(3 m^{2} n^{3}\right)^{-2}}{p^{4}} \div \frac{p^{-3}}{m}\)
|727247089|kitty|/Applications/kitty.app|0|
a \(2^{-1}\)
b \(5^{-1}\)
g \(2^{-4}\)
h \(3^{-3}\)
|727247132|kitty|/Applications/kitty.app|0|
b \(5^{-1}\)
g \(2^{-4}\)
h \(3^{-3}\)
|727247135|kitty|/Applications/kitty.app|0|
g \(2^{-4}\)
h \(3^{-3}\)
|727247136|kitty|/Applications/kitty.app|0|
h \(3^{-3}\)
|727247138|kitty|/Applications/kitty.app|0|
a \(\frac{1}{8}\)
b \(\frac{1}{9}\)
g \(\frac{1}{16}\)
h \(\frac{1}{64}\)
|727247149|kitty|/Applications/kitty.app|0|
b \(\frac{1}{9}\)
g \(\frac{1}{16}\)
h \(\frac{1}{64}\)
|727247150|kitty|/Applications/kitty.app|0|
g \(\frac{1}{16}\)
h \(\frac{1}{64}\)
|727247152|kitty|/Applications/kitty.app|0|
h \(\frac{1}{64}\)
|727247153|kitty|/Applications/kitty.app|0|
a \(a^{-3}\)
b \(x^{-7}\)
g \(\frac{3}{a^{-4}}\)
h \(\frac{5}{x^{-5}}\)
|727247180|kitty|/Applications/kitty.app|0|
b \(x^{-7}\)
g \(\frac{3}{a^{-4}}\)
h \(\frac{5}{x^{-5}}\)
|727247183|kitty|/Applications/kitty.app|0|
g \(\frac{3}{a^{-4}}\)
h \(\frac{5}{x^{-5}}\)
|727247184|kitty|/Applications/kitty.app|0|
h \(\frac{5}{x^{-5}}\)
|727247185|kitty|/Applications/kitty.app|0|
\(\mathbf{a}\left(\frac{1}{4}\right)^{-1}\)
\(\mathbf{b}\left(\frac{2}{5}\right)^{-2}\)
e \(3^{5} \times 3^{-2}\)
f \(5^{11} \times 5^{-8}\)
|727247194|kitty|/Applications/kitty.app|0|
\(\mathbf{b}\left(\frac{2}{5}\right)^{-2}\)
e \(3^{5} \times 3^{-2}\)
f \(5^{11} \times 5^{-8}\)
|727247196|kitty|/Applications/kitty.app|0|
e \(3^{5} \times 3^{-2}\)
f \(5^{11} \times 5^{-8}\)
|727247197|kitty|/Applications/kitty.app|0|
f \(5^{11} \times 5^{-8}\)
|727247198|kitty|/Applications/kitty.app|0|
e|727247201|kitty|/Applications/kitty.app|0|
a \(\frac{2^{3}}{2^{6}}\)
b \(\frac{4^{2}}{4^{4}}\)
g \(\frac{8^{6}}{8^{7}}\)
h \(\frac{20^{4}}{20^{6}}\)
|727247226|kitty|/Applications/kitty.app|0|
b \(\frac{4^{2}}{4^{4}}\)
g \(\frac{8^{6}}{8^{7}}\)
h \(\frac{20^{4}}{20^{6}}\)
|727247229|kitty|/Applications/kitty.app|0|
g \(\frac{8^{6}}{8^{7}}\)
h \(\frac{20^{4}}{20^{6}}\)
|727247230|kitty|/Applications/kitty.app|0|
h \(\frac{20^{4}}{20^{6}}\)
|727247231|kitty|/Applications/kitty.app|0|
a \(\frac{3}{x}\)
b \(\frac{5}{x^{2}}\)
c \(\frac{8}{x^{4}}\)
|727247243|kitty|/Applications/kitty.app|0|
b \(\frac{5}{x^{2}}\)
c \(\frac{8}{x^{4}}\)
|727247244|kitty|/Applications/kitty.app|0|
c \(\frac{8}{x^{4}}\)
|727247246|kitty|/Applications/kitty.app|0|
\(\mathbf{a}\left(\frac{1}{2}\right)^{-1}\)
\(\mathbf{b}\left(\frac{2}{3}\right)^{-1}\)
c \(\left(\frac{1}{2}\right)^{-2}\)
|727247257|kitty|/Applications/kitty.app|0|
\(\mathbf{b}\left(\frac{2}{3}\right)^{-1}\)
c \(\left(\frac{1}{2}\right)^{-2}\)
|727247258|kitty|/Applications/kitty.app|0|
c \(\left(\frac{1}{2}\right)^{-2}\)
|727247259|kitty|/Applications/kitty.app|0|
c|727247261|kitty|/Applications/kitty.app|0|
a \(x^{-6} y^{4} \times x^{2} y^{-2}\)
d \(2 a^{-1} b^{5} \times 7 a b^{-3}\)
g \(\frac{8 a^{-4}}{2 a^{6}}\)
l \(\frac{36 h^{-9}}{9 h^{-4}}\)
|727247282|kitty|/Applications/kitty.app|0|
d \(2 a^{-1} b^{5} \times 7 a b^{-3}\)
g \(\frac{8 a^{-4}}{2 a^{6}}\)
l \(\frac{36 h^{-9}}{9 h^{-4}}\)
|727247284|kitty|/Applications/kitty.app|0|
g \(\frac{8 a^{-4}}{2 a^{6}}\)
l \(\frac{36 h^{-9}}{9 h^{-4}}\)
|727247286|kitty|/Applications/kitty.app|0|
l \(\frac{36 h^{-9}}{9 h^{-4}}\)
|727247287|kitty|/Applications/kitty.app|0|
a \(6^{4} \times \ldots=6^{2}\)
d \(m^{5} \times \ldots=m^{-6}\)
g \(d^{-7} \div \ldots=d^{15}\)
j \(\left(a^{5}\right) \cdots=a^{-15}\)
|727247302|kitty|/Applications/kitty.app|0|
a \(6^{4} \times \ldots=6^{2}\)
d \(m^{5} \times \ldots=m^{-6}\)
g \(d^{-7} \div \ldots=d^{15}\)
j \(\left(a^{5}\right) \cdots=a^{-15}\)
\(\mathbf{m}(\ldots)^{-2}=\frac{m^{6}}{25}\)
\(\mathbf{p}(\ldots)^{-2}=p^{4} q^{-6}\)
|727247305|kitty|/Applications/kitty.app|0|
d \(m^{5} \times \ldots=m^{-6}\)
g \(d^{-7} \div \ldots=d^{15}\)
j \(\left(a^{5}\right) \cdots=a^{-15}\)
\(\mathbf{m}(\ldots)^{-2}=\frac{m^{6}}{25}\)
\(\mathbf{p}(\ldots)^{-2}=p^{4} q^{-6}\)
|727247308|kitty|/Applications/kitty.app|0|
g \(d^{-7} \div \ldots=d^{15}\)
j \(\left(a^{5}\right) \cdots=a^{-15}\)
\(\mathbf{m}(\ldots)^{-2}=\frac{m^{6}}{25}\)
\(\mathbf{p}(\ldots)^{-2}=p^{4} q^{-6}\)
|727247311|kitty|/Applications/kitty.app|0|
j \(\left(a^{5}\right) \cdots=a^{-15}\)
\(\mathbf{m}(\ldots)^{-2}=\frac{m^{6}}{25}\)
\(\mathbf{p}(\ldots)^{-2}=p^{4} q^{-6}\)
|727247313|kitty|/Applications/kitty.app|0|
\(\mathbf{m}(\ldots)^{-2}=\frac{m^{6}}{25}\)
\(\mathbf{p}(\ldots)^{-2}=p^{4} q^{-6}\)
|727247314|kitty|/Applications/kitty.app|0|
\(\mathbf{p}(\ldots)^{-2}=p^{4} q^{-6}\)
|727247315|kitty|/Applications/kitty.app|0|
a \(\left(3 a^{2} b^{-2}\right)^{3} \times\left(2 a^{4}\right)^{-2}\)
d \(\left(6 a^{5} b^{-4}\right)^{-3} \times 2\left(a^{3} b^{-3}\right)^{2}\)
g \(\frac{\left(2 a^{4} b^{-2}\right)^{3}}{c^{2}} \times \frac{\left(2^{2} a^{-3} b^{2}\right)^{-1}}{c}\)
\(\mathbf{j} \frac{\left(2 a^{4}\right)^{2}}{b^{7}} \div \frac{\left(a^{2}\right)^{-3}}{2 b}\)
|727247352|kitty|/Applications/kitty.app|0|
d \(\left(6 a^{5} b^{-4}\right)^{-3} \times 2\left(a^{3} b^{-3}\right)^{2}\)
g \(\frac{\left(2 a^{4} b^{-2}\right)^{3}}{c^{2}} \times \frac{\left(2^{2} a^{-3} b^{2}\right)^{-1}}{c}\)
\(\mathbf{j} \frac{\left(2 a^{4}\right)^{2}}{b^{7}} \div \frac{\left(a^{2}\right)^{-3}}{2 b}\)
|727247355|kitty|/Applications/kitty.app|0|
d|727247357|kitty|/Applications/kitty.app|0|
g \(\frac{\left(2 a^{4} b^{-2}\right)^{3}}{c^{2}} \times \frac{\left(2^{2} a^{-3} b^{2}\right)^{-1}}{c}\)
\(\mathbf{j} \frac{\left(2 a^{4}\right)^{2}}{b^{7}} \div \frac{\left(a^{2}\right)^{-3}}{2 b}\)
            \part
|727247358|kitty|/Applications/kitty.app|0|
g \(\frac{\left(2 a^{4} b^{-2}\right)^{3}}{c^{2}} \times \frac{\left(2^{2} a^{-3} b^{2}\right)^{-1}}{c}\)
\(\mathbf{j} \frac{\left(2 a^{4}\right)^{2}}{b^{7}} \div \frac{\left(a^{2}\right)^{-3}}{2 b}\)
|727247360|kitty|/Applications/kitty.app|0|
\(\mathbf{j} \frac{\left(2 a^{4}\right)^{2}}{b^{7}} \div \frac{\left(a^{2}\right)^{-3}}{2 b}\)
|727247362|kitty|/Applications/kitty.app|0|
            \end{solutionordottedlines}
|727247519|kitty|/Applications/kitty.app|0|
    \begin{parts}
|727249773|kitty|/Applications/kitty.app|0|
1|727249775|kitty|/Applications/kitty.app|0|
    \begin{part}
|727250079|kitty|/Applications/kitty.app|0|
 9 Copy and complete.|727250512|kitty|/Applications/kitty.app|0|
9|727250520|kitty|/Applications/kitty.app|0|
        \part \mathbf{k} \frac{\left(4 c^{4} d^{-3}\right)^{2}}{9} \div \frac{3 c^{-2}}{d}\)
|727250579|kitty|/Applications/kitty.app|0|
[1cm]|727250926|kitty|/Applications/kitty.app|0|
/\\fillin\[\]/\\fillin\[\]\[1cm\]/g
|727251252|kitty|/Applications/kitty.app|0|
ldots|727251468|kitty|/Applications/kitty.app|0|
ldo1cmts|727251476|kitty|/Applications/kitty.app|0|
ldts|727251495|kitty|/Applications/kitty.app|0|
cdots|727251502|kitty|/Applications/kitty.app|0|
\begin{questions}
    \Question[1] What is the fourth power of two equal to?
    \begin{solutionordottedlines}[2cm]
        $2^4 = 16$
    \end{solutionordottedlines}
    \Question[4] How do you interpret this?
    \begin{solutionordottedlines}[2cm]
        $2^4 = 2\times 2\times 2\times 2 = 16$
    \end{solutionordottedlines}
    \Question[4] But then what about $2^{-2}$? Does this have an answer? What is it's interpretation?!
    \begin{solutionorbox}[1in]
        $2^4 = 2\times 2\times 2\times 2 = 16$
    \end{solutionorbox}
    \Question[1] Why are learning about these powers?
    \begin{solutionordottedlines}[1.5in]
        Because they express small quantities and large quantites very well. And the world around us expresses itself to us in that way: there are $10^{22}$ stars in the universe (thereabouts) and the mass of an electron has the mass of $9.1 \times 10^{-31}$. These quantities would be dreadful to write out in full. Also, if we study these well we can manipulate them and make our mathematical expressions simpler.
    \end{solutionordottedlines}
\end{questions}
|727317239|kitty|/Applications/kitty.app|0|
Let us begin with a definition:
\begin{boxlaw}
    Let
|727317242|kitty|/Applications/kitty.app|0|
The Index Laws|727317326|kitty|/Applications/kitty.app|0|
Negative Indices|727317379|kitty|/Applications/kitty.app|0|
3-neg-ind|727317387|kitty|/Applications/kitty.app|0|
\section{Scientific Notation}
    \setcounter{secmarks}{0}
    \input{3-sci-not}
    \setcounter{sec3marks}{\thesecmarks}
|727317396|kitty|/Applications/kitty.app|0|
Scientific Notation|727317399|kitty|/Applications/kitty.app|0|
3-sci-not
\|727317456|kitty|/Applications/kitty.app|0|
algebra. pythag. consumer. factor. linear. formulas. index
|727317484|kitty|/Applications/kitty.app|0|
emph|727317514|kitty|/Applications/kitty.app|0|
newpage|727317527|kitty|/Applications/kitty.app|0|
             In the formula \(E = mc^2\):                                                                   
             \begin{itemize}                                                                                
                 \item \(E\) stands for energy.                                                             
                 \item \(m\) stands for mass.                                                               
                 \item \(c\) stands for the speed of light in a vacuum.                                     
             \end{itemize}                                                                                  
|727317620|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                        
     \begin{questions}                                                                                      
         \Question The profit \(\$ P\) made each day by a store owner who sells \(C D\) s is given by the   
 formula \(P=5 n-150\), where \(n\) is the number of CDs sold.                                              
         \begin{parts}                                                                                      
             \Part[1] What profit is made if the store owner sells 60 CDs?                                  
             \begin{solutionordottedlines}                                                                  
                 P = 5n - 150 \\                                                                            
                 P = 5(60) - 150 \\                                                                         
                 P = 300 - 150 \\                                                                           
                 P = \$150                                                                                  
             \end{solutionordottedlines}                                                                    
             \Part[1] Make \(n\) the subject of the formula.                                                
             \begin{solutionordottedlines}                                                                  
                 P = 5n - 150 \\                                                                            
                 P + 150 = 5n \\                                                                            
                 n = \frac{P + 150}{5}                                                                      
             \end{solutionordottedlines}                                                                    
             \Part[2] How many CDs were sold if the store made:                                             
             \begin{subparts}                                                                               
                 \subpart a profit of \(\$ 275\) ?                                                          
                 \begin{solutionordottedlines}                                                              
                     n = \frac{P + 150}{5} \\                                                               
                     n = \frac{275 + 150}{5} \\                                                             
                     n = \frac{425}{5} \\                                                                   
                     n = 85                                                                                 
                 \end{solutionordottedlines}                                                                
                 \subpart a profit of \(\$ 400\) ?                                                          
                 \begin{solutionordottedlines}                                                              
                     n = \frac{P + 150}{5} \\                                                               
                     n = \frac{400 + 150}{5} \\                                                             
                     n = \frac{550}{5} \\                                                                   
                     n = 110                                                                                
                 \end{solutionordottedlines}                                                                
                 \subpart a loss of \(\$ 100\) ?                                                            
                 \begin{solutionordottedlines}                                                              
                     n = \frac{P + 150}{5} \\                                                               
                     n = \frac{-100 + 150}{5} \\                                                            
                     n = \frac{50}{5} \\                                                                    
                     n = 10                                                                                 
                 \end{solutionordottedlines}                                                                
                 \subpart no profit?                                                                        
                 \begin{solutionordottedlines}                                                              
                     n = \frac{P + 150}{5} \\                                                               
                     n = \frac{0 + 150}{5} \\                                                               
                     n = \frac{150}{5} \\                                                                   
                     n = 30                                                                                 
                 \end{solutionordottedlines}                                                                
             \end{subparts}                                                                                 
         \end{parts}                                                                                        
         \Question The cost \(\$ C\) of hiring a reception room for a function is given by the formula \(C= 
 n+250\), where \(n\) is the number of people attending the function.                                       
         \begin{parts}                                                                                      
             \Part[1] Rearrange the formula to make \(n\) the subject.                                      
             \begin{solutionordottedlines}                                                                  
                 C = 12n + 250 \\                                                                           
                 C - 250 = 12n \\                                                                           
                 n = \frac{C - 250}{12}                                                                     
             \end{solutionordottedlines}                                                                    
             \Part[2] How many people attended the function if the cost of hiring the reception room was:   
             \begin{subparts}                                                                               
                 \subpart \(\$ 730\) ?                                                                      
                 \begin{solutionordottedlines}                                                              
                     n = \frac{C - 250}{12} \\                                                              
                     n = \frac{730 - 250}{12} \\                                                            
                     n = \frac{480}{12} \\                                                                  
                     n = 40                                                                                 
                 \end{solutionordottedlines}                                                                
                 \subpart \(\$ 1090\) ?                                                                     
                 \begin{solutionordottedlines}                                                              
                     n = \frac{C - 250}{12} \\                                                              
                     n = \frac{1090 - 250}{12} \\                                                           
                     n = \frac{840}{12} \\                                                                  
                     n = 70                                                                                 
                 \end{solutionordottedlines}                                                                
                 \subpart \(\$ 1210\) ?                                                                     
                 \begin{solutionordottedlines}                                                              
                     n = \frac{C - 250}{12} \\                                                              
                     n = \frac{1210 - 250}{12} \\                                                           
                     n = \frac{960}{12} \\                                                                  
                     n = 80                                                                                 
                 \end{solutionordottedlines}                                                                
                 \subpart \(\$ 1690\) ?                                                                     
                 \begin{solutionordottedlines}                                                              
                     n = \frac{C - 250}{12} \\                                                              
                     n = \frac{1690 - 250}{12} \\                                                           
                     n = \frac{1440}{12} \\                                                                 
                     n = 120                                                                                
                 \end{solutionordottedlines}                                                                
             \end{subparts}                                                                                 
         \end{parts}                                                                                        
         \Question Given the formula \(t=a+(n-1) d\) :                                                      
         \begin{parts}                                                                                      
             \Part[1] rearrange the formula to make \(a\) the subject                                       
             \begin{solutionordottedlines}                                                                  
                 t = a + (n - 1)d \\                                                                        
                 a = t - (n - 1)d                                                                           
             \end{solutionordottedlines}                                                                    
             \Part[2] find the value of \(a\) when:                                                         
             \begin{subparts}                                                                               
                 \subpart \(t=11, n=4\) and \(d=3\)                                                         
                 \begin{solutionordottedlines}                                                              
                     a = t - (n - 1)d \\                                                                    
                     a = 11 - (4 - 1) \cdot 3 \\                                                            
                     a = 11 - 3 \cdot 3 \\                                                                  
                     a = 11 - 9 \\                                                                          
                     a = 2                                                                                  
                 \end{solutionordottedlines}                                                                
                 \subpart \(t=8, n=5\) and \(d=-3\)                                                         
                 \begin{solutionordottedlines}                                                              
                     a = t - (n - 1)d \\                                                                    
                     a = 8 - (5 - 1) \cdot (-3) \\                                                          
                     a = 8 - 4 \cdot (-3) \\                                                                
                     a = 8 + 12 \\                                                                          
                     a = 20                                                                                 
                 \end{solutionordottedlines}                                                                
             \end{subparts}                                                                                 
             \Part[1] rearrange the formula to make \(d\) the subject                                       
             \begin{solutionordottedlines}                                                                  
                 t = a + (n - 1)d \\                                                                        
                 d = \frac{t - a}{n - 1}                                                                    
             \end{solutionordottedlines}                                                                    
             \Part[2] find the value of \(d\) when:                                                         
             \begin{subparts}                                                                               
                 \subpart \(t=48, a=3\) and \(n=16\)                                                        
                 \begin{solutionordottedlines}                                                              
                     d = \frac{t - a}{n - 1} \\                                                             
                     d = \frac{48 - 3}{16 - 1} \\                                                           
                     d = \frac{45}{15} \\                                                                   
                     d = 3                                                                                  
                 \end{solutionordottedlines}                                                                
                 \subpart \(t=120, a=-30\) and \(n=101\)                                                    
                 \begin{solutionordottedlines}                                                              
                     d = \frac{t - a}{n - 1} \\                                                             
                     d = \frac{120 - (-30)}{101 - 1} \\                                                     
                     d = \frac{150}{100} \\                                                                 
                     d = 1.5                                                                                
                 \end{solutionordottedlines}                                                                
             \end{subparts}                                                                                 
             \Part[1] rearrange the formula to make \(n\) the subject and find the value of \(n\) when      
 \(t=150, a=5\) and \(d=5\)                                                                                 
             \begin{solutionordottedlines}                                                                  
                 t = a + (n - 1)d \\                                                                        
                 n = \frac{t - a}{d} + 1 \\                                                                 
                 n = \frac{150 - 5}{5} + 1 \\                                                               
                 n = \frac{145}{5} + 1 \\                                                                   
                 n = 29 + 1 \\                                                                              
                 n = 30                                                                                     
             \end{solutionordottedlines}                                                                    
         \end{parts}                                                                                        
         \Question Rearrange each of these formulas to make the pronumeral in brackets the subject.         
         \begin{parts}                                                                                      
             \part \(y=m x+c\) \hfill\((c)\)                                                                
             \begin{solutionordottedlines}                                                                  
                 c = y - mx                                                                                 
             \end{solutionordottedlines}                                                                    
             \part \(A=\frac{1}{2} b h\)\hfill\((x)\)                                                       
             \begin{solutionordottedlines}                                                                  
                 % This part seems to have a typo, as there is no 'x' in the given formula.                 
                 % Assuming the subject to be made is 'b':                                                  
                 b = \frac{2A}{h}                                                                           
             \end{solutionordottedlines}                                                                    
             \part \(P=A+2 \ell h\)\hfill\((\ell)\)                                                         
             \begin{solutionordottedlines}                                                                  
                 \ell = \frac{P - A}{2h}                                                                    
             \end{solutionordottedlines}                                                                    
             \part \(A=2\pi r^2 + 2\pi r h\hfill{}(h)\)                                                     
             \begin{solutionordottedlines}                                                                  
                 h = \frac{A - 2\pi r^2}{2\pi r}                                                            
             \end{solutionordottedlines}                                                                    
             \part \(s=\frac{n}{2}(a+\ell)\)\hfill\((a)\)                                                   
             \begin{solutionordottedlines}                                                                  
                 a = \frac{2s}{n} - \ell                                                                    
             \end{solutionordottedlines}                                                                    
             \part \(V=\pi r^{2}+\pi r s\)\hfill\((s)\)                                                     
             \begin{solutionordottedlines}                                                                  
                 s = \frac{V}{\pi r} - r                                                                    
             \end{solutionordottedlines}                                                                    
         \end{parts}                                                                                        
         \Question[3] The formula for the sum \(S\) of the interior angles in a convex \(n\)-sided polygon  
 \(S=180(n-2)\). Rearrange the formula to make \(n\) the subject and use this to find the number of sides i 
 the polygon if the sum of the interior angles is:                                                          
         \begin{parts}\begin{multicols}{3}                                                                  
             \part \(1080^{\circ}\)                                                                         
             \begin{solutionordottedlines}                                                                  
                 n = \frac{S}{180} + 2 \\                                                                   
                 n = \frac{1080}{180} + 2 \\                                                                
                 n = 6 + 2 \\                                                                               
                 n = 8                                                                                      
             \end{solutionordottedlines}                                                                    
             \part \(1800^{\circ}\)                                                                         
             \begin{solutionordottedlines}                                                                  
                 n = \frac{S}{180} + 2 \\                                                                   
                 n = \frac{1800}{180} + 2 \\                                                                
                 n = 10 + 2 \\                                                                              
                 n = 12                                                                                     
             \end{solutionordottedlines}                                                                    
             \part \(3240^{\circ}\)                                                                         
             \begin{solutionordottedlines}                                                                  
                 n = \frac{S}{180} + 2 \\                                                                   
                 n = \frac{3240}{180} + 2 \\                                                                
                 n = 18 + 2 \\                                                                              
                 n = 20                                                                                     
             \end{solutionordottedlines}                                                                    
         \end{multicols}\end{parts}                                                                         
         \Question[2] 8 When an object is shot up into the air with a speed of \(u\) metres per second, its 
 height above the ground \(h\) metres and time of flight \(t\) seconds are related (ignoring air resistance 
 by \(h=u t-4.9 t^{2}\).                                                                                    
                                                                                                            
         Find the speed at which an object was fired if it reached a height of 27.5 metres after 5 seconds. 
         \begin{solutionordottedlines}                                                                      
             h = ut - 4.9t^2 \\                                                                             
             27.5 = u(5) - 4.9(5)^2 \\                                                                      
             27.5 = 5u - 4.9(25) \\                                                                         
             27.5 = 5u - 122.5 \\                                                                           
             5u = 27.5 + 122.5 \\                                                                           
             5u = 150 \\                                                                                    
             u = \frac{150}{5} \\                                                                           
             u = 30 \text{ metres per second}                                                               
         \end{solutionordottedlines}                                                                        
         \Question[4] Rearrange each of these formulas to make the pronumeral in brackets the subject. (All 
 pronumerals represent positive numbers.)                                                                   
         \begin{parts}\begin{multicols}{2}                                                                  
             \part \(c=a^{2}+b^{2}\hfill{}(a)\)                                                             
             \begin{solutionordottedlines}                                                                  
                 a = \sqrt{c - b^2}                                                                         
             \end{solutionordottedlines}                                                                    
             \part \(x=\sqrt{a b}\hfill{}(b)\)                                                              
             \begin{solutionordottedlines}                                                                  
                 b = \frac{x^2}{a}                                                                          
             \end{solutionordottedlines}                                                                    
             \part \(T=\frac{2\pi{}}{n}\hfill{}(n)\)                                                        
             \begin{solutionordottedlines}                                                                  
                 n = \frac{2\pi}{T}                                                                         
             \end{solutionordottedlines}                                                                    
             \part \(E=\frac{m}{2r^2}\hfill{}(r)\)                                                          
             \begin{solutionordottedlines}                                                                  
                 r = \sqrt{\frac{m}{2E}}                                                                    
             \end{solutionordottedlines}                                                                    
         \end{multicols}\end{parts}                                                                         
     \end{questions}                                                                                        
 \end{exercisebox}                                                                                          
|727317834|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \begin{questions}
        \Question The profit \(\$ P\) made each day by a store owner who sells \(C D\) s is given by the formula \(P=5 n-150\), where \(n\) is the number of CDs sold.
        \begin{parts}
            \Part[1] What profit is made if the store owner sells 60 CDs?
            \Part[1] Make \(n\) the subject of the formula.
            \Part[2] How many CDs were sold if the store made:
            \begin{subparts}
                \subpart a profit of \(\$ 275\) ?
                \subpart a profit of \(\$ 400\) ?
                \subpart a loss of \(\$ 100\) ?
                \subpart no profit?
            \end{subparts}
        \end{parts}
        \Question The cost \(\$ C\) of hiring a reception room for a function is given by the formula \(C=12 n+250\), where \(n\) is the number of people attending the function.
        \begin{parts}
            \Part[1] Rearrange the formula to make \(n\) the subject.
            \Part[2] How many people attended the function if the cost of hiring the reception room was:
            \begin{subparts}
                \subpart \(\$ 730\) ?
                \subpart \(\$ 1090\) ?
                \subpart \(\$ 1210\) ?
                \subpart \(\$ 1690\) ?
            \end{subparts}
        \end{parts}
        \Question Given the formula \(t=a+(n-1) d\) :
        \begin{parts}
            \Part[1] rearrange the formula to make \(a\) the subject
            \Part[2] find the value of \(a\) when:
            \begin{subparts}
                \subpart \(t=11, n=4\) and \(d=3\)
                \subpart \(t=8, n=5\) and \(d=-3\)
            \end{subparts}
            \Part[1] rearrange the formula to make \(d\) the subject
            \Part[2] find the value of \(d\) when:
            \begin{subparts}
                \subpart \(t=48, a=3\) and \(n=16\)
                \subpart \(t=120, a=-30\) and \(n=101\)
            \end{subparts}
            \Part[1] rearrange the formula to make \(n\) the subject and find the value of \(n\) when \(t=150, a=5\) and \(d=5\)
        \end{parts}
        \Question Rearrange each of these formulas to make the pronumeral in brackets the subject.
        \begin{parts}
            \part \(y=m x+c\) \hfill\((c)\)
            \part \(A=\frac{1}{2} b h\)\hfill\((x)\)
            \part \(P=A+2 \ell h\)\hfill\((\ell)\)
            \part \(A=2\pi r^2 + 2\pi r h\hfill{}(h)\)
            \part \(s=\frac{n}{2}(a+\ell)\)\hfill\((a)\)
            \part \(V=\pi r^{2}+\pi r s\)\hfill\((s)\)
        \end{parts}
        \Question[3] The formula for the sum \(S\) of the interior angles in a convex \(n\)-sided polygon is \(S=180(n-2)\). Rearrange the formula to make \(n\) the subject and use this to find the number of sides in the polygon if the sum of the interior angles is:
        \begin{parts}\begin{multicols}{3}
            \part \(1080^{\circ}\)
            \part \(1800^{\circ}\)
            \part \(3240^{\circ}\)
        \end{multicols}\end{parts}
        \Question[2] 8 When an object is shot up into the air with a speed of \(u\) metres per second, its height above the ground \(h\) metres and time of flight \(t\) seconds are related (ignoring air resistance) by \(h=u t-4.9 t^{2}\).

        Find the speed at which an object was fired if it reached a height of 27.5 metres after 5 seconds.
        \Question[4] Rearrange each of these formulas to make the pronumeral in brackets the subject. (All pronumerals represent positive numbers.)
        \begin{parts}\begin{multicols}{2}
            \part \(c=a^{2}+b^{2}\hfill{}(a)\)
            \part \(x=\sqrt{a b}\hfill{}(b)\)
            \part \(T=\frac{2\pi{}}{n}\hfill{}(n)\)
            \part \(E=\frac{m}{2r^2}\hfill{}(r)\)
        \end{multicols}\end{parts}
    \end{questions}
\end{exercisebox}
|727317854|kitty|/Applications/kitty.app|0|
         \begin{parts}                                                                                      
             \part \(D\) in terms of \(n\), where \(D\) is the number of degrees in \(n\) right angles      
             \begin{solutionordottedlines}[1cm]                                                             
                 \(D = 90n\)                                                                                
             \end{solutionordottedlines}                                                                    
             \part \(c\) in terms of \(D\), where \(c\) is the number of cents in \(\$ D\)                  
             \begin{solutionordottedlines}[1cm]                                                             
                 \(c = 100D\)                                                                               
             \end{solutionordottedlines}                                                                    
             \part \(m\) in terms of \(h\), where \(m\) is the number of minutes in \(h\) hours             
             \begin{solutionordottedlines}[1cm]                                                             
                 \(m = 60h\)                                                                                
             \end{solutionordottedlines}                                                                    
             \part \(d\) in terms of \(m\), where \(d\) is the number of days in \(m\) weeks                
             \begin{solutionordottedlines}[1cm]                                                             
                 \(d = 7m\)                                                                                 
             \end{solutionordottedlines}                                                                    
         \end{parts}                                                                                        
|727317904|kitty|/Applications/kitty.app|0|
        \begin{parts}
            \part \(D\) in terms of \(n\), where \(D\) is the number of degrees in \(n\) right angles
            \part \(c\) in terms of \(D\), where \(c\) is the number of cents in \(\$ D\)
            \part \(m\) in terms of \(h\), where \(m\) is the number of minutes in \(h\) hours
            \part \(d\) in terms of \(m\), where \(d\) is the number of days in \(m\) weeks
        \end{parts}
|727317907|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                        
     \begin{questions}                                                                                      
         \Question[4] Construct a formula for:                                                              
         \begin{parts}                                                                                      
             \part the number of centimetres \(n\) in \(p\) metres                                          
             \begin{solutionordottedlines}[1cm]                                                             
                 \(n = 100p\)                                                                               
             \end{solutionordottedlines}                                                                    
             \part the number of millilitres \(s\) in \(t\) litres                                          
             \begin{solutionordottedlines}[1cm]                                                             
                 \(s = 1000t\)                                                                              
             \end{solutionordottedlines}                                                                    
             \part the number of centimetres \(q\) in \(5 p\) metres                                        
             \begin{solutionordottedlines}[1cm]                                                             
                 \(q = 500p\)                                                                               
             \end{solutionordottedlines}                                                                    
             \part the number of grams \(x\) in \(\frac{y}{2}\) kilograms                                   
             \begin{solutionordottedlines}[1cm]                                                             
                 \(x = 500y\)                                                                               
             \end{solutionordottedlines}                                                                    
         \end{parts}                                                                                        
         \Question[6] Find a formula relating \(x\) and \(y\) for each of these statements, making \(y\) th 
 subject.                                                                                                   
         \begin{parts}                                                                                      
             \part \(y\) is three less than \(x\).                                                          
             \begin{solutionordottedlines}[1cm]                                                             
                 \(y = x - 3\)                                                                              
             \end{solutionordottedlines}                                                                    
             \part \(y\) is four more than the square of \(x\).                                             
             \begin{solutionordottedlines}[1cm]                                                             
                 \(y = x^2 + 4\)                                                                            
             \end{solutionordottedlines}                                                                    
             \part \(y\) is eight times the square root of one-fifth of \(x\).                              
             \begin{solutionordottedlines}[1cm]                                                             
                 \(y = 8 \sqrt{\frac{x}{5}}\)                                                               
             \end{solutionordottedlines}                                                                    
             \part \(x\) and \(y\) are supplementary angles.                                                
             \begin{solutionordottedlines}[1cm]                                                             
                 \(y = 180 - x\)                                                                            
             \end{solutionordottedlines}                                                                    
             \part A car travelled \(80 \mathrm{~km}\) in \(x\) hours at an average speed of \(y \mathrm{~k 
 / \mathrm{h}\).                                                                                            
             \begin{solutionordottedlines}[1cm]                                                             
                 \(y = \frac{80}{x}\)                                                                       
             \end{solutionordottedlines}                                                                    
             \part A car used \(x\) litres of petrol on a trip of \(80 \mathrm{~km}\) and the fuel          
 consumption was \(y\) litres \(/ 100 \mathrm{~km}\).                                                       
             \begin{solutionordottedlines}[1cm]                                                             
                 \(y = \frac{x}{80} \times 100\)                                                            
             \end{solutionordottedlines}                                                                    
         \end{parts}                                                                                        
         \Question[6] Find a formula relating the given pronumerals for each of these statements.           
         \begin{parts}                                                                                      
             \part The number of square \(\mathrm{cm} x\) in \(y\) square metres                            
             \begin{solutionordottedlines}[1cm]                                                             
                 \(x = 10000y\)                                                                             
             \end{solutionordottedlines}                                                                    
             \part The selling price \(\$ S\) of an article with an original price of \(\$ m\) when a       
 discount of \(20 \%\) is given                                                                             
             \begin{solutionordottedlines}[1cm]                                                             
                 \(S = m - 0.2m\)                                                                           
             \end{solutionordottedlines}                                                                    
             \part The length \(c \mathrm{~cm}\) of the hypotenuse and the lengths \(a \mathrm{~cm}\) and \ 
 \mathrm{~cm}\) of the other two sides in a right-angled triangle                                           
             \begin{solutionordottedlines}[1cm]                                                             
                 \(c = \sqrt{a^2 + b^2}\)                                                                   
             \end{solutionordottedlines}                                                                    
             \part The area \(A \mathrm{~cm}^{2}\) of a sector of a circle with a radius of length \(r      
 \mathrm{~cm}\) and angle \(\theta\) at the centre of the circle                                            
             \begin{solutionordottedlines}[1cm]                                                             
                 \(A = \frac{1}{2} r^2 \theta\)                                                             
             \end{solutionordottedlines}                                                                    
             \part The distance \(d \mathrm{~km}\) travelled by a car in \(t\) hours at an average speed of 
 \(75 \mathrm{~km} / \mathrm{h}\)                                                                           
             \begin{solutionordottedlines}[1cm]                                                             
                 \(d = 75t\)                                                                                
             \end{solutionordottedlines}                                                                    
             \part The number of hectares \(h\) in a rectangular paddock of length \(400 \mathrm{~m}\) and  
 width \(w \mathrm{~m}\)                                                                                    
             \begin{solutionordottedlines}[1cm]                                                             
                 \(h = \frac{400w}{10000}\)                                                                 
             \end{solutionordottedlines}                                                                    
         \end{parts}                                                                                        
     \end{questions}                                                                                        
 \end{exercisebox}                                                                                          
|727317918|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \begin{questions}
        \Question[4] Construct a formula for:
        \begin{parts}
            \part the number of centimetres \(n\) in \(p\) metres
            \part the number of millilitres \(s\) in \(t\) litres
            \part the number of centimetres \(q\) in \(5 p\) metres
            \part the number of grams \(x\) in \(\frac{y}{2}\) kilograms
        \end{parts}
        \Question[6] Find a formula relating \(x\) and \(y\) for each of these statements, making \(y\) the subject.
        \begin{parts}
            \part \(y\) is three less than \(x\).
            \part \(y\) is four more than the square of \(x\).
            \part \(y\) is eight times the square root of one-fifth of \(x\).
            \part \(x\) and \(y\) are supplementary angles.
            \part A car travelled \(80 \mathrm{~km}\) in \(x\) hours at an average speed of \(y \mathrm{~km} / \mathrm{h}\).
            \part A car used \(x\) litres of petrol on a trip of \(80 \mathrm{~km}\) and the fuel consumption was \(y\) litres \(/ 100 \mathrm{~km}\).
        \end{parts}
        \Question[6] Find a formula relating the given pronumerals for each of these statements.
        \begin{parts}
            \part The number of square \(\mathrm{cm} x\) in \(y\) square metres
            \part The selling price \(\$ S\) of an article with an original price of \(\$ m\) when a discount of \(20 \%\) is given
            \part The length \(c \mathrm{~cm}\) of the hypotenuse and the lengths \(a \mathrm{~cm}\) and \(b \mathrm{~cm}\) of the other two sides in a right-angled triangle
            \part The area \(A \mathrm{~cm}^{2}\) of a sector of a circle with a radius of length \(r \mathrm{~cm}\) and angle \(\theta\) at the centre of the circle
            \part The distance \(d \mathrm{~km}\) travelled by a car in \(t\) hours at an average speed of \(75 \mathrm{~km} / \mathrm{h}\)
            \part The number of hectares \(h\) in a rectangular paddock of length \(400 \mathrm{~m}\) and width \(w \mathrm{~m}\)
        \end{parts}
    \end{questions}
\end{exercisebox}
|727317924|kitty|/Applications/kitty.app|0|
 \subsection{Substitution into formulas}                                                                    
 \begin{questions}                                                                                          
     \Question[3] For each part, find the value of the subject when the other pronumerals have the value    
 indicated.                                                                                                 
     \begin{parts}                                                                                          
         \part \(A=\frac{1}{2}(a+b) h\), where \(a=4, b=6, h=10\)                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(A=\frac{1}{2}(4+6) \cdot 10 = \frac{1}{2} \cdot 10 \cdot 10 = 50\)                       
             \end{solutionordottedlines}                                                                    
         \part \(t=a+(n-1) d\), where \(a=30, n=8, d=4\)                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(t=30+(8-1) \cdot 4 = 30+7 \cdot 4 = 30+28 = 58\)                                         
             \end{solutionordottedlines}                                                                    
         \part \(E=\frac{1}{2} m v^{2}\), where \(m=8, v=4\)                                                
             \begin{solutionordottedlines}[2cm]                                                             
                 \(E=\frac{1}{2} \cdot 8 \cdot 4^{2} = 4 \cdot 16 = 64\)                                    
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[4] For each part, find the value of the subject when the other pronumerals have the value    
 indicated. Calculate \(\mathbf{a}-\mathbf{c}\) correct to 3 decimal places and \(\mathbf{d}\) correct to 2 
     \begin{parts}                                                                                          
         \part \(x=\sqrt{a b}\), where \(a=40, b=50\)                                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(x=\sqrt{40 \cdot 50} = \sqrt{2000} \approx 44.721\)                                      
             \end{solutionordottedlines}                                                                    
         \part \(V=\pi r^{2} h\), where \(r=12, h=20\)                                                      
             \begin{solutionordottedlines}[2cm]                                                             
                 \(V=\pi \cdot 12^{2} \cdot 20 = 144 \pi \cdot 20 = 2880 \pi \approx 9047.78\)              
             \end{solutionordottedlines}                                                                    
         \part \(T=2 \pi \sqrt{\frac{\ell}{g}}\), where \(\ell=88.2, g=9.8\)                                
             \begin{solutionordottedlines}[2cm]                                                             
                 \(T=2 \pi \sqrt{\frac{88.2}{9.8}} \approx 2 \pi \sqrt{9} = 2 \pi \cdot 3 \approx 18.850\)  
             \end{solutionordottedlines}                                                                    
         \part \(A=P(1+R)^{n}\), where \(P=10000, R=0.065, n=10\)                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(A=10000(1+0.065)^{10} \approx 10000 \cdot 1.8194 \approx 18194.00\)                      
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[1] For the formula \(S=2(\ell w+\ell h+h w)\), find \(h\) if \(S=592, \ell=10\) and \(w=8\). 
             \begin{solutionordottedlines}[2cm]                                                             
                 \(592=2(10 \cdot 8+10h+8h)\)                                                               
                 \(296=80+18h\)                                                                             
                 \(216=18h\)                                                                                
                 \(h=12\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \Question[1] For the formula \(s=u t+\frac{1}{2} a t^{2}\), find \(a\) if \(s=1000, u=20\) and \(t=5\) 
             \begin{solutionordottedlines}[2cm]                                                             
                 \(1000=20 \cdot 5+\frac{1}{2} a \cdot 5^{2}\)                                              
                 \(1000=100+\frac{1}{2} a \cdot 25\)                                                        
                 \(900=\frac{1}{2} a \cdot 25\)                                                             
                 \(36=a\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \Question[1] For the formula \(t=a+(n-1) d\), find \(n\) if \(t=58, d=3\) and \(a=7\).                 
             \begin{solutionordottedlines}[2cm]                                                             
                 \(58=7+(n-1) \cdot 3\)                                                                     
                 \(51=(n-1) \cdot 3\)                                                                       
                 \(17=n-1\)                                                                                 
                 \(n=18\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \Question[2] Given \(v^{2}=u^{2}+2 a x\) and \(v>0\), find the value of \(v\) (correct to 1 decimal    
 place) when:                                                                                               
     \begin{parts}\begin{multicols}{2}                                                                      
         \part \(u=0, a=5\) and \(x=10\)                                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{0^{2}+2 \cdot 5 \cdot 10} = \sqrt{100} = 10.0\)                                  
             \end{solutionordottedlines}                                                                    
         \part \(u=2, a=9.8\) and \(x=22\)                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{2^{2}+2 \cdot 9.8 \cdot 22} \approx \sqrt{4+431.2} \approx \sqrt{435.2} \approx  
 20.9\)                                                                                                     
             \end{solutionordottedlines}                                                                    
     \end{multicols}\end{parts}                                                                             
     \Question[2] Given \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\), find the value of:                         
     \begin{parts}\begin{multicols}{2}                                                                      
         \part \(u\) when \(f=2\) and \(v=4\)                                                               
             \begin{solutionordottedlines}[2cm]                                                             
                 \(\frac{1}{2}=\frac{1}{u}+\frac{1}{4}\)                                                    
                 \(\frac{1}{2}-\frac{1}{4}=\frac{1}{u}\)                                                    
                 \(\frac{1}{4}=\frac{1}{u}\)                                                                
                 \(u=4\)                                                                                    
             \end{solutionordottedlines}                                                                    
         \part \(u\) when \(f=3\) and \(v=4\)                                                               
             \begin{solutionordottedlines}[2cm]                                                             
                 \(\frac{1}{3}=\frac{1}{u}+\frac{1}{4}\)                                                    
                 \(\frac{1}{3}-\frac{1}{4}=\frac{1}{u}\)                                                    
                 \(\frac{1}{12}=\frac{1}{u}\)                                                               
                 \(u=12\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \end{multicols}\end{parts}                                                                             
     \Question[6] The formula for finding the number of degrees Fahrenheit \((F)\) for a temperature given  
 a number of degrees Celsius \((C)\) is \(F=\frac{9}{5} C+32\).                                             
                                                                                                            
     Fahrenheit temperatures are still used in the USA, but in Australia we commonly use Celsius.           
                                                                                                            
     Calculate the Fahrenheit temperatures which people in the USA would recognise for:                     
     \begin{parts}                                                                                          
         \part the freezing point of water, \(0^{\circ} \mathrm{C}\)                                        
             \begin{solutionordottedlines}[2cm]                                                             
                 \(F=\frac{9}{5} \cdot 0+32 = 32\)                                                          
             \end{solutionordottedlines}                                                                    
         \part the boiling point of water, \(100^{\circ} \mathrm{C}\)                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(F=\frac{9}{5} \cdot 100+32 = 180+32 = 212\)                                              
             \end{solutionordottedlines}                                                                    
         \part a nice summer temperature of \(25^{\circ} \mathrm{C}\)                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(F=\frac{9}{5} \cdot 25+32 = 45+32 = 77\)                                                 
             \end{solutionordottedlines}                                                                    
         \hspace{-1cm}\item[] Now calculate the Celsius temperatures which people in Australia would        
 recognise for:                                                                                             
         \part \(50^{\circ} \mathrm{F}\)                                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(C=\frac{5}{9}(50-32) = \frac{5}{9} \cdot 18 \approx 10\)                                 
             \end{solutionordottedlines}                                                                    
         \part \(104^{\circ} \mathrm{F}\)                                                                   
             \begin{solutionordottedlines}[2cm]                                                             
                 \(C=\frac{5}{9}(104-32) = \frac{5}{9} \cdot 72 \approx 40\)                                
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[4] Sam throws a stone down to the ground from the top of a cliff \(s\) metres high, with an  
 initial speed of \(u \mathrm{~m} / \mathrm{s}\). It accelerates at \(a \mathrm{~m} / \mathrm{s}^{2}\). The 
 stone hits the ground with a speed of \(v \mathrm{~m} / \mathrm{s}\) given by the formula \(v^{2}=u^{2}+2  
 s\). Find the speed at which the stone hits the ground, correct to 2 decimal places, if:                   
     \begin{parts}                                                                                          
         \part \(u=0, a=9.8\) and \(s=50\)                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{0^{2}+2 \cdot 9.8 \cdot 50} = \sqrt{980} \approx 31.30\)                         
             \end{solutionordottedlines}                                                                    
         \part \(u=5, a=9.8\) and \(s=35\)                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{5^{2}+2 \cdot 9.8 \cdot 35} \approx \sqrt{25+686} \approx \sqrt{711} \approx     
 26.67\)                                                                                                    
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
 \end{questions}                                                                                            
                                                                                                            
 \subsection{Changing the subject of a formula}                                                             
                                                                                                            
 \begin{questions}                                                                                          
     \Question[] Given the formula \(v=u+a t\) :                                                            
     \begin{parts}                                                                                          
         \Part[1] rearrange the formula to make \(u\) the subject                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(u=v-a t\)                                                                                
             \end{solutionordottedlines}                                                                    
         \Part[2] find the value of \(u\) when:                                                             
             \begin{solutionordottedlines}[2cm]                                                             
             \end{solutionordottedlines}                                                                    
         \begin{subparts}                                                                                   
             \subpart \(v=20, a=2\) and \(t=5 \quad\)                                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(u=20-2 \cdot 5 = 20-10 = 10\)                                                            
             \end{solutionordottedlines}                                                                    
             \subpart \(v=40, a=-6\) and \(t=4\)                                                            
             \begin{solutionordottedlines}[2cm]                                                             
                 \(u=40-(-6) \cdot 4 = 40+24 = 64\)                                                         
             \end{solutionordottedlines}                                                                    
         \end{subparts}                                                                                     
         \Part[1] rearrange the formula to make \(a\) the subject                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{v-u}{t}\)                                                                        
             \end{solutionordottedlines}                                                                    
         \Part[2] find the value of \(a\) when:                                                             
             \begin{solutionordottedlines}[2cm]                                                             
             \end{solutionordottedlines}                                                                    
         \begin{subparts}                                                                                   
             \subpart \(v=20, u=15\) and \(t=2 \quad\)                                                      
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{20-15}{2} = \frac{5}{2} = 2.5\)                                                  
             \end{solutionordottedlines}                                                                    
             \subpart \(v=-26.8, u=-14.4\) and \(t=2\)                                                      
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{-26.8-(-14.4)}{2} = \frac{-26.8+14.4}{2} = \frac{-12.4}{2} = -6.2\)              
             \end{solutionordottedlines}                                                                    
             \subpart \(v=\frac{1}{2}, u=\frac{2}{3}\) and \(t=\frac{5}{6}\)                                
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{\frac{1}{2}-\frac{2}{3}}{\frac{5}{6}} =                                          
 \frac{\frac{3}{6}-\frac{4}{6}}{\frac{5}{6}} = \frac{-\frac{1}{6}}{\frac{5}{6}} = -\frac{1}{5}\)            
             \end{solutionordottedlines}                                                                    
         \end{subparts}                                                                                     
         \Part[2] rearrange the formula to make \(t\) the subject and find \(t\) when \(v=6, u=7\) and      
 \(a=-3\).                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(t=\frac{v-u}{a}\)                                                                        
                 \(t=\frac{6-7}{-3} = \frac{-1}{-3} = \frac{1}{3}\)                                         
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[6] Rearrange each of these formulas to make the pronumeral in brackets the subject.          
     \begin{parts}                                                                                          
         \part \(y=m x+c\hfill{}(x)\)                                                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(x=\frac{y-c}{m}\)                                                                        
             \end{solutionordottedlines}                                                                    
         \part \(C=2 \pi r\hfill{}(r)\)                                                                     
             \begin{solutionordottedlines}[2cm]                                                             
                 \(r=\frac{C}{2 \pi}\)                                                                      
             \end{solutionordottedlines}                                                                    
         \part \(s=u t+\frac{1}{2} a t^{2}\hfill{}(a)\)                                                     
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{2(s-u t)}{t^{2}}\)                                                               
             \end{solutionordottedlines}                                                                    
         \part \(V=\frac{1}{3} \pi r^{2} h\hfill{}(h)\)                                                     
             \begin{solutionordottedlines}[2cm]                                                             
                 \(h=\frac{3 V}{\pi r^{2}}\)                                                                
             \end{solutionordottedlines}                                                                    
         \part \(S=\frac{n}{2}(a+\ell)\hfill{}(n)\)                                                         
             \begin{solutionordottedlines}[2cm]                                                             
                 \(n=\frac{2 S}{a+\ell}\)                                                                   
             \end{solutionordottedlines}                                                                    
         \part \(E=m g h+\frac{1}{2} m v^{2}\hfill{}(h)\)                                                   
             \begin{solutionordottedlines}[2cm]                                                             
                 \(h=\frac{E-\frac{1}{2} m v^{2}}{m g}\)                                                    
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[4] The kinetic energy \(E\) joules of a moving object is given by \(E=\frac{1}{2} m v^{2}\), 
 where \(m \mathrm{~kg}\) is the mass of the object and \(v \mathrm{~m} / \mathrm{s}\) is its speed.        
                                                                                                            
     Rearrange the formula to make \(m\) the subject and use this to find the mass of the object when its   
 energy and speed are, respectively:                                                                        
     \begin{parts}                                                                                          
         \part 400 joules, \(10 \mathrm{~m} / \mathrm{s}\)                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(m=\frac{2 E}{v^{2}}\)                                                                    
                 \(m=\frac{2 \cdot 400}{10^{2}} = \frac{800}{100} = 8\)                                     
             \end{solutionordottedlines}                                                                    
         \part 28 joules, \(4 \mathrm{~m} / \mathrm{s}\)                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(m=\frac{2 \cdot 28}{4^{2}} = \frac{56}{16} = 3.5\)                                       
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
 \end{questions}                                                                                            
|727317965|kitty|/Applications/kitty.app|0|
\subsection{Substitution into formulas}
\begin{questions}
    \Question[3] For each part, find the value of the subject when the other pronumerals have the value indicated.
    \begin{parts}
        \part \(A=\frac{1}{2}(a+b) h\), where \(a=4, b=6, h=10\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(t=a+(n-1) d\), where \(a=30, n=8, d=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(E=\frac{1}{2} m v^{2}\), where \(m=8, v=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[4] For each part, find the value of the subject when the other pronumerals have the value indicated. Calculate \(\mathbf{a}-\mathbf{c}\) correct to 3 decimal places and \(\mathbf{d}\) correct to 2 .
    \begin{parts}
        \part \(x=\sqrt{a b}\), where \(a=40, b=50\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(V=\pi r^{2} h\), where \(r=12, h=20\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(T=2 \pi \sqrt{\frac{\ell}{g}}\), where \(\ell=88.2, g=9.8\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(A=P(1+R)^{n}\), where \(P=10000, R=0.065, n=10\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[1] For the formula \(S=2(\ell w+\ell h+h w)\), find \(h\) if \(S=592, \ell=10\) and \(w=8\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \Question[1] For the formula \(s=u t+\frac{1}{2} a t^{2}\), find \(a\) if \(s=1000, u=20\) and \(t=5\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \Question[1] For the formula \(t=a+(n-1) d\), find \(n\) if \(t=58, d=3\) and \(a=7\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \Question[2] Given \(v^{2}=u^{2}+2 a x\) and \(v>0\), find the value of \(v\) (correct to 1 decimal place) when:
    \begin{parts}\begin{multicols}{2}
        \part \(u=0, a=5\) and \(x=10\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(u=2, a=9.8\) and \(x=22\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[2] Given \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\), find the value of:
    \begin{parts}\begin{multicols}{2}
        \part \(u\) when \(f=2\) and \(v=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(u\) when \(f=3\) and \(v=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[6] The formula for finding the number of degrees Fahrenheit \((F)\) for a temperature given as a number of degrees Celsius \((C)\) is \(F=\frac{9}{5} C+32\).

    Fahrenheit temperatures are still used in the USA, but in Australia we commonly use Celsius.

    Calculate the Fahrenheit temperatures which people in the USA would recognise for:
    \begin{parts}
        \part the freezing point of water, \(0^{\circ} \mathrm{C}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the boiling point of water, \(100^{\circ} \mathrm{C}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part a nice summer temperature of \(25^{\circ} \mathrm{C}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \hspace{-1cm}\item[] Now calculate the Celsius temperatures which people in Australia would recognise for:
        \part \(50^{\circ} \mathrm{F}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(104^{\circ} \mathrm{F}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[4] Sam throws a stone down to the ground from the top of a cliff \(s\) metres high, with an initial speed of \(u \mathrm{~m} / \mathrm{s}\). It accelerates at \(a \mathrm{~m} / \mathrm{s}^{2}\). The stone hits the ground with a speed of \(v \mathrm{~m} / \mathrm{s}\) given by the formula \(v^{2}=u^{2}+2 a s\). Find the speed at which the stone hits the ground, correct to 2 decimal places, if:
    \begin{parts}
        \part \(u=0, a=9.8\) and \(s=50\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(u=5, a=9.8\) and \(s=35\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
\end{questions}
|727317983|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}
 \begin{questions}
 \Question[3] For each part, find the value of the subject when the other pronumerals have the value indicated.
 \begin{parts}
 \part \(A=\ell w\), where \(\ell=5, w=8\)
 \begin{solutionordottedlines}[2cm]
 $A = 5 \times 8 = 40$
 \end{solutionordottedlines}
 \part \(s=\frac{d}{t}\), where \(d=120, t=6\)
 \begin{solutionordottedlines}[2cm]
 $s = \frac{120}{6} = 20$
 \end{solutionordottedlines}
 \part \(A=\frac{1}{2} x y\), where \(x=10, y=7\)
 \begin{solutionordottedlines}[2cm]
 $A = \frac{1}{2} \times 10 \times 7 = 35$
 \end{solutionordottedlines}
 \end{parts}
 \Question[4] For the formula \(v=u+at\), find:
 \begin{parts}
 \begin{multicols}{2}
 \part \(v\) if \(u=6, a=3\) and \(t=5\)
 \begin{solutionordottedlines}[2cm]
 $v = 6 + 3 \times 5 = 21$
 \end{solutionordottedlines}
 \part \(u\) if \(v=40, a=5\) and \(t=2\)
 \begin{solutionordottedlines}[2cm]
 $u = 40 - 5 \times 2 = 30$
 \end{solutionordottedlines}
 \part \(a\) if \(v=60, u=0\) and \(t=5\)
 \begin{solutionordottedlines}[2cm]
 $a = \frac{60 - 0}{5} = 12$
 \end{solutionordottedlines}
 \part \(t\) if \(v=100, u=20\) and \(a=6\)
 \begin{solutionordottedlines}[2cm]
 $t = \frac{100 - 20}{6} \approx 13.33$
 \end{solutionordottedlines}
 \end{multicols}
 \end{parts}
 \Question[4]$\,$
 \begin{parts}
 \part For the formula \(S=2(a-b)\), find \(a\) if \(S=60\) and \(b=10\).
 \begin{solutionordottedlines}[2cm]
 $a = \frac{60}{2} + 10 = 40$
 \end{solutionordottedlines}
 \part For the formula \(I=\frac{180n-360}{n}\), find \(n\) if \(I=120\).
 \begin{solutionordottedlines}[2cm]
 $180n - 360 = 120n$
 $60n = 360$
 $n = 6$
 \end{solutionordottedlines}
 \part For the formula \(a=\frac{m+n}{2}\), find \(m\) if \(a=20\) and \(n=6\).
 \begin{solutionordottedlines}[2cm]
 $m = 2 \times 20 - 6 = 34$
 \end{solutionordottedlines}
 \part For the formula \(A=\frac{PRT}{100}\), find \(P\) if \(A=1600, R=4\) and \(T=10\).
 \begin{solutionordottedlines}[2cm]
 $P = \frac{1600 \times 100}{4 \times 10} = 4000$
 \end{solutionordottedlines}
 \end{parts}
 \Question[3] For the formula \(s=ut+\frac{1}{2}at^2\), find the value of:
 \begin{parts}
 \part \(u\), when \(s=10, t=20\) and \(a=2\)
 \begin{solutionordottedlines}[2cm]
 $10 = u \times 20 + \frac{1}{2} \times 2 \times 20^2$
 $10 = 20u + 400$
 $u = -19.5$
 \end{solutionordottedlines}
 \part \(a\), when \(s=20, u=5\) and \(t=2\)
 \begin{solutionordottedlines}[2cm]
 $20 = 5 \times 2 + \frac{1}{2} \times a \times 2^2$
 $20 = 10 + 2a$
 $a = 5$
 \end{solutionordottedlines}
 \end{parts}
 \Question[3] Given that \(P=\frac{M+m}{M-m}\), find the value of \(P\) when:
 \begin{parts}
 \part \(M=8\) and \(m=4\)
 \begin{solutionordottedlines}[2cm]
 $P = \frac{8+4}{8-4} = 3$
 \end{solutionordottedlines}
 \part \(M=26\) and \(m=17\)
 \begin{solutionordottedlines}[2cm]
 $P = \frac{26+17}{26-17} = \frac{43}{9}$
 \end{solutionordottedlines}
 \end{parts}
 \Question[3] The area \(A \mathrm{~cm}^{2}\) of a square with side length \(x \mathrm{~cm}\) is given by \(A=x^{2}\). If
 \(A=20\), find:
 \begin{parts}
 \part the value of \(x\)
 \begin{solutionordottedlines}[2cm]
 $x = \sqrt{20} \approx 4.47$
 \end{solutionordottedlines}
 \part the value of \(x\) correct to 2 decimal places.
 \begin{solutionordottedlines}[2cm]
 $x \approx 4.47$ (to 2 decimal places)
 \end{solutionordottedlines}
 \end{parts}
 \Question[2] For a rectangle of length \(\ell\, \mathrm{cm}\) and width \(w \mathrm{~cm}\), the perimeter \(P \mathrm{~cm}\) i
 given by \(P=2(\ell+w)\).
 Use this formula to calculate the length of a rectangle which has width \(15 \mathrm{~cm}\) and perimeter \(57 \mathrm{~cm}\).
 \begin{solutionordottedlines}[1in]
 $57 = 2(\ell + 15)$
 $\ell = \frac{57}{2} - 15 = 13.5$
 \end{solutionordottedlines}
 \Question[6] The area $A\,\text{cm}^2$ of a triangle with side lengths, $a$ cm, $b$ cm and $c$ cm is given by \textit{Heron's
 formula}:
 \begin{theorembox}
 \subsection*{Heron's Formula}
 \[A^2 = s(s-a)(s-b)(s-c)\]
 where $s = \frac{a+b+c}{2} = $ half the perimeter. This is helpful for finding the area of non-right-angled triangles for whic
 you do not even have an angle for.\ Try to determine the area of this shape yourself without the formula if you dare.
 \begin{tikzpicture}
 \end{tikzpicture}
 \end{theorembox}
 Find the exact areas of the triangles whose side lengths are given below.
 \begin{parts}
 \part \(6 \mathrm{~cm}, 8 \mathrm{~cm}\) and \(10 \mathrm{~cm}\)
 \begin{solutionordottedlines}[2cm]
 $s = \frac{6+8+10}{2} = 12$
 $A = \sqrt{12(12-6)(12-8)(12-10)} = \sqrt{12 \times 6 \times 4 \times 2} = \sqrt{576} = 24$
 \end{solutionordottedlines}
 \part \(5 \mathrm{~cm}, 12 \mathrm{~cm}\) and \(13 \mathrm{~cm}\)
 \begin{solutionordottedlines}[2cm]
 $s = \frac{5+12+13}{2} = 15$
 $A = \sqrt{15(15-5)(15-12)(15-13)} = \sqrt{15 \times 10 \times 3 \times 2} = \sqrt{900} = 30$
 \end{solutionordottedlines}
 \part \(8 \mathrm{~cm}, 10 \mathrm{~cm}\) and \(14 \mathrm{~cm}\)
 \begin{solutionordottedlines}[2cm]
 $s = \frac{8+10+14}{2} = 16$
 $A = \sqrt{16(16-8)(16-10)(16-14)} = \sqrt{16 \times 8 \times 6 \times 2} = \sqrt{1536} \approx 39.19$
 \end{solutionordottedlines}
 \part \(13 \mathrm{~cm}, 14 \mathrm{~cm}\) and \(15 \mathrm{~cm}\)
 \begin{solutionordottedlines}[2cm]
 $s = \frac{13+14+15}{2} = 21$
 $A = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84$
 \end{solutionordottedlines}
 \end{parts}
 \end{questions}
 \end{exercisebox}
|727318002|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \begin{questions}
        \Question[3] For each part, find the value of the subject when the other pronumerals have the value indicated.
        \begin{parts}
            \part \(A=\ell w\), where \(\ell=5, w=8\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(s=\frac{d}{t}\), where \(d=120, t=6\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(A=\frac{1}{2} x y\), where \(x=10, y=7\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[4] For the formula \(v=u+a t\), find:
        \begin{parts}\begin{multicols}{2}
            \part \(v\) if \(u=6, a=3\) and \(t=5\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(u\) if \(v=40, a=5\) and \(t=2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(a\) if \(v=60, u=0\) and \(t=5\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(t\) if \(v=100, u=20\) and \(a=6\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4]$\,$
        \begin{parts}
            \part For the formula \(S=2(a-b)\), find \(a\) if \(S=60\) and \(b=10\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part For the formula \(I=\frac{180 n-360}{n}\), find \(n\) if \(I=120\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part For the formula \(a=\frac{m+n}{2}\), find \(m\) if \(a=20\) and \(n=6\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part For the formula \(A=\frac{P R T}{100}\), find \(P\) if \(A=1600, R=4\) and \(T=10\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] For the formula \(s=u t+\frac{1}{2} a t^{2}\), find the value of:
        \begin{parts}
            \part \(u\), when \(s=10, t=20\) and \(a=2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(a\), when \(s=20, u=5\) and \(t=2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] Given that \(P=\frac{M+m}{M-m}\), find the value of \(P\) when:
        \begin{parts}
            \part \(M=8\) and \(m=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(M=26\) and \(m=17\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] The area \(A \mathrm{~cm}^{2}\) of a square with side length \(x \mathrm{~cm}\) is given by \(A=x^{2}\). If \(A=20\), find:
        \begin{parts}
            \part the value of \(x\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part the value of \(x\) correct to 2 decimal places.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] For a rectangle of length \(\ell\, \mathrm{cm}\) and width \(w \mathrm{~cm}\), the perimeter \(P \mathrm{~cm}\) is given by \(P=2(\ell\,+w)\).
        \\Use this formula to calculate the length of a rectangle which has width \(15 \mathrm{~cm}\) and perimeter \(57 \mathrm{~cm}\).
        \begin{solutionordottedlines}[1in]
        \end{solutionordottedlines}
        \Question[6] The area $A\,\text{cm}^2$ of a triangle with side lengths, $a$ cm, $b$ cm and $c$ cm is given by \textit{Heron's formula}:
        \begin{theorembox}
            \subsection*{Heron's Formula}
            \[A^2 = s(s-a)(s-b)(s-c)\]
            where $s = \frac{a+b+c}{2} = $ half the perimeter.
            This is helpful for finding the area of non-right-angled triangles for which you do not even have an angle for.\\
            Try to determine the area of this shape yourself without the formula if you dare.
            \begin{tikzpicture}
            \end{tikzpicture}
        \end{theorembox}
        Find the exact areas of the triangles whose side lengths are given below.
        \begin{parts}
            \part \(6 \mathrm{~cm}, 8 \mathrm{~cm}\) and \(10 \mathrm{~cm}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(5 \mathrm{~cm}, 12 \mathrm{~cm}\) and \(13 \mathrm{~cm}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(8 \mathrm{~cm}, 10 \mathrm{~cm}\) and \(14 \mathrm{~cm}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(13 \mathrm{~cm}, 14 \mathrm{~cm}\) and \(15 \mathrm{~cm}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{exercisebox}
|727318030|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
|727318033|kitty|/Applications/kitty.app|0|
 \subsection{Substitution into formulas}                                                                    
 \begin{questions}                                                                                          
     \Question[3] For each part, find the value of the subject when the other pronumerals have the value    
 indicated.                                                                                                 
     \begin{parts}                                                                                          
         \part \(A=\frac{1}{2}(a+b) h\), where \(a=4, b=6, h=10\)                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(A=\frac{1}{2}(4+6) \cdot 10 = \frac{1}{2} \cdot 10 \cdot 10 = 50\)                       
             \end{solutionordottedlines}                                                                    
         \part \(t=a+(n-1) d\), where \(a=30, n=8, d=4\)                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(t=30+(8-1) \cdot 4 = 30+7 \cdot 4 = 30+28 = 58\)                                         
             \end{solutionordottedlines}                                                                    
         \part \(E=\frac{1}{2} m v^{2}\), where \(m=8, v=4\)                                                
             \begin{solutionordottedlines}[2cm]                                                             
                 \(E=\frac{1}{2} \cdot 8 \cdot 4^{2} = 4 \cdot 16 = 64\)                                    
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[4] For each part, find the value of the subject when the other pronumerals have the value    
 indicated. Calculate \(\mathbf{a}-\mathbf{c}\) correct to 3 decimal places and \(\mathbf{d}\) correct to 2 
     \begin{parts}                                                                                          
         \part \(x=\sqrt{a b}\), where \(a=40, b=50\)                                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(x=\sqrt{40 \cdot 50} = \sqrt{2000} \approx 44.721\)                                      
             \end{solutionordottedlines}                                                                    
         \part \(V=\pi r^{2} h\), where \(r=12, h=20\)                                                      
             \begin{solutionordottedlines}[2cm]                                                             
                 \(V=\pi \cdot 12^{2} \cdot 20 = 144 \pi \cdot 20 = 2880 \pi \approx 9047.78\)              
             \end{solutionordottedlines}                                                                    
         \part \(T=2 \pi \sqrt{\frac{\ell}{g}}\), where \(\ell=88.2, g=9.8\)                                
             \begin{solutionordottedlines}[2cm]                                                             
                 \(T=2 \pi \sqrt{\frac{88.2}{9.8}} \approx 2 \pi \sqrt{9} = 2 \pi \cdot 3 \approx 18.850\)  
             \end{solutionordottedlines}                                                                    
         \part \(A=P(1+R)^{n}\), where \(P=10000, R=0.065, n=10\)                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(A=10000(1+0.065)^{10} \approx 10000 \cdot 1.8194 \approx 18194.00\)                      
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[1] For the formula \(S=2(\ell w+\ell h+h w)\), find \(h\) if \(S=592, \ell=10\) and \(w=8\). 
             \begin{solutionordottedlines}[2cm]                                                             
                 \(592=2(10 \cdot 8+10h+8h)\)                                                               
                 \(296=80+18h\)                                                                             
                 \(216=18h\)                                                                                
                 \(h=12\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \Question[1] For the formula \(s=u t+\frac{1}{2} a t^{2}\), find \(a\) if \(s=1000, u=20\) and \(t=5\) 
             \begin{solutionordottedlines}[2cm]                                                             
                 \(1000=20 \cdot 5+\frac{1}{2} a \cdot 5^{2}\)                                              
                 \(1000=100+\frac{1}{2} a \cdot 25\)                                                        
                 \(900=\frac{1}{2} a \cdot 25\)                                                             
                 \(36=a\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \Question[1] For the formula \(t=a+(n-1) d\), find \(n\) if \(t=58, d=3\) and \(a=7\).                 
             \begin{solutionordottedlines}[2cm]                                                             
                 \(58=7+(n-1) \cdot 3\)                                                                     
                 \(51=(n-1) \cdot 3\)                                                                       
                 \(17=n-1\)                                                                                 
                 \(n=18\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \Question[2] Given \(v^{2}=u^{2}+2 a x\) and \(v>0\), find the value of \(v\) (correct to 1 decimal    
 place) when:                                                                                               
     \begin{parts}\begin{multicols}{2}                                                                      
         \part \(u=0, a=5\) and \(x=10\)                                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{0^{2}+2 \cdot 5 \cdot 10} = \sqrt{100} = 10.0\)                                  
             \end{solutionordottedlines}                                                                    
         \part \(u=2, a=9.8\) and \(x=22\)                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{2^{2}+2 \cdot 9.8 \cdot 22} \approx \sqrt{4+431.2} \approx \sqrt{435.2} \approx  
 20.9\)                                                                                                     
             \end{solutionordottedlines}                                                                    
     \end{multicols}\end{parts}                                                                             
     \Question[2] Given \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\), find the value of:                         
     \begin{parts}\begin{multicols}{2}                                                                      
         \part \(u\) when \(f=2\) and \(v=4\)                                                               
             \begin{solutionordottedlines}[2cm]                                                             
                 \(\frac{1}{2}=\frac{1}{u}+\frac{1}{4}\)                                                    
                 \(\frac{1}{2}-\frac{1}{4}=\frac{1}{u}\)                                                    
                 \(\frac{1}{4}=\frac{1}{u}\)                                                                
                 \(u=4\)                                                                                    
             \end{solutionordottedlines}                                                                    
         \part \(u\) when \(f=3\) and \(v=4\)                                                               
             \begin{solutionordottedlines}[2cm]                                                             
                 \(\frac{1}{3}=\frac{1}{u}+\frac{1}{4}\)                                                    
                 \(\frac{1}{3}-\frac{1}{4}=\frac{1}{u}\)                                                    
                 \(\frac{1}{12}=\frac{1}{u}\)                                                               
                 \(u=12\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \end{multicols}\end{parts}                                                                             
     \Question[6] The formula for finding the number of degrees Fahrenheit \((F)\) for a temperature given  
 a number of degrees Celsius \((C)\) is \(F=\frac{9}{5} C+32\).                                             
                                                                                                            
     Fahrenheit temperatures are still used in the USA, but in Australia we commonly use Celsius.           
                                                                                                            
     Calculate the Fahrenheit temperatures which people in the USA would recognise for:                     
     \begin{parts}                                                                                          
         \part the freezing point of water, \(0^{\circ} \mathrm{C}\)                                        
             \begin{solutionordottedlines}[2cm]                                                             
                 \(F=\frac{9}{5} \cdot 0+32 = 32\)                                                          
             \end{solutionordottedlines}                                                                    
         \part the boiling point of water, \(100^{\circ} \mathrm{C}\)                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(F=\frac{9}{5} \cdot 100+32 = 180+32 = 212\)                                              
             \end{solutionordottedlines}                                                                    
         \part a nice summer temperature of \(25^{\circ} \mathrm{C}\)                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(F=\frac{9}{5} \cdot 25+32 = 45+32 = 77\)                                                 
             \end{solutionordottedlines}                                                                    
         \hspace{-1cm}\item[] Now calculate the Celsius temperatures which people in Australia would        
 recognise for:                                                                                             
         \part \(50^{\circ} \mathrm{F}\)                                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(C=\frac{5}{9}(50-32) = \frac{5}{9} \cdot 18 \approx 10\)                                 
             \end{solutionordottedlines}                                                                    
         \part \(104^{\circ} \mathrm{F}\)                                                                   
             \begin{solutionordottedlines}[2cm]                                                             
                 \(C=\frac{5}{9}(104-32) = \frac{5}{9} \cdot 72 \approx 40\)                                
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[4] Sam throws a stone down to the ground from the top of a cliff \(s\) metres high, with an  
 initial speed of \(u \mathrm{~m} / \mathrm{s}\). It accelerates at \(a \mathrm{~m} / \mathrm{s}^{2}\). The 
 stone hits the ground with a speed of \(v \mathrm{~m} / \mathrm{s}\) given by the formula \(v^{2}=u^{2}+2  
 s\). Find the speed at which the stone hits the ground, correct to 2 decimal places, if:                   
     \begin{parts}                                                                                          
         \part \(u=0, a=9.8\) and \(s=50\)                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{0^{2}+2 \cdot 9.8 \cdot 50} = \sqrt{980} \approx 31.30\)                         
             \end{solutionordottedlines}                                                                    
         \part \(u=5, a=9.8\) and \(s=35\)                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{5^{2}+2 \cdot 9.8 \cdot 35} \approx \sqrt{25+686} \approx \sqrt{711} \approx     
 26.67\)                                                                                                    
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
 \end{questions}                                                                                            
|727318055|kitty|/Applications/kitty.app|0|
 \subsection{Substitution into formulas}                                                                    
 \begin{questions}                                                                                          
     \Question[3] For each part, find the value of the subject when the other pronumerals have the value    
 indicated.                                                                                                 
     \begin{parts}                                                                                          
         \part \(A=\frac{1}{2}(a+b) h\), where \(a=4, b=6, h=10\)                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(A=\frac{1}{2}(4+6) \cdot 10 = \frac{1}{2} \cdot 10 \cdot 10 = 50\)                       
             \end{solutionordottedlines}                                                                    
         \part \(t=a+(n-1) d\), where \(a=30, n=8, d=4\)                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(t=30+(8-1) \cdot 4 = 30+7 \cdot 4 = 30+28 = 58\)                                         
             \end{solutionordottedlines}                                                                    
         \part \(E=\frac{1}{2} m v^{2}\), where \(m=8, v=4\)                                                
             \begin{solutionordottedlines}[2cm]                                                             
                 \(E=\frac{1}{2} \cdot 8 \cdot 4^{2} = 4 \cdot 16 = 64\)                                    
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[4] For each part, find the value of the subject when the other pronumerals have the value    
 indicated. Calculate \(\mathbf{a}-\mathbf{c}\) correct to 3 decimal places and \(\mathbf{d}\) correct to 2 
     \begin{parts}                                                                                          
         \part \(x=\sqrt{a b}\), where \(a=40, b=50\)                                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(x=\sqrt{40 \cdot 50} = \sqrt{2000} \approx 44.721\)                                      
             \end{solutionordottedlines}                                                                    
         \part \(V=\pi r^{2} h\), where \(r=12, h=20\)                                                      
             \begin{solutionordottedlines}[2cm]                                                             
                 \(V=\pi \cdot 12^{2} \cdot 20 = 144 \pi \cdot 20 = 2880 \pi \approx 9047.78\)              
             \end{solutionordottedlines}                                                                    
         \part \(T=2 \pi \sqrt{\frac{\ell}{g}}\), where \(\ell=88.2, g=9.8\)                                
             \begin{solutionordottedlines}[2cm]                                                             
                 \(T=2 \pi \sqrt{\frac{88.2}{9.8}} \approx 2 \pi \sqrt{9} = 2 \pi \cdot 3 \approx 18.850\)  
             \end{solutionordottedlines}                                                                    
         \part \(A=P(1+R)^{n}\), where \(P=10000, R=0.065, n=10\)                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(A=10000(1+0.065)^{10} \approx 10000 \cdot 1.8194 \approx 18194.00\)                      
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[1] For the formula \(S=2(\ell w+\ell h+h w)\), find \(h\) if \(S=592, \ell=10\) and \(w=8\). 
             \begin{solutionordottedlines}[2cm]                                                             
                 \(592=2(10 \cdot 8+10h+8h)\)                                                               
                 \(296=80+18h\)                                                                             
                 \(216=18h\)                                                                                
                 \(h=12\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \Question[1] For the formula \(s=u t+\frac{1}{2} a t^{2}\), find \(a\) if \(s=1000, u=20\) and \(t=5\) 
             \begin{solutionordottedlines}[2cm]                                                             
                 \(1000=20 \cdot 5+\frac{1}{2} a \cdot 5^{2}\)                                              
                 \(1000=100+\frac{1}{2} a \cdot 25\)                                                        
                 \(900=\frac{1}{2} a \cdot 25\)                                                             
                 \(36=a\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \Question[1] For the formula \(t=a+(n-1) d\), find \(n\) if \(t=58, d=3\) and \(a=7\).                 
             \begin{solutionordottedlines}[2cm]                                                             
                 \(58=7+(n-1) \cdot 3\)                                                                     
                 \(51=(n-1) \cdot 3\)                                                                       
                 \(17=n-1\)                                                                                 
                 \(n=18\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \Question[2] Given \(v^{2}=u^{2}+2 a x\) and \(v>0\), find the value of \(v\) (correct to 1 decimal    
 place) when:                                                                                               
     \begin{parts}\begin{multicols}{2}                                                                      
         \part \(u=0, a=5\) and \(x=10\)                                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{0^{2}+2 \cdot 5 \cdot 10} = \sqrt{100} = 10.0\)                                  
             \end{solutionordottedlines}                                                                    
         \part \(u=2, a=9.8\) and \(x=22\)                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{2^{2}+2 \cdot 9.8 \cdot 22} \approx \sqrt{4+431.2} \approx \sqrt{435.2} \approx  
 20.9\)                                                                                                     
             \end{solutionordottedlines}                                                                    
     \end{multicols}\end{parts}                                                                             
     \Question[2] Given \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\), find the value of:                         
     \begin{parts}\begin{multicols}{2}                                                                      
         \part \(u\) when \(f=2\) and \(v=4\)                                                               
             \begin{solutionordottedlines}[2cm]                                                             
                 \(\frac{1}{2}=\frac{1}{u}+\frac{1}{4}\)                                                    
                 \(\frac{1}{2}-\frac{1}{4}=\frac{1}{u}\)                                                    
                 \(\frac{1}{4}=\frac{1}{u}\)                                                                
                 \(u=4\)                                                                                    
             \end{solutionordottedlines}                                                                    
         \part \(u\) when \(f=3\) and \(v=4\)                                                               
             \begin{solutionordottedlines}[2cm]                                                             
                 \(\frac{1}{3}=\frac{1}{u}+\frac{1}{4}\)                                                    
                 \(\frac{1}{3}-\frac{1}{4}=\frac{1}{u}\)                                                    
                 \(\frac{1}{12}=\frac{1}{u}\)                                                               
                 \(u=12\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \end{multicols}\end{parts}                                                                             
     \Question[6] The formula for finding the number of degrees Fahrenheit \((F)\) for a temperature given  
 a number of degrees Celsius \((C)\) is \(F=\frac{9}{5} C+32\).                                             
                                                                                                            
     Fahrenheit temperatures are still used in the USA, but in Australia we commonly use Celsius.           
                                                                                                            
     Calculate the Fahrenheit temperatures which people in the USA would recognise for:                     
     \begin{parts}                                                                                          
         \part the freezing point of water, \(0^{\circ} \mathrm{C}\)                                        
             \begin{solutionordottedlines}[2cm]                                                             
                 \(F=\frac{9}{5} \cdot 0+32 = 32\)                                                          
             \end{solutionordottedlines}                                                                    
         \part the boiling point of water, \(100^{\circ} \mathrm{C}\)                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(F=\frac{9}{5} \cdot 100+32 = 180+32 = 212\)                                              
             \end{solutionordottedlines}                                                                    
         \part a nice summer temperature of \(25^{\circ} \mathrm{C}\)                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(F=\frac{9}{5} \cdot 25+32 = 45+32 = 77\)                                                 
             \end{solutionordottedlines}                                                                    
         \hspace{-1cm}\item[] Now calculate the Celsius temperatures which people in Australia would        
 recognise for:                                                                                             
         \part \(50^{\circ} \mathrm{F}\)                                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(C=\frac{5}{9}(50-32) = \frac{5}{9} \cdot 18 \approx 10\)                                 
             \end{solutionordottedlines}                                                                    
         \part \(104^{\circ} \mathrm{F}\)                                                                   
             \begin{solutionordottedlines}[2cm]                                                             
                 \(C=\frac{5}{9}(104-32) = \frac{5}{9} \cdot 72 \approx 40\)                                
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[4] Sam throws a stone down to the ground from the top of a cliff \(s\) metres high, with an  
 initial speed of \(u \mathrm{~m} / \mathrm{s}\). It accelerates at \(a \mathrm{~m} / \mathrm{s}^{2}\). The 
 stone hits the ground with a speed of \(v \mathrm{~m} / \mathrm{s}\) given by the formula \(v^{2}=u^{2}+2  
 s\). Find the speed at which the stone hits the ground, correct to 2 decimal places, if:                   
     \begin{parts}                                                                                          
         \part \(u=0, a=9.8\) and \(s=50\)                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{0^{2}+2 \cdot 9.8 \cdot 50} = \sqrt{980} \approx 31.30\)                         
             \end{solutionordottedlines}                                                                    
         \part \(u=5, a=9.8\) and \(s=35\)                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{5^{2}+2 \cdot 9.8 \cdot 35} \approx \sqrt{25+686} \approx \sqrt{711} \approx     
 26.67\)                                                                                                    
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
 \end{questions}                                                                                            
                                                                                                            
 \subsection{Changing the subject of a formula}                                                             
                                                                                                            
 \begin{questions}                                                                                          
     \Question[] Given the formula \(v=u+a t\) :                                                            
     \begin{parts}                                                                                          
         \Part[1] rearrange the formula to make \(u\) the subject                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(u=v-a t\)                                                                                
             \end{solutionordottedlines}                                                                    
         \Part[2] find the value of \(u\) when:                                                             
             \begin{solutionordottedlines}[2cm]                                                             
             \end{solutionordottedlines}                                                                    
         \begin{subparts}                                                                                   
             \subpart \(v=20, a=2\) and \(t=5 \quad\)                                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(u=20-2 \cdot 5 = 20-10 = 10\)                                                            
             \end{solutionordottedlines}                                                                    
             \subpart \(v=40, a=-6\) and \(t=4\)                                                            
             \begin{solutionordottedlines}[2cm]                                                             
                 \(u=40-(-6) \cdot 4 = 40+24 = 64\)                                                         
             \end{solutionordottedlines}                                                                    
         \end{subparts}                                                                                     
         \Part[1] rearrange the formula to make \(a\) the subject                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{v-u}{t}\)                                                                        
             \end{solutionordottedlines}                                                                    
         \Part[2] find the value of \(a\) when:                                                             
             \begin{solutionordottedlines}[2cm]                                                             
             \end{solutionordottedlines}                                                                    
         \begin{subparts}                                                                                   
             \subpart \(v=20, u=15\) and \(t=2 \quad\)                                                      
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{20-15}{2} = \frac{5}{2} = 2.5\)                                                  
             \end{solutionordottedlines}                                                                    
             \subpart \(v=-26.8, u=-14.4\) and \(t=2\)                                                      
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{-26.8-(-14.4)}{2} = \frac{-26.8+14.4}{2} = \frac{-12.4}{2} = -6.2\)              
             \end{solutionordottedlines}                                                                    
             \subpart \(v=\frac{1}{2}, u=\frac{2}{3}\) and \(t=\frac{5}{6}\)                                
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{\frac{1}{2}-\frac{2}{3}}{\frac{5}{6}} =                                          
 \frac{\frac{3}{6}-\frac{4}{6}}{\frac{5}{6}} = \frac{-\frac{1}{6}}{\frac{5}{6}} = -\frac{1}{5}\)            
             \end{solutionordottedlines}                                                                    
         \end{subparts}                                                                                     
         \Part[2] rearrange the formula to make \(t\) the subject and find \(t\) when \(v=6, u=7\) and      
 \(a=-3\).                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(t=\frac{v-u}{a}\)                                                                        
                 \(t=\frac{6-7}{-3} = \frac{-1}{-3} = \frac{1}{3}\)                                         
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[6] Rearrange each of these formulas to make the pronumeral in brackets the subject.          
     \begin{parts}                                                                                          
         \part \(y=m x+c\hfill{}(x)\)                                                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(x=\frac{y-c}{m}\)                                                                        
             \end{solutionordottedlines}                                                                    
         \part \(C=2 \pi r\hfill{}(r)\)                                                                     
             \begin{solutionordottedlines}[2cm]                                                             
                 \(r=\frac{C}{2 \pi}\)                                                                      
             \end{solutionordottedlines}                                                                    
         \part \(s=u t+\frac{1}{2} a t^{2}\hfill{}(a)\)                                                     
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{2(s-u t)}{t^{2}}\)                                                               
             \end{solutionordottedlines}                                                                    
         \part \(V=\frac{1}{3} \pi r^{2} h\hfill{}(h)\)                                                     
             \begin{solutionordottedlines}[2cm]                                                             
                 \(h=\frac{3 V}{\pi r^{2}}\)                                                                
             \end{solutionordottedlines}                                                                    
         \part \(S=\frac{n}{2}(a+\ell)\hfill{}(n)\)                                                         
             \begin{solutionordottedlines}[2cm]                                                             
                 \(n=\frac{2 S}{a+\ell}\)                                                                   
             \end{solutionordottedlines}                                                                    
         \part \(E=m g h+\frac{1}{2} m v^{2}\hfill{}(h)\)                                                   
             \begin{solutionordottedlines}[2cm]                                                             
                 \(h=\frac{E-\frac{1}{2} m v^{2}}{m g}\)                                                    
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[4] The kinetic energy \(E\) joules of a moving object is given by \(E=\frac{1}{2} m v^{2}\), 
 where \(m \mathrm{~kg}\) is the mass of the object and \(v \mathrm{~m} / \mathrm{s}\) is its speed.        
                                                                                                            
     Rearrange the formula to make \(m\) the subject and use this to find the mass of the object when its   
 energy and speed are, respectively:                                                                        
     \begin{parts}                                                                                          
         \part 400 joules, \(10 \mathrm{~m} / \mathrm{s}\)                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(m=\frac{2 E}{v^{2}}\)                                                                    
                 \(m=\frac{2 \cdot 400}{10^{2}} = \frac{800}{100} = 8\)                                     
             \end{solutionordottedlines}                                                                    
         \part 28 joules, \(4 \mathrm{~m} / \mathrm{s}\)                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(m=\frac{2 \cdot 28}{4^{2}} = \frac{56}{16} = 3.5\)                                       
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
 \end{questions}                                                                                            

\subsection{Changing the subject of a formula}

\begin{questions}
    \Question[] Given the formula \(v=u+a t\) :
    \begin{parts}
        \Part[1] rearrange the formula to make \(u\) the subject
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Part[2] find the value of \(u\) when:
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \begin{subparts}
            \subpart \(v=20, a=2\) and \(t=5 \quad\) 
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \subpart \(v=40, a=-6\) and \(t=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{subparts}
        \Part[1] rearrange the formula to make \(a\) the subject
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Part[2] find the value of \(a\) when:
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \begin{subparts}
            \subpart \(v=20, u=15\) and \(t=2 \quad\) 
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \subpart \(v=-26.8, u=-14.4\) and \(t=2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \subpart \(v=\frac{1}{2}, u=\frac{2}{3}\) and \(t=\frac{5}{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{subparts}
        \Part[2] rearrange the formula to make \(t\) the subject and find \(t\) when \(v=6, u=7\) and \(a=-3\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[6] Rearrange each of these formulas to make the pronumeral in brackets the subject.
    \begin{parts}
        \part \(y=m x+c\hfill{}(x)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(C=2 \pi r\hfill{}(r)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(s=u t+\frac{1}{2} a t^{2}\hfill{}(a)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(V=\frac{1}{3} \pi r^{2} h\hfill{}(h)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(S=\frac{n}{2}(a+\ell)\hfill{}(n)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(E=m g h+\frac{1}{2} m v^{2}\hfill{}(h)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[4] The kinetic energy \(E\) joules of a moving object is given by \(E=\frac{1}{2} m v^{2}\), where \(m \mathrm{~kg}\) is the mass of the object and \(v \mathrm{~m} / \mathrm{s}\) is its speed.

    Rearrange the formula to make \(m\) the subject and use this to find the mass of the object when its energy and speed are, respectively:
    \begin{parts}
        \part 400 joules, \(10 \mathrm{~m} / \mathrm{s}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part 28 joules, \(4 \mathrm{~m} / \mathrm{s}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
\end{questions}
|727318138|kitty|/Applications/kitty.app|0|
\subsection{Changing the subject of a formula}

\begin{questions}
    \Question[] Given the formula \(v=u+a t\) :
    \begin{parts}
        \Part[1] rearrange the formula to make \(u\) the subject
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Part[2] find the value of \(u\) when:
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \begin{subparts}
            \subpart \(v=20, a=2\) and \(t=5 \quad\) 
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \subpart \(v=40, a=-6\) and \(t=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{subparts}
        \Part[1] rearrange the formula to make \(a\) the subject
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Part[2] find the value of \(a\) when:
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \begin{subparts}
            \subpart \(v=20, u=15\) and \(t=2 \quad\) 
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \subpart \(v=-26.8, u=-14.4\) and \(t=2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \subpart \(v=\frac{1}{2}, u=\frac{2}{3}\) and \(t=\frac{5}{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{subparts}
        \Part[2] rearrange the formula to make \(t\) the subject and find \(t\) when \(v=6, u=7\) and \(a=-3\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[6] Rearrange each of these formulas to make the pronumeral in brackets the subject.
    \begin{parts}
        \part \(y=m x+c\hfill{}(x)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(C=2 \pi r\hfill{}(r)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(s=u t+\frac{1}{2} a t^{2}\hfill{}(a)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(V=\frac{1}{3} \pi r^{2} h\hfill{}(h)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(S=\frac{n}{2}(a+\ell)\hfill{}(n)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(E=m g h+\frac{1}{2} m v^{2}\hfill{}(h)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[4] The kinetic energy \(E\) joules of a moving object is given by \(E=\frac{1}{2} m v^{2}\), where \(m \mathrm{~kg}\) is the mass of the object and \(v \mathrm{~m} / \mathrm{s}\) is its speed.

    Rearrange the formula to make \(m\) the subject and use this to find the mass of the object when its energy and speed are, respectively:
    \begin{parts}
        \part 400 joules, \(10 \mathrm{~m} / \mathrm{s}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part 28 joules, \(4 \mathrm{~m} / \mathrm{s}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
\end{questions}

|727318177|kitty|/Applications/kitty.app|0|
 \subsection{Constructing Formulas}
 \begin{questions}
 \Question[8] Find a formula for:
 \begin{parts}
 \part the number of cents \(z\) in \(x\) dollars and \(y\) cents
 \begin{solutionordottedlines}[2cm]
 \( z = 100x + y \)
 \end{solutionordottedlines}
 \part the number of minutes \(x\) in \(y\) minutes and \(z\) seconds
 \begin{solutionordottedlines}[2cm]
 \( x = y + \frac{z}{60} \)
 \end{solutionordottedlines}
 \part the number of hours \(x\) in \(y\) minutes and \(z\) seconds
 \begin{solutionordottedlines}[2cm]
 \( x = \frac{y}{60} + \frac{z}{3600} \)
 \end{solutionordottedlines}
 \part the cost \(\$ m\) of 1 book if 20 books cost \(\$ c\)
 \begin{solutionordottedlines}[2cm]
 \( m = \frac{c}{20} \)
 \end{solutionordottedlines}
 \part the cost \(\$ n\) of 1 suit if 5 suits cost \(\$ m\)
 \begin{solutionordottedlines}[2cm]
 \( n = \frac{m}{5} \)
 \end{solutionordottedlines}
 \part the cost \(\$ m\) of 1 tyre if \(x\) tyres cost \(\$ y\)
 \begin{solutionordottedlines}[2cm]
 \( m = \frac{y}{x} \)
 \end{solutionordottedlines}
 \part the cost \(\$ p\) of \(n\) suits if 4 suits cost \(\$ k\)
 \begin{solutionordottedlines}[2cm]
 \( p = \frac{k}{4}n \)
 \end{solutionordottedlines}
 \part the cost \(\$ q\) of \(x\) cars if 8 cars cost \(\$ b\)
 \begin{solutionordottedlines}[2cm]
 \( q = \frac{b}{8}x \)
 \end{solutionordottedlines}
 \end{parts}
 \Question[7] In each part, find a formula from the information given.
 \begin{parts}
 \part A hire car firm charges \(\$ 20\) per day plus 40 cents per \(\mathrm{km}\). What is the total cost \(\$ C\) for a day in which \(x \mathrm{~km}\) was travelled?
 \begin{solutionordottedlines}[2cm]
 \( C = 20 + 0.40x \)
 \end{solutionordottedlines}
 \part If there are 50 litres of petrol in the tank of a car and petrol is used at the rate of 4 litres per day, what is the number of litres \(y\) that remains after \(x\)
 days?
 \begin{solutionordottedlines}[2cm]
 \( y = 50 - 4x \)
 \end{solutionordottedlines}
 \part Cooking instructions for a forequarter of lamb are as follows: preheat oven to \(220^{\circ}C\). Cook for 30 minutes, then reduce heat to \(180^{\circ}C\) and cook for
 further \(25\) minutes for every \(500\) grams. If the lamb weighs \(x\) grams, what is the total cooking time \(T\) in minutes?
 \begin{solutionordottedlines}[2cm]
 \( T = 30 + 25 \left(\frac{x}{500}\right) \)
 \end{solutionordottedlines}
 \end{parts}
 \end{questions}
|727318293|kitty|/Applications/kitty.app|0|
\begin{questions}
    \Question[8] Find a formula for:
    \begin{parts}
        \part the number of cents \(z\) in \(x\) dollars and \(y\) cents
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the number of minutes \(x\) in \(y\) minutes and \(z\) seconds
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the number of hours \(x\) in \(y\) minutes and \(z\) seconds
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the cost \(\$ m\) of 1 book if 20 books cost \(\$ c\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the cost \(\$ n\) of 1 suit if 5 suits cost \(\$ m\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the cost \(\$ m\) of 1 tyre if \(x\) tyres cost \(\$ y\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the cost \(\$ p\) of \(n\) suits if 4 suits cost \(\$ k\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the cost \(\$ q\) of \(x\) cars if 8 cars cost \(\$ b\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[7] In each part, find a formula from the information given.
    \begin{parts}
        \part A hire car firm charges \(\$ 20\) per day plus 40 cents per \(\mathrm{km}\). What is the total cost \(\$ C\) for a day in which \(x \mathrm{~km}\) was travelled?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part If there are 50 litres of petrol in the tank of a car and petrol is used at the rate of 4 litres per day, what is the number of litres \(y\) that remains after \(x\) days?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part Cooking instructions for a forequarter of lamb are as follows: preheat oven to \(220^{\circ} \mathrm{C}\) and cook for \(45 \mathrm{~min}\) per kg plus an additional \(20 \mathrm{~min}\). What is the formula relating the cooking time \(T\) minutes and weight \(w \mathrm{~kg}\) ?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part In a sequence of numbers the first number is 2 , the second number is 4 , the third is 8 , the fourth is 16, etc. Assuming the doubling pattern continues, what is the formula you would use to calculate \(t\), the \(n\)th number?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part A piece of wire of length \(x \mathrm{~cm}\) is bent into a circle of area \(A \mathrm{~cm}^{2}\). What is the formula relating \(A\) and \(x\) ?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[10] A cyclic quadrilateral has all its vertices on a circle. Its area \(A\) is given by Brahmagupta's formula \[ A^{2}=(s-a)(s-b)(s-c)(s-d) \] where \(a, b, c\) and \(d\) are the side lengths of the quadrilateral and \(s=\frac{a+b+c+d}{2}\) is the 'semi-perimeter'. Find the exact area of a cyclic quadrilateral with side lengths:
    \begin{parts}
        \part \(4,5,6,7\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(7,4,4,3\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(8,9,10,13\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(39,52,25,60\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(51,40,68,75\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
\end{questions}
|727318327|kitty|/Applications/kitty.app|0|
    \Question[10] A cyclic quadrilateral has all its vertices on a circle. Its area \(A\) is given by Brahmagupta's formula \[ A^{2}=(s-a)(s-b)(s-c)(s-d) \] where \(a, b, c\) and \(d\) are the side lengths of the quadrilateral and \(s=\frac{a+b+c+d}{2}\) is the 'semi-perimeter'. Find the exact area of a cyclic quadrilateral with side lengths:
    \begin{parts}
        \part \(4,5,6,7\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(7,4,4,3\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(8,9,10,13\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(39,52,25,60\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(51,40,68,75\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
|727318399|kitty|/Applications/kitty.app|0|
 \subsection{Substitution into formulas}                                                                    
 \begin{questions}                                                                                          
     \Question[3] For each part, find the value of the subject when the other pronumerals have the value    
 indicated.                                                                                                 
     \begin{parts}                                                                                          
         \part \(A=\frac{1}{2}(a+b) h\), where \(a=4, b=6, h=10\)                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(A=\frac{1}{2}(4+6) \cdot 10 = \frac{1}{2} \cdot 10 \cdot 10 = 50\)                       
             \end{solutionordottedlines}                                                                    
         \part \(t=a+(n-1) d\), where \(a=30, n=8, d=4\)                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(t=30+(8-1) \cdot 4 = 30+7 \cdot 4 = 30+28 = 58\)                                         
             \end{solutionordottedlines}                                                                    
         \part \(E=\frac{1}{2} m v^{2}\), where \(m=8, v=4\)                                                
             \begin{solutionordottedlines}[2cm]                                                             
                 \(E=\frac{1}{2} \cdot 8 \cdot 4^{2} = 4 \cdot 16 = 64\)                                    
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[4] For each part, find the value of the subject when the other pronumerals have the value    
 indicated. Calculate \(\mathbf{a}-\mathbf{c}\) correct to 3 decimal places and \(\mathbf{d}\) correct to 2 
     \begin{parts}                                                                                          
         \part \(x=\sqrt{a b}\), where \(a=40, b=50\)                                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(x=\sqrt{40 \cdot 50} = \sqrt{2000} \approx 44.721\)                                      
             \end{solutionordottedlines}                                                                    
         \part \(V=\pi r^{2} h\), where \(r=12, h=20\)                                                      
             \begin{solutionordottedlines}[2cm]                                                             
                 \(V=\pi \cdot 12^{2} \cdot 20 = 144 \pi \cdot 20 = 2880 \pi \approx 9047.78\)              
             \end{solutionordottedlines}                                                                    
         \part \(T=2 \pi \sqrt{\frac{\ell}{g}}\), where \(\ell=88.2, g=9.8\)                                
             \begin{solutionordottedlines}[2cm]                                                             
                 \(T=2 \pi \sqrt{\frac{88.2}{9.8}} \approx 2 \pi \sqrt{9} = 2 \pi \cdot 3 \approx 18.850\)  
             \end{solutionordottedlines}                                                                    
         \part \(A=P(1+R)^{n}\), where \(P=10000, R=0.065, n=10\)                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(A=10000(1+0.065)^{10} \approx 10000 \cdot 1.8194 \approx 18194.00\)                      
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[1] For the formula \(S=2(\ell w+\ell h+h w)\), find \(h\) if \(S=592, \ell=10\) and \(w=8\). 
             \begin{solutionordottedlines}[2cm]                                                             
                 \(592=2(10 \cdot 8+10h+8h)\)                                                               
                 \(296=80+18h\)                                                                             
                 \(216=18h\)                                                                                
                 \(h=12\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \Question[1] For the formula \(s=u t+\frac{1}{2} a t^{2}\), find \(a\) if \(s=1000, u=20\) and \(t=5\) 
             \begin{solutionordottedlines}[2cm]                                                             
                 \(1000=20 \cdot 5+\frac{1}{2} a \cdot 5^{2}\)                                              
                 \(1000=100+\frac{1}{2} a \cdot 25\)                                                        
                 \(900=\frac{1}{2} a \cdot 25\)                                                             
                 \(36=a\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \Question[1] For the formula \(t=a+(n-1) d\), find \(n\) if \(t=58, d=3\) and \(a=7\).                 
             \begin{solutionordottedlines}[2cm]                                                             
                 \(58=7+(n-1) \cdot 3\)                                                                     
                 \(51=(n-1) \cdot 3\)                                                                       
                 \(17=n-1\)                                                                                 
                 \(n=18\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \Question[2] Given \(v^{2}=u^{2}+2 a x\) and \(v>0\), find the value of \(v\) (correct to 1 decimal    
 place) when:                                                                                               
     \begin{parts}\begin{multicols}{2}                                                                      
         \part \(u=0, a=5\) and \(x=10\)                                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{0^{2}+2 \cdot 5 \cdot 10} = \sqrt{100} = 10.0\)                                  
             \end{solutionordottedlines}                                                                    
         \part \(u=2, a=9.8\) and \(x=22\)                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{2^{2}+2 \cdot 9.8 \cdot 22} \approx \sqrt{4+431.2} \approx \sqrt{435.2} \approx  
 20.9\)                                                                                                     
             \end{solutionordottedlines}                                                                    
     \end{multicols}\end{parts}                                                                             
     \Question[2] Given \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\), find the value of:                         
     \begin{parts}\begin{multicols}{2}                                                                      
         \part \(u\) when \(f=2\) and \(v=4\)                                                               
             \begin{solutionordottedlines}[2cm]                                                             
                 \(\frac{1}{2}=\frac{1}{u}+\frac{1}{4}\)                                                    
                 \(\frac{1}{2}-\frac{1}{4}=\frac{1}{u}\)                                                    
                 \(\frac{1}{4}=\frac{1}{u}\)                                                                
                 \(u=4\)                                                                                    
             \end{solutionordottedlines}                                                                    
         \part \(u\) when \(f=3\) and \(v=4\)                                                               
             \begin{solutionordottedlines}[2cm]                                                             
                 \(\frac{1}{3}=\frac{1}{u}+\frac{1}{4}\)                                                    
                 \(\frac{1}{3}-\frac{1}{4}=\frac{1}{u}\)                                                    
                 \(\frac{1}{12}=\frac{1}{u}\)                                                               
                 \(u=12\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \end{multicols}\end{parts}                                                                             
     \Question[6] The formula for finding the number of degrees Fahrenheit \((F)\) for a temperature given  
 a number of degrees Celsius \((C)\) is \(F=\frac{9}{5} C+32\).                                             
                                                                                                            
     Fahrenheit temperatures are still used in the USA, but in Australia we commonly use Celsius.           
                                                                                                            
     Calculate the Fahrenheit temperatures which people in the USA would recognise for:                     
     \begin{parts}                                                                                          
         \part the freezing point of water, \(0^{\circ} \mathrm{C}\)                                        
             \begin{solutionordottedlines}[2cm]                                                             
                 \(F=\frac{9}{5} \cdot 0+32 = 32\)                                                          
             \end{solutionordottedlines}                                                                    
         \part the boiling point of water, \(100^{\circ} \mathrm{C}\)                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(F=\frac{9}{5} \cdot 100+32 = 180+32 = 212\)                                              
             \end{solutionordottedlines}                                                                    
         \part a nice summer temperature of \(25^{\circ} \mathrm{C}\)                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(F=\frac{9}{5} \cdot 25+32 = 45+32 = 77\)                                                 
             \end{solutionordottedlines}                                                                    
         \hspace{-1cm}\item[] Now calculate the Celsius temperatures which people in Australia would        
 recognise for:                                                                                             
         \part \(50^{\circ} \mathrm{F}\)                                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(C=\frac{5}{9}(50-32) = \frac{5}{9} \cdot 18 \approx 10\)                                 
             \end{solutionordottedlines}                                                                    
         \part \(104^{\circ} \mathrm{F}\)                                                                   
             \begin{solutionordottedlines}[2cm]                                                             
                 \(C=\frac{5}{9}(104-32) = \frac{5}{9} \cdot 72 \approx 40\)                                
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[4] Sam throws a stone down to the ground from the top of a cliff \(s\) metres high, with an  
 initial speed of \(u \mathrm{~m} / \mathrm{s}\). It accelerates at \(a \mathrm{~m} / \mathrm{s}^{2}\). The 
 stone hits the ground with a speed of \(v \mathrm{~m} / \mathrm{s}\) given by the formula \(v^{2}=u^{2}+2  
 s\). Find the speed at which the stone hits the ground, correct to 2 decimal places, if:                   
     \begin{parts}                                                                                          
         \part \(u=0, a=9.8\) and \(s=50\)                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{0^{2}+2 \cdot 9.8 \cdot 50} = \sqrt{980} \approx 31.30\)                         
             \end{solutionordottedlines}                                                                    
         \part \(u=5, a=9.8\) and \(s=35\)                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{5^{2}+2 \cdot 9.8 \cdot 35} \approx \sqrt{25+686} \approx \sqrt{711} \approx     
 26.67\)                                                                                                    
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
 \end{questions}                                                                                            
                                                                                                            
 \subsection{Changing the subject of a formula}                                                             
                                                                                                            
 \begin{questions}                                                                                          
     \Question[] Given the formula \(v=u+a t\) :                                                            
     \begin{parts}                                                                                          
         \Part[1] rearrange the formula to make \(u\) the subject                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(u=v-a t\)                                                                                
             \end{solutionordottedlines}                                                                    
         \Part[2] find the value of \(u\) when:                                                             
             \begin{solutionordottedlines}[2cm]                                                             
             \end{solutionordottedlines}                                                                    
         \begin{subparts}                                                                                   
             \subpart \(v=20, a=2\) and \(t=5 \quad\)                                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(u=20-2 \cdot 5 = 20-10 = 10\)                                                            
             \end{solutionordottedlines}                                                                    
             \subpart \(v=40, a=-6\) and \(t=4\)                                                            
             \begin{solutionordottedlines}[2cm]                                                             
                 \(u=40-(-6) \cdot 4 = 40+24 = 64\)                                                         
             \end{solutionordottedlines}                                                                    
         \end{subparts}                                                                                     
         \Part[1] rearrange the formula to make \(a\) the subject                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{v-u}{t}\)                                                                        
             \end{solutionordottedlines}                                                                    
         \Part[2] find the value of \(a\) when:                                                             
             \begin{solutionordottedlines}[2cm]                                                             
             \end{solutionordottedlines}                                                                    
         \begin{subparts}                                                                                   
             \subpart \(v=20, u=15\) and \(t=2 \quad\)                                                      
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{20-15}{2} = \frac{5}{2} = 2.5\)                                                  
             \end{solutionordottedlines}                                                                    
             \subpart \(v=-26.8, u=-14.4\) and \(t=2\)                                                      
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{-26.8-(-14.4)}{2} = \frac{-26.8+14.4}{2} = \frac{-12.4}{2} = -6.2\)              
             \end{solutionordottedlines}                                                                    
             \subpart \(v=\frac{1}{2}, u=\frac{2}{3}\) and \(t=\frac{5}{6}\)                                
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{\frac{1}{2}-\frac{2}{3}}{\frac{5}{6}} =                                          
 \frac{\frac{3}{6}-\frac{4}{6}}{\frac{5}{6}} = \frac{-\frac{1}{6}}{\frac{5}{6}} = -\frac{1}{5}\)            
             \end{solutionordottedlines}                                                                    
         \end{subparts}                                                                                     
         \Part[2] rearrange the formula to make \(t\) the subject and find \(t\) when \(v=6, u=7\) and      
 \(a=-3\).                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(t=\frac{v-u}{a}\)                                                                        
                 \(t=\frac{6-7}{-3} = \frac{-1}{-3} = \frac{1}{3}\)                                         
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[6] Rearrange each of these formulas to make the pronumeral in brackets the subject.          
     \begin{parts}                                                                                          
         \part \(y=m x+c\hfill{}(x)\)                                                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(x=\frac{y-c}{m}\)                                                                        
             \end{solutionordottedlines}                                                                    
         \part \(C=2 \pi r\hfill{}(r)\)                                                                     
             \begin{solutionordottedlines}[2cm]                                                             
                 \(r=\frac{C}{2 \pi}\)                                                                      
             \end{solutionordottedlines}                                                                    
         \part \(s=u t+\frac{1}{2} a t^{2}\hfill{}(a)\)                                                     
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{2(s-u t)}{t^{2}}\)                                                               
             \end{solutionordottedlines}                                                                    
         \part \(V=\frac{1}{3} \pi r^{2} h\hfill{}(h)\)                                                     
             \begin{solutionordottedlines}[2cm]                                                             
                 \(h=\frac{3 V}{\pi r^{2}}\)                                                                
             \end{solutionordottedlines}                                                                    
         \part \(S=\frac{n}{2}(a+\ell)\hfill{}(n)\)                                                         
             \begin{solutionordottedlines}[2cm]                                                             
                 \(n=\frac{2 S}{a+\ell}\)                                                                   
             \end{solutionordottedlines}                                                                    
         \part \(E=m g h+\frac{1}{2} m v^{2}\hfill{}(h)\)                                                   
             \begin{solutionordottedlines}[2cm]                                                             
                 \(h=\frac{E-\frac{1}{2} m v^{2}}{m g}\)                                                    
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[4] The kinetic energy \(E\) joules of a moving object is given by \(E=\frac{1}{2} m v^{2}\), 
 where \(m \mathrm{~kg}\) is the mass of the object and \(v \mathrm{~m} / \mathrm{s}\) is its speed.        
                                                                                                            
     Rearrange the formula to make \(m\) the subject and use this to find the mass of the object when its   
 energy and speed are, respectively:                                                                        
     \begin{parts}                                                                                          
         \part 400 joules, \(10 \mathrm{~m} / \mathrm{s}\)                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(m=\frac{2 E}{v^{2}}\)                                                                    
                 \(m=\frac{2 \cdot 400}{10^{2}} = \frac{800}{100} = 8\)                                     
             \end{solutionordottedlines}                                                                    
         \part 28 joules, \(4 \mathrm{~m} / \mathrm{s}\)                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(m=\frac{2 \cdot 28}{4^{2}} = \frac{56}{16} = 3.5\)                                       
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
 \end{questions}                                                                                            


|727318437|kitty|/Applications/kitty.app|0|
                                                                                                                                                                                
 \subsection{Constructing Formulas}                                                                                                                                             
                                                                                                                                                                                
 \begin{questions}                                                                                                                                                              
     \Question[8] Find a formula for:                                                                                                                                           
     \begin{parts}                                                                                                                                                              
         \part the number of cents \(z\) in \(x\) dollars and \(y\) cents                                                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(z = 100x + y\)                                                                                                                                               
             \end{solutionordottedlines}                                                                                                                                        
         \part the number of minutes \(x\) in \(y\) minutes and \(z\) seconds                                                                                                   
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(x = y + \frac{z}{60}\)                                                                                                                                       
             \end{solutionordottedlines}                                                                                                                                        
         \part the number of hours \(x\) in \(y\) minutes and \(z\) seconds                                                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(x = \frac{y}{60} + \frac{z}{3600}\)                                                                                                                          
             \end{solutionordottedlines}                                                                                                                                        
         \part the cost \(\$ m\) of 1 book if 20 books cost \(\$ c\)                                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(m = \frac{c}{20}\)                                                                                                                                           
             \end{solutionordottedlines}                                                                                                                                        
         \part the cost \(\$ n\) of 1 suit if 5 suits cost \(\$ m\)                                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(n = \frac{m}{5}\)                                                                                                                                            
             \end{solutionordottedlines}                                                                                                                                        
         \part the cost \(\$ m\) of 1 tyre if \(x\) tyres cost \(\$ y\)                                                                                                         
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(m = \frac{y}{x}\)                                                                                                                                            
             \end{solutionordottedlines}                                                                                                                                        
         \part the cost \(\$ p\) of \(n\) suits if 4 suits cost \(\$ k\)                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(p = \frac{k}{4}n\)                                                                                                                                           
             \end{solutionordottedlines}                                                                                                                                        
         \part the cost \(\$ q\) of \(x\) cars if 8 cars cost \(\$ b\)                                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(q = \frac{b}{8}x\)                                                                                                                                           
             \end{solutionordottedlines}                                                                                                                                        
     \end{parts}                                                                                                                                                                
     \Question[7] In each part, find a formula from the information given.                                                                                                      
     \begin{parts}                                                                                                                                                              
         \part A hire car firm charges \(\$ 20\) per day plus 40 cents per \(\mathrm{km}\). What is the total cost \(\$ C\) for a day in which \(x \mathrm{~km}\) was travelled 
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(C = 20 + 0.40x\)                                                                                                                                             
             \end{solutionordottedlines}                                                                                                                                        
         \part If there are 50 litres of petrol in the tank of a car and petrol is used at the rate of 4 litres per day, what is the number of litres \(y\) that remains after  
 \(x\) days?                                                                                                                                                                    
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(y = 50 - 4x\)                                                                                                                                                
             \end{solutionordottedlines}                                                                                                                                        
         \part Cooking instructions for a forequarter of lamb are as follows: preheat oven to \(220^{\circ} \mathrm{C}\) and cook for \(45 \mathrm{~min}\) per kg plus an       
 additional \(20 \mathrm{~min}\). What is the formula relating the cooking time \(T\) minutes and weight \(w \mathrm{~kg}\) ?                                                   
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(T = 45w + 20\)                                                                                                                                               
             \end{solutionordottedlines}                                                                                                                                        
         \part In a sequence of numbers the first number is 2 , the second number is 4 , the third is 8 , the fourth is 16, etc. Assuming the doubling pattern continues, what  
 the formula you would use to calculate \(t\), the \(n\)th number?                                                                                                              
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(t = 2^n\)                                                                                                                                                    
             \end{solutionordottedlines}                                                                                                                                        
         \part A piece of wire of length \(x \mathrm{~cm}\) is bent into a circle of area \(A \mathrm{~cm}^{2}\). What is the formula relating \(A\) and \(x\) ?                
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(A = \frac{\pi}{4\pi^2}x^2 = \frac{x^2}{4\pi}\)                                                                                                               
             \end{solutionordottedlines}                                                                                                                                        
     \end{parts}                                                                                                                                                                
     \Question[10] A cyclic quadrilateral has all its vertices on a circle. Its area \(A\) is given by Brahmagupta's formula \[ A^{2}=(s-a)(s-b)(s-c)(s-d) \] where \(a, b, c\) 
 and \(d\) are the side lengths of the quadrilateral and \(s=\frac{a+b+c+d}{2}\) is the 'semi-perimeter'. Find the exact area of a cyclic quadrilateral with side lengths:      
     \begin{parts}                                                                                                                                                              
         \part \(4,5,6,7\)                                                                                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(s = \frac{4+5+6+7}{2} = 11\) \\                                                                                                                              
                 \(A^{2} = (11-4)(11-5)(11-6)(11-7) = 7 \cdot 6 \cdot 5 \cdot 4\) \\                                                                                            
                 \(A = \sqrt{7 \cdot 6 \cdot 5 \cdot 4} = \sqrt{840} = 2\sqrt{210}\)                                                                                            
             \end{solutionordottedlines}                                                                                                                                        
         \part \(7,4,4,3\)                                                                                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(s = \frac{7+4+4+3}{2} = 9\) \\                                                                                                                               
                 \(A^{2} = (9-7)(9-4)(9-4)(9-3) = 2 \cdot 5 \cdot 5 \cdot 6\) \\                                                                                                
                 \(A = \sqrt{2 \cdot 5 \cdot 5 \cdot 6} = \sqrt{300} = 10\sqrt{3}\)                                                                                             
             \end{solutionordottedlines}                                                                                                                                        
         \part \(8,9,10,13\)                                                                                                                                                    
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(s = \frac{8+9+10+13}{2} = 20\) \\                                                                                                                            
                 \(A^{2} = (20-8)(20-9)(20-10)(20-13) = 12 \cdot 11 \cdot 10 \cdot 7\) \\                                                                                       
                 \(A = \sqrt{12 \cdot 11 \cdot 10 \cdot 7} = \sqrt{9240} = 2\sqrt{2310}\)                                                                                       
             \end{solutionordottedlines}                                                                                                                                        
         \part \(39,52,25,60\)                                                                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(s = \frac{39+52+25+60}{2} = 88\) \\                                                                                                                          
                 \(A^{2} = (88-39)(88-52)(88-25)(88-60) = 49 \cdot 36 \cdot 63 \cdot 28\) \\                                                                                    
                 \(A = \sqrt{49 \cdot 36 \cdot 63 \cdot 28} = \sqrt{3895584} = 1974\)                                                                                           
             \end{solutionordottedlines}                                                                                                                                        
         \part \(51,40,68,75\)                                                                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(s = \frac{51+40+68+75}{2} = 117\) \\                                                                                                                         
                 \(A^{2} = (117-51)(117-40)(117-68)(117-75) = 66 \cdot 77 \cdot 49 \cdot 42\) \\                                                                                
                 \(A = \sqrt{66 \cdot 77 \cdot 49 \cdot 42} = \sqrt{11388348} = 3372\)                                                                                          
             \end{solutionordottedlines}                                                                                                                                        
     \end{parts}                                                                                                                                                                
 \end{questions}                                                                                                                                                                
                                                                                                                                                                                
|727318570|kitty|/Applications/kitty.app|0|
\subsection{Constructing Formulas}

\begin{questions}
    \Question[8] Find a formula for:
    \begin{parts}
        \part the number of cents \(z\) in \(x\) dollars and \(y\) cents
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the number of minutes \(x\) in \(y\) minutes and \(z\) seconds
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the number of hours \(x\) in \(y\) minutes and \(z\) seconds
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the cost \(\$ m\) of 1 book if 20 books cost \(\$ c\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the cost \(\$ n\) of 1 suit if 5 suits cost \(\$ m\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the cost \(\$ m\) of 1 tyre if \(x\) tyres cost \(\$ y\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the cost \(\$ p\) of \(n\) suits if 4 suits cost \(\$ k\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the cost \(\$ q\) of \(x\) cars if 8 cars cost \(\$ b\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[7] In each part, find a formula from the information given.
    \begin{parts}
        \part A hire car firm charges \(\$ 20\) per day plus 40 cents per \(\mathrm{km}\). What is the total cost \(\$ C\) for a day in which \(x \mathrm{~km}\) was travelled?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part If there are 50 litres of petrol in the tank of a car and petrol is used at the rate of 4 litres per day, what is the number of litres \(y\) that remains after \(x\) days?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part Cooking instructions for a forequarter of lamb are as follows: preheat oven to \(220^{\circ} \mathrm{C}\) and cook for \(45 \mathrm{~min}\) per kg plus an additional \(20 \mathrm{~min}\). What is the formula relating the cooking time \(T\) minutes and weight \(w \mathrm{~kg}\) ?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part In a sequence of numbers the first number is 2 , the second number is 4 , the third is 8 , the fourth is 16, etc. Assuming the doubling pattern continues, what is the formula you would use to calculate \(t\), the \(n\)th number?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part A piece of wire of length \(x \mathrm{~cm}\) is bent into a circle of area \(A \mathrm{~cm}^{2}\). What is the formula relating \(A\) and \(x\) ?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[10] A cyclic quadrilateral has all its vertices on a circle. Its area \(A\) is given by Brahmagupta's formula \[ A^{2}=(s-a)(s-b)(s-c)(s-d) \] where \(a, b, c\) and \(d\) are the side lengths of the quadrilateral and \(s=\frac{a+b+c+d}{2}\) is the 'semi-perimeter'. Find the exact area of a cyclic quadrilateral with side lengths:
    \begin{parts}
        \part \(4,5,6,7\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(7,4,4,3\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(8,9,10,13\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(39,52,25,60\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(51,40,68,75\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
\end{questions}
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 \subsection{Constructing Formulas}                                                                                                                                             
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Up until now it has been trivial to substitute values into \textit{formulas} and have the answer pop out quite simply. But now we shall need to do some work to find the answers - we will need to \textbf{rearrange} the \textbf{expression}.

\begin{boxdef}
    \large{\textbf{Trivial: }}
    \begin{solutionordottedlines}[1cm]
        Something that is easy, simple or self-evident.
    \end{solutionordottedlines}
\end{boxdef}


\begin{examplebox}
    \begin{questions}
        \Question[1] The manager of a bed-and-breakfast guest house finds that the weekly profit \(\$ P\) is given by the formula \[ P=40 G-600 \] where \(G\) is the number of guests who stay during the week. Make \(G\) the subject of the formula and use the result to find the number of guests needed to make a profit of \(\$ 800\).
        \begin{solutionordottedlines}[3cm]
            \[
            \begin{aligned}
            P & =40 G-600 \\
            P+600 & =40 G \\
            G & =\frac{P+600}{40} \\
            \text { When } P & =800, \\
            G & =\frac{800+600}{40} \\
            & =\frac{1400}{40}=35
            \end{aligned}
            \]

            Thirty-five guests are required to make a profit of \(\$ 800\).
        \end{solutionordottedlines}
        \Question[3] Given the formula \(v^{2}=u^{2}+2 a s\) :
        \begin{parts}
            \part rearrange the formula to make \(s\) the subject
            \begin{solutionordottedlines}[2cm]
                \(\quad v^{2}=u^{2}+2 a s\)

                \[
                \begin{array}{rlrl}
                v^{2}-u^{2} & =2 a s & & \text { (Subtract } u^{2} \text { from both sides of the formula.) } \\
                s & =\frac{v^{2}-u^{2}}{2 a} & \text { (Divide both sides of the equation by } 2 a .)
                \end{array}
                \]
            \end{solutionordottedlines}
            \part find the value of \(s\) when \(u=4, v=10\) and \(a=2\)
            \begin{solutionordottedlines}[2cm]
                When \(u=4, v=10\) and \(a=2\).

                \[
                \begin{aligned}
                s & =\frac{10^{2}-4^{2}}{2 \times 2} \\
                & =\frac{100-16}{4} \\
                & =\frac{84}{4} \\
                & =21
                \end{aligned}
                \]
            \end{solutionordottedlines}
            \part find the value of \(s\) when \(u=4, v=12\) and \(a=3\)
            \begin{solutionordottedlines}[2cm]
                When \(u=4, v=12\) and \(a=3\).

                \[
                \begin{aligned}
                s & =\frac{12^{2}-4^{2}}{2 \times 3} \\
                & =\frac{144-16}{6} \\
                & =\frac{128}{6} \\
                & =21 \frac{1}{3}
                \end{aligned}
                \]
            \end{solutionordottedlines}
        \end{parts}
        \Question[4] Rearrange each of these formulas to make the pronumeral in brackets the subject.
        \begin{parts}
            \part \(E=\frac{p^{2}}{2 m}\) \hfill\((m)\)
            \begin{solutionordottedlines}[2cm]
                \(\quad E=\frac{p^{2}}{2 m}\)

                \[
                \begin{aligned}
                m E & =\frac{p^{2}}{2} & & \text { (Multiply both sides of the equation by } m .) \\
                m & =\frac{p^{2}}{2 E} & & \text { (Divide both sides by } E .)
                \end{aligned}
                \]
            \end{solutionordottedlines}
            \part \(T=2 \pi \sqrt{\frac{\ell}{g}}\) \hfill\((\ell)\)
            \begin{solutionordottedlines}[2cm]
                \[
                \begin{array}{rlr}
                T & =2 \pi \sqrt{\frac{\ell}{g}} & \\
                \frac{T}{2 \pi} & \left.=\sqrt{\frac{\ell}{g}} \quad \text { (Divide both sides of the equation by } 2 \pi .\right) \\
                \frac{\ell}{g} & =\left(\frac{T}{2 \pi}\right)^{2} & \text { (Square both sides of the equation.) } \\
                & =\frac{T^{2}}{4 \pi^{2}} & \\
                \ell & \left.=\frac{T^{2}}{4 \pi^{2}} \times g \quad \text { (Multiply both sides of the equation by } g .\right)
                \end{array}
                \]

                That is, \(\ell=\frac{T^{2} g}{4 \pi^{2}}\)
            \end{solutionordottedlines}
            \part \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\) \hfill\((u)\)
            \begin{solutionordottedlines}[2cm]
                \(\quad \frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
                \(\frac{1}{f}-\frac{1}{v}=\frac{1}{u}\)
                (Subtract \(\frac{1}{v}\) from both sides.)
                \(\frac{v-f}{f v}=\frac{1}{u}\)
                (common denominator on LHS of equation)

                \[
                u=\frac{f v}{v-f}
                \]

                (Take reciprocals of both sides.)

                Note: \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\) does not imply \(f=u+v\).
            \end{solutionordottedlines}
            \part \(P=\sqrt{h+c}-a\)\hfill\((h)\)
            \begin{solutionordottedlines}[2cm]
                \[
                P=\sqrt{h+c}-a
                \]

                \[
                \begin{array}{rlrl}
                P+a & =\sqrt{h+c} \quad & \text { (Add } a \text { to both sides of the equation. }) \\
                (P+a)^{2} & =h+c \quad \quad \text { (Square both sides. }) \\
                h & =(P+a)^{2}-c
                \end{array}
                \]

                The previous example shows some of the techniques that can be used to rearrange a formula.
            \end{solutionordottedlines}
            \part \(\frac{3 p}{4}-\frac{5}{q}=\frac{p^{2}}{3 q}\)\hfill\((q)\)
            \begin{solutionordottedlines}[2cm]
                \[
                \begin{gathered}
                \frac{3 p}{4}-\frac{5}{q}=\frac{p^{2}}{3 q} \\
                12 q\left(\frac{3 p}{4}-\frac{5}{q}\right)=12 q \times \frac{p^{2}}{3 q}
                \end{gathered}
                \]

                \(3 p \times 3 q-5 \times 12=p^{2} \times 4 \quad\) (Multiply both sides by the lowest common denominator, \(12 q\).)

                \[
                \begin{aligned}
                9 p q-60 & =4 p^{2} \\
                9 p q & =4 p^{2}+60 \\
                q & =\frac{4 p^{2}+60}{9 p}
                \end{aligned}
                \]
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{examplebox}


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    & & & &\\
    Year 9 Mathematics Linear Topic Test Booklet & \centering\$100 & \centering 1 & \centering -\$0 & \$100\\[2.5ex]\hline
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 \begin{exercisebox}                                                                                        
     \begin{questions}                                                                                      
         \Question The profit \(\$ P\) made each day by a store owner who sells \(C D\) s is given by the   
 formula \(P=5 n-150\), where \(n\) is the number of CDs sold.                                              
         \begin{parts}                                                                                      
             \Part[1] What profit is made if the store owner sells 60 CDs?                                  
             \begin{solutionordottedlines}                                                                  
                 P = 5n - 150 \\                                                                            
                 P = 5(60) - 150 \\                                                                         
                 P = 300 - 150 \\                                                                           
                 P = \$150                                                                                  
             \end{solutionordottedlines}                                                                    
             \Part[1] Make \(n\) the subject of the formula.                                                
             \begin{solutionordottedlines}                                                                  
                 P = 5n - 150 \\                                                                            
                 P + 150 = 5n \\                                                                            
                 \(n = \frac{P + 150}{5}\)
             \end{solutionordottedlines}                                                                    
             \Part[2] How many CDs were sold if the store made:                                             
             \begin{subparts}                                                                               
                 \subpart a profit of \(\$ 275\) ?                                                          
                 \begin{solutionordottedlines}                                                              
                     \(n = \frac{P + 150}{5}\) \\                                                               
                     n = \frac{275 + 150}{5} \\                                                             
                     n = \frac{425}{5} \\                                                                   
                     n = 85                                                                                 
                 \end{solutionordottedlines}                                                                
                 \subpart a profit of \(\$ 400\) ?                                                          
                 \begin{solutionordottedlines}                                                              
                     n = \frac{P + 150}{5} \\                                                               
                     n = \frac{400 + 150}{5} \\                                                             
                     n = \frac{550}{5} \\                                                                   
                     n = 110                                                                                
                 \end{solutionordottedlines}                                                                
                 \subpart a loss of \(\$ 100\) ?                                                            
                 \begin{solutionordottedlines}                                                              
                     n = \frac{P + 150}{5} \\                                                               
                     n = \frac{-100 + 150}{5} \\                                                            
                     n = \frac{50}{5} \\                                                                    
                     n = 10                                                                                 
                 \end{solutionordottedlines}                                                                
                 \subpart no profit?                                                                        
                 \begin{solutionordottedlines}                                                              
                     n = \frac{P + 150}{5} \\                                                               
                     n = \frac{0 + 150}{5} \\                                                               
                     n = \frac{150}{5} \\                                                                   
                     n = 30                                                                                 
                 \end{solutionordottedlines}                                                                
             \end{subparts}                                                                                 
         \end{parts}                                                                                        
         \Question The cost \(\$ C\) of hiring a reception room for a function is given by the formula \(C= 
 n+250\), where \(n\) is the number of people attending the function.                                       
         \begin{parts}                                                                                      
             \Part[1] Rearrange the formula to make \(n\) the subject.                                      
             \begin{solutionordottedlines}                                                                  
                 C = 12n + 250 \\                                                                           
                 C - 250 = 12n \\                                                                           
                 n = \frac{C - 250}{12}                                                                     
             \end{solutionordottedlines}                                                                    
             \Part[2] How many people attended the function if the cost of hiring the reception room was:   
             \begin{subparts}                                                                               
                 \subpart \(\$ 730\) ?                                                                      
                 \begin{solutionordottedlines}                                                              
                     n = \frac{C - 250}{12} \\                                                              
                     n = \frac{730 - 250}{12} \\                                                            
                     n = \frac{480}{12} \\                                                                  
                     n = 40                                                                                 
                 \end{solutionordottedlines}                                                                
                 \subpart \(\$ 1090\) ?                                                                     
                 \begin{solutionordottedlines}                                                              
                     n = \frac{C - 250}{12} \\                                                              
                     n = \frac{1090 - 250}{12} \\                                                           
                     n = \frac{840}{12} \\                                                                  
                     n = 70                                                                                 
                 \end{solutionordottedlines}                                                                
                 \subpart \(\$ 1210\) ?                                                                     
                 \begin{solutionordottedlines}                                                              
                     n = \frac{C - 250}{12} \\                                                              
                     n = \frac{1210 - 250}{12} \\                                                           
                     n = \frac{960}{12} \\                                                                  
                     n = 80                                                                                 
                 \end{solutionordottedlines}                                                                
                 \subpart \(\$ 1690\) ?                                                                     
                 \begin{solutionordottedlines}                                                              
                     n = \frac{C - 250}{12} \\                                                              
                     n = \frac{1690 - 250}{12} \\                                                           
                     n = \frac{1440}{12} \\                                                                 
                     n = 120                                                                                
                 \end{solutionordottedlines}                                                                
             \end{subparts}                                                                                 
         \end{parts}                                                                                        
         \Question Given the formula \(t=a+(n-1) d\) :                                                      
         \begin{parts}                                                                                      
             \Part[1] rearrange the formula to make \(a\) the subject                                       
             \begin{solutionordottedlines}                                                                  
                 t = a + (n - 1)d \\                                                                        
                 a = t - (n - 1)d                                                                           
             \end{solutionordottedlines}                                                                    
             \Part[2] find the value of \(a\) when:                                                         
             \begin{subparts}                                                                               
                 \subpart \(t=11, n=4\) and \(d=3\)                                                         
                 \begin{solutionordottedlines}                                                              
                     a = t - (n - 1)d \\                                                                    
                     a = 11 - (4 - 1) \cdot 3 \\                                                            
                     a = 11 - 3 \cdot 3 \\                                                                  
                     a = 11 - 9 \\                                                                          
                     a = 2                                                                                  
                 \end{solutionordottedlines}                                                                
                 \subpart \(t=8, n=5\) and \(d=-3\)                                                         
                 \begin{solutionordottedlines}                                                              
                     a = t - (n - 1)d \\                                                                    
                     a = 8 - (5 - 1) \cdot (-3) \\                                                          
                     a = 8 - 4 \cdot (-3) \\                                                                
                     a = 8 + 12 \\                                                                          
                     a = 20                                                                                 
                 \end{solutionordottedlines}                                                                
             \end{subparts}                                                                                 
             \Part[1] rearrange the formula to make \(d\) the subject                                       
             \begin{solutionordottedlines}                                                                  
                 t = a + (n - 1)d \\                                                                        
                 d = \frac{t - a}{n - 1}                                                                    
             \end{solutionordottedlines}                                                                    
             \Part[2] find the value of \(d\) when:                                                         
             \begin{subparts}                                                                               
                 \subpart \(t=48, a=3\) and \(n=16\)                                                        
                 \begin{solutionordottedlines}                                                              
                     d = \frac{t - a}{n - 1} \\                                                             
                     d = \frac{48 - 3}{16 - 1} \\                                                           
                     d = \frac{45}{15} \\                                                                   
                     d = 3                                                                                  
                 \end{solutionordottedlines}                                                                
                 \subpart \(t=120, a=-30\) and \(n=101\)                                                    
                 \begin{solutionordottedlines}                                                              
                     d = \frac{t - a}{n - 1} \\                                                             
                     d = \frac{120 - (-30)}{101 - 1} \\                                                     
                     d = \frac{150}{100} \\                                                                 
                     d = 1.5                                                                                
                 \end{solutionordottedlines}                                                                
             \end{subparts}                                                                                 
             \Part[1] rearrange the formula to make \(n\) the subject and find the value of \(n\) when      
 \(t=150, a=5\) and \(d=5\)                                                                                 
             \begin{solutionordottedlines}                                                                  
                 t = a + (n - 1)d \\                                                                        
                 n = \frac{t - a}{d} + 1 \\                                                                 
                 n = \frac{150 - 5}{5} + 1 \\                                                               
                 n = \frac{145}{5} + 1 \\                                                                   
                 n = 29 + 1 \\                                                                              
                 n = 30                                                                                     
             \end{solutionordottedlines}                                                                    
         \end{parts}                                                                                        
         \Question Rearrange each of these formulas to make the pronumeral in brackets the subject.         
         \begin{parts}                                                                                      
             \part \(y=m x+c\) \hfill\((c)\)                                                                
             \begin{solutionordottedlines}                                                                  
                 c = y - mx                                                                                 
             \end{solutionordottedlines}                                                                    
             \part \(A=\frac{1}{2} b h\)\hfill\((x)\)                                                       
             \begin{solutionordottedlines}                                                                  
                 % This part seems to have a typo, as there is no 'x' in the given formula.                 
                 % Assuming the subject to be made is 'b':                                                  
                 b = \frac{2A}{h}                                                                           
             \end{solutionordottedlines}                                                                    
             \part \(P=A+2 \ell h\)\hfill\((\ell)\)                                                         
             \begin{solutionordottedlines}                                                                  
                 \ell = \frac{P - A}{2h}                                                                    
             \end{solutionordottedlines}                                                                    
             \part \(A=2\pi r^2 + 2\pi r h\hfill{}(h)\)                                                     
             \begin{solutionordottedlines}                                                                  
                 h = \frac{A - 2\pi r^2}{2\pi r}                                                            
             \end{solutionordottedlines}                                                                    
             \part \(s=\frac{n}{2}(a+\ell)\)\hfill\((a)\)                                                   
             \begin{solutionordottedlines}                                                                  
                 a = \frac{2s}{n} - \ell                                                                    
             \end{solutionordottedlines}                                                                    
             \part \(V=\pi r^{2}+\pi r s\)\hfill\((s)\)                                                     
             \begin{solutionordottedlines}                                                                  
                 s = \frac{V}{\pi r} - r                                                                    
             \end{solutionordottedlines}                                                                    
         \end{parts}                                                                                        
         \Question[3] The formula for the sum \(S\) of the interior angles in a convex \(n\)-sided polygon  
 \(S=180(n-2)\). Rearrange the formula to make \(n\) the subject and use this to find the number of sides i 
 the polygon if the sum of the interior angles is:                                                          
         \begin{parts}\begin{multicols}{3}                                                                  
             \part \(1080^{\circ}\)                                                                         
             \begin{solutionordottedlines}                                                                  
                 n = \frac{S}{180} + 2 \\                                                                   
                 n = \frac{1080}{180} + 2 \\                                                                
                 n = 6 + 2 \\                                                                               
                 n = 8                                                                                      
             \end{solutionordottedlines}                                                                    
             \part \(1800^{\circ}\)                                                                         
             \begin{solutionordottedlines}                                                                  
                 n = \frac{S}{180} + 2 \\                                                                   
                 n = \frac{1800}{180} + 2 \\                                                                
                 n = 10 + 2 \\                                                                              
                 n = 12                                                                                     
             \end{solutionordottedlines}                                                                    
             \part \(3240^{\circ}\)                                                                         
             \begin{solutionordottedlines}                                                                  
                 n = \frac{S}{180} + 2 \\                                                                   
                 n = \frac{3240}{180} + 2 \\                                                                
                 n = 18 + 2 \\                                                                              
                 n = 20                                                                                     
             \end{solutionordottedlines}                                                                    
         \end{multicols}\end{parts}                                                                         
         \Question[2] 8 When an object is shot up into the air with a speed of \(u\) metres per second, its 
 height above the ground \(h\) metres and time of flight \(t\) seconds are related (ignoring air resistance 
 by \(h=u t-4.9 t^{2}\).                                                                                    
                                                                                                            
         Find the speed at which an object was fired if it reached a height of 27.5 metres after 5 seconds. 
         \begin{solutionordottedlines}                                                                      
             h = ut - 4.9t^2 \\                                                                             
             27.5 = u(5) - 4.9(5)^2 \\                                                                      
             27.5 = 5u - 4.9(25) \\                                                                         
             27.5 = 5u - 122.5 \\                                                                           
             5u = 27.5 + 122.5 \\                                                                           
             5u = 150 \\                                                                                    
             u = \frac{150}{5} \\                                                                           
             u = 30 \text{ metres per second}                                                               
         \end{solutionordottedlines}                                                                        
         \Question[4] Rearrange each of these formulas to make the pronumeral in brackets the subject. (All 
 pronumerals represent positive numbers.)                                                                   
         \begin{parts}\begin{multicols}{2}                                                                  
             \part \(c=a^{2}+b^{2}\hfill{}(a)\)                                                             
             \begin{solutionordottedlines}                                                                  
                 a = \sqrt{c - b^2}                                                                         
             \end{solutionordottedlines}                                                                    
             \part \(x=\sqrt{a b}\hfill{}(b)\)                                                              
             \begin{solutionordottedlines}                                                                  
                 b = \frac{x^2}{a}                                                                          
             \end{solutionordottedlines}                                                                    
             \part \(T=\frac{2\pi{}}{n}\hfill{}(n)\)                                                        
             \begin{solutionordottedlines}                                                                  
                 n = \frac{2\pi}{T}                                                                         
             \end{solutionordottedlines}                                                                    
             \part \(E=\frac{m}{2r^2}\hfill{}(r)\)                                                          
             \begin{solutionordottedlines}                                                                  
                 r = \sqrt{\frac{m}{2E}}                                                                    
             \end{solutionordottedlines}                                                                    
         \end{multicols}\end{parts}                                                                         
     \end{questions}                                                                                        
 \end{exercisebox}                                                                                          
|727320176|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                        
     \begin{questions}                                                                                      
         \Question The profit \(\$ P\) made each day by a store owner who sells \(C D\) s is given by the   
 formula \(P=5 n-150\), where \(n\) is the number of CDs sold.                                              
         \begin{parts}                                                                                      
             \Part[1] What profit is made if the store owner sells 60 CDs?                                  
             \begin{solutionordottedlines}                                                                  
                 P = 5n - 150 \\                                                                            
                 P = 5(60) - 150 \\                                                                         
                 P = 300 - 150 \\                                                                           
                 P = \$150                                                                                  
             \end{solutionordottedlines}                                                                    
             \Part[1] Make \(n\) the subject of the formula.                                                
             \begin{solutionordottedlines}                                                                  
                 P = 5n - 150 \\                                                                            
                 P + 150 = 5n \\                                                                            
                 \(n = \frac{P + 150}{5}\)
             \end{solutionordottedlines}                                                                    
             \Part[2] How many CDs were sold if the store made:                                             
             \begin{subparts}                                                                               
                 \subpart a profit of \(\$ 275\) ?                                                          
                 \begin{solutionordottedlines}                                                              
                     \(n = \frac{P + 150}{5}\) \\                                                               
                     n = \frac{275 + 150}{5} \\                                                             
                     n = \frac{425}{5} \\                                                                   
                     n = 85                                                                                 
                 \end{solutionordottedlines}                                                                
                 \subpart a profit of \(\$ 400\) ?                                                          
                 \begin{solutionordottedlines}                                                              
                     n = \frac{P + 150}{5} \\                                                               
                     n = \frac{400 + 150}{5} \\                                                             
                     n = \frac{550}{5} \\                                                                   
                     n = 110                                                                                
                 \end{solutionordottedlines}                                                                
                 \subpart a loss of \(\$ 100\) ?                                                            
                 \begin{solutionordottedlines}                                                              
                     n = \frac{P + 150}{5} \\                                                               
                     n = \frac{-100 + 150}{5} \\                                                            
                     n = \frac{50}{5} \\                                                                    
                     n = 10                                                                                 
                 \end{solutionordottedlines}                                                                
                 \subpart no profit?                                                                        
                 \begin{solutionordottedlines}                                                              
                     n = \frac{P + 150}{5} \\                                                               
                     n = \frac{0 + 150}{5} \\                                                               
                     n = \frac{150}{5} \\                                                                   
                     n = 30                                                                                 
                 \end{solutionordottedlines}                                                                
             \end{subparts}                                                                                 
         \end{parts}                                                                                        
         \Question The cost \(\$ C\) of hiring a reception room for a function is given by the formula \(C= 
 n+250\), where \(n\) is the number of people attending the function.                                       
         \begin{parts}                                                                                      
             \Part[1] Rearrange the formula to make \(n\) the subject.                                      
             \begin{solutionordottedlines}                                                                  
                 C = 12n + 250 \\                                                                           
                 C - 250 = 12n \\                                                                           
                 n = \frac{C - 250}{12}                                                                     
             \end{solutionordottedlines}                                                                    
             \Part[2] How many people attended the function if the cost of hiring the reception room was:   
             \begin{subparts}                                                                               
                 \subpart \(\$ 730\) ?                                                                      
                 \begin{solutionordottedlines}                                                              
                     n = \frac{C - 250}{12} \\                                                              
                     n = \frac{730 - 250}{12} \\                                                            
                     n = \frac{480}{12} \\                                                                  
                     n = 40                                                                                 
                 \end{solutionordottedlines}                                                                
                 \subpart \(\$ 1090\) ?                                                                     
                 \begin{solutionordottedlines}                                                              
                     n = \frac{C - 250}{12} \\                                                              
                     n = \frac{1090 - 250}{12} \\                                                           
                     n = \frac{840}{12} \\                                                                  
                     n = 70                                                                                 
                 \end{solutionordottedlines}                                                                
                 \subpart \(\$ 1210\) ?                                                                     
                 \begin{solutionordottedlines}                                                              
                     n = \frac{C - 250}{12} \\                                                              
                     n = \frac{1210 - 250}{12} \\                                                           
                     n = \frac{960}{12} \\                                                                  
                     n = 80                                                                                 
                 \end{solutionordottedlines}                                                                
                 \subpart \(\$ 1690\) ?                                                                     
                 \begin{solutionordottedlines}                                                              
                     n = \frac{C - 250}{12} \\                                                              
                     n = \frac{1690 - 250}{12} \\                                                           
                     n = \frac{1440}{12} \\                                                                 
                     n = 120                                                                                
                 \end{solutionordottedlines}                                                                
             \end{subparts}                                                                                 
         \end{parts}                                                                                        
         \Question Given the formula \(t=a+(n-1) d\) :                                                      
         \begin{parts}                                                                                      
             \Part[1] rearrange the formula to make \(a\) the subject                                       
             \begin{solutionordottedlines}                                                                  
                 t = a + (n - 1)d \\                                                                        
                 a = t - (n - 1)d                                                                           
             \end{solutionordottedlines}                                                                    
             \Part[2] find the value of \(a\) when:                                                         
             \begin{subparts}                                                                               
                 \subpart \(t=11, n=4\) and \(d=3\)                                                         
                 \begin{solutionordottedlines}                                                              
                     a = t - (n - 1)d \\                                                                    
                     a = 11 - (4 - 1) \cdot 3 \\                                                            
                     a = 11 - 3 \cdot 3 \\                                                                  
                     a = 11 - 9 \\                                                                          
                     a = 2                                                                                  
                 \end{solutionordottedlines}                                                                
                 \subpart \(t=8, n=5\) and \(d=-3\)                                                         
                 \begin{solutionordottedlines}                                                              
                     a = t - (n - 1)d \\                                                                    
                     a = 8 - (5 - 1) \cdot (-3) \\                                                          
                     a = 8 - 4 \cdot (-3) \\                                                                
                     a = 8 + 12 \\                                                                          
                     a = 20                                                                                 
                 \end{solutionordottedlines}                                                                
             \end{subparts}                                                                                 
             \Part[1] rearrange the formula to make \(d\) the subject                                       
             \begin{solutionordottedlines}                                                                  
                 t = a + (n - 1)d \\                                                                        
                 d = \frac{t - a}{n - 1}                                                                    
             \end{solutionordottedlines}                                                                    
             \Part[2] find the value of \(d\) when:                                                         
             \begin{subparts}                                                                               
                 \subpart \(t=48, a=3\) and \(n=16\)                                                        
                 \begin{solutionordottedlines}                                                              
                     d = \frac{t - a}{n - 1} \\                                                             
                     d = \frac{48 - 3}{16 - 1} \\                                                           
                     d = \frac{45}{15} \\                                                                   
                     d = 3                                                                                  
                 \end{solutionordottedlines}                                                                
                 \subpart \(t=120, a=-30\) and \(n=101\)                                                    
                 \begin{solutionordottedlines}                                                              
                     d = \frac{t - a}{n - 1} \\                                                             
                     d = \frac{120 - (-30)}{101 - 1} \\                                                     
                     d = \frac{150}{100} \\                                                                 
                     d = 1.5                                                                                
                 \end{solutionordottedlines}                                                                
             \end{subparts}                                                                                 
             \Part[1] rearrange the formula to make \(n\) the subject and find the value of \(n\) when      
 \(t=150, a=5\) and \(d=5\)                                                                                 
             \begin{solutionordottedlines}                                                                  
                 t = a + (n - 1)d \\                                                                        
                 n = \frac{t - a}{d} + 1 \\                                                                 
                 n = \frac{150 - 5}{5} + 1 \\                                                               
                 n = \frac{145}{5} + 1 \\                                                                   
                 n = 29 + 1 \\                                                                              
                 n = 30                                                                                     
             \end{solutionordottedlines}                                                                    
         \end{parts}                                                                                        
         \Question Rearrange each of these formulas to make the pronumeral in brackets the subject.         
         \begin{parts}                                                                                      
             \part \(y=m x+c\) \hfill\((c)\)                                                                
             \begin{solutionordottedlines}                                                                  
                 c = y - mx                                                                                 
             \end{solutionordottedlines}                                                                    
             \part \(A=\frac{1}{2} b h\)\hfill\((x)\)                                                       
             \begin{solutionordottedlines}                                                                  
                 % This part seems to have a typo, as there is no 'x' in the given formula.                 
                 % Assuming the subject to be made is 'b':                                                  
                 b = \frac{2A}{h}                                                                           
             \end{solutionordottedlines}                                                                    
             \part \(P=A+2 \ell h\)\hfill\((\ell)\)                                                         
             \begin{solutionordottedlines}                                                                  
                 \ell = \frac{P - A}{2h}                                                                    
             \end{solutionordottedlines}                                                                    
             \part \(A=2\pi r^2 + 2\pi r h\hfill{}(h)\)                                                     
             \begin{solutionordottedlines}                                                                  
                 h = \frac{A - 2\pi r^2}{2\pi r}                                                            
             \end{solutionordottedlines}                                                                    
             \part \(s=\frac{n}{2}(a+\ell)\)\hfill\((a)\)                                                   
             \begin{solutionordottedlines}                                                                  
                 a = \frac{2s}{n} - \ell                                                                    
             \end{solutionordottedlines}                                                                    
             \part \(V=\pi r^{2}+\pi r s\)\hfill\((s)\)                                                     
             \begin{solutionordottedlines}                                                                  
                 s = \frac{V}{\pi r} - r                                                                    
             \end{solutionordottedlines}                                                                    
         \end{parts}                                                                                        
         \Question[3] The formula for the sum \(S\) of the interior angles in a convex \(n\)-sided polygon  
 \(S=180(n-2)\). Rearrange the formula to make \(n\) the subject and use this to find the number of sides i 
 the polygon if the sum of the interior angles is:                                                          
         \begin{parts}\begin{multicols}{3}                                                                  
             \part \(1080^{\circ}\)                                                                         
             \begin{solutionordottedlines}                                                                  
                 n = \frac{S}{180} + 2 \\                                                                   
                 n = \frac{1080}{180} + 2 \\                                                                
                 n = 6 + 2 \\                                                                               
                 n = 8                                                                                      
             \end{solutionordottedlines}                                                                    
             \part \(1800^{\circ}\)                                                                         
             \begin{solutionordottedlines}                                                                  
                 n = \frac{S}{180} + 2 \\                                                                   
                 n = \frac{1800}{180} + 2 \\                                                                
                 n = 10 + 2 \\                                                                              
                 n = 12                                                                                     
             \end{solutionordottedlines}                                                                    
             \part \(3240^{\circ}\)                                                                         
             \begin{solutionordottedlines}                                                                  
                 n = \frac{S}{180} + 2 \\                                                                   
                 n = \frac{3240}{180} + 2 \\                                                                
                 n = 18 + 2 \\                                                                              
                 n = 20                                                                                     
             \end{solutionordottedlines}                                                                    
         \end{multicols}\end{parts}                                                                         
         \Question[2] 8 When an object is shot up into the air with a speed of \(u\) metres per second, its 
 height above the ground \(h\) metres and time of flight \(t\) seconds are related (ignoring air resistance 
 by \(h=u t-4.9 t^{2}\).                                                                                    
                                                                                                            
         Find the speed at which an object was fired if it reached a height of 27.5 metres after 5 seconds. 
         \begin{solutionordottedlines}                                                                      
             h = ut - 4.9t^2 \\                                                                             
             27.5 = u(5) - 4.9(5)^2 \\                                                                      
             27.5 = 5u - 4.9(25) \\                                                                         
             27.5 = 5u - 122.5 \\                                                                           
             5u = 27.5 + 122.5 \\                                                                           
             5u = 150 \\                                                                                    
             u = \frac{150}{5} \\                                                                           
             u = 30 \text{ metres per second}                                                               
         \end{solutionordottedlines}                                                                        
         \Question[4] Rearrange each of these formulas to make the pronumeral in brackets the subject. (All 
 pronumerals represent positive numbers.)                                                                   
         \begin{parts}\begin{multicols}{2}                                                                  
             \part \(c=a^{2}+b^{2}\hfill{}(a)\)                                                             
             \begin{solutionordottedlines}                                                                  
                 a = \sqrt{c - b^2}                                                                         
             \end{solutionordottedlines}                                                                    
             \part \(x=\sqrt{a b}\hfill{}(b)\)                                                              
             \begin{solutionordottedlines}                                                                  
                 b = \frac{x^2}{a}                                                                          
             \end{solutionordottedlines}                                                                    
             \part \(T=\frac{2\pi{}}{n}\hfill{}(n)\)                                                        
             \begin{solutionordottedlines}                                                                  
                 n = \frac{2\pi}{T}                                                                         
             \end{solutionordottedlines}                                                                    
             \part \(E=\frac{m}{2r^2}\hfill{}(r)\)                                                          
             \begin{solutionordottedlines}                                                                  
                 r = \sqrt{\frac{m}{2E}}                                                                    
             \end{solutionordottedlines}                                                                    
         \end{multicols}\end{parts}                                                                         
     \end{questions}                                                                                        
 \end{exercisebox}                                                                                          

|727320210|kitty|/Applications/kitty.app|0|
      \end{questions}
  \end{exercisebox}
|727320218|kitty|/Applications/kitty.app|0|

          \Question[4] Rearrange each of these formulas to make the pronumeral in brackets the subject. (All
  pronumerals represent positive numbers.)
          \begin{parts}\begin{multicols}{2}
              \part \( c=a^{2}+b^{2} \)\hfill\(( a )\)
              \begin{solutionordottedlines}
                  \( a = \sqrt{c - b^2} \)
              \end{solutionordottedlines}
              \part \( x=\sqrt{ab} \)\hfill\(( b|727320238|kitty|/Applications/kitty.app|0|
              \part \( 3240^\circ \)
              \begin{solutionordottedlines}
                  \( n = \frac{S}{180} + 2 \) \\
                  \( n = \frac{3240}{180} + 2 \) \\
                  \( n = 18 + 2 \) \\
                  \( n = 20 \)
              \end{solutionordottedlines}
          \end{multicols}\end{parts}
          \Question[2] 8 When an object is shot up into the air with a speed of \( u \) metres per second, its
  height above the ground \( h \) metres and time of flight \( t \) seconds are related (ignoring air resistance
  by \( h=ut-4.9t^{2} \).

          Find the speed at which an object was fired if it reached a height of 27.5 metres after 5 seconds.
          \begin{solutionordottedlines}
              \( h = ut - 4.9t^2 \) \\
              \( 27.5 = u(5) - 4.9(5)^2 \) \\
              \( 27.5 = 5u - 4.9(25) \) \\
              \( 27.5 = 5u - 122.5 \) \\
              \( 5u = 27.5 + 122.5 \) \\
              \( 5u = 150 \) \\
              \( u = \frac{150}{5} \) \\
              \( u = 30 \text{ metres per second} \)
          \end{solutionordottedlines}
          \Question[4] Rearrange each of these formulas to make the pronumeral in brackets the subject. (All
  pronumerals represent positive numbers.)
          \begin{parts}\begin{multicols}{2}
              \part \( c=a^{2}+b^{2} \)\hfill\(( a )\)
              \begin{solutionordottedlines}
                  \( a = \sqrt{c - b^2} \)
              \end{solutionordottedlines}
              \part \( x=\sqrt{ab} \)\hfill\(( b )\)
              \begin{solutionordottedlines}
                  \( b = \frac{x^2}{a} \)
              \end{solutionordottedlines}
              \part \( T=\frac{2\pi}{n} \)\hfill\(( n )\)
              \begin{solutionordottedlines}
                  \( n = \frac{2\pi}{T} \)
              \end{solutionordottedlines}
              \part \( E=\frac{m}{2r^2} \)\hfill\(( r )\)
              \begin{solutionordottedlines}
                  \( r = \sqrt{\frac{m}{2E}} \)
              \end{solutionordottedlines}
          \end{multicols}\end{parts}
      \end{questions}
  \end{exercisebox}

|727320245|kitty|/Applications/kitty.app|0|
(n-2) \). Rearrange the formula to make \( n \) the subject and use this to find the number of sides in
  the polygon if the sum of the interior angles is:
          \begin{parts}\begin{multicols}{3}
              \part \( 1080^\circ \)
              \begin{solutionordottedlines}
                  \( n = \frac{S}{180} + 2 \) \\
                  \( n = \frac{1080}{180} + 2 \) \\
                  \( n = 6 + 2 \) \\
                  \( n = 8 \)
              \end{solutionordottedlines}
              \part \( 1800^\circ \)
              \begin{solutionordottedlines}
                  \( n = \frac{S}{180} + 2 \) \\
                  \( n = \frac{1800}{180} + 2 \) \\
                  \( n = 10 + 2 \) \\
                  \( n = 12 \)
              \end{solutionordottedlines}
              \part \( 3240^\circ \)
              \begin{solutionordottedlines}
                  \( n = \frac{S}{180} + 2 \) \\
                  \( n = \frac{3240}{180} + 2 \) \\
                  \( n = 18 + 2 \) \\
                  \( n = 20 \)
              \end{solutionordottedlines}
          \end{multicols}\end{parts}
          \Question[2] 8 When an object is shot up into the air with a speed of \( u \) metres per second, its
  height above the ground \( h \) metres and time of flight \( t \) seconds are related (ignoring air resistance
  by \( h=ut-4.9t^{2} \).

          Find the speed at which an object was fired if it reached a height of 27.5 metres after 5 seconds.
          \begin{solutionordottedlines}
              \( h = ut - 4.9t^2 \) \\
              \( 27.5 = u(5) - 4.9(5)^2 \) \\
              \( 27.5 = 5u - 4.9(25) \) \\
              \( 27.5 = 5u - 122.5 \) \\
              \( 5u = 27.5 + 122.5 \) \\
              \( 5u = 150 \) \\
              \( u = \frac{150}{5} \) \\
              \( u = 30 \text{ metres per second} \)
          \end{solutionordottedlines}
          \Question[4] Rearrange each of these formulas to make the pronumeral in brackets the subject. (All
  pronumerals represent positive numbers.)
          \begin{parts}\begin{multicols}{2}
              \part \( c=a^{2}+b^{2} \)\hfill\(( a )\)
              \begin{solutionordottedlines}
                  \( a = \sqrt{c - b^2} \)
              \end{solutionordottedlines}
              \part \( x=\sqrt{ab} \)\hfill\(( b )\)
              \begin{solutionordottedlines}
                  \( b = \frac{x^2}{a} \)
              \end{solutionordottedlines}
              \part \( T=\frac{2\pi}{n} \)\hfill\(( n )\)
              \begin{solutionordottedlines}
                  \( n = \frac{2\pi}{T} \)
              \end{solutionordottedlines}
              \part \( E=\frac{m}{2r^2} \)\hfill\(( r )\)
              \begin{solutionordottedlines}
                  \( r = \sqrt{\frac{m}{2E}} \)
              \end{solutionordottedlines}
          \end{multicols}\end{parts}
      \end{questions}
  \end{exercisebox}

|727320278|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
      \begin{questions}
          \Question The profit \( P \) made each day by a store owner who sells CDs is given by the
  formula \( P=5n-150 \), where \( n \) is the number of CDs sold.
          \begin{parts}
              \Part[1] What profit is made if the store owner sells 60 CDs?
              \begin{solutionordottedlines}
                  \( P = 5n - 150 \) \\
                  \( P = 5(60) - 150 \) \\
                  \( P = 300 - 150 \) \\
                  \( P = \$150 \)
              \end{solutionordottedlines}
              \Part[1] Make \( n \) the subject of the formula.
              \begin{solutionordottedlines}
                  \( P = 5n - 150 \) \\
                  \( P + 150 = 5n \) \\
                  \( n = \frac{P + 150}{5} \)
              \end{solutionordottedlines}
              \Part[2] How many CDs were sold if the store made:
              \begin{subparts}
                  \subpart a profit of \( \$275 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{P + 150}{5} \) \\
                      \( n = \frac{275 + 150}{5} \) \\
                      \( n = \frac{425}{5} \) \\
                      \( n = 85 \)
                  \end{solutionordottedlines}
                  \subpart a profit of \( \$400 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{P + 150}{5} \) \\
                      \( n = \frac{400 + 150}{5} \) \\
                      \( n = \frac{550}{5} \) \\
                      \( n = 110 \)
                  \end{solutionordottedlines}
                  \subpart a loss of \( \$100 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{P + 150}{5} \) \\
                      \( n = \frac{-100 + 150}{5} \) \\
                      \( n = \frac{50}{5} \) \\
                      \( n = 10 \)
                  \end{solutionordottedlines}
                  \subpart no profit?
                  \begin{solutionordottedlines}
                      \( n = \frac{P + 150}{5} \) \\
                      \( n = \frac{0 + 150}{5} \) \\
                      \( n = \frac{150}{5} \) \\
                      \( n = 30 \)
                  \end{solutionordottedlines}
              \end{subparts}
          \end{parts}
          \Question The cost \( C \) of hiring a reception room for a function is given by the formula \( C=
  n+250 \), where \( n \) is the number of people attending the function.
          \begin{parts}
              \Part[1] Rearrange the formula to make \(|727320328|kitty|/Applications/kitty.app|0|

                  \( n = \frac{145}{5} + 1 \) \\
                  \( n = 29 + 1 \) \\
                  \( n = 30 \)
              \end{solutionordottedlines}
          \end{parts}
          \Question Rearrange each of these formulas to make the pronumeral in brackets the subject.
          \begin{parts}
              \part \( y=mx+c \) \hfill\(( c )\)
              \begin{solutionordottedlines}
                  \( c = y - mx \)
              \end{solutionordottedlines}
              \part \( A=\frac{1}{2}bh \)\hfill\(( x )\)
              \begin{solutionordottedlines}
                  % This part seems to have a typo, as there is no 'x' in the given formula.
                  % Assuming the subject to be made is 'b':
                  \( b = \frac{2A}{h} \)
              \end{solutionordottedlines}
              \part \( P=A+2\ell h \)\hfill\(( \ell )\)
              \begin{solutionordottedlines}
                  \( \ell = \frac{P - A}{2h} \)
              \end{solutionordottedlines}
              \part \( A=2\pi r^2 + 2\pi rh \)\hfill\(( h )\)
              \begin{solutionordottedlines}
                  \( h = \frac{A - 2\pi r^2}{2\pi r} \)
              \end{solutionordottedlines}
              \part \( s=\frac{n}{2}(a+\ell) \)\hfill\(( a )\)
              \begin{solutionordottedlines}
                  \( a = \frac{2s}{n} - \ell \)
              \end{solutionordottedlines}
              \part \( V=\pi r^{2}+\pi rs \)\hfill\(( s )\)
              \begin{solutionordottedlines}
                  \( s = \frac{V}{\pi r} - r \)
              \end{solutionordottedlines}
          \end{parts}
          \Question[3] The formula for the sum \( S \) of the interior angles in a convex \( n \)-sided polygon
  \( S=180(n-2) \). Rearrange the formula to make \( n \) the subject and use this to find the number of sides in
  the polygon if the sum of the interior angles is:
          \begin{parts}\begin{multicols}{3}
              \part \( 1080^\circ \)
              \begin{solutionordottedlines}
                  \( n = \frac{S}{180} + 2 \) \\
                  \( n = \frac{1080}{180} + 2 \) \\
                  \( n = 6 + 2 \) \\
                  \( n = 8 \)
              \end{solutionordottedlines}
              \part \( 1800^\circ \)
              \begin{solutionordottedlines}
                  \( n = \frac{S}{180} + 2 \) \\
                  \( n = \frac{1800}{180} + 2 \) \\
                  \( n = 10 + 2 \) \\
                  \( n = 12 \)
              \end{solutionordottedlines}
              \part \( 3240^\circ \)
              \begin{solutionordottedlines}
                  \( n = \frac{S}{180} + 2 \) \\
                  \( n = \frac{3240}{180} + 2 \) \\
                  \( n = 18 + 2 \) \\
                  \( n = 20 \)
              \end{solutionordottedlines}
          \end{multicols}\end{parts}
          \Question[2] 8 When an object is shot up into the air with a speed of \( u \) metres per second, its
  height above the ground \( h \) metres and time of flight \( t \) seconds are related (ignoring air resistance
  by \( h=ut-4.9t^{2} \).

          Find the speed at which an object was fired if it reached a height of 27.5 metres after 5 seconds.
          \begin{solutionordottedlines}
              \( h = ut - 4.9t^2 \) \\
              \( 27.5 = u(5) - 4.9(5)^2 \) \\
              \( 27.5 = 5u - 4.9(25) \) \\
              \( 27.5 = 5u - 122.5 \) \\
              \( 5u = 27.5 + 122.5 \) \\
              \( 5u = 150 \) \\
              \( u = \frac{150}{5} \) \\
              \( u = 30 \text{ metres per second} \)
          \end{solutionordottedlines}
          \Question[4] Rearrange each of these formulas to make the pronumeral in brackets the subject. (All
  pronumerals represent positive numbers.)
          \begin{parts}\begin{multicols}{2}
              \part \( c=a^{2}+b^{2} \)\hfill\(( a )\)
              \begin{solutionordottedlines}
                  \( a = \sqrt{c - b^2} \)
              \end{solutionordottedlines}
              \part \( x=\sqrt{ab} \)\hfill\(( b )\)
              \begin{solutionordottedlines}
                  \( b = \frac{x^2}{a} \)
              \end{solutionordottedlines}
              \part \( T=\frac{2\pi}{n} \)\hfill\(( n )\)
              \begin{solutionordottedlines}
                  \( n = \frac{2\pi}{T} \)
              \end{solutionordottedlines}
              \part \( E=\frac{m}{2r^2} \)\hfill\(( r )\)
              \begin{solutionordottedlines}
                  \( r = \sqrt{\frac{m}{2E}} \)
              \end{solutionordottedlines}
          \end{multicols}\end{parts}
      \end{questions}
  \end{exercisebox}
|727320343|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
      \begin{questions}
          \Question The profit \( P \) made each day by a store owner who sells CDs is given by the
  formula \( P=5n-150 \), where \( n \) is the number of CDs sold.
          \begin{parts}
              \Part[1] What profit is made if the store owner sells 60 CDs?
              \begin{solutionordottedlines}
                  \( P = 5n - 150 \) \\
                  \( P = 5(60) - 150 \) \\
                  \( P = 300 - 150 \) \\
                  \( P = \$150 \)
              \end{solutionordottedlines}
              \Part[1] Make \( n \) the subject of the formula.
              \begin{solutionordottedlines}
                  \( P = 5n - 150 \) \\
                  \( P + 150 = 5n \) \\
                  \( n = \frac{P + 150}{5} \)
              \end{solutionordottedlines}
              \Part[2] How many CDs were sold if the store made:
              \begin{subparts}
                  \subpart a profit of \( \$275 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{P + 150}{5} \) \\
                      \( n = \frac{275 + 150}{5} \) \\
                      \( n = \frac{425}{5} \) \\
                      \( n = 85 \)
                  \end{solutionordottedlines}
                  \subpart a profit of \( \$400 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{P + 150}{5} \) \\
                      \( n = \frac{400 + 150}{5} \) \\
                      \( n = \frac{550}{5} \) \\
                      \( n = 110 \)
                  \end{solutionordottedlines}
                  \subpart a loss of \( \$100 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{P + 150}{5} \) \\
                      \( n = \frac{-100 + 150}{5} \) \\
                      \( n = \frac{50}{5} \) \\
                      \( n = 10 \)
                  \end{solutionordottedlines}
                  \subpart no profit?
                  \begin{solutionordottedlines}
                      \( n = \frac{P + 150}{5} \) \\
                      \( n = \frac{0 + 150}{5} \) \\
                      \( n = \frac{150}{5} \) \\
                      \( n = 30 \)
                  \end{solutionordottedlines}
              \end{subparts}
          \end{parts}
          \Question The cost \( C \) of hiring a reception room for a function is given by the formula \( C=
  n+250 \), where \( n \) is the number of people attending the function.
          \begin{parts}
              \Part[1] Rearrange the formula to make \( n \) the subject.
              \begin{solutionordottedlines}
                  \( C = 12n + 250 \) \\
                  \( C - 250 = 12n \) \\
                  \( n = \frac{C - 250}{12} \)
              \end{solutionordottedlines}
              \Part[2] How many people attended the function if the cost of hiring the reception room was:
              \begin{subparts}
                  \subpart \( \$730 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{C - 250}{12} \) \\
                      \( n = \frac{730 - 250}{12} \) \\
                      \( n = \frac{480}{12} \) \\
                      \( n = 40 \)
                  \end{solutionordottedlines}
                  \subpart \( \$1090 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{C - 250}{12} \) \\
                      \( n = \frac{1090 - 250}{12} \) \\
                      \( n = \frac{840}{12} \) \\
                      \( n = 70 \)
                  \end{solutionordottedlines}
                  \subpart \( \$1210 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{C - 250}{12} \) \\
                      \( n = \frac{1210 - 250}{12} \) \\
                      \( n = \frac{960}{12} \) \\
                      \( n = 80 \)
                  \end{solutionordottedlines}
                  \subpart \( \$1690 \) ?
                  \begin{solutionordottedlines}
                      \( n = \frac{C - 250}{12} \) \\
                      \( n = \frac{1690 - 250}{12} \) \\
                      \( n = \frac{1440}{12} \) \\
                      \( n = 120 \)
                  \end{solutionordottedlines}
              \end{subparts}
          \end{parts}
|727320369|kitty|/Applications/kitty.app|0|
         \Question Given the formula \( t=a+(n-1)d \) :
          \begin{parts}
              \Part[1] rearrange the formula to make \( a \) the subject
              \begin{solutionordottedlines}
                  \( t = a + (n - 1)d \) \\
                  \( a = t - (n - 1)d \)
              \end{solutionordottedlines}
              \Part[2] find the value of \( a \) when:
              \begin{subparts}
                  \subpart \( t=11, n=4 \) and \( d=3 \)
                  \begin{solutionordottedlines}
                      \( a = t - (n - 1)d \) \\
                      \( a = 11 - (4 - 1) \cdot 3 \) \\
                      \( a = 11 - 3 \cdot 3 \) \\
                      \( a = 11 - 9 \) \\
                      \( a = 2 \)
                  \end{solutionordottedlines}
                  \subpart \( t=8, n=5 \) and \( d=-3 \)
                  \begin{solutionordottedlines}
                      \( a = t - (n - 1)d \) \\
                      \( a = 8 - (5 - 1) \cdot (-3) \) \\
                      \( a = 8 - 4 \cdot (-3) \) \\
                      \( a = 8 + 12 \) \\
                      \( a = 20 \)
                  \end{solutionordottedlines}
              \end{subparts}
              \Part[1] rearrange the formula to make \( d \) the subject
              \begin{solutionordottedlines}
                  \( t = a + (n - 1)d \) \\
                  \( d = \frac{t - a}{n - 1} \)
              \end{solutionordottedlines}
              \Part[2] find the value of \( d \) when:
              \begin{subparts}
                  \subpart \( t=48, a=3 \) and \( n=16 \)
                  \begin{solutionordottedlines}
                      \( d = \frac{t - a}{n - 1} \) \\
                      \( d = \frac{48 - 3}{16 - 1} \) \\
                      \( d = \frac{45}{15} \) \\
                      \( d = 3 \)
                  \end{solutionordottedlines}
                  \subpart \( t=120, a=-30 \) and \( n=101 \)
                  \begin{solutionordottedlines}
                      \( d = \frac{t - a}{n - 1} \) \\
                      \( d = \frac{120 - (-30)}{101 - 1} \) \\
                      \( d = \frac{150}{100} \) \\
                      \( d = 1.5 \)
                  \end{solutionordottedlines}
              \end{subparts}
              \Part[1] rearrange the formula to make \( n \) the subject and find the value of \( n \) when
  \( t=150, a=5 \) and \( d=5 \)
              \begin{solutionordottedlines}
                  \( t = a + (n - 1)d \) \\
                  \( n = \frac{t - a}{d} + 1 \) \\
                  \( n = \frac{150 - 5}{5} + 1 \) \\
                  \( n = \frac{145}{5} + 1 \) \\
                  \( n = 29 + 1 \) \\
                  \( n = 30 \)
              \end{solutionordottedlines}
          \end{parts}
          \Question Rearrange each of these formulas to make the pronumeral in brackets the subject.
          \begin{parts}
              \part \( y=mx+c \) \hfill\(( c )\)
              \begin{solutionordottedlines}
                  \( c = y - mx \)
              \end{solutionordottedlines}
              \part \( A=\frac{1}{2}bh \)\hfill\(( x )\)
              \begin{solutionordottedlines}
                  % This part seems to have a typo, as there is no 'x' in the given formula.
                  % Assuming the subject to be made is 'b':
                  \( b = \frac{2A}{h} \)
              \end{solutionordottedlines}
              \part \( P=A+2\ell h \)\hfill\(( \ell )\)
              \begin{solutionordottedlines}
                  \( \ell = \frac{P - A}{2h} \)
              \end{solutionordottedlines}
              \part \( A=2\pi r^2 + 2\pi rh \)\hfill\(( h )\)
              \begin{solutionordottedlines}
                  \( h = \frac{A - 2\pi r^2}{2\pi r} \)
              \end{solutionordottedlines}
              \part \( s=\frac{n}{2}(a+\ell) \)\hfill\(( a )\)
              \begin{solutionordottedlines}
                  \( a = \frac{2s}{n} - \ell \)
              \end{solutionordottedlines}
              \part \( V=\pi r^{2}+\pi rs \)\hfill\(( s )\)
              \begin{solutionordottedlines}
                  \( s = \frac{V}{\pi r} - r \)
              \end{solutionordottedlines}
          \end{parts}
          \Question[3] The formula for the sum \( S \) of the interior angles in a convex \( n \)-sided polygon
  \( S=180(n-2) \). Rearrange the formula to make \( n \) the subject and use this to find the number of sides in
  the polygon if the sum of the interior angles is:
          \begin{parts}\begin{multicols}{3}
              \part \( 1080^\circ \)
              \begin{solutionordottedlines}
                  \( n = \frac{S}{180} + 2 \) \\
                  \( n = \frac{1080}{180} + 2 \) \\
                  \( n = 6 + 2 \) \\
                  \( n = 8 \)
              \end{solutionordottedlines}
              \part \( 1800^\circ \)
              \begin{solutionordottedlines}
                  \( n = \frac{S}{180} + 2 \) \\
                  \( n = \frac{1800}{180} + 2 \) \\
                  \( n = 10 + 2 \) \\
                  \( n = 12 \)
              \end{solutionordottedlines}
              \part \( 3240^\circ \)
              \begin{solutionordottedlines}
                  \( n = \frac{S}{180} + 2 \) \\
                  \( n = \frac{3240}{180} + 2 \) \\
                  \( n = 18 + 2 \) \\
                  \( n = 20 \)
              \end{solutionordottedlines}
          \end{multicols}\end{parts}
|727320391|kitty|/Applications/kitty.app|0|
      \Question[2] 8 When an object is shot up into the air with a speed of \( u \) metres per second, its
  height above the ground \( h \) metres and time of flight \( t \) seconds are related (ignoring air resistance
  by \( h=ut-4.9t^{2} \).

          Find the speed at which an object was fired if it reached a height of 27.5 metres after 5 seconds.
          \begin{solutionordottedlines}
              \( h = ut - 4.9t^2 \) \\
              \( 27.5 = u(5) - 4.9(5)^2 \) \\
              \( 27.5 = 5u - 4.9(25) \) \\
              \( 27.5 = 5u - 122.5 \) \\
              \( 5u = 27.5 + 122.5 \) \\
              \( 5u = 150 \) \\
              \( u = \frac{150}{5} \) \\
              \( u = 30 \text{ metres per second} \)
          \end{solutionordottedlines}
          \Question[4] Rearrange each of these formulas to make the pronumeral in brackets the subject. (All
  pronumerals represent positive numbers.)
          \begin{parts}\begin{multicols}{2}
              \part \( c=a^{2}+b^{2} \)\hfill\(( a )\)
              \begin{solutionordottedlines}
                  \( a = \sqrt{c - b^2} \)
              \end{solutionordottedlines}
              \part \( x=\sqrt{ab} \)\hfill\(( b )\)
              \begin{solutionordottedlines}
                  \( b = \frac{x^2}{a} \)
              \end{solutionordottedlines}
              \part \( T=\frac{2\pi}{n} \)\hfill\(( n )\)
              \begin{solutionordottedlines}
                  \( n = \frac{2\pi}{T} \)
              \end{solutionordottedlines}
              \part \( E=\frac{m}{2r^2} \)\hfill\(( r )\)
              \begin{solutionordottedlines}
                  \( r = \sqrt{\frac{m}{2E}} \)
              \end{solutionordottedlines}
          \end{multicols}\end{parts}
      \end{questions}
  \end{exercisebox}

|727320423|kitty|/Applications/kitty.app|0|
Can't locate YAML/Tiny.pm in @INC (you may need to install the YAML::Tiny module)|727321937|kitty|/Applications/kitty.app|0|
sudo cpan Unicode::GCString
sudo cpan App::cpanminus
sudo cpan YAML::Tiny
sudo perl -MCPAN -e 'install "File::HomeDir"'|727321954|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
 \begin{examplebox}                                                                                                                                         
|727322150|kitty|/Applications/kitty.app|0|
This section is tricky as negative numbers always tend to be. The most important thing you will learn here is:

\begin{lawbox}
    \large
    \[a^{-n} = \frac{1}{a^n}\]
\end{lawbox}

I want you to take a second to realise that this \fuzzyfont{breaks} your interpretation of $a$ being raised to some power $n$ as being $n$ successive multiplications of the number $a$.
\[a^n = \underbrace{a \times a \times a \ldots}_{n} \]

Suddenly what does it mean to multiply $a$ by itself \textit{$-n$} times?!

Here is some working out space for you to reconcile with this fact.
\begin{solutionorbox}[2in]
 \textbf{Proof:}

 Consider the product of \( a^n \) and \( a^{-n} \):
 \[ a^n \cdot a^{-n} = a^{n + (-n)} = a^0 \]

 Since anything raised to the power of 0 is 1, we have:
 \[ a^0 = 1 \]

 Therefore:
 \[ a^n \cdot a^{-n} = 1 \]

 Dividing both sides by \( a^n \), we get:
 \[ \frac{a^n \cdot a^{-n}}{a^n} = \frac{1}{a^n} \]

 Simplifying the left side, we have:
 \[ a^{-n} = \frac{1}{a^n} \]

 Hence proved.
\end{solutionorbox}

\begin{examplebox}
    \subsection{Examples:}
    \begin{questions}
        \item[Worked Example:] \[2^{-3} = \frac{1}{2^3} = \frac{1}{8}\]
        \Question[4] Evaluate
        \begin{parts}\begin{multicols}{2}
            \part \(6^{-2}\)
            \begin{solutionordottedlines}[2cm]
                \(6^{-2}=\frac{1}{6^{2}}\)
                \(=\frac{1}{36}\)
            \end{solutionordottedlines}
            \part \(4^{-3}\)
            \begin{solutionordottedlines}[2cm]
                \(4^{-3}=\frac{1}{4^{3}}\)
                \(=\frac{1}{64}\)
            \end{solutionordottedlines}
            \part \(2^{-7}\)
            \begin{solutionordottedlines}[2cm]
                \(2^{-7}=\frac{1}{2^{7}}\)
                \(=\frac{1}{128}\)
            \end{solutionordottedlines}
            \part \(10^{-3}\)
            \begin{solutionordottedlines}[2cm]
                \(10^{-3}=\frac{1}{10^{3}}\)
                \(=\frac{1}{1000}\)
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[3] These ones just reciprocate!
        \begin{parts}\begin{multicols}{3}
            \part \(\left(\frac{1}{3}\right)^{-1}\)
            \begin{solutionordottedlines}[3cm]
                \(\left(\frac{1}{3}\right)^{-1}=\frac{3}{1}\)
                \(=3\)
            \end{solutionordottedlines}
            \part \(\left(\frac{2}{7}\right)^{-2}\)
            \begin{solutionordottedlines}[3cm]
                \(\left(\frac{2}{7}\right)^{-2}=\left(\frac{7}{2}\right)^{2}\)
                \(=\frac{49}{4}\)
            \end{solutionordottedlines}
            \part \(\left(4 \frac{1}{4}\right)^{-2}\)
            \begin{solutionordottedlines}[3cm]
                \(\left(4 \frac{1}{4}\right)^{-2}=\left(\frac{17}{4}\right)^{-2}\)
                \(=\left(\frac{4}{17}\right)^{2}\)
                \(=\frac{16}{289}\)
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[8] Write as a single power and then evaluate:
        \begin{parts}
            \part \(3^{4} \times 3^{-2}\)
            \begin{solutionordottedlines}[2cm]
                \(3^{4} \times 3^{-2}=3^{2}\)
            \end{solutionordottedlines}
            \part \(5^{7} \times 5^{-8}\)
            \begin{solutionordottedlines}[2cm]
                \(5^{7} \times 5^{-8}=5^{-1}\)

                \[
                =9
                \]
            \end{solutionordottedlines}
            \part \(13^{-8} \times 13^{15} \times 13^{-7}\)
            \begin{solutionordottedlines}[2cm]
                \(\begin{aligned} 13^{-8} \times 13^{15} \times 13^{-7} & =13^{0} \\ & =1\end{aligned}\)

                \[
                =1
                \]

                \[
                =\frac{1}{5}
                \]
            \end{solutionordottedlines}
            \part \(\left(\frac{2}{3}\right)^{-6} \times\left(\frac{2}{3}\right)^{4}\)
            \begin{solutionordottedlines}[2cm]
                \(\left(\frac{2}{3}\right)^{-6} \times\left(\frac{2}{3}\right)^{4}=\left(\frac{2}{3}\right)^{-2}\)

                \[
                =\left(\frac{3}{2}\right)^{2}
                \]

                \[
                =\frac{9}{4}
                \]
            \end{solutionordottedlines}
            \begin{multicols}{2}
            \part \[\frac{2^{4}}{2^{5}}\]
            \begin{solutionordottedlines}[4cm]
                \(\frac{2^{4}}{2^{5}}=2^{-1}\)
                \(=\frac{1}{2}\)
            \end{solutionordottedlines}
            \part \[\frac{3^{4}}{3^{7}}\]
            \begin{solutionordottedlines}[4cm]
                \(\frac{3^{4}}{3^{7}}=3^{-3}\)
                \(=\frac{1}{3^{3}}\)
                \(=\frac{1}{27}\)
            \end{solutionordottedlines}
            \part \[\frac{5}{5^{3}}\]
            \begin{solutionordottedlines}[4cm]
                \(\frac{5}{5^{3}}=5^{-2}\)
                \(=\frac{1}{5^{2}}\)
                \(=\frac{1}{25}\)
            \end{solutionordottedlines}
            \part \[\frac{3^{4}}{3^{6}}\]
            \begin{solutionordottedlines}[4cm]
                \(\frac{3^{4}}{3^{6}}=3^{-2}\)
                \(=\frac{1}{3^{2}}\)
                \(=\frac{1}{9}\)
            \end{solutionordottedlines}
            \end{multicols}
        \end{parts}
        \Question[4] Simplify, expressing the answers with positive indices
        \begin{parts}\begin{multicols}{2}
            \part \(a^{2} b^{-3} \times a^{-4} b^{5}\)
            \begin{solutionordottedlines}[3cm]
                \(a^{2} b^{-3} \times a^{-4} b^{5}=a^{2-4} \times b^{-3+5}\)
                \(=\frac{1}{a^{2}} \times b^{2}\)
                \(=\frac{b^{2}}{a^{2}}\)
            \end{solutionordottedlines}
            \part \(\frac{x^{2} y^{3}}{x^{3} y^{2}}\)
            \begin{solutionordottedlines}[3cm]
                \(\frac{x^{2} y^{3}}{x^{3} y^{2}}=x^{-1} y^{1}\)
                \(=\frac{1}{x} \times y\)
                \(=\frac{y}{x}\)
            \end{solutionordottedlines}
            \part \(\left(2 a^{-2} b^{3}\right)^{-2}\)
            \begin{solutionordottedlines}[3cm]
                \(\left(2 a^{-2} b^{3}\right)^{-2}=2^{-2} \times a^{4} \times b^{-6}\)
                \(=\frac{1}{2^{2}} \times a^{4} \times \frac{1}{b^{6}}\)
                \(=\frac{a^{4}}{4 b^{6}}\)
            \end{solutionordottedlines}
            \part \(\left(\frac{3 m^{2}}{n}\right)^{-4}\)
            \begin{solutionordottedlines}[3cm]
                \(\left(\frac{3 m^{2}}{n}\right)^{-4}=\left(\frac{n}{3 m^{2}}\right)^{4}\)
                \(=\frac{n^{4}}{81 m^{8}}\)
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \end{questions}
\end{examplebox}

\fbox{\fbox{\parbox{\textwidth}{Note that in general \(\left(\frac{a}{b}\right)^{-1}=\frac{b}{a}\).
}}}

|727322311|kitty|/Applications/kitty.app|0|
Let us begin simply with the following:

\begin{examplebox}
    \subsection{Examples:}
    \begin{questions}
        \Question[1] Express as a power or as a product of powers.
        \begin{parts}
            \part \(5 \times 5 \times 5\)
            \begin{solutionordottedlines}
                \(5 \times 5 \times 5=5^{3} \quad\) 
            \end{solutionordottedlines}
            \part \(3 \times 3 \times 7 \times 7 \times 7 \times 7\)
            \begin{solutionordottedlines}
                \(3 \times 3 \times 7 \times 7 \times 7 \times 7=3^{2} \times 7^{4}\)
            \end{solutionordottedlines}
            \part
        \end{parts}
        \Question[1] Express each number as a power of a prime.
        \begin{parts}
            \part 81
            \begin{solutionordottedlines}[2cm]
                \(\quad 81=3 \times 3 \times 3 \times 3\)
            \end{solutionordottedlines}
            \part 128
            \begin{solutionordottedlines}[2cm]
                \(128=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\)
                \(=3^{4}\)
                \(=2^{7}\)
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{examplebox}

Now some definitions:
\begin{boxdef}
    \textbf{Base:}
    \begin{solutionordottedlines}[1cm]
        The number 2 in \(2^{4}\) is called the base.
    \end{solutionordottedlines}
\end{boxdef}
\begin{boxdef}
    \textbf{Index:}
    \begin{solutionordottedlines}[1cm]
        The number 4 in \(2^{4}\) is called the index or exponent.
    \end{solutionordottedlines}
\end{boxdef}

\begin{boxlaw}
    \subsection*{Index Laws}
    \subsubsection*{Product of Powers}
        To multiply powers of the same base, add the indices.

        \[
        a^{m} a^{n}=a^{m+n}
        \]

    \subsubsection*{Division of Powers}
        To divide powers of the same base, subtract the indices.

        \[
        \frac{a^{m}}{a^{n}}=a^{m-n} \quad \text { where } m>n \text { and } a \neq 0
        \]
    \subsubsection*{Powers of Powers}
        To raise a power to a power, multiply the indices.

        \[
        \left(a^{m}\right)^{n}=a^{m n}
        \]
    \subsubsection*{Power of Products}

        A power of a product is the product of the powers.

        \[(a b)^{m}=a^{m} b^{m}\]

    \subsubsection*{Power of Quotient}
        A power of a quotient is the quotient of the powers.

        \[
        \left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}} \quad \text { where } b \neq 0
        \]
    \subsubsection*{Zero Exponent Law}
        Any base raised to the power of $0$ is equal to $1$.
        \[a^0 = 1\]
    \subsubsection*{Negative Exponent Law (next lesson)}
         \[ a^{-n} = \frac{1}{a^n} \] 
         A negative exponent indicates the reciprocal of the base raised to the corresponding positive exponent.
\end{boxlaw}

Now with these laws in mind, try the following 10 examples:
\begin{examplebox}
    \subsection{Examples}
    \begin{questions}
        \Question[10] Simplify, expressing the answer in index form.
        \begin{parts}\begin{multicols}{2}
            \part \(3^{2} \times 3^{4}\)
                \begin{solutionordottedlines}[2cm]
                    \(3^{2} \times 3^{4}=3^{6}\)
                \end{solutionordottedlines}
            \part \(a^{3} \times a^{5}\)
                \begin{solutionordottedlines}[2cm]
                    \(a^{3} \times a^{5}=a^{8}\)
                \end{solutionordottedlines}
            \part \(3 x^{2} \times x^{3}\)
                \begin{solutionordottedlines}[2cm]
                    \(3 x^{2} \times x^{3}=3 x^{5}\)
                \end{solutionordottedlines}
            \part \(2 a^{2} b^{3} \times 5 a b^{2}\)
                \begin{solutionordottedlines}[2cm]
                    \(2 a^{2} b^{3} \times 5 a b^{2}=10 \times a^{2+1} \times b^{3+2}\)

                    \[
                    =10 a^{3} b^{5}
                    \]
                \end{solutionordottedlines}
            \part \(\frac{3^{5}}{3^{2}}\)
                \begin{solutionordottedlines}[2cm]
                    \(\frac{3^{5}}{3^{2}}=3^{5-2}\)
                \end{solutionordottedlines}
            \part \(\frac{9^{5}}{9^{4}}\)
                \begin{solutionordottedlines}[2cm]
                    \(\frac{9^{5}}{9^{4}}=9^{1}\)
                \end{solutionordottedlines}
            \part \(10^{6} \div 10^{4}\)
                \begin{solutionordottedlines}[2cm]
                    \(10^{6} \div 10^{4}=10^{6-4}\)
                    \(=3^{3}\)
                    \(=9\)
                    \(=10^{2}\)
                    \(=27\)
                    \(=100\)
                \end{solutionordottedlines}
            \part \(a^{7} \div a^{4}\)
                \begin{solutionordottedlines}[2cm]
                    \(\begin{aligned} a^{7} \div a^{4} & =a^{7-4} \\ & =a^{3}\end{aligned}\)
                \end{solutionordottedlines}
            \part \(\frac{3 y^{4}}{y}\)
                \begin{solutionordottedlines}[2cm]
                    \(\frac{3 y^{4}}{y}=3 \times \frac{y^{4}}{y}\)
                \end{solutionordottedlines}
            \part \(\frac{6 x^{5}}{2 x^{3}}\)
                \begin{solutionordottedlines}[2cm]
                    \(\frac{6 x^{5}}{2 x^{3}}=\frac{6}{2} \times \frac{x^{5}}{x^{3}}\)
                    \(=3 \times y^{4-1}\)
                    \(=3 \times x^{5-3}\)
                    \(=3 y^{3}\)
                    \(=3 x^{2}\)
                \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4] Double marks for each of these:
        \begin{parts}
            \part \[\frac{3 x^{3} y^{2}}{4 x y} \times \frac{6 x^{2} y^{3}}{x^{3} y^{2}}\]
            \begin{solutionordottedlines}[2cm]
                \(\frac{3 x^{3} y^{2}}{4 x y} \times \frac{6 x^{2} y^{3}}{x^{3} y^{2}}=\frac{18 x^{5} y^{5}}{4 x^{4} y^{3}}\)
                \(=\frac{9 x y^{2}}{2}\)
            \end{solutionordottedlines}
            \part \[\frac{8 a^{2} b^{3}}{3 a^{3} b} \div \frac{4 a b^{2}}{9 a^{3} b^{5}}\]
            \begin{solutionordottedlines}[2cm]
                \(\frac{8 a^{2} b^{3}}{3 a^{3} b} \div \frac{4 a b^{2}}{9 a^{3} b^{5}}=\frac{8 a^{2} b^{3}}{3 a^{3} b} \times \frac{9 a^{3} b^{5}}{4 a b^{2}}\)

                \[
                =\frac{72 \times a^{5} \times b^{8}}{12 \times a^{4} \times b^{3}}
                \]

                \[
                =6 a b^{5}
                \]
            \end{solutionordottedlines}
        \end{parts}
        \Question[6] And a $1.5\times$ multiplier for these:
        \begin{parts}
            \part \(\left(\frac{2}{3}\right)^{2}\)
            \begin{solutionordottedlines}[2cm]
                \(\begin{aligned}\left(\frac{2}{3}\right)^{3} & =\frac{2^{3}}{3^{3}} \\ & =\frac{8}{27}\end{aligned}\)
            \end{solutionordottedlines}
            \part \(\left(\frac{m}{n}\right)^{5}\)
            \begin{solutionordottedlines}[2cm]
                \(\left(\frac{m}{n}\right)^{5}=\frac{m^{5}}{n^{5}}\)
            \end{solutionordottedlines}
            \part \(\left(\frac{x^{3}}{y^{2}}\right)^{2} \times\left(\frac{y}{x}\right)^{4}\)
            \begin{solutionordottedlines}[2cm]
                \(\left(\frac{x^{3}}{y^{2}}\right)^{2} \times\left(\frac{y}{x}\right)^{4}=\frac{x^{6}}{y^{4}} \times \frac{y^{4}}{x^{4}}\)
            \end{solutionordottedlines}
            \part \(\left(\frac{2 x^{2}}{3}\right)^{2} \div \frac{4 x^{3}}{9}\)
            \begin{solutionordottedlines}[2cm]
                \(\left(\frac{2 x^{2}}{3}\right)^{2} \div \frac{4 x^{3}}{9}=\frac{4 x^{4}}{9} \times \frac{9}{4 x^{3}}\)
                \(=x^{2}\)
                \(=x\)
            \end{solutionordottedlines}
        \end{parts}
        \begin{parts}
            \part \(\left(5 a^{3}\right)^{0}\)
            \begin{solutionordottedlines}[2cm]
                \(\left(5 a^{3}\right)^{0}=1\)
            \end{solutionordottedlines}
            \part \(\frac{6 x^{2} y}{x y^{2}} \times \frac{y^{3} x}{2 y^{2} x^{2}}\)
            \begin{solutionordottedlines}[2cm]
                \[
                \begin{aligned}
                \frac{6 x^{2} y}{x y^{2}} \times \frac{y^{3} x}{2 y^{2} x^{2}} & =\frac{6}{2} \times \frac{x^{3}}{x^{3}} \times \frac{y^{4}}{y^{4}} \\
                & =3 x^{0} y^{0} \\
                & =3 \times 1 \times 1 \\
                & =3
                \end{aligned}
                \]
            \end{solutionordottedlines}
            \part \(\left(m n^{2}\right)^{0}\)
            \begin{solutionordottedlines}[2cm]
                \(\left(m n^{2}\right)^{0}=1\)
            \end{solutionordottedlines}
            \part \(\left(a^{4} b^{2}\right)^{3}\)
            \begin{solutionordottedlines}[2cm]
                \(\left(a^{4} b^{2}\right)^{3}=a^{12} b^{6}\)
            \end{solutionordottedlines}
            \part \(\left(2 a^{4}\right)^{3}\)
            \begin{solutionordottedlines}[2cm]
                \(\left(2 a^{4}\right)^{3}=2^{3} \times a^{12}\)
            \end{solutionordottedlines}
            \part \(2\left(x^{2} y\right)^{0} \times\left(x^{2} y^{3}\right)^{3}\)
            \begin{solutionordottedlines}[2cm]
                \(2\left(x^{2} y\right)^{0} \times\left(x^{2} y^{3}\right)^{3}=2 \times 1 \times x^{6} y^{9}\)
                \(=8 a^{12}\)
                \(=2 x^{6} y^{9}\)
            \end{solutionordottedlines}
            \part
        \end{parts}
    \end{questions}
\end{examplebox}

\fbox{\fbox{\parbox{\textwidth}{\textbf{Note: }There are different possible interpretations of the word 'simplify'. There may be more than one acceptable simplified form.
}}}

|727322486|kitty|/Applications/kitty.app|0|
                                                                                                                                                            
 \begin{exercisebox}                                                                                                                                        
     \subsection{Exercises:}                                                                                                                                
     \begin{questions}                                                                                                                                      
         \Question[4] Express with a positive index and then evaluate.                                                                                      
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(2^{-1}\)                                                                                                                               
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(2^{-1} = \frac{1}{2}\)\\                                                                                                                 
                 \(2^{-1} = 0.5\)                                                                                                                           
             \end{solutionordottedlines}                                                                                                                    
             \part \(5^{-1}\)                                                                                                                               
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(5^{-1} = \frac{1}{5}\)\\                                                                                                                 
                 \(5^{-1} = 0.2\)                                                                                                                           
             \end{solutionordottedlines}                                                                                                                    
             \part \(2^{-4}\)                                                                                                                               
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(2^{-4} = \frac{1}{2^4}\)\\                                                                                                               
                 \(2^{-4} = 0.0625\)                                                                                                                        
             \end{solutionordottedlines}                                                                                                                    
             \part \(3^{-3}\)                                                                                                                               
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(3^{-3} = \frac{1}{3^3}\)\\                                                                                                               
                 \(3^{-3} = \frac{1}{27}\)                                                                                                                  
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[4] Write each fraction as a power of a prime with a negative index.                                                                      
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(\frac{1}{8}\)                                                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{1}{8} = 2^{-3}\)                                                                                                                   
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{1}{9}\)                                                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{1}{9} = 3^{-2}\)                                                                                                                   
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{1}{16}\)                                                                                                                         
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{1}{16} = 2^{-4}\)                                                                                                                  
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{1}{64}\)                                                                                                                         
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{1}{64} = 2^{-6}\)                                                                                                                  
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[4] Express with positive indices, evaluating where possible.                                                                             
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(a^{-3}\)                                                                                                                               
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(a^{-3} = \frac{1}{a^3}\)                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
             \part \(x^{-7}\)                                                                                                                               
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(x^{-7} = \frac{1}{x^7}\)                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{3}{a^{-4}}\)                                                                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{3}{a^{-4}} = 3a^4\)                                                                                                                
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{5}{x^{-5}}\)                                                                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{5}{x^{-5}} = 5x^5\)                                                                                                                
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[4] Simplify and then evaluate                                                                                                            
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(\left(\frac{1}{4}\right)^{-1}\)                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\left(\frac{1}{4}\right)^{-1} = 4\)                                                                                                      
             \end{solutionordottedlines}                                                                                                                    
             \part \(\left(\frac{2}{5}\right)^{-2}\)                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\left(\frac{2}{5}\right)^{-2} = \left(\frac{5}{2}\right)^2\)\\                                                                           
                 \(\left(\frac{2}{5}\right)^{-2} = \frac{25}{4}\)                                                                                           
             \end{solutionordottedlines}                                                                                                                    
             \part \(3^{5} \times 3^{-2}\)                                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(3^{5} \times 3^{-2} = 3^{5-2}\)\\                                                                                                        
                 \(3^{5} \times 3^{-2} = 3^3\)\\                                                                                                            
                 \(3^{5} \times 3^{-2} = 27\)                                                                                                               
             \end{solutionordottedlines}                                                                                                                    
             \part \(5^{11} \times 5^{-8}\)                                                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(5^{11} \times 5^{-8} = 5^{11-8}\)\\                                                                                                      
                 \(5^{11} \times 5^{-8} = 5^3\)\\                                                                                                           
                 \(5^{11} \times 5^{-8} = 125\)                                                                                                             
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[4]$\,$                                                                                                                                   
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(\frac{2^{3}}{2^{6}}\)                                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{2^{3}}{2^{6}} = 2^{3-6}\)\\                                                                                                        
                 \(\frac{2^{3}}{2^{6}} = 2^{-3}\)\\                                                                                                         
                 \(\frac{2^{3}}{2^{6}} = \frac{1}{8}\)                                                                                                      
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{4^{2}}{4^{4}}\)                                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{4^{2}}{4^{4}} = 4^{2-4}\)\\                                                                                                        
                 \(\frac{4^{2}}{4^{4}} = 4^{-2}\)\\                                                                                                         
                 \(\frac{4^{2}}{4^{4}} = \frac{1}{16}\)                                                                                                     
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{8^{6}}{8^{7}}\)                                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{8^{6}}{8^{7}} = 8^{6-7}\)\\                                                                                                        
                 \(\frac{8^{6}}{8^{7}} = 8^{-1}\)\\                                                                                                         
                 \(\frac{8^{6}}{8^{7}} = \frac{1}{8}\)                                                                                                      
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{20^{4}}{20^{6}}\)                                                                                                                
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{20^{4}}{20^{6}} = 20^{4-6}\)\\                                                                                                     
                 \(\frac{20^{4}}{20^{6}} = 20^{-2}\)\\                                                                                                      
                 \(\frac{20^{4}}{20^{6}} = \frac{1}{400}\)                                                                                                  
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[3] Express with negative index.                                                                                                          
         \begin{parts}\begin{multicols}{3}                                                                                                                  
             \part \(\frac{3}{x}\)                                                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{3}{x} = 3x^{-1}\)                                                                                                                  
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{5}{x^{2}}\)                                                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{5}{x^{2}} = 5x^{-2}\)                                                                                                              
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{8}{x^{4}}\)                                                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{8}{x^{4}} = 8x^{-4}\)                                                                                                              
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[3] Evaluate.                                                                                                                             
         \begin{parts}\begin{multicols}{3}                                                                                                                  
             \part \(\left(\frac{1}{2}\right)^{-1}\)                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\left(\frac{1}{2}\right)^{-1} = 2\)                                                                                                      
             \end{solutionordottedlines}                                                                                                                    
             \part \(\left(\frac{2}{3}\right)^{-1}\)                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\left(\frac{2}{3}\right)^{-1} = \frac{3}{2}\)                                                                                            
             \end{solutionordottedlines}                                                                                                                    
             \part \(\left(\frac{1}{2}\right)^{-2}\)                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\left(\frac{1}{2}\right)^{-2} = 2^2\)\\                                                                                                  
                 \(\left(\frac{1}{2}\right)^{-2} = 4\)                                                                                                      
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[4] Simplify, expressing the answer with positive indices.                                                                                
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(x^{-6} y^{4} \times x^{2} y^{-2}\)                                                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(x^{-6} y^{4} \times x^{2} y^{-2} = x^{-6+2} y^{4-2}\)\\                                                                                  
                 \(x^{-6} y^{4} \times x^{2} y^{-2} = x^{-4} y^{2}\)\\                                                                                      
                 \(x^{-6} y^{4} \times x^{2} y^{-2} = \frac{y^2}{x^4}\)                                                                                     
             \end{solutionordottedlines}                                                                                                                    
             \part \(2 a^{-1} b^{5} \times 7 a b^{-3}\)                                                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(2 a^{-1} b^{5} \times 7 a b^{-3} = 14 a^{-1+1} b^{5-3}\)\\                                                                               
                 \(2 a^{-1} b^{5} \times 7 a b^{-3} = 14 b^{2}\)                                                                                            
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{8 a^{-4}}{2 a^{6}}\)                                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{8 a^{-4}}{2 a^{6}} = 4 a^{-4-6}\)\\                                                                                                
                 \(\frac{8 a^{-4}}{2 a^{6}} = 4 a^{-10}\)\\                                                                                                 
                 \(\frac{8 a^{-4}}{2 a^{6}} = \frac{4}{a^{10}}\)                                                                                            
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{36 h^{-9}}{9 h^{-4}}\)                                                                                                           
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{36 h^{-9}}{9 h^{-4}} = 4 h^{-9+4}\)\\                                                                                              
                 \(\frac{36 h^{-9}}{9 h^{-4}} = 4 h^{-5}\)\\                                                                                                
                 \(\frac{36 h^{-9}}{9 h^{-4}} = \frac{4}{h^{5}}\)                                                                                           
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[6] Fill in the missing term                                                                                                              
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(6^{4} \times \ldots=6^{2}\)                                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(6^{4} \times 6^{-2} = 6^{2}\)                                                                                                            
             \end{solutionordottedlines}                                                                                                                    
             \part \(m^{5} \times \ldots=m^{-6}\)                                                                                                           
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(m^{5} \times m^{-11} = m^{-6}\)                                                                                                          
             \end{solutionordottedlines}                                                                                                                    
             \part \(d^{-7} \div \ldots=d^{15}\)                                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(d^{-7} \div d^{-22} = d^{15}\)                                                                                                           
             \end{solutionordottedlines}                                                                                                                    
             \part \(\left(a^{5}\right) \cdots=a^{-15}\)                                                                                                    
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\left(a^{5}\right)^{-3} = a^{-15}\)                                                                                                      
             \end{solutionordottedlines}                                                                                                                    
             \part \((\ldots)^{-2}=\frac{m^{6}}{25}\)                                                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \((\frac{m^{3}}{5})^{-2} = \frac{m^{6}}{25}\)                                                                                              
             \end{solutionordottedlines}                                                                                                                    
             \part \((\ldots)^{-2}=p^{4} q^{-6}\)                                                                                                           
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \((p^{2} q^{-3})^{-2} = p^{4} q^{-6}\)                                                                                                     
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[4] Simplify, expressing the answers with positive indices. Evaluate powers where possible.                                               
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(\left(3 a^{2} b^{-2}\right)^{3} \times\left(2 a^{4}\right)^{-2}\)                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\left(3 a^{2} b^{-2}\right)^{3} \times\left(2 a^{4}\right)^{-2} = 27 a^{6} b^{-6} \times \frac{1}{4 a^{8}}\)\\                           
                 \(\left(3 a^{2} b^{-2}\right)^{3} \times\left(2 a^{4}\right)^{-2} = \frac{27}{4} a^{-2} b^{-6}\)\\                                         
                 \(\left(3 a^{2} b^{-2}\right)^{3} \times\left(2 a^{4}\right)^{-2} = \frac{27}{4 a^{2} b^{6}}\)                                             
             \end{solutionordottedlines}                                                                                                                    
             \part \(\left(6 a^{5} b^{-4}\right)^{-3} \times 2\left(a^{3} b^{-3}\right)^{2}\)                                                               
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\left(6 a^{5} b^{-4}\right)^{-3} \times 2\left(a^{3} b^{-3}\right)^{2} = \frac{1}{216 a^{15} b^{-12}} \times 2 a^{6} b^{-6}\)\\          
                 \(\left(6 a^{5} b^{-4}\right)^{-3} \times 2\left(a^{3} b^{-3}\right)^{2} = \frac{2}{216} a^{-9} b^{6}\)\\                                  
                 \(\left(6 a^{5} b^{-4}\right)^{-3} \times 2\left(a^{3} b^{-3}\right)^{2} = \frac{1}{108 a^{9} b^{6}}\)                                     
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{\left(2 a^{4} b^{-2}\right)^{3}}{c^{2}} \times \frac{\left(2^{2} a^{-3} b^{2}\right)^{-1}}{c}\)                                  
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{\left(2 a^{4} b^{-2}\right)^{3}}{c^{2}} \times \frac{\left(2^{2} a^{-3} b^{2}\right)^{-1}}{c} = \frac{8 a^{12} b^{-6}}{c^{2}} \tim 
 \frac{c}{4 a^{3} b^{-2}}\)\\                                                                                                                               
                 \(\frac{\left(2 a^{4} b^{-2}\right)^{3}}{c^{2}} \times \frac{\left(2^{2} a^{-3} b^{2}\right)^{-1}}{c} = \frac{2 a^{9} b^{-4}}{c^{3}}\)     
             \end{solutionordottedlines}                                                                                                                    
             \part \(\mathbf{j} \frac{\left(2 a^{4}\right)^{2}}{b^{7}} \div \frac{\left(a^{2}\right)^{-3}}{2 b}\)                                           
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\mathbf{j} \frac{\left(2 a^{4}\right)^{2}}{b^{7}} \div \frac{\left(a^{2}\right)^{-3}}{2 b} = \mathbf{j} \frac{4 a^{8}}{b^{7}} \times     
 \frac{2 b}{a^{-6}}\)\\                                                                                                                                     
                 \(\mathbf{j} \frac{\left(2 a^{4}\right)^{2}}{b^{7}} \div \frac{\left(a^{2}\right)^{-3}}{2 b} = \mathbf{j} \frac{8 a^{14}}{b^{6}}\)         
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
     \end{questions}                                                                                                                                        
 \end{exercisebox}                                                                                                                                          
                                                                                                                                                            
|727322968|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \Question[4] Express with a positive index and then evaluate.
        \begin{parts}\begin{multicols}{2}
            \part \(2^{-1}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(5^{-1}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(2^{-4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(3^{-3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4] Write each fraction as a power of a prime with a negative index.
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{1}{8}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{1}{9}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{1}{16}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{1}{64}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4] Express with positive indices, evaluating where possible.
        \begin{parts}\begin{multicols}{2}
            \part \(a^{-3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(x^{-7}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{3}{a^{-4}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{5}{x^{-5}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4] Simplify and then evaluate
        \begin{parts}\begin{multicols}{2}
            \part \(\left(\frac{1}{4}\right)^{-1}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(\frac{2}{5}\right)^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(3^{5} \times 3^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(5^{11} \times 5^{-8}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4]$\,$ 
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{2^{3}}{2^{6}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{4^{2}}{4^{4}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{8^{6}}{8^{7}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{20^{4}}{20^{6}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[3] Express with negative index.
        \begin{parts}\begin{multicols}{3}
            \part \(\frac{3}{x}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{5}{x^{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{8}{x^{4}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[3] Evaluate.
        \begin{parts}\begin{multicols}{3}
            \part \(\left(\frac{1}{2}\right)^{-1}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(\frac{2}{3}\right)^{-1}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(\frac{1}{2}\right)^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4] Simplify, expressing the answer with positive indices.
        \begin{parts}\begin{multicols}{2}
            \part \(x^{-6} y^{4} \times x^{2} y^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(2 a^{-1} b^{5} \times 7 a b^{-3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{8 a^{-4}}{2 a^{6}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{36 h^{-9}}{9 h^{-4}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[6] Fill in the missing term
        \begin{parts}\begin{multicols}{2}
            \part \(6^{4} \times \ldots=6^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(m^{5} \times \ldots=m^{-6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(d^{-7} \div \ldots=d^{15}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(a^{5}\right) \cdots=a^{-15}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((\ldots)^{-2}=\frac{m^{6}}{25}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((\ldots)^{-2}=p^{4} q^{-6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4] Simplify, expressing the answers with positive indices. Evaluate powers where possible.
        \begin{parts}\begin{multicols}{2}
            \part \(\left(3 a^{2} b^{-2}\right)^{3} \times\left(2 a^{4}\right)^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(6 a^{5} b^{-4}\right)^{-3} \times 2\left(a^{3} b^{-3}\right)^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{\left(2 a^{4} b^{-2}\right)^{3}}{c^{2}} \times \frac{\left(2^{2} a^{-3} b^{2}\right)^{-1}}{c}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\mathbf{j} \frac{\left(2 a^{4}\right)^{2}}{b^{7}} \div \frac{\left(a^{2}\right)^{-3}}{2 b}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \end{questions}
\end{exercisebox}



|727322983|kitty|/Applications/kitty.app|0|
                                                                                                                                                            
 \begin{exercisebox}                                                                                                                                        
     \subsection{Exercises:}                                                                                                                                
     \begin{questions}                                                                                                                                      
         \Question[3] State the base and index of:                                                                                                          
         \begin{parts}\begin{multicols}{3}                                                                                                                  
             \part \(6^{4}=\fillin[6]\)                                                                                                                     
             \begin{solutionordottedlines}                                                                                                                  
                 Base: 6, Index: 4                                                                                                                          
             \end{solutionordottedlines}                                                                                                                    
             \part \(7^{3}=\fillin[7]\)                                                                                                                     
             \begin{solutionordottedlines}                                                                                                                  
                 Base: 7, Index: 3                                                                                                                          
             \end{solutionordottedlines}                                                                                                                    
             \part \(8^{2}=\fillin[8]\)                                                                                                                     
             \begin{solutionordottedlines}                                                                                                                  
                 Base: 8, Index: 2                                                                                                                          
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[3] Express as a power of a prime number.                                                                                                 
         \begin{parts}\begin{multicols}{3}                                                                                                                  
             \part $8 = \fillin[2^3]$                                                                                                                       
             \part $27 = \fillin[3^3]$                                                                                                                      
             \part $64 = \fillin[2^6]$                                                                                                                      
         \end{multicols}\end{parts}                                                                                                                         
         \Question[3] Evaluate:                                                                                                                             
         \begin{parts}\begin{multicols}{3}                                                                                                                  
             \part \(3^{4}=\fillin[81]\)                                                                                                                    
             \part \(2^{7}=\fillin[128]\)                                                                                                                   
             \part \(5^{5}=\fillin[3125]\)                                                                                                                  
         \end{multicols}\end{parts}                                                                                                                         
         \Question[3] Express as a product of powers of prime numbers.                                                                                      
         \begin{parts}\begin{multicols}{3}                                                                                                                  
             \part $18=\fillin[2 \times 3^2]$                                                                                                               
             \part $24=\fillin[2^3 \times 3]$                                                                                                               
             \part $144=\fillin[2^4 \times 3^2]$                                                                                                            
         \end{multicols}\end{parts}                                                                                                                         
         \Question[3] Simplify:                                                                                                                             
         \begin{parts}\begin{multicols}{3}                                                                                                                  
             \part \(2^{7} \times 2^{3}\)                                                                                                                   
             \begin{solutionordottedlines}                                                                                                                  
                 \(2^{10}\)                                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
             \part \(3^{3} \times 3^{4} \times 3^{5}\)                                                                                                      
             \begin{solutionordottedlines}                                                                                                                  
                 \(3^{12}\)                                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
             \part \(3 x^{2} \times 4 x^{3}\)                                                                                                               
             \begin{solutionordottedlines}                                                                                                                  
                 \(12 x^{5}\)                                                                                                                               
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[4] Simplify:                                                                                                                             
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(a^{2} b^{3} \times b^{2}\)                                                                                                             
             \begin{solutionordottedlines}                                                                                                                  
                 \(a^{2} b^{5}\)                                                                                                                            
             \end{solutionordottedlines}                                                                                                                    
             \part \(a^{3} b \times a^{2} b^{3}\)                                                                                                           
             \begin{solutionordottedlines}                                                                                                                  
                 \(a^{5} b^{4}\)                                                                                                                            
             \end{solutionordottedlines}                                                                                                                    
             \part \(2 x y^{2} \times 3 x^{2} y\)                                                                                                           
             \begin{solutionordottedlines}                                                                                                                  
                 \(6 x^{3} y^{3}\)                                                                                                                          
             \end{solutionordottedlines}                                                                                                                    
             \part \(4 a^{3} b^{2} \times a^{2} b^{4}\)                                                                                                     
             \begin{solutionordottedlines}                                                                                                                  
                 \(4 a^{5} b^{6}\)                                                                                                                          
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[6]$\,$                                                                                                                                   
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(\frac{3^7}{3^2}\)                                                                                                                      
             \begin{solutionordottedlines}                                                                                                                  
                 \(3^5\)                                                                                                                                    
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{2^{6}}{2^{2}}\)                                                                                                                  
             \begin{solutionordottedlines}                                                                                                                  
                 \(2^4\)                                                                                                                                    
             \end{solutionordottedlines}                                                                                                                    
             \part \(10^{7} \div 10^{2}\)                                                                                                                   
             \begin{solutionordottedlines}                                                                                                                  
                 \(10^5\)                                                                                                                                   
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{10^{12}}{10^{4}}\)                                                                                                               
             \begin{solutionordottedlines}                                                                                                                  
                 \(10^8\)                                                                                                                                   
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{2 x^{3}}{x^{2}}\)                                                                                                                
             \begin{solutionordottedlines}                                                                                                                  
                 \(2 x\)                                                                                                                                    
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{6 x^{5}}{2 x^{2}}\)                                                                                                              
             \begin{solutionordottedlines}                                                                                                                  
                 \(3 x^{3}\)                                                                                                                                
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[4]$\,$                                                                                                                                   
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(\frac{a^{3} b^{2}}{a^{2}}\)                                                                                                            
             \begin{solutionordottedlines}                                                                                                                  
                 \(a b^{2}\)                                                                                                                                
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{x^{3} y^{2}}{x y}\)                                                                                                              
             \begin{solutionordottedlines}                                                                                                                  
                 \(x^{2} y\)                                                                                                                                
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)                                                                                                     
             \begin{solutionordottedlines}                                                                                                                  
                 \(3 a^{4} b\)                                                                                                                              
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{15 x y^{3}}{3 y^{2}}\)                                                                                                           
             \begin{solutionordottedlines}                                                                                                                  
                 \(5 x y\)                                                                                                                                  
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[4]$\,$                                                                                                                                   
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)                                                                                     
             \begin{solutionordottedlines}                                                                                                                  
                 \(a^{4} b^{2}\)                                                                                                                            
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)                                                                             
             \begin{solutionordottedlines}                                                                                                                  
                 \(x^{4} y^{4}\)                                                                                                                            
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)                                                                           
             \begin{solutionordottedlines}                                                                                                                  
                 \(3 a^{2} b\)                                                                                                                              
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)                                                            
             \begin{solutionordottedlines}                                                                                                                  
                 \(2 x^{2} y\)                                                                                                                              
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[6]$\,$                                                                                                                                   
         \begin{parts}\begin{multicols}{3}                                                                                                                  
             \part \(a^{4} \times \fillin[a^{6}]=a^{10}\)                                                                                                   
             \begin{solutionordottedlines}                                                                                                                  
                 \(a^{6}\)                                                                                                                                  
             \end{solutionordottedlines}                                                                                                                    
             \part \(9 d^{5} \times \fillin[3]=27 d^{6}\)                                                                                                   
             \begin{solutionordottedlines}                                                                                                                  
                 \(3 d\)                                                                                                                                    
             \end{solutionordottedlines}                                                                                                                    
             \part \(15 d^{7} \div \fillin[5 d^{5}]=3 d^{2}\)                                                                                               
             \begin{solutionordottedlines}                                                                                                                  
                 \(5 d^{5}\)                                                                                                                                
             \end{solutionordottedlines}                                                                                                                    
             \part \(8 a b^{4} \times \fillin[3 a b^{2}]=24 a^{2} b^{6}\)                                                                                   
             \begin{solutionordottedlines}                                                                                                                  
                 \(3 a b^{2}\)                                                                                                                              
             \end{solutionordottedlines}                                                                                                                    
             \part \(\ell^{6} m^{7} \div \fillin[\ell^{4} m^{2}]=\ell^{2} m^{5}\)                                                                           
             \begin{solutionordottedlines}                                                                                                                  
                 \(\ell^{4} m^{2}\)                                                                                                                         
             \end{solutionordottedlines}                                                                                                                    
             \part \(b^{7} \times \fillin[b^{9}]=b^{16}\)                                                                                                   
             \begin{solutionordottedlines}                                                                                                                  
                 \(b^{9}\)                                                                                                                                  
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[2]$\,$                                                                                                                                   
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(a^{0}\)                                                                                                                                
             \begin{solutionordottedlines}                                                                                                                  
                 \(1\)                                                                                                                                      
             \end{solutionordottedlines}                                                                                                                    
             \part \(2 x^{0}\)                                                                                                                              
             \begin{solutionordottedlines}                                                                                                                  
                 \(2\)                                                                                                                                      
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[6]$\,$                                                                                                                                   
         \begin{parts}\begin{multicols}{3}                                                                                                                  
             \part \(3 a^{0}\)                                                                                                                              
             \begin{solutionordottedlines}                                                                                                                  
                 \(3\)                                                                                                                                      
             \end{solutionordottedlines}                                                                                                                    
             \part \(6 a^{0}\)                                                                                                                              
             \begin{solutionordottedlines}                                                                                                                  
                 \(6\)                                                                                                                                      
             \end{solutionordottedlines}                                                                                                                    
             \part \(4 a^{0}+3 b^{0}\)                                                                                                                      
             \begin{solutionordottedlines}                                                                                                                  
                 \(7\)                                                                                                                                      
             \end{solutionordottedlines}                                                                                                                    
             \part \(6 a^{0}+7 m^{0}\)                                                                                                                      
             \begin{solutionordottedlines}                                                                                                                  
                 \(13\)                                                                                                                                     
             \end{solutionordottedlines}                                                                                                                    
             \part \((4 b)^{0}+2 b^{0}\)                                                                                                                    
             \begin{solutionordottedlines}                                                                                                                  
                 \(3\)                                                                                                                                      
             \end{solutionordottedlines}                                                                                                                    
             \part \((3 b)^{0}-5 d^{0}\)                                                                                                                    
             \begin{solutionordottedlines}                                                                                                                  
                 \(-4\)                                                                                                                                     
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[2] Simplify, leaving the answer as a power.                                                                                              
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(\left(2^{3}\right)^{4}\)                                                                                                               
             \begin{solutionordottedlines}                                                                                                                  
                 \(2^{12}\)                                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
             \part \(\left(3^{2}\right)^{3}\)                                                                                                               
             \begin{solutionordottedlines}                                                                                                                  
                 \(3^{6}\)                                                                                                                                  
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[4] $\,$                                                                                                                                  
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)                                                                                  
             \begin{solutionordottedlines}                                                                                                                  
                 \(a^{14}\)                                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
             \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)                                                                                  
             \begin{solutionordottedlines}                                                                                                                  
                 \(x^{17}\)                                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
             \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)                                                                                           
             \begin{solutionordottedlines}                                                                                                                  
                 \(6 a^{2} b^{8}\)                                                                                                                          
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)                                                                      
             \begin{solutionordottedlines}                                                                                                                  
                 \(4 b^{2}\)                                                                                                                                
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[4]$\,$                                                                                                                                   
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \((3 a)^{2}\)                                                                                                                            
             \begin{solutionordottedlines}                                                                                                                  
                 \(9 a^{2}\)                                                                                                                                
             \end{solutionordottedlines}                                                                                                                    
             \part \((2 x)^{3}\)                                                                                                                            
             \begin{solutionordottedlines}                                                                                                                  
                 \(8 x^{3}\)                                                                                                                                
             \end{solutionordottedlines}                                                                                                                    
             \part \(\left(\frac{a}{5}\right)^{2}\)                                                                                                         
             \begin{solutionordottedlines}                                                                                                                  
                 \(\frac{a^{2}}{25}\)                                                                                                                       
             \end{solutionordottedlines}                                                                                                                    
             \part \(\left(\frac{2}{x}\right)^{3}\)                                                                                                         
             \begin{solutionordottedlines}                                                                                                                  
                 \(\frac{8}{x^{3}}\)                                                                                                                        
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[4]$\,$                                                                                                                                   
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)                                                                                          
             \begin{solutionordottedlines}                                                                                                                  
                 \(12 a^{5} b^{5}\)                                                                                                                         
             \end{solutionordottedlines}                                                                                                                    
             \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)                                                                            
             \begin{solutionordottedlines}                                                                                                                  
                 \(27 x^{5} y^{8}\)                                                                                                                         
             \end{solutionordottedlines}                                                                                                                    
             \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)                                                                        
             \begin{solutionordottedlines}                                                                                                                  
                 \(125 x^{7} y^{13}\)                                                                                                                       
             \end{solutionordottedlines}                                                                                                                    
             \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)                                                                                            
             \begin{solutionordottedlines}                                                                                                                  
                 \(24 a^{9} b^{3}\)                                                                                                                         
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
         \Question[4]$\,$                                                                                                                                   
         \begin{parts}\begin{multicols}{2}                                                                                                                  
             \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)                                                              
             \begin{solutionordottedlines}                                                                                                                  
                 \(\frac{x y^{4}}{y^{2}}\)                                                                                                                  
             \end{solutionordottedlines}                                                                                                                    
             \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)                                                              
             \begin{solutionordottedlines}                                                                                                                  
                 \(\frac{8 a b}{b^{2}}\)                                                                                                                    
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)                                       
             \begin{solutionordottedlines}                                                                                                                  
                 \(x^{2} y^{4}\)                                                                                                                            
             \end{solutionordottedlines}                                                                                                                    
             \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)                                                
             \begin{solutionordottedlines}                                                                                                                  
                 \(\frac{3 a^{2} b^{2}}{4}\)                                                                                                                
             \end{solutionordottedlines}                                                                                                                    
         \end{multicols}\end{parts}                                                                                                                         
     \end{questions}                                                                                                                                        
 \end{exercisebox}                                                                                                                                          
                                                                                                                                                            
|727323191|kitty|/Applications/kitty.app|0|

\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \Question[3] State the base and index of:
        \begin{parts}\begin{multicols}{3}
            \part \(6^{4}=\fillin[]\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(7^{3}=\fillin[]\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(8^{2}=\fillin[]\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[3] Express as a power of a prime number.
        \begin{parts}\begin{multicols}{3}
            \part $8 = \fillin[]$
            \part $27 = \fillin[]$
            \part $64 = \fillin[]$
        \end{multicols}\end{parts}
        \Question[3] Evaluate:
        \begin{parts}\begin{multicols}{3}
            \part \(3^{4}=\fillin[]\)
            \part \(2^{7}=\fillin[]\)
            \part \(5^{5}=\fillin[]\)
        \end{multicols}\end{parts}
        \Question[3] Express as a product of powers of prime numbers.
        \begin{parts}\begin{multicols}{3}
            \part $18=\fillin[]$
            \part $24=\fillin[]$
            \part $144=\fillin[]$
        \end{multicols}\end{parts}
        \Question[3] Simplify:
        \begin{parts}\begin{multicols}{3}
            \part \(2^{7} \times 2^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(3^{3} \times 3^{4} \times 3^{5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(3 x^{2} \times 4 x^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4] Simplify:
        \begin{parts}\begin{multicols}{2}
            \part \(a^{2} b^{3} \times b^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(a^{3} b \times a^{2} b^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(2 x y^{2} \times 3 x^{2} y\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(4 a^{3} b^{2} \times a^{2} b^{4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[6]$\,$
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{3^7}{3^2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{2^{6}}{2^{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(10^{7} \div 10^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{10^{12}}{10^{4}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{2 x^{3}}{x^{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{6 x^{5}}{2 x^{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4]$\,$
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a^{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{x^{3} y^{2}}{x y}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{12 a^{6} b^{2}}{4 a^{2} b}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{15 x y^{3}}{3 y^{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4]$\,$
        \begin{parts}\begin{multicols}{2}
            \part \(\frac{a^{3} b^{2}}{a b} \times \frac{a^{2} b}{a}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{x^{3} y}{x y^{2}} \times \frac{x^{4} y^{5}}{x^{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{6 a b^{2}}{5 a^{3} b} \div \frac{12 a b}{15 a^{5} b}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{7 x^{3} y^{4}}{2 x y^{2}} \div \frac{21 x^{2} y^{3}}{4 x^{3} y^{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[6]$\,$
        \begin{parts}\begin{multicols}{3}
            \part \(a^{4} \times \fillin[][1cm]=a^{10}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(9 d^{5} \times \fillin[][1cm]=27 d^{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(15 d^{7} \div \fillin[][1cm]=3 d^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(8 a b^{4} \times \fillin[][1cm]=24 a^{2} b^{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\ell^{6} m^{7} \div \fillin[][1cm]=\ell^{2} m^{5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(b^{7} \times \fillin[][1cm]=b^{16}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[2]$\,$
        \begin{parts}\begin{multicols}{2}
            \part \(a^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(2 x^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[6]$\,$
        \begin{parts}\begin{multicols}{3}
            \part \(3 a^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(6 a^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(4 a^{0}+3 b^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(6 a^{0}+7 m^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((4 b)^{0}+2 b^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((3 b)^{0}-5 d^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[2] Simplify, leaving the answer as a power.
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2^{3}\right)^{4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(3^{2}\right)^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4] $\,$ 
        \begin{parts}\begin{multicols}{2}
            \part \(\left(a^{3}\right)^{2} \times\left(a^{3}\right)^{4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(x^{4}\right)^{2} \times\left(x^{3}\right)^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(2 a b^{2} \times 3 a\left(b^{3}\right)^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{3 a b}{\left(b^{2}\right)^{3}} \times \frac{4 b^{7}}{3 a}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4]$\,$
        \begin{parts}\begin{multicols}{2}
            \part \((3 a)^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((2 x)^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(\frac{a}{5}\right)^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(\frac{2}{x}\right)^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4]$\,$
        \begin{parts}\begin{multicols}{2}
            \part \(\left(2 a^{2} b\right)^{2} \times 3 a b^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(3 x y^{2}\right)^{3} \times\left(x^{2} y\right)^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(5 x y^{2}\right)^{3} \times\left(x^{2} y^{3}\right)^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(2 a^{3} b\right)^{3} \times 3 a^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4]$\,$
        \begin{parts}\begin{multicols}{2}
            \part \(\left(\frac{x^{2}}{y}\right)^{2} \times\left(\frac{y^{2}}{x}\right)^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(\frac{4 a^{2}}{b}\right)^{2} \times\left(\frac{b}{2 a}\right)^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{\left(3 x y^{2}\right)^{2} \times\left(2 x^{2} y\right)^{3}}{\left(6 x^{2} y\right)^{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{3 a^{2} b^{4} \times\left(2 a b^{2}\right)^{3}}{\left(4 a^{2} b^{3}\right)^{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \end{questions}
\end{exercisebox}
|727323210|kitty|/Applications/kitty.app|0|
\subsection*{The Index Laws}

\begin{questions}
    \Question[3] State the base and index of:
    \begin{parts}\begin{multicols}{3}
        \part \(10^{4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part $5$
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(6^{0}\)
    \end{multicols}\end{parts}
    \Question[3] Express as a power of a prime number.
    \begin{parts}\begin{multicols}{3}
        \part 243
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part 125
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part 81
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[3] Evaluate:
    \begin{parts}\begin{multicols}{3}
        \part \(7^{4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(2^{3} \times 3^{5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(6^{4} \times 3^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[3] Express as a product of powers of prime numbers.
    \begin{parts}\begin{multicols}{3}
        \part 90
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part 700
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part 84
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[6] Simplify, leaving the answer as a power or a product of powers.
    \begin{parts}\begin{multicols}{3}
        \part \(3^{4} \times 3^{5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(3^{4} \times 3^{7}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(a^{3} \times a^{8}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(b^{7} \times b^{12}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(2 y \times 3 y^{4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(4 b^{2} \times 3 b^{4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[2] Simplify:
    \begin{parts}\begin{multicols}{2}
        \part \(x^{2} y \times x^{3} y\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(5 a^{4} b \times 2 a b^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[6] Simplify, leaving the answer as a power or a product of powers.
    \begin{parts}\begin{multicols}{2}
        \part \(5^{4} \div 5\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(7^{5} \div 7^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{a^{4}}{a}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{a^{5}}{a^{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{10 y^{12}}{5 y^{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{27 p^{4}}{9 p}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[4] Simplify:
    \begin{parts}\begin{multicols}{2}
        \part \[\frac{a^{5} b^{3}}{a^{4} b}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{x^{4} y^{7}}{x^{3} y^{2}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{16 a^{4} b^{3}}{12 a^{2} b^{2}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{27 x^{2} y^{3}}{18 x y^{2}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[4] Simplify:
    \begin{parts}\begin{multicols}{2}
        \part \[\frac{2 a b^{2}}{3 a^{2} b^{4}} \times \frac{6 a^{4} b^{5}}{a b}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{12 x^{4} y^{3}}{3 x^{2} y} \times \frac{x^{2} y^{4}}{x^{3} y^{5}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{14 a^{4} b^{3}}{3 a b^{2}} \div \frac{7 a^{5} b^{4}}{6 a^{3} b^{5}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{12 x^{2} y}{x^{3} y^{4}} \div \frac{6 x y^{2}}{x^{6} y^{7}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[9] Copy and complete.
    \begin{parts}\begin{multicols}{3}
        \part \(4 a^{3} \times \fillin[][1cm]=12 a^{7}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(a^{8} \div \fillin[][1cm]=a^{4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(x^{10} \div \fillin[][1cm]=x^{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(9 d^{6} \div \fillin[][1cm]=3 d\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(m^{4} n^{5} \times \fillin[][1cm]=m^{10} n^{7}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(a^{7} b^{4} \div \fillin[][1cm]=a^{2} b\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(14 x^{5} y^{2} \times \fillin[][1cm]=42 x^{10} y^{5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(9 m^{7} n^{4} \div \fillin[][1cm]=3 m^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(18 p^{2} q^{6} \div \fillin[][1cm]=3 p q\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[2] Simplify:
    \begin{parts}\begin{multicols}{2}
        \part \(x y^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(7 x^{0} y^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[6] Simplify:
    \begin{parts}\begin{multicols}{2}
        \part \((4 a)^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \((3 b)^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \((2 a+1)^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \((4 a+3 b)^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(5 m^{0}+7 b\right)^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(6 m-2 c^{0}\right)^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[2] Simplify, leaving the answer as a power.
    \begin{parts}\begin{multicols}{2}
        \part \(\left(a^{2}\right)^{5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(y^{5}\right)^{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[1] Simplify \[\frac{\left(y^{3}\right)^{4}}{\left(y^{4}\right)^{2}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \Question[1] Simplify \[\frac{3\left(x^{3} y\right)^{2}}{\left(x^{2} y\right)^{2}} \div \frac{12 x^{4} y^{2}}{\left(2 x^{3} y\right)^{2}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \Question[6] Copy and complete (using index law 3).
    \begin{parts}\begin{multicols}{2}
        \part \(\left(a^{6}\right)^{\fillin[][1cm]}=a^{24}\)
        \part \(\left(b^{3}\right)^{\fillin[][1cm]}=b^{21}\)
        \part \(\left(m^{6}\right) \fillin[][1cm]=m^{30}\)
        \part \((\fillin[][1cm])^{6}=p^{36}\)
        \part \((\fillin[][1cm])^{4}=a^{8}\)
        \part \((\fillin[][1cm])^{3}=m^{15}\)
    \end{multicols}\end{parts}
    \Question[6]$\,$
    \begin{parts}
        \part Is it true that \(\left(a^{2}\right)^{6}=\left(a^{6}\right)^{2}\) ? 
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part Is it true that \(\left(b^{4}\right)^{7}=\left(b^{7}\right)^{4}\) ?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part Generalise your result.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
|727323271|kitty|/Applications/kitty.app|0|

\subsection*{The Index Laws}

\begin{questions}
\Question[3] State the base and index of:
\begin{parts}\begin{multicols}{3}
		\part \(10^{4}\)
		\begin{solutionordottedlines}[2cm]
			Base: 10, Index: 4
		\end{solutionordottedlines}
		\part $5$
		\begin{solutionordottedlines}[2cm]
			Base: 5, Index: 1
		\end{solutionordottedlines}
		\part \(6^{0}\)
		\begin{solutionordottedlines}[2cm]
			Base: 6, Index: 0
		\end{solutionordottedlines}
	\end{multicols}\end{parts}
\Question[3] Express as a power of a prime number.
\begin{parts}\begin{multicols}{3}
		\part 243
		\begin{solutionordottedlines}[2cm]
			$243 = 3^{5}$
		\end{solutionordottedlines}
		\part 125
		\begin{solutionordottedlines}[2cm]
			$125 = 5^{3}$
		\end{solutionordottedlines}
		\part 81
		\begin{solutionordottedlines}[2cm]
			$81 = 3^{4}$
		\end{solutionordottedlines}
	\end{multicols}\end{parts}
\Question[3] Evaluate:
\begin{parts}\begin{multicols}{3}
		\part \(7^{4}\)
		\begin{solutionordottedlines}[2cm]
			$2401$
		\end{solutionordottedlines}
		\part \(2^{3} \times 3^{5}\)
		\begin{solutionordottedlines}[2cm]
			$8 \times 243 = 1944$
		\end{solutionordottedlines}
		\part \(6^{4} \times 3^{2}\)
		\begin{solutionordottedlines}[2cm]
			$1296 \times 9 = 11664$
		\end{solutionordottedlines}
	\end{multicols}\end{parts}
\Question[3] Express as a product of powers of prime numbers.
\begin{parts}\begin{multicols}{3}
		\part 90
		\begin{solutionordottedlines}[2cm]
			$90 = 2 \times 3^{2} \times 5$
		\end{solutionordottedlines}
		\part 700
		\begin{solutionordottedlines}[2cm]
			$700 = 2^{2} \times 5^{2} \times 7$
		\end{solutionordottedlines}
		\part 84
		\begin{solutionordottedlines}[2cm]
			$84 = 2^{2} \times 3 \times 7$
		\end{solutionordottedlines}
	\end{multicols}\end{parts}
\Question[6] Simplify, leaving the answer as a power or a product of powers.
\begin{parts}\begin{multicols}{3}
		\part \(3^{4} \times 3^{5}\)
		\begin{solutionordottedlines}[2cm]
			$3^{9}$
		\end{solutionordottedlines}
		\part \(3^{4} \times 3^{7}\)
		\begin{solutionordottedlines}[2cm]
			$3^{11}$
		\end{solutionordottedlines}
		\part \(a^{3} \times a^{8}\)
		\begin{solutionordottedlines}[2cm]
			$a^{11}$
		\end{solutionordottedlines}
		\part \(b^{7} \times b^{12}\)
		\begin{solutionordottedlines}[2cm]
			$b^{19}$
		\end{solutionordottedlines}
		\part \(2 y \times 3 y^{4}\)
		\begin{solutionordottedlines}[2cm]
			$6 y^{5}$
		\end{solutionordottedlines}
		\part \(4 b^{2} \times 3 b^{4}\)
		\begin{solutionordottedlines}[2cm]
			$12 b^{6}$
		\end{solutionordottedlines}
	\end{multicols}\end{parts}
\Question[2] Simplify:
\begin{parts}\begin{multicols}{2}
		\part \(x^{2} y \times x^{3} y\)
		\begin{solutionordottedlines}[2cm]
			$x^{5} y^{2}$
		\end{solutionordottedlines}
		\part \(5 a^{4} b \times 2 a b^{3}\)
		\begin{solutionordottedlines}[2cm]
			$10 a^{5} b^{4}$
		\end{solutionordottedlines}
	\end{multicols}\end{parts}
\Question[6] Simplify, leaving the answer as a power or a product of powers.
\begin{parts}\begin{multicols}{2}
		\part \(5^{4} \div 5\)
		\begin{solutionordottedlines}[2cm]
			$5^{3}$
		\end{solutionordottedlines}
		\part \(7^{5} \div 7^{3}\)
		\begin{solutionordottedlines}[2cm]
			$7^{2}$
		\end{solutionordottedlines}
		\part \(\frac{a^{4}}{a}\)
		\begin{solutionordottedlines}[2cm]
			$a^{3}$
		\end{solutionordottedlines}
		\part \(\frac{a^{5}}{a^{3}}\)
		\begin{solutionordottedlines}[2cm]
			$a^{2}$
		\end{solutionordottedlines}
		\part \(\frac{10 y^{12}}{5 y^{3}}\)
		\begin{solutionordottedlines}[2cm]
			$2 y^{9}$
		\end{solutionordottedlines}
		\part \(\frac{27 p^{4}}{9 p}\)
		\begin{solutionordottedlines}[2cm]
			$3 p^{3}$
		\end{solutionordottedlines}
	\end{multicols}\end{parts}
\Question[4] Simplify:
\begin{parts}\begin{multicols}{2}
		\part \[\frac{a^{5} b^{3}}{a^{4} b}\]
		\begin{solutionordottedlines}[2cm]
			$a b^{2}$
		\end{solutionordottedlines}
		\part \[\frac{x^{4} y^{7}}{x^{3} y^{2}}\]
		\begin{solutionordottedlines}[2cm]
			$x y^{5}$
		\end{solutionordottedlines}
		\part \[\frac{16 a^{4} b^{3}}{12 a^{2} b^{2}}\]
		\begin{solutionordottedlines}[2cm]
			$\frac{4}{3} a^{2} b$
		\end{solutionordottedlines}
		\part \[\frac{27 x^{2} y^{3}}{18 x y^{2}}\]
		\begin{solutionordottedlines}[2cm]
			$\frac{3}{2} x y$
		\end{solutionordottedlines}
	\end{multicols}\end{parts}
\Question[4] Simplify:
\begin{parts}\begin{multicols}{2}
		\part \[\frac{2 a b^{2}}{3 a^{2} b^{4}} \times \frac{6 a^{4} b^{5}}{a b}\]
		\begin{solutionordottedlines}[2cm]
			$4 a b^{-1}$
		\end{solutionordottedlines}
		\part \[\frac{12 x^{4} y^{3}}{3 x^{2} y} \times \frac{x^{2} y^{4}}{x^{3} y^{5}}\]
		\begin{solutionordottedlines}[2cm]
			$4 x y$
		\end{solutionordottedlines}
		\part \[\frac{14 a^{4} b^{3}}{3 a b^{2}} \div \frac{7 a^{5} b^{4}}{6 a^{3} b^{5}}\]
		\begin{solutionordottedlines}[2cm]
			$8 a^{2} b^{-4}$
		\end{solutionordottedlines}
		\part \[\frac{12 x^{2} y}{x^{3} y^{4}} \div \frac{6 x y^{2}}{x^{6} y^{7}}\]
		\begin{solutionordottedlines}[2cm]
			$2 x^{-7} y^{-8}$
		\end{solutionordottedlines}
	\end{multicols}\end{parts}
\Question[9] Copy and complete.
\begin{parts}\begin{multicols}{3}
		\part \(4 a^{3} \times \fillin[][1cm]=12 a^{7}\)
		\begin{solutionordottedlines}[2cm]
			$3 a^{4}$
		\end{solutionordottedlines}
		\part \(a^{8} \div \fillin[][1cm]=a^{4}\)
		\begin{solutionordottedlines}[2cm]
			$a^{4}$
		\end{solutionordottedlines}
		\part \(x^{10} \div \fillin[][1cm]=x^{6}\)
		\begin{solutionordottedlines}[2cm]
			$x^{4}$
		\end{solutionordottedlines}
		\part \(9 d^{6} \div \fillin[][1cm]=3 d\)
		\begin{solutionordottedlines}[2cm]
			$3 d^{5}$
		\end{solutionordottedlines}
		\part \(m^{4} n^{5} \times \fillin[][1cm]=m^{10} n^{7}\)
		\begin{solutionordottedlines}[2cm]
			$m^{6} n^{2}$
		\end{solutionordottedlines}
		\part \(a^{7} b^{4} \div \fillin[][1cm]=a^{2} b\)
		\begin{solutionordottedlines}[2cm]
			$a^{5} b^{3}$
		\end{solutionordottedlines}
		\part \(14 x^{5} y^{2} \times \fillin[][1cm]=42 x^{10} y^{5}\)
		\begin{solutionordottedlines}[2cm]
			$3 x^{5} y^{3}$
		\end{solutionordottedlines}
		\part \(9 m^{7} n^{4} \div \fillin[][1cm]=3 m^{2}\)
		\begin{solutionordottedlines}[2cm]
			$3 m^{5} n^{4}$
		\end{solutionordottedlines}
		\part \(18 p^{2} q^{6} \div \fillin[][1cm]=3 p q\)
		\begin{solutionordottedlines}[2cm]
			$6 p q^{5}$
		\end{solutionordottedlines}
	\end{multicols}\end{parts}
\Question[2] Simplify:
\begin{parts}\begin{multicols}{2}
		\part \(x y^{0}\)
		\begin{solutionordottedlines}[2cm]
			$x$
		\end{solutionordottedlines}
		\part \(7 x^{0} y^{0}\)
		\begin{solutionordottedlines}[2cm]
			$7$
		\end{solutionordottedlines}
	\end{multicols}\end{parts}
\Question[6] Simplify:
\begin{parts}\begin{multicols}{2}
		\part \((4 a)^{0}\)
		\begin{solutionordottedlines}[2cm]
			$1$
		\end{solutionordottedlines}
		\part \((3 b)^{0}\)
		\begin{solutionordottedlines}[2cm]
			$1$
		\end{solutionordottedlines}
		\part \((2 a+1)^{0}\)
		\begin{solutionordottedlines}[2cm]
			$1$
		\end{solutionordottedlines}
		\part \((4 a+3 b)^{0}\)
		\begin{solutionordottedlines}[2cm]
			$1$
		\end{solutionordottedlines}
		\part \(\left(5 m^{0}+7 b\right)^{0}\)
		\begin{solutionordottedlines}[2cm]
			$1$
		\end{solutionordottedlines}
		\part \(\left(6 m-2 c^{0}\right)^{0}\)
		\begin{solutionordottedlines}[2cm]
			$1$
		\end{solutionordottedlines}
	\end{multicols}\end{parts}
\Question[2] Simplify, leaving the answer as a power.
\begin{parts}\begin{multicols}{2}
		\part \(\left(a^{2}\right)^{5}\)
		\begin{solutionordottedlines}[2cm]
			$a^{10}$
		\end{solutionordottedlines}
		\part \(\left(y^{5}\right)^{6}\)
		\begin{solutionordottedlines}[2cm]
			$y^{30}$
		\end{solutionordottedlines}
	\end{multicols}\end{parts}
\Question[1] Simplify \[\frac{\left(y^{3}\right)^{4}}{\left(y^{4}\right)^{2}}\]
\begin{solutionordottedlines}[2cm]
	$y^{4}$
\end{solutionordottedlines}
\Question[1] Simplify \[\frac{3\left(x^{3} y\right)^{2}}{\left(x^{2} y\right)^{2}} \div \frac{12 x^{4} y^{2}}{\left(2 x^{3} y\right)^{2}}\]
\begin{solutionordottedlines}[2cm]
	$\frac{3}{4}$
\end{solutionordottedlines}
\Question[6] Copy and complete (using index law 3).
\begin{parts}\begin{multicols}{2}
		\part \(\left(a^{6}\right)^{\fillin[][1cm]}=a^{24}\)
		\begin{solutionordottedlines}[2cm]
			$4$
		\end{solutionordottedlines}
		\part \(\left(b^{3}\right)^{\fillin[][1cm]}=b^{21}\)
		\begin{solutionordottedlines}[2cm]
			$7$
		\end{solutionordottedlines}
		\part \(\left(m^{6}\right) \fillin[][1cm]=m^{30}\)
		\begin{solutionordottedlines}[2cm]
			$5$
		\end{solutionordottedlines}
		\part \((\fillin[][1cm])^{6}=p^{36}\)
		\begin{solutionordottedlines}[2cm]
			$p^{6}$
		\end{solutionordottedlines}
		\part \((\fillin[][1cm])^{4}=a^{8}\)
		\begin{solutionordottedlines}[2cm]
			$a^{2}$
		\end{solutionordottedlines}
		\part \((\fillin[][1cm])^{3}=m^{15}\)
		\begin{solutionordottedlines}[2cm]
			$m^{5}$
		\end{solutionordottedlines}
	\end{multicols}\end{parts}
\Question[6]$\,$
\begin{parts}
	\part Is it true that \(\left(a^{2}\right)^{6}=\left(a^{6}\right)^{2}\) ?
	\begin{solutionordottedlines}[2cm]
		Yes, both equal $a^{12}$.
	\end{solutionordottedlines}
	\part Is it true that \(\left(b^{4}\right)^{7}=\left(b^{7}\right)^{4}\) ?
	\begin{solutionordottedlines}[2cm]
		Yes, both equal $b^{28}$.
	\end{solutionordottedlines}
	\part Generalise your result.
	\begin{solutionordottedlines}[2cm]
		\(\left(a^{m}\right)^{n}=\left(a^{n}\right)^{m}=a^{mn}\)
	\end{solutionordottedlines}
\end{parts}
|727323291|kitty|/Applications/kitty.app|0|

    \Question[4] Simplify by expanding the brackets.
    \begin{parts}\begin{multicols}{2}
        \part \(\left(x y^{3}\right)^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(a^{2} b\right)^{4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(\frac{a}{b}\right)^{5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(\frac{x^{2}}{y}\right)^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[2] Simplify:
    \begin{parts}
        \part \[\left(\frac{x^{3}}{y^{2}}\right)^{2} \div\left(\frac{x}{y^{2}}\right)^{3}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\left(\frac{2 x^{4}}{y}\right)^{5} \div\left(\frac{4 x^{3}}{y^{3}}\right)^{2}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[2] Simplify:
    \begin{parts}
        \part \[\frac{\left(2 x^{2} y^{3}\right)^{3} \times\left(5 x y^{2}\right)^{2}}{\left(10 x^{2} y\right)^{2} \times(x y)^{3}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{(6 a b)^{3} \times 2 a^{7} b^{4}}{(2 a b)^{4} \times\left(3 a^{2} b\right)^{2}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[9] Copy and complete.
    \begin{parts}\begin{multicols}{3}
        \part \((\fillin[][1cm])^{4}=a^{8} b^{12}\)
        \part \((\fillin[][1cm])^{6}=m^{30} n^{24}\)
        \part \(\left(p^{3} q\right) \fillin[][1cm]=p^{9} q^{3}\)
        \part \(\left(x^{4} y^{7}\right) \fillin[][1cm]=1\)
        \part \((\fillin[][1cm])^{4}=16 a^{8}\)
        \part \((\fillin[][1cm])^{3}=27 q^{9}\)
        \part \((\fillin[][1cm])^{2}=49 m^{6}\)
        \part \((\fillin[][1cm])^{3}=64 \ell^{9} m^{3}\)
        \part \((\fillin[][1cm])^{2}=25 m^{10} n^{6}\)
    \end{multicols}\end{parts}
\end{questions}


\subsection*{Negative Indices}

\begin{questions}
    \Question[6] Express with a positive index and then evaluate.
    \begin{parts}\begin{multicols}{2}
        \part \(3^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(6^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(9^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(5^{-3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(3^{-4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(10^{-5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[6] Write each fraction as a power of a prime with a negative index.
    \begin{parts}\begin{multicols}{2}
        \part \(\frac{1}{27}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{1}{49}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{1}{121}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{1}{125}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{1}{169}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{1}{81}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[6] Express with positive indices, evaluating where possible.
    \begin{parts}\begin{multicols}{2}
        \part \(3 a^{-4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(5 x^{-7}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(4 a^{-5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{1}{x^{-3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(3^{-2} a^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(4^{-2} x^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[4] Simplify where possible and then evaluate.
    \begin{parts}\begin{multicols}{2}
        \part \(\left(3 \frac{1}{3}\right)^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(\frac{2}{3}\right)^{-3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(7^{3} \times 7^{-5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(4^{3} \times 4^{-5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[6] Write as a single power and then evaluate.
    \begin{parts}\begin{multicols}{2}
        \part \[\frac{3^{8}}{3^{9}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{6^{5}}{6^{8}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{7^{1}}{7^{3}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{5^{7}}{5^{10}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{3^{5}}{3^{9}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{12^{12}}{12^{14}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[3] Express with negative index.
    \begin{parts}\begin{multicols}{3}
        \part \(\frac{3}{2 x^{4}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{4}{3 x^{7}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{2}{3 x^{5}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[7] Evaluate.
    \begin{parts}\begin{multicols}{3}
        \part \(\left(\frac{4}{5}\right)^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(2 \frac{1}{4}\right)^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(1 \frac{1}{5}\right)^{-3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[9] Simplify, expressing the answer with positive indices.
    \begin{parts}\begin{multicols}{3}
        \part \(a^{-3} b^{-5} \times a^{5} b^{-3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(3 x^{-2} y^{5} \times 5 x^{-7} y^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(7 a^{3} m^{-4} \times 8 a^{-5} m^{-3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(3 r^{2} s^{3} \times 4 r^{-3} s^{-5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{16 a^{-4}}{8 a^{5}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{18 a^{-4}}{4 a^{5}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{27 m^{-3}}{9 m^{-2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{72 a^{4} b^{-3}}{36 a b^{-2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{7 a^{2} b^{-3} c^{-4}}{21 a^{5} b^{-7} c^{-9}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[9] Copy and complete.
    \begin{parts}\begin{multicols}{3}
        \part \(9^{5} \times \fillin[][1cm]=9^{4}\)
        \part \(b^{9} \times \fillin[][1cm]=b^{7}\)
        \part \(a^{11} \div \fillin[][1cm]=a^{14}\)
        \part \(b^{7} \div \fillin[][1cm]=b^{15}\)
        \part \(e^{-7} \div \fillin[][1cm]=e^{-5}\)
        \part \(\left(m^{-2}\right) \fillin[][1cm]=m^{10}\)
        \part \((\fillin[][1cm])^{-3}=\frac{1}{27 a^{9}}\)
        \part \((\fillin[][1cm])^{-3}=\frac{a^{6}}{b^{9}}\)
        \part \((\fillin[][1cm])^{-6}=\frac{m^{12} n^{18}}{p^{6}}\)
    \end{multicols}\end{parts}
    \Question[6] Simplify, expressing the answers with positive indices. Evaluate powers where possible.
    \begin{parts}\begin{multicols}{2}
        \part \[\left(5 x^{4} y^{6}\right)^{-3} \times\left(5^{2} x y^{-1}\right)^{3}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\left(5 m^{2} n^{-3}\right)^{-2} \times 2\left(m^{-2} n^{3}\right)^{2}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{\left(x^{2}\right)^{2}}{y} \times \frac{\left(y^{2}\right)^{-3}}{x^{3}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{\left(2 x^{3}\right)^{-2}}{y^{4}} \times \frac{\left(2 x^{7}\right)^{2}}{3 y^{5}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{\left(m^{2} n^{3}\right)^{2}}{p^{-3}} \times\left(m n p^{-2}\right)^{-3}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{\left(a^{2}\right)^{3}}{b^{3}} \div\left(\frac{a}{b^{2}}\right)^{-2}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \[\frac{\left(3 m^{2} n^{3}\right)^{-2}}{p^{4}} \div \frac{p^{-3}}{m}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
\end{questions}
|727323306|kitty|/Applications/kitty.app|0|
     \Question[4] Simplify by expanding the brackets.                                                                                                       
     \begin{parts}\begin{multicols}{2}                                                                                                                      
         \part \(\left(x y^{3}\right)^{2}\)                                                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(x^2 y^6\)                                                                                                                                
             \end{solutionordottedlines}                                                                                                                    
         \part \(\left(a^{2} b\right)^{4}\)                                                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(a^8 b^4\)                                                                                                                                
             \end{solutionordottedlines}                                                                                                                    
         \part \(\left(\frac{a}{b}\right)^{5}\)                                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{a^5}{b^5}\)                                                                                                                        
             \end{solutionordottedlines}                                                                                                                    
         \part \(\left(\frac{x^{2}}{y}\right)^{3}\)                                                                                                         
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{x^6}{y^3}\)                                                                                                                        
             \end{solutionordottedlines}                                                                                                                    
     \end{multicols}\end{parts}                                                                                                                             
     \Question[2] Simplify:                                                                                                                                 
     \begin{parts}                                                                                                                                          
         \part \[\left(\frac{x^{3}}{y^{2}}\right)^{2} \div\left(\frac{x}{y^{2}}\right)^{3}\]                                                                
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(x^3\)                                                                                                                                    
             \end{solutionordottedlines}                                                                                                                    
         \part \[\left(\frac{2 x^{4}}{y}\right)^{5} \div\left(\frac{4 x^{3}}{y^{3}}\right)^{2}\]                                                            
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{32 x^{8}}{y}\)                                                                                                                     
             \end{solutionordottedlines}                                                                                                                    
     \end{parts}                                                                                                                                            
     \Question[2] Simplify:                                                                                                                                 
     \begin{parts}                                                                                                                                          
         \part \[\frac{\left(2 x^{2} y^{3}\right)^{3} \times\left(5 x y^{2}\right)^{2}}{\left(10 x^{2} y\right)^{2} \times(x y)^{3}}\]                      
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(25 y\)                                                                                                                                   
             \end{solutionordottedlines}                                                                                                                    
         \part \[\frac{(6 a b)^{3} \times 2 a^{7} b^{4}}{(2 a b)^{4} \times\left(3 a^{2} b\right)^{2}}\]                                                    
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(3 a^2\)                                                                                                                                  
             \end{solutionordottedlines}                                                                                                                    
     \end{parts}                                                                                                                                            
     \Question[9] Copy and complete.                                                                                                                        
     \begin{parts}\begin{multicols}{3}                                                                                                                      
         \part \((a^2 b^3)^{4}=a^{8} b^{12}\)                                                                                                               
         \part \((m^5 n^4)^{6}=m^{30} n^{24}\)                                                                                                              
         \part \(\left(p^{3} q\right)^3=p^{9} q^{3}\)                                                                                                       
         \part \(\left(x^{4} y^{7}\right)^{-1}=1\)                                                                                                          
         \part \((2 a^2)^{4}=16 a^{8}\)                                                                                                                     
         \part \((3 q^3)^{3}=27 q^{9}\)                                                                                                                     
         \part \((7 m^3)^{2}=49 m^{6}\)                                                                                                                     
         \part \((4 \ell^3 m)^{3}=64 \ell^{9} m^{3}\)                                                                                                       
         \part \((5 m^5 n^3)^{2}=25 m^{10} n^{6}\)                                                                                                          
     \end{multicols}\end{parts}                                                                                                                             
 \end{questions}                                                                                                                                            
                                                                                                                                                            
 \subsection*{Negative Indices}                                                                                                                             
                                                                                                                                                            
 \begin{questions}                                                                                                                                          
     \Question[6] Express with a positive index and then evaluate.                                                                                          
     \begin{parts}\begin{multicols}{2}                                                                                                                      
         \part \(3^{-2}\)                                                                                                                                   
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{1}{9}\)                                                                                                                            
             \end{solutionordottedlines}                                                                                                                    
         \part \(6^{-2}\)                                                                                                                                   
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{1}{36}\)                                                                                                                           
             \end{solutionordottedlines}                                                                                                                    
         \part \(9^{-2}\)                                                                                                                                   
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{1}{81}\)                                                                                                                           
             \end{solutionordottedlines}                                                                                                                    
         \part \(5^{-3}\)                                                                                                                                   
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{1}{125}\)                                                                                                                          
             \end{solutionordottedlines}                                                                                                                    
         \part \(3^{-4}\)                                                                                                                                   
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{1}{81}\)                                                                                                                           
             \end{solutionordottedlines}                                                                                                                    
         \part \(10^{-5}\)                                                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{1}{100000}\)                                                                                                                       
             \end{solutionordottedlines}                                                                                                                    
     \end{multicols}\end{parts}                                                                                                                             
     \Question[6] Write each fraction as a power of a prime with a negative index.                                                                          
     \begin{parts}\begin{multicols}{2}                                                                                                                      
         \part \(\frac{1}{27}\)                                                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(3^{-3}\)                                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
         \part \(\frac{1}{49}\)                                                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(7^{-2}\)                                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
         \part \(\frac{1}{121}\)                                                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(11^{-2}\)                                                                                                                                
             \end{solutionordottedlines}                                                                                                                    
         \part \(\frac{1}{125}\)                                                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(5^{-3}\)                                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
         \part \(\frac{1}{169}\)                                                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(13^{-2}\)                                                                                                                                
             \end{solutionordottedlines}                                                                                                                    
         \part \(\frac{1}{81}\)                                                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(3^{-4}\)                                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
     \end{multicols}\end{parts}                                                                                                                             
     \Question[6] Express with positive indices, evaluating where possible.                                                                                 
     \begin{parts}\begin{multicols}{2}                                                                                                                      
         \part \(3 a^{-4}\)                                                                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{3}{a^4}\)                                                                                                                          
             \end{solutionordottedlines}                                                                                                                    
         \part \(5 x^{-7}\)                                                                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{5}{x^7}\)                                                                                                                          
             \end{solutionordottedlines}                                                                                                                    
         \part \(4 a^{-5}\)                                                                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{4}{a^5}\)                                                                                                                          
             \end{solutionordottedlines}                                                                                                                    
         \part \(\frac{1}{x^{-3}}\)                                                                                                                         
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(x^3\)                                                                                                                                    
             \end{solutionordottedlines}                                                                                                                    
         \part \(3^{-2} a^{-2}\)                                                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{1}{9 a^2}\)                                                                                                                        
             \end{solutionordottedlines}                                                                                                                    
         \part \(4^{-2} x^{-2}\)                                                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{1}{16 x^2}\)                                                                                                                       
             \end{solutionordottedlines}                                                                                                                    
     \end{multicols}\end{parts}                                                                                                                             
     \Question[4] Simplify where possible and then evaluate.                                                                                                
     \begin{parts}\begin{multicols}{2}                                                                                                                      
         \part \(\left(3 \frac{1}{3}\right)^{-2}\)                                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{1}{11.11}\)                                                                                                                        
             \end{solutionordottedlines}                                                                                                                    
         \part \(\left(\frac{2}{3}\right)^{-3}\)                                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{27}{8}\)                                                                                                                           
             \end{solutionordottedlines}                                                                                                                    
         \part \(7^{3} \times 7^{-5}\)                                                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(7^{-2}\)                                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
         \part \(4^{3} \times 4^{-5}\)                                                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(4^{-2}\)                                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
     \end{multicols}\end{parts}                                                                                                                             
     \Question[6] Write as a single power and then evaluate.                                                                                                
     \begin{parts}\begin{multicols}{2}                                                                                                                      
         \part \[\frac{3^{8}}{3^{9}}\]                                                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(3^{-1}\)                                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
         \part \[\frac{6^{5}}{6^{8}}\]                                                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(6^{-3}\)                                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
         \part \[\frac{7^{1}}{7^{3}}\]                                                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(7^{-2}\)                                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
         \part \[\frac{5^{7}}{5^{10}}\]                                                                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(5^{-3}\)                                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
         \part \[\frac{3^{5}}{3^{9}}\]                                                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(3^{-4}\)                                                                                                                                 
             \end{solutionordottedlines}                                                                                                                    
         \part \[\frac{12^{12}}{12^{14}}\]                                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(12^{-2}\)                                                                                                                                
             \end{solutionordottedlines}                                                                                                                    
     \end{multicols}\end{parts}                                                                                                                             
     \Question[3] Express with negative index.                                                                                                              
     \begin{parts}\begin{multicols}{3}                                                                                                                      
         \part \(\frac{3}{2 x^{4}}\)                                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(3 \cdot 2^{-1} x^{-4}\)                                                                                                                  
             \end{solutionordottedlines}                                                                                                                    
         \part \(\frac{4}{3 x^{7}}\)                                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(4 \cdot 3^{-1} x^{-7}\)                                                                                                                  
             \end{solutionordottedlines}                                                                                                                    
         \part \(\frac{2}{3 x^{5}}\)                                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(2 \cdot 3^{-1} x^{-5}\)                                                                                                                  
             \end{solutionordottedlines}                                                                                                                    
     \end{multicols}\end{parts}                                                                                                                             
     \Question[7] Evaluate.                                                                                                                                 
     \begin{parts}\begin{multicols}{3}                                                                                                                      
         \part \(\left(\frac{4}{5}\right)^{-2}\)                                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{25}{16}\)                                                                                                                          
             \end{solutionordottedlines}                                                                                                                    
         \part \(\left(2 \frac{1}{4}\right)^{-2}\)                                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{16}{81}\)                                                                                                                          
             \end{solutionordottedlines}                                                                                                                    
         \part \(\left(1 \frac{1}{5}\right)^{-3}\)                                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{125}{216}\)                                                                                                                        
             \end{solutionordottedlines}                                                                                                                    
     \end{multicols}\end{parts}                                                                                                                             
     \Question[9] Simplify, expressing the answer with positive indices.                                                                                    
     \begin{parts}\begin{multicols}{3}                                                                                                                      
         \part \(a^{-3} b^{-5} \times a^{5} b^{-3}\)                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(a^2 b^{-8}\)                                                                                                                             
             \end{solutionordottedlines}                                                                                                                    
         \part \(3 x^{-2} y^{5} \times 5 x^{-7} y^{-2}\)                                                                                                    
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(15 x^{-9} y^3\)                                                                                                                          
             \end{solutionordottedlines}                                                                                                                    
         \part \(7 a^{3} m^{-4} \times 8 a^{-5} m^{-3}\)                                                                                                    
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(56 a^{-2} m^{-7}\)                                                                                                                       
             \end{solutionordottedlines}                                                                                                                    
         \part \(3 r^{2} s^{3} \times 4 r^{-3} s^{-5}\)                                                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(12 r^{-1} s^{-2}\)                                                                                                                       
             \end{solutionordottedlines}                                                                                                                    
         \part \(\frac{16 a^{-4}}{8 a^{5}}\)                                                                                                                
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(2 a^{-9}\)                                                                                                                               
             \end{solutionordottedlines}                                                                                                                    
         \part \(\frac{18 a^{-4}}{4 a^{5}}\)                                                                                                                
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(4.5 a^{-9}\)                                                                                                                             
             \end{solutionordottedlines}                                                                                                                    
         \part \(\frac{27 m^{-3}}{9 m^{-2}}\)                                                                                                               
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(3 m^{-1}\)                                                                                                                               
             \end{solutionordottedlines}                                                                                                                    
         \part \(\frac{72 a^{4} b^{-3}}{36 a b^{-2}}\)                                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(2 a^3 b^{-1}\)                                                                                                                           
             \end{solutionordottedlines}                                                                                                                    
         \part \(\frac{7 a^{2} b^{-3} c^{-4}}{21 a^{5} b^{-7} c^{-9}}\)                                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{1}{3} a^{-3} b^4 c^5\)                                                                                                             
             \end{solutionordottedlines}                                                                                                                    
     \end{multicols}\end{parts}                                                                                                                             
     \Question[9] Copy and complete.                                                                                                                        
     \begin{parts}\begin{multicols}{3}                                                                                                                      
         \part \(9^{5} \times 9^{-1}=9^{4}\)                                                                                                                
         \part \(b^{9} \times b^{-2}=b^{7}\)                                                                                                                
         \part \(a^{11} \div a^{-3}=a^{14}\)                                                                                                                
         \part \(b^{7} \div b^{-8}=b^{15}\)                                                                                                                 
         \part \(e^{-7} \div e^{2}=e^{-5}\)                                                                                                                 
         \part \(\left(m^{-2}\right)^{-5}=m^{10}\)                                                                                                          
         \part \((3 a^3)^{-3}=\frac{1}{27 a^{9}}\)                                                                                                          
         \part \((\frac{a^2}{b^3})^{-3}=\frac{a^{6}}{b^{9}}\)                                                                                               
         \part \((\frac{m^2 n^3}{p})^{-6}=\frac{m^{12} n^{18}}{p^{6}}\)                                                                                     
     \end{multicols}\end{parts}                                                                                                                             
     \Question[6] Simplify, expressing the answers with positive indices. Evaluate powers where possible.                                                   
     \begin{parts}\begin{multicols}{2}                                                                                                                      
         \part \[\left(5 x^{4} y^{6}\right)^{-3} \times\left(5^{2} x y^{-1}\right)^{3}\]                                                                    
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(x^{-9} y^{-9}\)                                                                                                                          
             \end{solutionordottedlines}                                                                                                                    
         \part \[\left(5 m^{2} n^{-3}\right)^{-2} \times 2\left(m^{-2} n^{3}\right)^{2}\]                                                                   
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(2 m^{-2} n^{6}\)                                                                                                                         
             \end{solutionordottedlines}                                                                                                                    
         \part \[\frac{\left(x^{2}\right)^{2}}{y} \times \frac{\left(y^{2}\right)^{-3}}{x^{3}}\]                                                            
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(x y^{-5}\)                                                                                                                               
             \end{solutionordottedlines}                                                                                                                    
         \part \[\frac{\left(2 x^{3}\right)^{-2}}{y^{4}} \times \frac{\left(2 x^{7}\right)^{2}}{3 y^{5}}\]                                                  
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{8 x^{10}}{3 y^{9}}\)                                                                                                               
             \end{solutionordottedlines}                                                                                                                    
         \part \[\frac{\left(m^{2} n^{3}\right)^{2}}{p^{-3}} \times\left(m n p^{-2}\right)^{-3}\]                                                           
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(m^{-1} n^{-6} p^3\)                                                                                                                      
             \end{solutionordottedlines}                                                                                                                    
         \part \[\frac{\left(a^{2}\right)^{3}}{b^{3}} \div\left(\frac{a}{b^{2}}\right)^{-2}\]                                                               
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(a^7 b\)                                                                                                                                  
             \end{solutionordottedlines}                                                                                                                    
         \part \[\frac{\left(3 m^{2} n^{3}\right)^{-2}}{p^{4}} \div \frac{p^{-3}}{m}\]                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                                             
                 \(\frac{m^5}{9 n^6 p}\)                                                                                                                    
             \end{solutionordottedlines}                                                                                                                    
     \end{multicols}\end{parts}                                                                                                                             
 \end{questions}                                                                                                                                            
                                                                                                                                                            
|727323588|kitty|/Applications/kitty.app|0|
2dy48%!rSa*4UJd|727761727|kitty|/Applications/kitty.app|0|
	- [ ] [COMP2511](COMP2511/index.md)
|727760702|kitty|/Applications/kitty.app|0|
@T3rr0rbyt3!|727766245|kitty|/Applications/kitty.app|0|
Statement from Australian Taxation Office|727766368|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
174201004|727766535|kitty|/Applications/kitty.app|0|
34 300 938 877|727766845|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
636df9b3-4e98-4d8e-8207-ad94002e0f56
|727767550|kitty|/Applications/kitty.app|0|
429a7d3d-bb3b-4e4e-b951-ac1c0122e920|727767560|kitty|/Applications/kitty.app|0|
V1DWLrq!yWwao3B!|727767570|kitty|/Applications/kitty.app|0|
5 Gallagher place coffs|727825238|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
💸️|727826095|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
75 493 363 262|727852365|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
DOC-20240119-WA0002..pdf|727855167|Finder|/System/Library/CoreServices/Finder.app|0|
Screenshot_20240124-155951.jpg|727855566|Finder|/System/Library/CoreServices/Finder.app|0|
Scott Pilgrim vs The World The Game (eShop)
Scott Pilgrim vs The World The Game [0100394011C30800] [v65536].nsp|727939104|Finder|/System/Library/CoreServices/Finder.app|0|
Awoo Installer|727939997|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
tar: RAR solid archive support unavailable.|728120145|kitty|/Applications/kitty.app|0|
https://drive.google.com/file/d/1fBXTM07xbNodXZoK87pnvWhWetnr7fC-/view?usp=sharing|728125584|Finder|/System/Library/CoreServices/Finder.app|0|
Microsoft Basic Data ⁨BAJAJ_DUO⁩               256.4 GB   disk4s2

|728128110|kitty|/Applications/kitty.app|0|
Animal Crossing New Horizons [01006F8002326000][v0](eShop).nsp|728128121|Finder|/System/Library/CoreServices/Finder.app|0|
hekate_ctcaer_6.0.7.bin|728131545|Finder|/System/Library/CoreServices/Finder.app|0|
ModuleNotFoundError: No module named 'usb'
|728131705|kitty|/Applications/kitty.app|0|
pycurl~=7.43.0
|728132771|kitty|/Applications/kitty.app|0|
SUBSYSTEM=="usb", ATTRS{idVendor}=="16c0", ATTRS{idProduct}=="27e2", GROUP="plugdev"
|728132815|kitty|/Applications/kitty.app|0|
ModuleNotFoundError: No module named 'PyQt5'|728132945|kitty|/Applications/kitty.app|0|
 No module named 'PyQt5'
|728133125|kitty|/Applications/kitty.app|0|
/usr/local/opt/python@3.11/bin/python3.11 -m pip install --upgrade pip|728133163|kitty|/Applications/kitty.app|0|
requests~=2.25.0
tqdm~=4.53.0
colorama~=0.4.4
beautifulsoup4~=4.9.3
urllib3~=1.26.0
pyusb~=1.1.0
pyqt5~=5.15.1
google-api-python-client~=1.12.7
google-auth-oauthlib~=0.4.2
pycryptodome~=3.9.9
pycryptoplus~=0.0.1
pyopenssl~=19.1.0
Flask~=1.1.2
Pillow~=8.0.1
zstandard~=0.14.0
|728133310|kitty|/Applications/kitty.app|0|
pip install --install-option="--with-openssl" --install-option="--openssl-dir=/usr/local/opt/openssl" pycurl
|728133330|kitty|/Applications/kitty.app|0|
ImportError: No module named PyQt5.QtGui
|728133401|kitty|/Applications/kitty.app|0|
PYCURL_SSL_LIBRARY=openssl LDFLAGS="-L/usr/local/opt/openssl/lib" CPPFLAGS="-I/usr/local/opt/openssl/include" pip install --no-cache-dir pycurl
|728133420|kitty|/Applications/kitty.app|0|
pip install pyqt5|728133438|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
python3 -m pip uninstall PyQt5
python3 -m pip uninstall PyQt5-sip
python3 -m pip uninstall PyQtWebEngine

# reinstall
python3 -m pip install PyQt5
python3 -m pip install PyQt5-sip
python3 -m pip install PyQtWebEngine|728133459|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
oduleNotFoundError: No module named 'colorama'|728133524|kitty|/Applications/kitty.app|0|
ModuleNotFoundError: No module named 'Crypto'
|728133743|kitty|/Applications/kitty.app|0|
pip uninstall pycrypto
easy_install pycrypto|728133756|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
pip3 uninstall crypto 
pip3 uninstall pycrypto 
pip3 install pycryptodome|728133795|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
https://www.ziperto.com/adventure-time-pirates-of-the-enchiridion-switch/|728183014|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
## Trimester 1
    - [ ] [COMP3411](COMP3411/index.md)
	- [ ] [COMP2511](COMP2511/index.md)

## Trimester 2
	- [ ] [COMP3900](COMP3900/index.md)
    - [ ] [MATH2121](MATH2121/index.md)
    - [ ] [ARTS1631](ARTS1630/index.md)

## Trimester 3
	- [ ] [COMP4920](COMP4920/index.md)
    - [ ] [COMP4121](COMP4121/index.md)
    - [F] [MATH2521](MATH2521/index.md)
    - [ ] [MATH3411](math3411/index.md)
|728445991|kitty|/Applications/kitty.app|0|
    - [ ] [COMP3411](COMP3411/index.md)
	- [ ] [COMP2511](COMP2511/index.md)
|728445997|kitty|/Applications/kitty.app|0|
	- [ ] [COMP6080](COMP6080/index.md)
|728446004|kitty|/Applications/kitty.app|0|
COMP6080|728446005|kitty|/Applications/kitty.app|0|
ARTS1631|728446047|kitty|/Applications/kitty.app|0|
	- [ ] [COMP4920](COMP4920/index.md)
    - [ ] [COMP4121](COMP4121/index.md)
    - [F] [MATH2521](MATH2521/index.md)
    - [ ] [MATH3411](math3411/index.md)
|728446067|kitty|/Applications/kitty.app|0|
    - [ ] [MATH3801](MATH3801/index.md)
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	- [ ] [COMP3900](COMP3900/index.md)
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    - [F] [MATH2521](MATH2521/index.md)
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    - [ ] [MATH3411](math3411/index.md)
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    - [ ] [ARTS1631](ARTS1630/index.md)
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	- WAM: 80*
    - Total WAM: 69
	- Units: 12
	- Total Units: 108 + 24 = 136
        - need to double count MATH1A and MATH1B from minor.
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70*|728446266|kitty|/Applications/kitty.app|0|
 65|728446268|kitty|/Applications/kitty.app|0|
48|728446269|kitty|/Applications/kitty.app|0|
66 + 42 = 108|728446271|kitty|/Applications/kitty.app|0|
 ] [COMP2511](COMP2511/index.md)|728446274|kitty|/Applications/kitty.app|0|
    - [ ] [MATH2121](MATH2121/index.md)
    - [ ] [MATH3801](MATH3801/index.md)
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\colorlet{clarboxcolor}{yellow!20}
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\newmdenv[backgroundcolor=hintboxcolor]{hintbox}
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subsection|728448119|kitty|/Applications/kitty.app|0|
    \begin{}
|728448120|kitty|/Applications/kitty.app|0|
    This is saying that if you took 2 and cubed it, you would get 8. Or in our newfound index notation: $8^\frac{1}{3} = 2$
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\begin{examplebox}
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            = \fillinlinelength[\frac{1}{8}]$
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=
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\end{multicols}|728451942|kitty|/Applications/kitty.app|0|
        \Question[3]$\,$
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\begin{multicols}{3}|728452014|kitty|/Applications/kitty.app|0|
\begin{multicols}{2}|728452045|kitty|/Applications/kitty.app|0|
\faExclamationCircle|728452368|kitty|/Applications/kitty.app|0|
Image: 155x146 (357.9 KB)|728453012|Brave Browser Beta|/Applications/Brave Browser Beta.app|1|1c825dd87e40740c796803f3f6a0c85f3360a812.tiff
\subsection{Substitution into formulas}
\begin{questions}
    \Question[3] For each part, find the value of the subject when the other pronumerals have the value indicated.
    \begin{parts}
        \part \(A=\frac{1}{2}(a+b) h\), where \(a=4, b=6, h=10\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(t=a+(n-1) d\), where \(a=30, n=8, d=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(E=\frac{1}{2} m v^{2}\), where \(m=8, v=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[4] For each part, find the value of the subject when the other pronumerals have the value indicated. Calculate \(\mathbf{a}-\mathbf{c}\) correct to 3 decimal places and \(\mathbf{d}\) correct to 2 .
    \begin{parts}
        \part \(x=\sqrt{a b}\), where \(a=40, b=50\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(V=\pi r^{2} h\), where \(r=12, h=20\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(T=2 \pi \sqrt{\frac{\ell}{g}}\), where \(\ell=88.2, g=9.8\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(A=P(1+R)^{n}\), where \(P=10000, R=0.065, n=10\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[1] For the formula \(S=2(\ell w+\ell h+h w)\), find \(h\) if \(S=592, \ell=10\) and \(w=8\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \Question[1] For the formula \(s=u t+\frac{1}{2} a t^{2}\), find \(a\) if \(s=1000, u=20\) and \(t=5\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \Question[1] For the formula \(t=a+(n-1) d\), find \(n\) if \(t=58, d=3\) and \(a=7\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \Question[2] Given \(v^{2}=u^{2}+2 a x\) and \(v>0\), find the value of \(v\) (correct to 1 decimal place) when:
    \begin{parts}\begin{multicols}{2}
        \part \(u=0, a=5\) and \(x=10\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(u=2, a=9.8\) and \(x=22\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[2] Given \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\), find the value of:
    \begin{parts}\begin{multicols}{2}
        \part \(u\) when \(f=2\) and \(v=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(u\) when \(f=3\) and \(v=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \Question[6] The formula for finding the number of degrees Fahrenheit \((F)\) for a temperature given as a number of degrees Celsius \((C)\) is \(F=\frac{9}{5} C+32\).

    Fahrenheit temperatures are still used in the USA, but in Australia we commonly use Celsius.

    Calculate the Fahrenheit temperatures which people in the USA would recognise for:
    \begin{parts}
        \part the freezing point of water, \(0^{\circ} \mathrm{C}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the boiling point of water, \(100^{\circ} \mathrm{C}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part a nice summer temperature of \(25^{\circ} \mathrm{C}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \hspace{-1cm}\item[] Now calculate the Celsius temperatures which people in Australia would recognise for:
        \part \(50^{\circ} \mathrm{F}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(104^{\circ} \mathrm{F}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[4] Sam throws a stone down to the ground from the top of a cliff \(s\) metres high, with an initial speed of \(u \mathrm{~m} / \mathrm{s}\). It accelerates at \(a \mathrm{~m} / \mathrm{s}^{2}\). The stone hits the ground with a speed of \(v \mathrm{~m} / \mathrm{s}\) given by the formula \(v^{2}=u^{2}+2 a s\). Find the speed at which the stone hits the ground, correct to 2 decimal places, if:
    \begin{parts}
        \part \(u=0, a=9.8\) and \(s=50\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(u=5, a=9.8\) and \(s=35\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
\end{questions}

\subsection{Changing the subject of a formula}

\begin{questions}
    \Question[] Given the formula \(v=u+a t\) :
    \begin{parts}
        \Part[1] rearrange the formula to make \(u\) the subject
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Part[2] find the value of \(u\) when:
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \begin{subparts}
            \subpart \(v=20, a=2\) and \(t=5 \quad\) 
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \subpart \(v=40, a=-6\) and \(t=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{subparts}
        \Part[1] rearrange the formula to make \(a\) the subject
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Part[2] find the value of \(a\) when:
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \begin{subparts}
            \subpart \(v=20, u=15\) and \(t=2 \quad\) 
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \subpart \(v=-26.8, u=-14.4\) and \(t=2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \subpart \(v=\frac{1}{2}, u=\frac{2}{3}\) and \(t=\frac{5}{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{subparts}
        \Part[2] rearrange the formula to make \(t\) the subject and find \(t\) when \(v=6, u=7\) and \(a=-3\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[6] Rearrange each of these formulas to make the pronumeral in brackets the subject.
    \begin{parts}
        \part \(y=m x+c\hfill{}(x)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(C=2 \pi r\hfill{}(r)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(s=u t+\frac{1}{2} a t^{2}\hfill{}(a)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(V=\frac{1}{3} \pi r^{2} h\hfill{}(h)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(S=\frac{n}{2}(a+\ell)\hfill{}(n)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(E=m g h+\frac{1}{2} m v^{2}\hfill{}(h)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[4] The kinetic energy \(E\) joules of a moving object is given by \(E=\frac{1}{2} m v^{2}\), where \(m \mathrm{~kg}\) is the mass of the object and \(v \mathrm{~m} / \mathrm{s}\) is its speed.

    Rearrange the formula to make \(m\) the subject and use this to find the mass of the object when its energy and speed are, respectively:
    \begin{parts}
        \part 400 joules, \(10 \mathrm{~m} / \mathrm{s}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part 28 joules, \(4 \mathrm{~m} / \mathrm{s}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
\end{questions}


\subsection{Constructing Formulas}

\begin{questions}
    \Question[8] Find a formula for:
    \begin{parts}
        \part the number of cents \(z\) in \(x\) dollars and \(y\) cents
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the number of minutes \(x\) in \(y\) minutes and \(z\) seconds
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the number of hours \(x\) in \(y\) minutes and \(z\) seconds
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the cost \(\$ m\) of 1 book if 20 books cost \(\$ c\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the cost \(\$ n\) of 1 suit if 5 suits cost \(\$ m\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the cost \(\$ m\) of 1 tyre if \(x\) tyres cost \(\$ y\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the cost \(\$ p\) of \(n\) suits if 4 suits cost \(\$ k\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part the cost \(\$ q\) of \(x\) cars if 8 cars cost \(\$ b\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[7] In each part, find a formula from the information given.
    \begin{parts}
        \part A hire car firm charges \(\$ 20\) per day plus 40 cents per \(\mathrm{km}\). What is the total cost \(\$ C\) for a day in which \(x \mathrm{~km}\) was travelled?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part If there are 50 litres of petrol in the tank of a car and petrol is used at the rate of 4 litres per day, what is the number of litres \(y\) that remains after \(x\) days?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part Cooking instructions for a forequarter of lamb are as follows: preheat oven to \(220^{\circ} \mathrm{C}\) and cook for \(45 \mathrm{~min}\) per kg plus an additional \(20 \mathrm{~min}\). What is the formula relating the cooking time \(T\) minutes and weight \(w \mathrm{~kg}\) ?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part In a sequence of numbers the first number is 2 , the second number is 4 , the third is 8 , the fourth is 16, etc. Assuming the doubling pattern continues, what is the formula you would use to calculate \(t\), the \(n\)th number?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part A piece of wire of length \(x \mathrm{~cm}\) is bent into a circle of area \(A \mathrm{~cm}^{2}\). What is the formula relating \(A\) and \(x\) ?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[10] A cyclic quadrilateral has all its vertices on a circle. Its area \(A\) is given by Brahmagupta's formula \[ A^{2}=(s-a)(s-b)(s-c)(s-d) \] where \(a, b, c\) and \(d\) are the side lengths of the quadrilateral and \(s=\frac{a+b+c+d}{2}\) is the 'semi-perimeter'. Find the exact area of a cyclic quadrilateral with side lengths:
    \begin{parts}
        \part \(4,5,6,7\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(7,4,4,3\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(8,9,10,13\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(39,52,25,60\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(51,40,68,75\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
\end{questions}
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\section{Fractional Indices}
    \setcounter{secmarks}{0}
    \input{2-ind-frac}
    \setcounter{sec2marks}{\thesecmarks}
\section{Scientific Notation}
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    \input{3-sci-not}
    \setcounter{sec3marks}{\thesecmarks}
\section{Significant Figures}
    \setcounter{secmarks}{0}
    \input{4-sig-fig}
    \setcounter{sec4marks}{\thesecmarks}
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    \setcounter{secmarks}{0}
    \input{2-ind-frac}
    \setcounter{sec2marks}{\thesecmarks}
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    \setcounter{secmarks}{0}
    \input{3-sci-not}
    \setcounter{sec3marks}{\thesecmarks}
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    \setcounter{secmarks}{0}
    \input{4-sig-fig}
    \setcounter{sec4marks}{\thesecmarks}
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\faUniversity
testing: \faExclamationCircle
\faGavel
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        \begin{parts}\begin{multicols}{3}
        \end{multicols}\end{parts}
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        \begin{parts}\begin{multicols}{2}
        \end{multicols}\end{parts}
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a \(\sqrt[3]{8}\)
b \(\sqrt[5]{32}\)
c \(\sqrt[3]{216}\)
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1 Evaluate:
d \(\sqrt[4]{81}\)
e \(\sqrt[3]{64}\)
f \(\sqrt[5]{2^{10}}\)
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a \(\sqrt{14}\)
b \(\sqrt[4]{64}\)
c \(\sqrt[5]{7}\)
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2 Write using fractional indices. Evaluate, correct to 4 decimal places.
d \(\sqrt[7]{11}\)
e \(\sqrt[3]{2^{7}}\)
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2 Write using fractional indices. Evaluate, correct to 4 decimal places.
|728454028|kitty|/Applications/kitty.app|0|
2|728454035|kitty|/Applications/kitty.app|0|
a \(4^{\frac{1}{2}}\)
b \(27^{\frac{1}{3}}\)
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a \(4^{\frac{1}{2}}\)
b \(27^{\frac{1}{3}}\)
e \(64^{\frac{1}{2}}\)
f \(25^{\frac{1}{2}}\)
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a \(4^{\frac{1}{2}}\)
b \(27^{\frac{1}{3}}\)
e \(64^{\frac{1}{2}}\)
f \(25^{\frac{1}{2}}\)
i \(32^{\frac{1}{5}}\)
j \(625^{\frac{1}{4}}\)
|728454051|kitty|/Applications/kitty.app|0|
Example 16 3 Evaluate:
c \(243^{\frac{1}{5}}\)
d \(81^{\frac{1}{4}}\)
g \(125^{\frac{1}{3}}\)
h \(64^{\frac{1}{3}}\)
k \(216^{\frac{1}{3}}\)
l \(49^{\frac{1}{2}}\)
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a \(4^{\frac{5}{2}}\)
b \(25^{\frac{3}{2}}\)
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a \(4^{\frac{5}{2}}\)
b \(25^{\frac{3}{2}}\)
e \(32^{\frac{2}{5}}\)
f \(81^{\frac{3}{4}}\)
|728454081|kitty|/Applications/kitty.app|0|
a \(4^{\frac{5}{2}}\)
b \(25^{\frac{3}{2}}\)
e \(32^{\frac{2}{5}}\)
f \(81^{\frac{3}{4}}\)
i \((\sqrt[4]{16})^{3}\)
j \((\sqrt[3]{27})^{2}\)
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4 Evaluate:
c \(125^{\frac{2}{3}}\)
d \(64 \frac{5}{6}\)
g \(216^{\frac{2}{3}}\)
h \(243^{\overline{5}}\)
k \(\sqrt[5]{32^{4}}\)
l \(\sqrt[3]{2^{6}}\)
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a \(\left(a^{\frac{1}{2}}\right)^{2}\)
b \(\left(b^{\frac{1}{3}}\right)^{6}\)
|728454124|kitty|/Applications/kitty.app|0|
a \(\left(a^{\frac{1}{2}}\right)^{2}\)
b \(\left(b^{\frac{1}{3}}\right)^{6}\)
e \(x^{\frac{1}{2}} \times x^{\frac{3}{2}}\)

f \(y^{\frac{1}{3}} \times y^{\frac{2}{3}}\)
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a \(\left(a^{\frac{1}{2}}\right)^{2}\)
b \(\left(b^{\frac{1}{3}}\right)^{6}\)
e \(x^{\frac{1}{2}} \times x^{\frac{3}{2}}\)

f \(y^{\frac{1}{3}} \times y^{\frac{2}{3}}\)
i \(x^{\frac{3}{2}} \div x^{\frac{1}{2}}\)

j \(y^{\frac{2}{3}} \div y^{\frac{1}{3}}\)
|728454129|kitty|/Applications/kitty.app|0|
\(\mathbf{m}\left(4 m^{6}\right)^{\frac{1}{2}}\)

n \(\left(27 n^{12}\right)^{\frac{1}{3}}\)
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t|728454174|kitty|/Applications/kitty.app|0|
h|728454174|kitty|/Applications/kitty.app|0|
b|728454175|kitty|/Applications/kitty.app|0|
m|728454176|kitty|/Applications/kitty.app|0|
5 Simplify:
c \(\left(c^{12}\right)^{\frac{1}{4}}\)
d \(\left(c^{10}\right)^{\frac{1}{5}}\)


g \(p^{\frac{3}{4}} \times p^{\frac{2}{5}}\)

h \(q^{\frac{3}{2}} \times q^{\frac{2}{3}}\)


k \(p^{\frac{3}{4}} \div p^{\frac{2}{5}}\)

l \(q^{\frac{3}{2}} \div q^{\frac{2}{3}}\)


o \(\left(2 x^{\frac{2}{3}}\right)^{3}\)

p \(\left(3 y^{\frac{1}{2}}\right)^{4}\)
|728454185|kitty|/Applications/kitty.app|0|
a \(4^{-\frac{1}{2}}\)
b \(25^{-\frac{1}{2}}\)
|728454244|kitty|/Applications/kitty.app|0|
a \(4^{-\frac{1}{2}}\)
b \(25^{-\frac{1}{2}}\)
f \(\left(\frac{1}{81}\right)^{-\frac{1}{4}}\)
g \(81^{-\frac{1}{4}}\)
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c \(\left(\frac{8}{125}\right)^{-\frac{1}{3}}\)
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h \(\left(\frac{1}{25}\right)^{-\frac{1}{2}}\)
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6 Evaluate:
d \(\left(\frac{64}{27}\right)^{-\frac{1}{3}}\)
e \(32^{-\frac{2}{5}}\)
i \(\left(\frac{16}{81}\right)^{-\frac{1}{4}}\)
j \(\left(\frac{32}{243}\right)^{-\frac{1}{5}}\)
|728454273|kitty|/Applications/kitty.app|0|
c \(\left(2 x^{\frac{2}{3}}\right)^{-3}\)
|728454291|kitty|/Applications/kitty.app|0|
h \(q^{\frac{3}{2}} \times q^{-\frac{2}{3}}\)
|728454292|kitty|/Applications/kitty.app|0|
a \(\left(a^{\frac{1}{2}}\right)^{-2}\)
b \(\left(b^{-\frac{2}{3}}\right)^{6}\)
|728454297|kitty|/Applications/kitty.app|0|
a \(\left(a^{\frac{1}{2}}\right)^{-2}\)
b \(\left(b^{-\frac{2}{3}}\right)^{6}\)
f \(y^{\frac{1}{3}} \times y^{-\frac{2}{3}}\)
g \(p^{\frac{3}{4}} \times p^{-\frac{2}{5}}\)
|728454299|kitty|/Applications/kitty.app|0|
a \(\left(a^{\frac{1}{2}}\right)^{-2}\)
b \(\left(b^{-\frac{2}{3}}\right)^{6}\)
f \(y^{\frac{1}{3}} \times y^{-\frac{2}{3}}\)
g \(p^{\frac{3}{4}} \times p^{-\frac{2}{5}}\)
k \(p^{\frac{3}{4}} \div p^{-\frac{2}{5}}\)
l \(q^{\frac{3}{2}} \div q^{-\frac{2}{3}}\)
|728454455|kitty|/Applications/kitty.app|0|
7 Simplify, expressing the answer with positive indices.
d \(\left(3 y^{\frac{1}{2}}\right)^{-4}\)
e \(x^{\frac{1}{2}} \times x^{-\frac{3}{2}}\)
i \(x^{\frac{3}{2}} \div x^{-\frac{1}{2}}\)
j \(y^{\frac{2}{3}} \div y^{-\frac{1}{3}}\)
\(\mathbf{m}\left(4 m^{-6}\right)^{\frac{1}{2}}\)
n \(\left(27 n^{-12}\right)^{\frac{1}{3}}\)
o \(\left(2 x^{-\frac{2}{5}}\right)^{5}\)
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7 Simplify, expressing the answer with positive indices.
|728454478|kitty|/Applications/kitty.app|0|
7|728454485|kitty|/Applications/kitty.app|0|
\begin{centering}
|728454535|kitty|/Applications/kitty.app|0|
\end{centering}
|728454537|kitty|/Applications/kitty.app|0|
\begin{mu}
|728454666|kitty|/Applications/kitty.app|0|
\
r|728454701|kitty|/Applications/kitty.app|0|
p|728454702|kitty|/Applications/kitty.app|0|
\u
|728454702|kitty|/Applications/kitty.app|0|
\begin{defbox}
    \textbf{Tedious}: 
    \begin{solutionordottedlines}[1cm]
    \end{solutionordottedlines}
\end{defbox}
|728455428|kitty|/Applications/kitty.app|0|
Tedious|728455438|kitty|/Applications/kitty.app|0|
a 610
b 21000
e 0.0067
f 0.00002
c 46000000
d 81
g 0.07
h 8.17
|728455558|kitty|/Applications/kitty.app|0|
a \(2.1 \times 10^{3}\)
b \(6.3 \times 10^{5}\)
c \(5 \times 10^{-4}\)
d \(8.12 \times 10^{-2}\)
|728455949|kitty|/Applications/kitty.app|0|
a|728455951|kitty|/Applications/kitty.app|0|
Simplify and write in scientific notation.
a \(\left(3 \times 10^{4}\right) \times\left(2 \times 10^{6}\right)\)
b \(\left(9 \times 10^{7}\right) \div\left(3 \times 10^{4}\right)\)
c \(\left(4.1 \times 10^{4}\right)^{2}\)
d \(\left(2 \times 10^{5}\right)^{-2}\)

\section*{Solution}
a \(\left(3 \times 10^{4}\right) \times\left(2 \times 10^{6}\right)=3 \times 10^{4} \times 2 \times 10^{6}\)

\[
\begin{aligned}
& =3 \times 2 \times 10^{4} \times 10^{6} \\
& =6 \times 10^{10}
\end{aligned}
\]

b \(\left(9 \times 10^{7}\right) \div\left(3 \times 10^{4}\right)=\frac{9 \times 10^{7}}{3 \times 10^{4}}\)

\[
\begin{aligned}
& =\frac{9}{3} \times \frac{10^{7}}{10^{4}} \\
& =3 \times 10^{3}
\end{aligned}
\]

c \(\left(4.1 \times 10^{4}\right)^{2}=4.1^{2} \times\left(10^{4}\right)^{2}\)

d \(\left(2 \times 10^{5}\right)^{-2}=2^{-2} \times\left(10^{5}\right)^{-2}\)

\[
\begin{aligned}
& =16.81 \times 10^{8} \\
& =1.681 \times 10^{9}
\end{aligned}
\]

\[
\begin{aligned}
& =\frac{1}{2^{2}} \times 10^{-10} \\
& =0.25 \times 10^{-10} \\
& =2.5 \times 10^{-11}
\end{aligned}
\]
|728456114|kitty|/Applications/kitty.app|0|
a \(\left(3 \times 10^{4}\right) \times\left(2 \times 10^{6}\right)\)
b \(\left(9 \times 10^{7}\right) \div\left(3 \times 10^{4}\right)\)
c \(\left(4.1 \times 10^{4}\right)^{2}\)
d \(\left(2 \times 10^{5}\right)^{-2}\)
|728456120|kitty|/Applications/kitty.app|0|
Simplify and write in scientific notation.
|728456132|kitty|/Applications/kitty.app|0|
            \begin{solutionordottedlines}
            \end{solutionordottedlines}
|728456140|kitty|/Applications/kitty.app|0|
a \(\left(3 \times 10^{4}\right) \times\left(2 \times 10^{6}\right)=3 \times 10^{4} \times 2 \times 10^{6}\)

\[
\begin{aligned}
& =3 \times 2 \times 10^{4} \times 10^{6} \\
& =6 \times 10^{10}
\end{aligned}
\]
|728456155|kitty|/Applications/kitty.app|0|
b \(\left(9 \times 10^{7}\right) \div\left(3 \times 10^{4}\right)=\frac{9 \times 10^{7}}{3 \times 10^{4}}\)

\[
\begin{aligned}
& =\frac{9}{3} \times \frac{10^{7}}{10^{4}} \\
& =3 \times 10^{3}
\end{aligned}
\]
|728456172|kitty|/Applications/kitty.app|0|
\[
\begin{aligned}
& =16.81 \times 10^{8} \\
& =1.681 \times 10^{9}
\end{aligned}
\]
|728456214|kitty|/Applications/kitty.app|0|
d \(\left(2 \times 10^{5}\right)^{-2}=2^{-2} \times\left(10^{5}\right)^{-2}\)

\[
\begin{aligned}
& =\frac{1}{2^{2}} \times 10^{-10} \\
& =0.25 \times 10^{-10} \\
& =2.5 \times 10^{-11}
\end{aligned}
\]
|728456221|kitty|/Applications/kitty.app|0|
c \(\left(4.1 \times 10^{4}\right)^{2}=4.1^{2} \times\left(10^{4}\right)^{2}\)

\[
\begin{aligned}
& =16.81 \times 10^{8} \\
& =1.681 \times 10^{9}
\end{aligned}
\]
|728456229|kitty|/Applications/kitty.app|0|
\section*{Solution}
|728456264|kitty|/Applications/kitty.app|0|
[2in]|728456290|kitty|/Applications/kitty.app|0|
big|728456392|kitty|/Applications/kitty.app|0|
}|728456457|kitty|/Applications/kitty.app|0|
defboxdefbox|728456615|kitty|/Applications/kitty.app|0|
defbox|728456622|kitty|/Applications/kitty.app|0|
        \Question[1]
|728456856|kitty|/Applications/kitty.app|0|
        \Question[2]
        \Question[8]
|728456896|kitty|/Applications/kitty.app|0|
a \(\frac{1}{10}\)
b \(\frac{1}{100}\)
c \(\frac{1}{1000}\)
d 1 trillionth
e \(\frac{1}{100000}\)
f 1 millionth
|728456923|kitty|/Applications/kitty.app|0|
c 26000
|728456976|kitty|/Applications/kitty.app|0|
c 26000
f 4000000000000
|728456978|kitty|/Applications/kitty.app|0|
c 26000
f 4000000000000
i 0.00072
|728456980|kitty|/Applications/kitty.app|0|
c 26000
f 4000000000000
i 0.00072
l 0.000000206
|728456982|kitty|/Applications/kitty.app|0|
3 Write in scientific notation.
c 26000
f 4000000000000
i 0.00072
l 0.000000206
|728456985|kitty|/Applications/kitty.app|0|
3 Write in scientific notation.
|728456991|kitty|/Applications/kitty.app|0|
        \end{parts}\end{multicols}
|728457001|kitty|/Applications/kitty.app|0|
        \begin{multicols}{2}\begin{parts}
|728457003|kitty|/Applications/kitty.app|0|
a 510
b 5300
d 796000000
e 576000000000
g 0.008
h 0.06
j 0.000041
k 0.000000006
|728457017|kitty|/Applications/kitty.app|0|
c \(8.6 \times 10^{2}\)
|728457044|kitty|/Applications/kitty.app|0|
c \(8.6 \times 10^{2}\)
f \(7.2 \times 10^{1}\)
|728457046|kitty|/Applications/kitty.app|0|
c \(8.6 \times 10^{2}\)
f \(7.2 \times 10^{1}\)
i \(8.72 \times 10^{-4}\)
|728457047|kitty|/Applications/kitty.app|0|
4 Write in decimal form:
c \(8.6 \times 10^{2}\)
f \(7.2 \times 10^{1}\)
i \(8.72 \times 10^{-4}\)
l \(2.6 \times 10^{-7}\)
|728457053|kitty|/Applications/kitty.app|0|
c \(8.6 \times 10^{2}\)
f \(7.2 \times 10^{1}\)
i \(8.72 \times 10^{-4}\)
l \(2.6 \times 10^{-7}\)
|728457059|kitty|/Applications/kitty.app|0|
4 Write in decimal form:
a \(3.24 \times 10^{4}\)
b \(7.2 \times 10^{3}\)
d \(2.7 \times 10^{6}\)
e \(5.1 \times 10^{0}\)
g \(5.6 \times 10^{-2}\)
h \(1.7 \times 10^{-3}\)
j \(2.01 \times 10^{-3}\)
k \(9.7 \times 10^{-1}\)
|728457061|kitty|/Applications/kitty.app|0|
4 Write in decimal form:|728457067|kitty|/Applications/kitty.app|0|
6 Light travels approximately \(299000 \mathrm{~km}\) in a second. Express this in scientific notation.

7 The mass of a copper sample is \(0.0089 \mathrm{~kg}\). Express this in scientific notation.

8 The distance between interconnecting lines on a silicon chip for a computer is approximately \(0.00000004 \mathrm{~m}\). Express this in scientific notation.
|728457130|kitty|/Applications/kitty.app|0|
8 The distance between interconnecting lines on a silicon chip for a computer is approximately \(0.00000004 \mathrm{~m}\). Express this in scientific notation.
|728457140|kitty|/Applications/kitty.app|0|
7 The mass of a copper sample is \(0.0089 \mathrm{~kg}\). Express this in scientific notation.
|728457142|kitty|/Applications/kitty.app|0|
a \(\left(4 \times 10^{5}\right) \times\left(2 \times 10^{6}\right)\)
b \(\left(2.1 \times 10^{6}\right) \times\left(3 \times 10^{7}\right)\)
c \(\left(4 \times 10^{2}\right) \times\left(5 \times 10^{-7}\right)\)
d \(\left(3 \times 10^{6}\right) \times\left(8 \times 10^{-3}\right)\)
e \(\left(5 \times 10^{4}\right) \div\left(2 \times 10^{3}\right)\)
f \(\left(8 \times 10^{9}\right) \div\left(4 \times 10^{3}\right)\)
g \(\left(6 \times 10^{-4}\right) \div\left(8 \times 10^{-5}\right)\)
h \(\left(1.2 \times 10^{6}\right) \div\left(4 \times 10^{7}\right)\)
|728457175|kitty|/Applications/kitty.app|0|
i \(\left(2.1 \times 10^{2}\right)^{4}\)
j \(\left(3 \times 10^{-2}\right)^{3}\)
k \(\frac{\left(2 \times 10^{5}\right) \times\left(4 \times 10^{4}\right)}{1.6 \times 10^{3}}\)
l \(\frac{\left(8 \times 10^{6}\right) \times\left(4 \times 10^{3}\right)}{5 \times 10^{7}}\)
|728457194|kitty|/Applications/kitty.app|0|
9 Simplify, expressing the answer in scientific notation.
m \(\left(4 \times 10^{-2}\right)^{2} \times\left(5 \times 10^{7}\right)\)
n \(\left(6 \times 10^{-3}\right) \times\left(4 \times 10^{7}\right)^{2}\)
o \(\frac{\left(4 \times 10^{5}\right)^{3}}{\left(8 \times 10^{4}\right)^{2}}\)
p \(\frac{\left(2 \times 10^{-1}\right)^{5}}{\left(4 \times 10^{-2}\right)^{3}}\)
|728457196|kitty|/Applications/kitty.app|0|
9 Simplify, expressing the answer in scientific notation.
|728457201|kitty|/Applications/kitty.app|0|
10 If light travels at \(3 \times 10^{5} \mathrm{~km} / \mathrm{s}\) and our galaxy is approximately 80000 light years across, how many kilometres is it across? (A light year is the distance light travels in a year.)

11 The mass of a hydrogen atom is approximately \(1.674 \times 10^{-27} \mathrm{~kg}\) and the mass of an electron is approximately \(9.1 \times 10^{-31} \mathrm{~kg}\). How many electrons, correct to the nearest whole number, will have the same mass as a single hydrogen atom?
|728457272|kitty|/Applications/kitty.app|0|
12 If the average distance from the Earth to the Sun is \(1.4951 \times 10^{8} \mathrm{~km}\) and light travels at \(3 \times 10^{5} \mathrm{~km} / \mathrm{s}\), how long does it take light to travel from the Sun to the Earth?

13 The furthest galaxy detected by optical telescopes is approximately \(4.6 \times 10^{9}\) light years from us. How far is this in kilometres? (Light travels at \(3 \times 10^{5} \mathrm{~km} / \mathrm{s}\).)
|728457278|kitty|/Applications/kitty.app|0|

13 The furthest galaxy detected by optical telescopes is approximately \(4.6 \times 10^{9}\) light years from us. How far is this in kilometres? (Light travels at \(3 \times 10^{5} \mathrm{~km} / \mathrm{s}\).)
|728457283|kitty|/Applications/kitty.app|0|
        \begin{multicols}{2}\begin{parts}
        \end{parts}\end{multicols}
|728457293|kitty|/Applications/kitty.app|0|
14 In a lottery there are \(\frac{45 \times 44 \times 43 \times 42 \times 41 \times 40}{720}\) different possible outcomes. If I mark each outcome on an entry form one at a time, and it takes me an average of 1 minute to mark each outcome, how long will it take me to cover all different possible outcomes?
|728457298|kitty|/Applications/kitty.app|0|
 = \fillin[]|728457497|kitty|/Applications/kitty.app|0|
For example, \(0.00034061=3.4061 \times 10^{-4}\)

\[
\begin{aligned}
& \approx 3 \times 10^{-4} \quad(\text { correct to } 1 \text { significant figure }) \\
& \approx 3.4 \times 10^{-4} \quad(\text { correct to } 2 \text { significant figures }) \\
& \approx 3.41 \times 10^{-4} \quad(\text { correct to } 3 \text { significant figures }) \\
& \approx 3.406 \times 10^{-4} \quad(\text { correct to } 4 \text { significant figures })
\end{aligned}
\]
|728459702|kitty|/Applications/kitty.app|0|
Write in scientific notation and then round correct to 3 significant figures.
|728459755|kitty|/Applications/kitty.app|0|
        \
|728459769|kitty|/Applications/kitty.app|0|
        \begin{multicols}{2}
|728459769|kitty|/Applications/kitty.app|0|
Write in scientific notation and then round correct to 2 significant figures.
a 276000000
b 0.000000654
|728459840|kitty|/Applications/kitty.app|0|
Write in scientific notation and then round correct to 2 significant figures.|728459853|kitty|/Applications/kitty.app|0|
a 2.7043
b 634.96
c 8764.37
d 256412
e 0.003612
f 0.024186
|728460060|kitty|/Applications/kitty.app|0|
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
 & 4 sig. figs & 3 sig. figs & 2 sig. figs & 1 sig. fig. \\
\hline
274.62 &  &  &  &  \\
\hline
0.041236 &  &  &  &  \\
\hline
1704.28 &  &  &  &  \\
\hline
\(1.9925 \times 10^{27}\) &  &  &  &  \\
\hline
\end{tabular}
\end{center}
|728460135|kitty|/Applications/kitty.app|0|
2 Write in scientific notation, correct to 2 significant figures.
a 368.2
b 278000
c 0.004321
d 0.000021906
|728460215|kitty|/Applications/kitty.app|0|
b \(6.24 \div 0.026\)
|728460228|kitty|/Applications/kitty.app|0|
d \(\left(5.43 \times 10^{-6}\right) \div\left(6.24 \times 10^{-4}\right)\)
|728460229|kitty|/Applications/kitty.app|0|
f \(\frac{17.364 \times 24.32 \times 5.4^{2}}{3.6 \times 7.31^{2}}\)
|728460229|kitty|/Applications/kitty.app|0|
h \(\frac{6.283 \times 10^{8} \times 5.24 \times 10^{6}}{\left(4.37 \times 10^{7}\right)^{2}}\)
|728460230|kitty|/Applications/kitty.app|0|
4 Use a calculator to evaluate the following, giving the answer in scientific notation correct to 3 significant figures.
a \(3.24 \times 0.067\)
c \(4.736 \times 10^{13} \times 2.34 \times 10^{-6}\)
e \(0.0276^{2} \times \sqrt{0.723}\)
g \(\frac{6.54\left(5.26^{2}+3.24\right)}{5.4+\sqrt{6.34}}\)
|728460232|kitty|/Applications/kitty.app|0|
5 Use a calculator to evaluate, giving the answer in scientific notation correct to 4 significant figures.
a \(1.234 \times 0.1988\)
b \(1.234 \div 0.1988\)
c \(1.9346^{3}\)
d \(\left(7.919 \times 10^{21}\right)^{2}\)
|728460239|kitty|/Applications/kitty.app|0|
 1 Evaluate:|728460364|kitty|/Applications/kitty.app|0|
 2 Write using fractional indices. Evaluate, correct to 4 decimal places.|728460395|kitty|/Applications/kitty.app|0|
k|728461018|kitty|/Applications/kitty.app|0|
g|728461019|kitty|/Applications/kitty.app|0|
        \part \mathbf{m}\left(4 m^{-6}\right)^{\frac{1}{2}}\)
|728461075|kitty|/Applications/kitty.app|0|
{|728461162|kitty|/Applications/kitty.app|0|
this|728538829|kitty|/Applications/kitty.app|0|

This topic test following quiz has deliberately interleaved the subtopics together to force your brain to change gears between the different types of problems: thus working harder and enabling you to learn more. Goodluck!
|728538930|kitty|/Applications/kitty.app|0|
\fbox{%
\makebox[\textwidth]{
\randomword{formula}%
\hspace{\fill}
\randomword{subject}%
\hspace{\fill}
\randomword{pronumeral}%
\hspace{\fill}
\randomword{substitution}%
\hspace{\fill}
\randomword{expression}%
\hspace{\fill}
\randomword{construction}%
\hspace{\fill}
\randomword{equation}%
\hspace{\fill}
\randomword{rearrange}%
\hspace{\fill}
}
}
|728538943|kitty|/Applications/kitty.app|0|
\fbox{%
\makebox[\textwidth]{
\randomword{index}%
\hspace{\fill}
\randomword{indices}%
\hspace{\fill}
\randomword{power}%
\hspace{\fill}
\randomword{root}%
\hspace{\fill}
\randomword{exponent}%
\hspace{\fill}
\randomword{reciprocal}%
\hspace{\fill}
\randomword{sig fig}%
\hspace{\fill}
\randomword{base}%
\hspace{\fill}
\randomword{product}%
\hspace{\fill}
}
}
|728538964|kitty|/Applications/kitty.app|0|
Fractional Indices|728538979|kitty|/Applications/kitty.app|0|
\section{Scientific Notation}
    \setcounter{secmarks}{0}
    \input{3-sci-not}
    \setcounter{sec3marks}{\thesecmarks}
\section{Significant Figures}
    \setcounter{secmarks}{0}
    \input{4-sig-fig}
    \setcounter{sec4marks}{\thesecmarks}
|728539007|kitty|/Applications/kitty.app|0|
\title{\textsc{Year 9 Mathematics\\ Topic Test 3\\Linear Equations \& Inequalities}}
|728539113|kitty|/Applications/kitty.app|0|
Linear|728539120|kitty|/Applications/kitty.app|0|
\title{\textsc{Year 9 Mathematics\\ Topic 7b: Index Laws}}
|728539131|kitty|/Applications/kitty.app|0|
f|728540526|kitty|/Applications/kitty.app|0|
hspace|728540603|kitty|/Applications/kitty.app|0|
l|728540619|kitty|/Applications/kitty.app|0|
        \Questio
|728540686|kitty|/Applications/kitty.app|0|
.|728540776|kitty|/Applications/kitty.app|0|
 \small{\textsc{each mark is doubled}}()}|728540779|kitty|/Applications/kitty.app|0|




\section*{Challenge exercise}
|728540785|kitty|/Applications/kitty.app|0|
\includegraphics[width=\textwidth]{2023_12_09_5c1d855de2ec0db78c30g-19}
|728540804|kitty|/Applications/kitty.app|0|
\end{testbox}
|728540908|kitty|/Applications/kitty.app|0|
    \end{questions
|728540938|kitty|/Applications/kitty.app|0|




--- index

|728540942|kitty|/Applications/kitty.app|0|
a \(120^{2}\)
|728541771|kitty|/Applications/kitty.app|0|
j \(\left(b^{6}\right)^{5}\)
|728541782|kitty|/Applications/kitty.app|0|
        \begin{}
|728541919|kitty|/Applications/kitty.app|0|
        \begin{parts}
|728541987|kitty|/Applications/kitty.app|0|
blue|728542164|kitty|/Applications/kitty.app|0|
testbox|728542170|kitty|/Applications/kitty.app|0|
testbproblemboxox|728542180|kitty|/Applications/kitty.app|0|
\|728542218|kitty|/Applications/kitty.app|0|
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
|728542306|kitty|/Applications/kitty.app|0|
January 21, 2024|728983001|kitty|/Applications/kitty.app|0|
Formulas|728983030|kitty|/Applications/kitty.app|0|
 \subsection{Fractional Indices}                                                                                
 \begin{questions}                                                                                              
     \Question[3] Evaluate:                                                                                     
     \begin{multicols}{3}                                                                                       
         \begin{parts}                                                                                          
             \part \(\sqrt[4]{81}\)                                                                             
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( \sqrt[4]{81} = 3 \)                                                                         
             \end{solutionordottedlines}                                                                        
             \part \(\sqrt[3]{64}\)                                                                             
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( \sqrt[3]{64} = 4 \)                                                                         
             \end{solutionordottedlines}                                                                        
             \part \(\sqrt[5]{2^{10}}\)                                                                         
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( \sqrt[5]{2^{10}} = 2^{2} = 4 \)                                                             
             \end{solutionordottedlines}                                                                        
         \end{parts}                                                                                            
     \end{multicols}                                                                                            
     \Question[2] Write using fractional indices. Evaluate, correct to 4 decimal places.                        
     \begin{multicols}{2}\begin{parts}                                                                          
         \part \(\sqrt[7]{11}\)                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( 11^{\frac{1}{7}} \approx 1.4407 \)                                                          
             \end{solutionordottedlines}                                                                        
         \part \(\sqrt[3]{2^{7}}\)                                                                              
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( (2^{7})^{\frac{1}{3}} = 2^{\frac{7}{3}} \approx 5.0397 \)                                   
             \end{solutionordottedlines}                                                                        
     \end{parts}\end{multicols}                                                                                 
     \Question[6] Evaluate:                                                                                     
     \begin{multicols}{2}\begin{parts}                                                                          
         \part \(243^{\frac{1}{5}}\)                                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( 243^{\frac{1}{5}} = 3 \)                                                                    
             \end{solutionordottedlines}                                                                        
         \part \(81^{\frac{1}{4}}\)                                                                             
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( 81^{\frac{1}{4}} = 3 \)                                                                     
             \end{solutionordottedlines}                                                                        
         \part \(125^{\frac{1}{3}}\)                                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( 125^{\frac{1}{3}} = 5 \)                                                                    
             \end{solutionordottedlines}                                                                        
         \part \(64^{\frac{1}{3}}\)                                                                             
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( 64^{\frac{1}{3}} = 4 \)                                                                     
             \end{solutionordottedlines}                                                                        
         \part \(216^{\frac{1}{3}}\)                                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( 216^{\frac{1}{3}} = 6 \)                                                                    
             \end{solutionordottedlines}                                                                        
         \part \(49^{\frac{1}{2}}\)                                                                             
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( 49^{\frac{1}{2}} = 7 \)                                                                     
             \end{solutionordottedlines}                                                                        
     \end{parts}\end{multicols}                                                                                 
     \Question[6] Evaluate:                                                                                     
     \begin{multicols}{2}\begin{parts}                                                                          
         \part \(125^{\frac{2}{3}}\)                                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( 125^{\frac{2}{3}} = 25 \)                                                                   
             \end{solutionordottedlines}                                                                        
         \part \(64 \frac{5}{6}\)                                                                               
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( 64^{\frac{5}{6}} \approx 32 \)                                                              
             \end{solutionordottedlines}                                                                        
         \part \(216^{\frac{2}{3}}\)                                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( 216^{\frac{2}{3}} = 36 \)                                                                   
             \end{solutionordottedlines}                                                                        
         \part \(243^{\overline{5}}\)                                                                           
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( 243^{-\frac{1}{5}} \approx 0.4822 \)                                                        
             \end{solutionordottedlines}                                                                        
         \part \(\sqrt[5]{32^{4}}\)                                                                             
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( \sqrt[5]{32^{4}} = 32^{\frac{4}{5}} = 16 \)                                                 
             \end{solutionordottedlines}                                                                        
         \part \(\sqrt[3]{2^{6}}\)                                                                              
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( \sqrt[3]{2^{6}} = 2^{2} = 4 \)                                                              
             \end{solutionordottedlines}                                                                        
     \end{parts}\end{multicols}                                                                                 
     \Question[8] Simplify:                                                                                     
     \begin{multicols}{2}\begin{parts}                                                                          
         \part \(\left(c^{12}\right)^{\frac{1}{4}}\)                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( \left(c^{12}\right)^{\frac{1}{4}} = c^{3} \)                                                
             \end{solutionordottedlines}                                                                        
         \part \(\left(c^{10}\right)^{\frac{1}{5}}\)                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( \left(c^{10}\right)^{\frac{1}{5}} = c^{2} \)                                                
             \end{solutionordottedlines}                                                                        
         \part \(p^{\frac{3}{4}} \times p^{\frac{2}{5}}\)                                                       
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( p^{\frac{3}{4}} \times p^{\frac{2}{5}} = p^{\frac{3}{4}+\frac{2}{5}} =                      
 p^{\frac{15}{20}+\frac{8}{20}} = p^{\frac{23}{20}} \)                                                          
             \end{solutionordottedlines}                                                                        
         \part \(q^{\frac{3}{2}} \times q^{\frac{2}{3}}\)                                                       
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( q^{\frac{3}{2}} \times q^{\frac{2}{3}} = q^{\frac{3}{2}+\frac{2}{3}} =                      
 q^{\frac{9}{6}+\frac{4}{6}} = q^{\frac{13}{6}} \)                                                              
             \end{solutionordottedlines}                                                                        
         \part \(p^{\frac{3}{4}} \div p^{\frac{2}{5}}\)                                                         
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( p^{\frac{3}{4}} \div p^{\frac{2}{5}} = p^{\frac{3}{4}-\frac{2}{5}} =                        
 p^{\frac{15}{20}-\frac{8}{20}} = p^{\frac{7}{20}} \)                                                           
             \end{solutionordottedlines}                                                                        
         \part \(q^{\frac{3}{2}} \div q^{\frac{2}{3}}\)                                                         
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( q^{\frac{3}{2}} \div q^{\frac{2}{3}} = q^{\frac{3}{2}-\frac{2}{3}} =                        
 q^{\frac{9}{6}-\frac{4}{6}} = q^{\frac{5}{6}} \)                                                               
             \end{solutionordottedlines}                                                                        
         \part \(\left(2 x^{\frac{2}{3}}\right)^{3}\)                                                           
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( \left(2 x^{\frac{2}{3}}\right)^{3} = 2^{3} \cdot x^{2} = 8x^{2} \)                          
             \end{solutionordottedlines}                                                                        
         \part \(\left(3 y^{\frac{1}{2}}\right)^{4}\)                                                           
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( \left(3 y^{\frac{1}{2}}\right)^{4} = 3^{4} \cdot y^{2} = 81y^{2} \)                         
             \end{solutionordottedlines}                                                                        
     \end{parts}\end{multicols}                                                                                 
     \Question[4] Evaluate:                                                                                     
     \begin{parts}                                                                                              
         \part \(\left(\frac{64}{27}\right)^{-\frac{1}{3}}\)                                                    
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( \left(\frac{64}{27}\right)^{-\frac{1}{3}} = \left(\frac{27}{64}\right)^{\frac{1}{3}} =      
 \frac{3}{4} \)                                                                                                 
             \end{solutionordottedlines}                                                                        
         \part \(32^{-\frac{2}{5}}\)                                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( 32^{-\frac{2}{5}} \approx 0.2795 \)                                                         
             \end{solutionordottedlines}                                                                        
         \part \(\left(\frac{16}{81}\right)^{-\frac{1}{4}}\)                                                    
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( \left(\frac{16}{81}\right)^{-\frac{1}{4}} = \left(\frac{81}{16}\right)^{\frac{1}{4}} \appro 
 2.3784 \)                                                                                                      
             \end{solutionordottedlines}                                                                        
         \part \(\left(\frac{32}{243}\right)^{-\frac{1}{5}}\)                                                   
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( \left(\frac{32}{243}\right)^{-\frac{1}{5}} = \left(\frac{243}{32}\right)^{\frac{1}{5}}      
 \approx 1.9036 \)                                                                                              
             \end{solutionordottedlines}                                                                        
     \end{parts}                                                                                                
     \Question[6] Simplify, expressing the answer with positive indices.                                        
     \begin{parts}                                                                                              
         \part \(\left(3 y^{\frac{1}{2}}\right)^{-4}\)                                                          
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( \left(3 y^{\frac{1}{2}}\right)^{-4} = \left(\frac{1}{3}\right)^{4} \cdot y^{-2} =           
 \frac{1}{81y^{2}} \)                                                                                           
             \end{solutionordottedlines}                                                                        
         \part \(x^{\frac{1}{2}} \times x^{-\frac{3}{2}}\)                                                      
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( x^{\frac{1}{2}} \times x^{-\frac{3}{2}} = x^{\frac{1}{2}-\frac{3}{2}} = x^{-1} = \frac{1}{x 
 \)                                                                                                             
             \end{solutionordottedlines}                                                                        
         \part \(x^{\frac{3}{2}} \div x^{-\frac{1}{2}}\)                                                        
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( x^{\frac{3}{2}} \div x^{-\frac{1}{2}} = x^{\frac{3}{2}+\frac{1}{2}} = x^{2} \)              
             \end{solutionordottedlines}                                                                        
         \part \(y^{\frac{2}{3}} \div y^{-\frac{1}{3}}\)                                                        
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( y^{\frac{2}{3}} \div y^{-\frac{1}{3}} = y^{\frac{2}{3}+\frac{1}{3}} = y^{1} = y \)          
             \end{solutionordottedlines}                                                                        
         \part \(\left(27 n^{-12}\right)^{\frac{1}{3}}\)                                                        
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( \left(27 n^{-12}\right)^{\frac{1}{3}} = 27^{\frac{1}{3}} \cdot n^{-4} = 3n^{-4} =           
 \frac{3}{n^{4}} \)                                                                                             
             \end{solutionordottedlines}                                                                        
         \part \(\left(2 x^{-\frac{2}{5}}\right)^{5}\)                                                          
             \begin{solutionordottedlines}[2cm]                                                                 
                 \( \left(2 x^{-\frac{2}{5}}\right)^{5} = 2^{5} \cdot x^{-2} = 32x^{-2} = \frac{32}{x^{2}} \)   
             \end{solutionordottedlines}                                                                        
     \end{parts}                                                                                                
 \end{questions}                                                                                                
|728983733|kitty|/Applications/kitty.app|0|
\subsection{Fractional Indices}
\begin{questions}
    \Question[3] Evaluate:
    \begin{multicols}{3}
        \begin{parts}
            \part \(\sqrt[4]{81}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\sqrt[3]{64}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\sqrt[5]{2^{10}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
    \end{multicols}
    \Question[2] Write using fractional indices. Evaluate, correct to 4 decimal places.
    \begin{multicols}{2}\begin{parts}
        \part \(\sqrt[7]{11}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\sqrt[3]{2^{7}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}\end{multicols}
    \Question[6] Evaluate:
    \begin{multicols}{2}\begin{parts}
        \part \(243^{\frac{1}{5}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(81^{\frac{1}{4}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(125^{\frac{1}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(64^{\frac{1}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(216^{\frac{1}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(49^{\frac{1}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}\end{multicols}
    \Question[6] Evaluate:
    \begin{multicols}{2}\begin{parts}
        \part \(125^{\frac{2}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(64 \frac{5}{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(216^{\frac{2}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(243^{\overline{5}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\sqrt[5]{32^{4}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\sqrt[3]{2^{6}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}\end{multicols}
    \Question[8] Simplify:
    \begin{multicols}{2}\begin{parts}
        \part \(\left(c^{12}\right)^{\frac{1}{4}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(c^{10}\right)^{\frac{1}{5}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(p^{\frac{3}{4}} \times p^{\frac{2}{5}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(q^{\frac{3}{2}} \times q^{\frac{2}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(p^{\frac{3}{4}} \div p^{\frac{2}{5}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(q^{\frac{3}{2}} \div q^{\frac{2}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(2 x^{\frac{2}{3}}\right)^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(3 y^{\frac{1}{2}}\right)^{4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}\end{multicols}
    \Question[4] Evaluate:
    \begin{parts}
        \part \(\left(\frac{64}{27}\right)^{-\frac{1}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(32^{-\frac{2}{5}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(\frac{16}{81}\right)^{-\frac{1}{4}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(\frac{32}{243}\right)^{-\frac{1}{5}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[6] Simplify, expressing the answer with positive indices.
    \begin{parts}
        \part \(\left(3 y^{\frac{1}{2}}\right)^{-4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(x^{\frac{1}{2}} \times x^{-\frac{3}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(x^{\frac{3}{2}} \div x^{-\frac{1}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(y^{\frac{2}{3}} \div y^{-\frac{1}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(27 n^{-12}\right)^{\frac{1}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(2 x^{-\frac{2}{5}}\right)^{5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
\end{questions}
|728983737|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                            
     \subsection{Exercises:}                                                                                    
     \begin{questions}                                                                                          
         \Question[3] Evaluate:                                                                                 
         \begin{parts}                                                                                          
             \begin{multicols}{3}                                                                               
                 \part \(\sqrt[3]{8}\)                                                                          
                 \begin{solutionordottedlines}[2cm]                                                             
                     2                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(\sqrt[5]{32}\)                                                                         
                 \begin{solutionordottedlines}[2cm]                                                             
                     2                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(\sqrt[3]{216}\)                                                                        
                 \begin{solutionordottedlines}[2cm]                                                             
                     6                                                                                          
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
         \Question[3] Write using fractional indices. Evaluate, correct to 4 decimal places.                    
         \begin{parts}                                                                                          
             \begin{multicols}{3}                                                                               
                 \part \(\sqrt{14}\)                                                                            
                 \begin{solutionordottedlines}[2cm]                                                             
                     14^{\frac{1}{2}} \approx 3.7417                                                            
                 \end{solutionordottedlines}                                                                    
                 \part \(\sqrt[4]{64}\)                                                                         
                 \begin{solutionordottedlines}[2cm]                                                             
                     64^{\frac{1}{4}} = 2.8284                                                                  
                 \end{solutionordottedlines}                                                                    
                 \part \(\sqrt[5]{7}\)                                                                          
                 \begin{solutionordottedlines}[2cm]                                                             
                     7^{\frac{1}{5}} \approx 1.4758                                                             
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
         \Question[6] Evaluate:                                                                                 
         \begin{parts}                                                                                          
             \begin{multicols}{3}                                                                               
                 \part \(4^{\frac{1}{2}}\)                                                                      
                 \begin{solutionordottedlines}[2cm]                                                             
                     2                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(27^{\frac{1}{3}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     3                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(64^{\frac{1}{2}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     8                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(25^{\frac{1}{2}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     5                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(32^{\frac{1}{5}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     2                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(625^{\frac{1}{4}}\)                                                                    
                 \begin{solutionordottedlines}[2cm]                                                             
                     5                                                                                          
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
         \Question[6] Evaluate:                                                                                 
         \begin{parts}                                                                                          
             \begin{multicols}{3}                                                                               
                 \part \(4^{\frac{5}{2}}\)                                                                      
                 \begin{solutionordottedlines}[2cm]                                                             
                     32                                                                                         
                 \end{solutionordottedlines}                                                                    
                 \part \(25^{\frac{3}{2}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     125                                                                                        
                 \end{solutionordottedlines}                                                                    
                 \part \(32^{\frac{2}{5}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     4                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(81^{\frac{3}{4}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     27                                                                                         
                 \end{solutionordottedlines}                                                                    
                 \part \((\sqrt[4]{16})^{3}\)                                                                   
                 \begin{solutionordottedlines}[2cm]                                                             
                     8                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \((\sqrt[3]{27})^{2}\)                                                                   
                 \begin{solutionordottedlines}[2cm]                                                             
                     9                                                                                          
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
         \Question[8] Simplify:                                                                                 
         \begin{parts}                                                                                          
             \begin{multicols}{2}                                                                               
                 \part \(\left(a^{\frac{1}{2}}\right)^{2}\)                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     a                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(\left(b^{\frac{1}{3}}\right)^{6}\)                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     b^2                                                                                        
                 \end{solutionordottedlines}                                                                    
                 \part \(x^{\frac{1}{2}} \times x^{\frac{3}{2}}\)                                               
                 \begin{solutionordottedlines}[2cm]                                                             
                     x^2                                                                                        
                 \end{solutionordottedlines}                                                                    
                 \part \(y^{\frac{1}{3}} \times y^{\frac{2}{3}}\)                                               
                 \begin{solutionordottedlines}[2cm]                                                             
                     y                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(x^{\frac{3}{2}} \div x^{\frac{1}{2}}\)                                                 
                 \begin{solutionordottedlines}[2cm]                                                             
                     x                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(y^{\frac{2}{3}} \div y^{\frac{1}{3}}\)                                                 
                 \begin{solutionordottedlines}[2cm]                                                             
                     y^{\frac{1}{3}}                                                                            
                 \end{solutionordottedlines}                                                                    
                 \part \(\left(4 m^{6}\right)^{\frac{1}{2}}\)                                                   
                 \begin{solutionordottedlines}[2cm]                                                             
                     2m^3                                                                                       
                 \end{solutionordottedlines}                                                                    
                 \part \(\left(27 n^{12}\right)^{\frac{1}{3}}\)                                                 
                 \begin{solutionordottedlines}[2cm]                                                             
                     3n^4                                                                                       
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
         \Question[4] Evaluate:                                                                                 
         \begin{parts}                                                                                          
             \begin{multicols}{2}                                                                               
                 \part \(4^{-\frac{1}{2}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     \frac{1}{2}                                                                                
                 \end{solutionordottedlines}                                                                    
                 \part \(25^{-\frac{1}{2}}\)                                                                    
                 \begin{solutionordottedlines}[2cm]                                                             
                     \frac{1}{5}                                                                                
                 \end{solutionordottedlines}                                                                    
                 \part \(\left(\frac{1}{81}\right)^{-\frac{1}{4}}\)                                             
                 \begin{solutionordottedlines}[2cm]                                                             
                     3                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(81^{-\frac{1}{4}}\)                                                                    
                 \begin{solutionordottedlines}[2cm]                                                             
                     \frac{1}{3}                                                                                
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
         \Question[6] Simplify, expressing the answer with positive indices.                                    
         \begin{parts}                                                                                          
             \begin{multicols}{3}                                                                               
                 \part \(\left(a^{\frac{1}{2}}\right)^{-2}\)                                                    
                 \begin{solutionordottedlines}[2cm]                                                             
                     a^{-1}                                                                                     
                 \end{solutionordottedlines}                                                                    
                 \part \(\left(b^{-\frac{2}{3}}\right)^{6}\)                                                    
                 \begin{solutionordottedlines}[2cm]                                                             
                     b^{-4}                                                                                     
                 \end{solutionordottedlines}                                                                    
                 \part \(y^{\frac{1}{3}} \times y^{-\frac{2}{3}}\)                                              
                 \begin{solutionordottedlines}[2cm]                                                             
                     y^{-\frac{1}{3}}                                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \(p^{\frac{3}{4}} \times p^{-\frac{2}{5}}\)                                              
                 \begin{solutionordottedlines}[2cm]                                                             
                     p^{\frac{7}{20}}                                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \(p^{\frac{3}{4}} \div p^{-\frac{2}{5}}\)                                                
                 \begin{solutionordottedlines}[2cm]                                                             
                     p^{\frac{19}{20}}                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(q^{\frac{3}{2}} \div q^{-\frac{2}{3}}\)                                                
                 \begin{solutionordottedlines}[2cm]                                                             
                     q^{\frac{13}{6}}                                                                           
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
     \end{questions}                                                                                            
 \end{exercisebox}                                                                                              
|728983863|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \Question[3] Evaluate:
        \begin{parts}\begin{multicols}{3}
            \part \(\sqrt[3]{8}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\sqrt[5]{32}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\sqrt[3]{216}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[3] Write using fractional indices. Evaluate, correct to 4 decimal places.
        \begin{parts}\begin{multicols}{3}
            \part \(\sqrt{14}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\sqrt[4]{64}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\sqrt[5]{7}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[6] Evaluate:
        \begin{parts}\begin{multicols}{3}
            \part \(4^{\frac{1}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(27^{\frac{1}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(64^{\frac{1}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(25^{\frac{1}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(32^{\frac{1}{5}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(625^{\frac{1}{4}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[6] Evaluate:
        \begin{parts}\begin{multicols}{3}
            \part \(4^{\frac{5}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(25^{\frac{3}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(32^{\frac{2}{5}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(81^{\frac{3}{4}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((\sqrt[4]{16})^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \((\sqrt[3]{27})^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[8] Simplify:
        \begin{parts}\begin{multicols}{2}
            \part \(\left(a^{\frac{1}{2}}\right)^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(b^{\frac{1}{3}}\right)^{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(x^{\frac{1}{2}} \times x^{\frac{3}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(y^{\frac{1}{3}} \times y^{\frac{2}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(x^{\frac{3}{2}} \div x^{\frac{1}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(y^{\frac{2}{3}} \div y^{\frac{1}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(4 m^{6}\right)^{\frac{1}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(27 n^{12}\right)^{\frac{1}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[4] Evaluate:
        \begin{parts}\begin{multicols}{2}
            \part \(4^{-\frac{1}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(25^{-\frac{1}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(\frac{1}{81}\right)^{-\frac{1}{4}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(81^{-\frac{1}{4}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \Question[6] Simplify, expressing the answer with positive indices.
        \begin{parts}\begin{multicols}{3}
            \part \(\left(a^{\frac{1}{2}}\right)^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(b^{-\frac{2}{3}}\right)^{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(y^{\frac{1}{3}} \times y^{-\frac{2}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(p^{\frac{3}{4}} \times p^{-\frac{2}{5}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(p^{\frac{3}{4}} \div p^{-\frac{2}{5}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(q^{\frac{3}{2}} \div q^{-\frac{2}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \end{questions}
\end{exercisebox}
|728983873|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                            
     \subsection{Exercises:}                                                                                    
     \begin{questions}                                                                                          
         \Question[6] Write as a power of 10.                                                                   
         \begin{multicols}{2}\begin{parts}                                                                      
             \part \(\frac{1}{10} = \fillin[10^{-1}]\)                                                          
             \part \(\frac{1}{100} = \fillin[10^{-2}]\)                                                         
             \part \(\frac{1}{1000} = \fillin[10^{-3}]\)                                                        
             \part 1 trillionth = \fillin[10^{-12}]\)                                                           
             \part \(\frac{1}{100000} = \fillin[10^{-5}]\)                                                      
             \part 1 millionth = \fillin[10^{-6}]\)                                                             
         \end{parts}\end{multicols}                                                                             
         \Question[8] Write in scientific notation.                                                             
         \begin{multicols}{2}\begin{parts}                                                                      
             \part 510 = \fillin[$5.1 \times 10^2$]                                                             
             \part 5300 = \fillin[$5.3 \times 10^3$]                                                            
             \part 796000000 = \fillin[$7.96 \times 10^8$]                                                      
             \part 576000000000 = \fillin[$5.76 \times 10^{11}$]                                                
             \part 0.008 = \fillin[$8 \times 10^{-3}$]                                                          
             \part 0.06 = \fillin[$6 \times 10^{-2}$]                                                           
             \part 0.000041 = \fillin[$4.1 \times 10^{-5}$]                                                     
             \part 0.000000006 = \fillin[$6 \times 10^{-9}$]                                                    
         \end{parts}\end{multicols}                                                                             
         \Question[8] Write in decimal form:                                                                    
         \begin{multicols}{2}\begin{parts}                                                                      
             \part \(3.24 \times 10^{4} = \fillin[32400]\)                                                      
             \part \(7.2 \times 10^{3} = \fillin[7200]\)                                                        
             \part \(2.7 \times 10^{6} = \fillin[2700000]\)                                                     
             \part \(5.1 \times 10^{0} = \fillin[5.1]\)                                                         
             \part \(5.6 \times 10^{-2} = \fillin[0.056]\)                                                      
             \part \(1.7 \times 10^{-3} = \fillin[0.0017]\)                                                     
             \part \(2.01 \times 10^{-3} = \fillin[0.00201]\)                                                   
             \part \(9.7 \times 10^{-1} = \fillin[0.97]\)                                                       
         \end{parts}\end{multicols}                                                                             
         \Question[1] Light travels approximately \(299000 \mathrm{~km}\) in a second. Express this in scientif 
 notation.                                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(2.99 \times 10^5 \mathrm{~km/s}\)                                                            
             \end{solutionordottedlines}                                                                        
         \Question[2] The mass of a copper sample is \(0.0089 \mathrm{~kg}\). Express this in scientific        
 notation.                                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(8.9 \times 10^{-3} \mathrm{~kg}\)                                                            
             \end{solutionordottedlines}                                                                        
         \Question[2] The distance between interconnecting lines on a silicon chip for a computer is            
 approximately \(0.00000004 \mathrm{~m}\). Express this in scientific notation.                                 
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(4 \times 10^{-8} \mathrm{~m}\)                                                               
             \end{solutionordottedlines}                                                                        
         \Question[8] Simplify, expressing the answer in scientific notation.                                   
         \begin{multicols}{2}\begin{parts}                                                                      
             \part \(\left(4 \times 10^{5}\right) \times\left(2 \times 10^{6}\right)\)                          
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(8 \times 10^{11}\)                                                                           
             \end{solutionordottedlines}                                                                        
             \part \(\left(2.1 \times 10^{6}\right) \times\left(3 \times 10^{7}\right)\)                        
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(6.3 \times 10^{13}\)                                                                         
             \end{solutionordottedlines}                                                                        
             \part \(\left(4 \times 10^{2}\right) \times\left(5 \times 10^{-7}\right)\)                         
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(2 \times 10^{-4}\)                                                                           
             \end{solutionordottedlines}                                                                        
             \part \(\left(3 \times 10^{6}\right) \times\left(8 \times 10^{-3}\right)\)                         
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(2.4 \times 10^{4}\)                                                                          
             \end{solutionordottedlines}                                                                        
             \part \(\left(5 \times 10^{4}\right) \div\left(2 \times 10^{3}\right)\)                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(2.5 \times 10^{1}\)                                                                          
             \end{solutionordottedlines}                                                                        
             \part \(\left(8 \times 10^{9}\right) \div\left(4 \times 10^{3}\right)\)                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(2 \times 10^{6}\)                                                                            
             \end{solutionordottedlines}                                                                        
             \part \(\left(6 \times 10^{-4}\right) \div\left(8 \times 10^{-5}\right)\)                          
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(7.5 \times 10^{0}\)                                                                          
             \end{solutionordottedlines}                                                                        
             \part \(\left(1.2 \times 10^{6}\right) \div\left(4 \times 10^{7}\right)\)                          
             \begin{solutionordottedlines}[2cm]                                                                 
                 \(3 \times 10^{-2}\)                                                                           
             \end{solutionordottedlines}                                                                        
         \end{parts}\end{multicols}                                                                             
         \Question[3] If the average distance from the Earth to the Sun is \(1.4951 \times 10^{8} \mathrm{~km}\ 
 and light travels at \(3 \times 10^{5} \mathrm{~km} / \mathrm{s}\), how long does it take light to travel from 
 the Sun to the Earth?                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                 
                 \[                                                                                             
                 \begin{aligned}                                                                                
                 & \text{Time} = \frac{\text{Distance}}{\text{Speed}} \\                                        
                 & = \frac{1.4951 \times 10^{8} \mathrm{~km}}{3 \times 10^{5} \mathrm{~km/s}} \\                
                 & = 4.98367 \times 10^{2} \mathrm{~s} \\                                                       
                 & = 4.98367 \times 10^{2} \mathrm{~s} \approx 498.367 \mathrm{~s}                              
                 \end{aligned}                                                                                  
                 \]                                                                                             
             \end{solutionordottedlines}                                                                        
         \Question[3] The furthest galaxy detected by optical telescopes is approximately \(4.6 \times 10^{9}\) 
 light years from us. How far is this in kilometres? (Light travels at \(3 \times 10^{5} \mathrm{~km} /         
 \mathrm{s}\).)                                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                 
                 \[                                                                                             
                 \begin{aligned}                                                                                
                 & \text{Distance in km} = 4.6 \times 10^{9} \text{ light years} \times \frac{9.461 \times      
 10^{12} \mathrm{~km}}{\text{light year}} \\                                                                    
                 & = 4.6 \times 9.461 \times 10^{9+12} \mathrm{~km} \\                                          
                 & = 43.5206 \times 10^{21} \mathrm{~km} \\                                                     
                 & = 4.35206 \times 10^{22} \mathrm{~km}                                                        
                 \end{aligned}                                                                                  
                 \]                                                                                             
             \end{solutionordottedlines}                                                                        
     \end{questions}                                                                                            
 \end{exercisebox}                                                                                              
|728983899|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \Question[6] Write as a power of 10.
        \begin{multicols}{2}\begin{parts}
            \part \(\frac{1}{10} = \fillin[]\)
            \part \(\frac{1}{100} = \fillin[]\)
            \part \(\frac{1}{1000} = \fillin[]\)
            \part 1 trillionth = \fillin[]
            \part \(\frac{1}{100000} = \fillin[]\)
            \part 1 millionth = \fillin[]
        \end{parts}\end{multicols}
        \Question[8] Write in scientific notation.
        \begin{multicols}{2}\begin{parts}
            \part 510 = \fillin[]
            \part 5300 = \fillin[]
            \part 796000000 = \fillin[]
            \part 576000000000 = \fillin[]
            \part 0.008 = \fillin[]
            \part 0.06 = \fillin[]
            \part 0.000041 = \fillin[]
            \part 0.000000006 = \fillin[]
        \end{parts}\end{multicols}
        \Question[8] Write in decimal form:
        \begin{multicols}{2}\begin{parts}
            \part \(3.24 \times 10^{4} = \fillin[]\)
            \part \(7.2 \times 10^{3} = \fillin[]\)
            \part \(2.7 \times 10^{6} = \fillin[]\)
            \part \(5.1 \times 10^{0} = \fillin[]\)
            \part \(5.6 \times 10^{-2} = \fillin[]\)
            \part \(1.7 \times 10^{-3} = \fillin[]\)
            \part \(2.01 \times 10^{-3} = \fillin[]\)
            \part \(9.7 \times 10^{-1} = \fillin[]\)
        \end{parts}\end{multicols}
        \Question[1] Light travels approximately \(299000 \mathrm{~km}\) in a second. Express this in scientific notation.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Question[2] The mass of a copper sample is \(0.0089 \mathrm{~kg}\). Express this in scientific notation.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Question[2] The distance between interconnecting lines on a silicon chip for a computer is approximately \(0.00000004 \mathrm{~m}\). Express this in scientific notation.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Question[8] Simplify, expressing the answer in scientific notation.
        \begin{multicols}{2}\begin{parts}
            \part \(\left(4 \times 10^{5}\right) \times\left(2 \times 10^{6}\right)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(2.1 \times 10^{6}\right) \times\left(3 \times 10^{7}\right)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(4 \times 10^{2}\right) \times\left(5 \times 10^{-7}\right)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(3 \times 10^{6}\right) \times\left(8 \times 10^{-3}\right)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(5 \times 10^{4}\right) \div\left(2 \times 10^{3}\right)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(8 \times 10^{9}\right) \div\left(4 \times 10^{3}\right)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(6 \times 10^{-4}\right) \div\left(8 \times 10^{-5}\right)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(1.2 \times 10^{6}\right) \div\left(4 \times 10^{7}\right)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}\end{multicols}
        \Question[3] If the average distance from the Earth to the Sun is \(1.4951 \times 10^{8} \mathrm{~km}\) and light travels at \(3 \times 10^{5} \mathrm{~km} / \mathrm{s}\), how long does it take light to travel from the Sun to the Earth?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Question[3] The furthest galaxy detected by optical telescopes is approximately \(4.6 \times 10^{9}\) light years from us. How far is this in kilometres? (Light travels at \(3 \times 10^{5} \mathrm{~km} / \mathrm{s}\).)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{questions}
\end{exercisebox}
|728983909|kitty|/Applications/kitty.app|0|
 \begin{exercisebox}                                                                                            
     \subsection{Exercises:}                                                                                    
     \begin{questions}                                                                                          
         \Question[6] Write in scientific notation, correct to 3 significant figures.                           
         \begin{multicols}{2}\begin{parts}                                                                      
             \part 2.7043                                                                                       
             \begin{solutionordottedlines}[2cm]                                                                 
                 $2.70 \times 10^0$                                                                             
             \end{solutionordottedlines}                                                                        
             \part 634.96                                                                                       
             \begin{solutionordottedlines}[2cm]                                                                 
                 $6.35 \times 10^2$                                                                             
             \end{solutionordottedlines}                                                                        
             \part 8764.37                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                 
                 $8.76 \times 10^3$                                                                             
             \end{solutionordottedlines}                                                                        
             \part 256412                                                                                       
             \begin{solutionordottedlines}[2cm]                                                                 
                 $2.56 \times 10^5$                                                                             
             \end{solutionordottedlines}                                                                        
             \part 0.003612                                                                                     
             \begin{solutionordottedlines}[2cm]                                                                 
                 $3.61 \times 10^{-3}$                                                                          
             \end{solutionordottedlines}                                                                        
             \part 0.024186                                                                                     
             \begin{solutionordottedlines}[2cm]                                                                 
                 $2.42 \times 10^{-2}$                                                                          
             \end{solutionordottedlines}                                                                        
         \end{parts}\end{multicols}                                                                             
         \Question[16]$\,$                                                                                      
             \begin{center}                                                                                     
             \begin{tabular}{|l|l|l|l|l|}                                                                       
             \hline                                                                                             
              & 4 sig. figs & 3 sig. figs & 2 sig. figs & 1 sig. fig. \\                                        
             \hline                                                                                             
             274.62 & $2.746 \times 10^2$ & $2.75 \times 10^2$ & $2.7 \times 10^2$ & $3 \times 10^2$ \\         
             \hline                                                                                             
             0.041236 & $4.124 \times 10^{-2}$ & $4.12 \times 10^{-2}$ & $4.1 \times 10^{-2}$ & $4 \times       
 10^{-2}$ \\                                                                                                    
             \hline                                                                                             
             1704.28 & $1.704 \times 10^3$ & $1.70 \times 10^3$ & $1.7 \times 10^3$ & $2 \times 10^3$ \\        
             \hline                                                                                             
             \(1.9925 \times 10^{27}\) & $1.993 \times 10^{27}$ & $1.99 \times 10^{27}$ & $2.0 \times 10^{27}$  
 $2 \times 10^{27}$ \\                                                                                          
             \hline                                                                                             
             \end{tabular}                                                                                      
             \end{center}                                                                                       
     \end{questions}                                                                                            
 \end{exercisebox}                                                                                              
|728983927|kitty|/Applications/kitty.app|0|
\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \Question[6] Write in scientific notation, correct to 3 significant figures.
        \begin{multicols}{2}\begin{parts}
            \part 2.7043
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 634.96
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 8764.37
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 256412
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 0.003612
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 0.024186
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}\end{multicols}
        \Question[16]$\,$
            \begin{center}
            \begin{tabular}{|l|l|l|l|l|}
            \hline
             & 4 sig. figs & 3 sig. figs & 2 sig. figs & 1 sig. fig. \\
            \hline
            274.62 &  &  &  &  \\
            \hline
            0.041236 &  &  &  &  \\
            \hline
            1704.28 &  &  &  &  \\
            \hline
            \(1.9925 \times 10^{27}\) &  &  &  &  \\
            \hline
            \end{tabular}
            \end{center}
    \end{questions}
\end{exercisebox}
|728983932|kitty|/Applications/kitty.app|0|
 \subsection{Scientific Notation}                                                                               
 \begin{questions}                                                                                              
     \Question[4] Write in scientific notation.                                                                 
     \begin{multicols}{2}\begin{parts}                                                                          
         \part 26000                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 $2.6 \times 10^{4}$                                                                            
             \end{solutionordottedlines}                                                                        
         \part 4000000000000                                                                                    
             \begin{solutionordottedlines}[2cm]                                                                 
                 $4 \times 10^{12}$                                                                             
             \end{solutionordottedlines}                                                                        
         \part 0.00072                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                 
                 $7.2 \times 10^{-4}$                                                                           
             \end{solutionordottedlines}                                                                        
         \part 0.000000206                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                 
                 $2.06 \times 10^{-7}$                                                                          
             \end{solutionordottedlines}                                                                        
     \end{parts}\end{multicols}                                                                                 
     \Question[4] Write in decimal form:                                                                        
     \begin{multicols}{2}\begin{parts}                                                                          
         \part \(8.6 \times 10^{2}\)                                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 860                                                                                            
             \end{solutionordottedlines}                                                                        
         \part \(7.2 \times 10^{1}\)                                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 72                                                                                             
             \end{solutionordottedlines}                                                                        
         \part \(8.72 \times 10^{-4}\)                                                                          
             \begin{solutionordottedlines}[2cm]                                                                 
                 0.000872                                                                                       
             \end{solutionordottedlines}                                                                        
         \part \(2.6 \times 10^{-7}\)                                                                           
             \begin{solutionordottedlines}[2cm]                                                                 
                 0.00000026                                                                                     
             \end{solutionordottedlines}                                                                        
     \end{parts}\end{multicols}                                                                                 
     \Question[6] Simplify, expressing the answer in scientific notation.                                       
     \begin{parts}                                                                                              
         \part \(\left(4 \times 10^{-2}\right)^{2} \times\left(5 \times 10^{7}\right)\)                         
             \begin{solutionordottedlines}[2cm]                                                                 
                 $1.6 \times 10^{-4} \times 5 \times 10^{7} = 8 \times 10^{3}$                                  
             \end{solutionordottedlines}                                                                        
         \part \(\left(6 \times 10^{-3}\right) \times\left(4 \times 10^{7}\right)^{2}\)                         
             \begin{solutionordottedlines}[2cm]                                                                 
                 $6 \times 10^{-3} \times 16 \times 10^{14} = 9.6 \times 10^{12}$                               
             \end{solutionordottedlines}                                                                        
         \part \(\frac{\left(4 \times 10^{5}\right)^{3}}{\left(8 \times 10^{4}\right)^{2}}\)                    
             \begin{solutionordottedlines}[2cm]                                                                 
                 $\frac{64 \times 10^{15}}{64 \times 10^{8}} = 10^{7}$                                          
             \end{solutionordottedlines}                                                                        
         \part \(\frac{\left(2 \times 10^{-1}\right)^{5}}{\left(4 \times 10^{-2}\right)^{3}}\)                  
             \begin{solutionordottedlines}[2cm]                                                                 
                 $\frac{32 \times 10^{-5}}{64 \times 10^{-6}} = 0.5 \times 10^{1} = 5 \times 10^{0}$            
             \end{solutionordottedlines}                                                                        
     \end{parts}                                                                                                
     \Question[2] If light travels at \(3 \times 10^{5} \mathrm{~km} / \mathrm{s}\) and our galaxy is           
 approximately 80000 light years across, how many kilometres is it across? (A light year is the distance light  
 travels in a year.)                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 $3 \times 10^{5} \mathrm{~km/s} \times 60 \times 60 \times 24 \times 365.25 \times 80000 =     
 7.5696 \times 10^{17} \mathrm{~km}$                                                                            
             \end{solutionordottedlines}                                                                        
     \Question[2] The mass of a hydrogen atom is approximately \(1.674 \times 10^{-27} \mathrm{~kg}\) and the   
 mass of an electron is approximately \(9.1 \times 10^{-31} \mathrm{~kg}\). How many electrons, correct to the  
 nearest whole number, will have the same mass as a single hydrogen atom?                                       
             \begin{solutionordottedlines}[2cm]                                                                 
                 $\frac{1.674 \times 10^{-27}}{9.1 \times 10^{-31}} \approx 1839$ electrons                     
             \end{solutionordottedlines}                                                                        
     \Question[2] In a lottery there are \(\frac{45 \times 44 \times 43 \times 42 \times 41 \times 40}{720}\)   
 different possible outcomes. If I mark each outcome on an entry form one at a time, and it takes me an average 
 of 1 minute to mark each outcome, how long will it take me to cover all different possible outcomes?           
             \begin{solutionordottedlines}[2cm]                                                                 
                 $\frac{45 \times 44 \times 43 \times 42 \times 41 \times 40}{720} = 8,145,060$ minutes         
             \end{solutionordottedlines}                                                                        
 \end{questions}                                                                                                
                                                                                                                
 \subsection{Significant Figures}                                                                               
                                                                                                                
 \begin{questions}                                                                                              
     \Question[4] Write in scientific notation, correct to 2 significant figures.                               
     \begin{multicols}{2}\begin{parts}                                                                          
         \part 368.2                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 $3.7 \times 10^{2}$                                                                            
             \end{solutionordottedlines}                                                                        
         \part 278000                                                                                           
             \begin{solutionordottedlines}[2cm]                                                                 
                 $2.8 \times 10^{5}$                                                                            
             \end{solutionordottedlines}                                                                        
         \part 0.004321                                                                                         
             \begin{solutionordottedlines}[2cm]                                                                 
                 $4.3 \times 10^{-3}$                                                                           
             \end{solutionordottedlines}                                                                        
         \part 0.000021906                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                 
                 $2.2 \times 10^{-5}$                                                                           
             \end{solutionordottedlines}                                                                        
     \end{parts}\end{multicols}                                                                                 
     \Question[4] Use a calculator to evaluate the following, giving the answer in scientific notation correct  
 3 significant figures.                                                                                         
     \begin{multicols}{2}\begin{parts}                                                                          
         \part \(3.24 \times 0.067\)                                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 $2.17 \times 10^{-1}$                                                                          
             \end{solutionordottedlines}                                                                        
         \part \(4.736 \times 10^{13} \times 2.34 \times 10^{-6}\)                                              
             \begin{solutionordottedlines}[2cm]                                                                 
                 $1.11 \times 10^{8}$                                                                           
             \end{solutionordottedlines}                                                                        
         \part \(0.0276^{2} \times \sqrt{0.723}\)                                                               
             \begin{solutionordottedlines}[2cm]                                                                 
                 $5.48 \times 10^{-4}$                                                                          
             \end{solutionordottedlines}                                                                        
         \part \(\frac{6.54\left(5.26^{2}+3.24\right)}{5.4+\sqrt{6.34}}\)                                       
             \begin{solutionordottedlines}[2cm]                                                                 
                 $2.97 \times 10^{1}$                                                                           
             \end{solutionordottedlines}                                                                        
     \end{parts}\end{multicols}                                                                                 
     \Question[4] Use a calculator to evaluate, giving the answer in scientific notation correct to 4 significa 
 figures.                                                                                                       
     \begin{multicols}{2}\begin{parts}                                                                          
         \part \(1.234 \times 0.1988\)                                                                          
             \begin{solutionordottedlines}[2cm]                                                                 
                 $2.453 \times 10^{-1}$                                                                         
             \end{solutionordottedlines}                                                                        
         \part \(1.234 \div 0.1988\)                                                                            
             \begin{solutionordottedlines}[2cm]                                                                 
                 $6.206 \times 10^{0}$                                                                          
             \end{solutionordottedlines}                                                                        
         \part \(1.9346^{3}\)                                                                                   
             \begin{solutionordottedlines}[2cm]                                                                 
                 $7.238 \times 10^{0}$                                                                          
             \end{solutionordottedlines}                                                                        
         \part \(\left(7.919 \times 10^{21}\right)^{2}\)                                                        
             \begin{solutionordottedlines}[2cm]                                                                 
                 $6.271 \times 10^{43}$                                                                         
             \end{solutionordottedlines}                                                                        
     \end{parts}\end{multicols}                                                                                 
 \end{questions}                                                                                                
|728983955|kitty|/Applications/kitty.app|0|
\subsection{Scientific Notation}
\begin{questions}
    \Question[4] Write in scientific notation.
    \begin{multicols}{2}\begin{parts}
        \part 26000
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part 4000000000000
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part 0.00072
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part 0.000000206
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}\end{multicols}
    \Question[4] Write in decimal form:
    \begin{multicols}{2}\begin{parts}
        \part \(8.6 \times 10^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(7.2 \times 10^{1}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(8.72 \times 10^{-4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(2.6 \times 10^{-7}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}\end{multicols}
    \Question[6] Simplify, expressing the answer in scientific notation.
    \begin{parts}
        \part \(\left(4 \times 10^{-2}\right)^{2} \times\left(5 \times 10^{7}\right)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(6 \times 10^{-3}\right) \times\left(4 \times 10^{7}\right)^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{\left(4 \times 10^{5}\right)^{3}}{\left(8 \times 10^{4}\right)^{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{\left(2 \times 10^{-1}\right)^{5}}{\left(4 \times 10^{-2}\right)^{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}
    \Question[2] If light travels at \(3 \times 10^{5} \mathrm{~km} / \mathrm{s}\) and our galaxy is approximately 80000 light years across, how many kilometres is it across? (A light year is the distance light travels in a year.)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \Question[2] The mass of a hydrogen atom is approximately \(1.674 \times 10^{-27} \mathrm{~kg}\) and the mass of an electron is approximately \(9.1 \times 10^{-31} \mathrm{~kg}\). How many electrons, correct to the nearest whole number, will have the same mass as a single hydrogen atom?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \Question[2] In a lottery there are \(\frac{45 \times 44 \times 43 \times 42 \times 41 \times 40}{720}\) different possible outcomes. If I mark each outcome on an entry form one at a time, and it takes me an average of 1 minute to mark each outcome, how long will it take me to cover all different possible outcomes?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
\end{questions}

\subsection{Significant Figures}

\begin{questions}
    \Question[4] Write in scientific notation, correct to 2 significant figures.
    \begin{multicols}{2}\begin{parts}
        \part 368.2
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part 278000
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part 0.004321
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part 0.000021906
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}\end{multicols}
    \Question[4] Use a calculator to evaluate the following, giving the answer in scientific notation correct to 3 significant figures.
    \begin{multicols}{2}\begin{parts}
        \part \(3.24 \times 0.067\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(4.736 \times 10^{13} \times 2.34 \times 10^{-6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(0.0276^{2} \times \sqrt{0.723}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\frac{6.54\left(5.26^{2}+3.24\right)}{5.4+\sqrt{6.34}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}\end{multicols}
    \Question[4] Use a calculator to evaluate, giving the answer in scientific notation correct to 4 significant figures.
    \begin{multicols}{2}\begin{parts}
        \part \(1.234 \times 0.1988\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(1.234 \div 0.1988\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(1.9346^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \part \(\left(7.919 \times 10^{21}\right)^{2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{parts}\end{multicols}
\end{questions}
|728983973|kitty|/Applications/kitty.app|0|
s/^.*$/\\(&\\)/

|728984085|kitty|/Applications/kitty.app|0|
s/^\s*\(.\{-}\)\s*$/\\(\1\\)/

|728984125|kitty|/Applications/kitty.app|0|
\)|728984301|kitty|/Applications/kitty.app|0|
approx|728984341|kitty|/Applications/kitty.app|0|
         \Question[3] Write using fractional indices. Evaluate, correct to 4 decimal places.                    
         \begin{parts}                                                                                          
             \begin{multicols}{3}                                                                               
                 \part \(\sqrt{14}\)                                                                            
                 \begin{solutionordottedlines}[2cm]                                                             
                    \(14^{\frac{1}{2}} \approxeq 3.7417\)
                 \end{solutionordottedlines}                                                                    
                 \part \(\sqrt[4]{64}\)                                                                         
                 \begin{solutionordottedlines}[2cm]                                                             
                    \(64^{\frac{1}{4}} = 2.8284\)
                 \end{solutionordottedlines}                                                                    
                 \part \(\sqrt[5]{7}\)                                                                          
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(7^{\frac{1}{5}} \approxeq 1.4758\)                                                             
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
         \Question[6] Evaluate:                                                                                 
         \begin{parts}                                                                                          
             \begin{multicols}{3}                                                                               
                 \part \(4^{\frac{1}{2}}\)                                                                      
                 \begin{solutionordottedlines}[2cm]                                                             
                     2                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(27^{\frac{1}{3}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     3                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(64^{\frac{1}{2}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     8                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(25^{\frac{1}{2}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     5                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(32^{\frac{1}{5}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     2                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(625^{\frac{1}{4}}\)                                                                    
                 \begin{solutionordottedlines}[2cm]                                                             
                     5                                                                                          
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
         \Question[6] Evaluate:                                                                                 
         \begin{parts}                                                                                          
             \begin{multicols}{3}                                                                               
                 \part \(4^{\frac{5}{2}}\)                                                                      
                 \begin{solutionordottedlines}[2cm]                                                             
                     32                                                                                         
                 \end{solutionordottedlines}                                                                    
                 \part \(25^{\frac{3}{2}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     125                                                                                        
                 \end{solutionordottedlines}                                                                    
                 \part \(32^{\frac{2}{5}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     4                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(81^{\frac{3}{4}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     27                                                                                         
                 \end{solutionordottedlines}                                                                    
                 \part \((\sqrt[4]{16})^{3}\)                                                                   
                 \begin{solutionordottedlines}[2cm]                                                             
                     8                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \((\sqrt[3]{27})^{2}\)                                                                   
                 \begin{solutionordottedlines}[2cm]                                                             
                     9                                                                                          
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
         \Question[8] Simplify:                                                                                 
         \begin{parts}                                                                                          
             \begin{multicols}{2}                                                                               
                 \part \(\left(a^{\frac{1}{2}}\right)^{2}\)                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     a                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(\left(b^{\frac{1}{3}}\right)^{6}\)                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(b^2\)                                                                                        
                 \end{solutionordottedlines}                                                                    
                 \part \(x^{\frac{1}{2}} \times x^{\frac{3}{2}}\)                                               
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(x^2\)                                                                                        
                 \end{solutionordottedlines}                                                                    
                 \part \(y^{\frac{1}{3}} \times y^{\frac{2}{3}}\)                                               
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(y\)                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(x^{\frac{3}{2}} \div x^{\frac{1}{2}}\)                                                 
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(x\)                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(y^{\frac{2}{3}} \div y^{\frac{1}{3}}\)                                                 
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(y^{\frac{1}{3}}\)                                                                            
                 \end{solutionordottedlines}                                                                    
                 \part \(\left(4 m^{6}\right)^{\frac{1}{2}}\)                                                   
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(2m^3\)                                                                                       
                 \end{solutionordottedlines}                                                                    
                 \part \(\left(27 n^{12}\right)^{\frac{1}{3}}\)                                                 
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(3n^4\)                                                                                       
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
         \Question[4] Evaluate:                                                                                 
         \begin{parts}                                                                                          
             \begin{multicols}{2}                                                                               
                 \part \(4^{-\frac{1}{2}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(\frac{1}{2}\)                                                                                
                 \end{solutionordottedlines}                                                                    
                 \part \(25^{-\frac{1}{2}}\)                                                                    
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(\frac{1}{5}\)                                                                                
                 \end{solutionordottedlines}                                                                    
                 \part \(\left(\frac{1}{81}\right)^{-\frac{1}{4}}\)                                             
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(3\)                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(81^{-\frac{1}{4}}\)                                                                    
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(\frac{1}{3}\)                                                                                
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
         \Question[6] Simplify, expressing the answer with positive indices.                                    
         \begin{parts}                                                                                          
             \begin{multicols}{3}                                                                               
                 \part \(\left(a^{\frac{1}{2}}\right)^{-2}\)                                                    
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(a^{-1}\)                                                                                     
                 \end{solutionordottedlines}                                                                    
                 \part \(\left(b^{-\frac{2}{3}}\right)^{6}\)                                                    
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(b^{-4}\)                                                                                     
                 \end{solutionordottedlines}                                                                    
                 \part \(y^{\frac{1}{3}} \times y^{-\frac{2}{3}}\)                                              
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(y^{-\frac{1}{3}}\)                                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \(p^{\frac{3}{4}} \times p^{-\frac{2}{5}}\)                                              
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(p^{\frac{7}{20}}\)                                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \(p^{\frac{3}{4}} \div p^{-\frac{2}{5}}\)                                                
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(p^{\frac{19}{20}}\)                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(q^{\frac{3}{2}} \div q^{-\frac{2}{3}}\)                                                
                 \begin{solutionordottedlines}[2cm]                                                             
                     \(q^{\frac{13}{6}}\)                                                                           
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
     \end{questions}                                                                                            
 \end{exercisebox}                                                                                              
|728984431|kitty|/Applications/kitty.app|0|
         \Question[3] Write using fractional indices. Evaluate, correct to 4 decimal places.                    
         \begin{parts}                                                                                          
             \begin{multicols}{3}                                                                               
                 \part \(\sqrt{14}\)                                                                            
                 \begin{solutionordottedlines}[2cm]                                                             
                     14^{\frac{1}{2}} \approx 3.7417                                                            
                 \end{solutionordottedlines}                                                                    
                 \part \(\sqrt[4]{64}\)                                                                         
                 \begin{solutionordottedlines}[2cm]                                                             
                     64^{\frac{1}{4}} = 2.8284                                                                  
                 \end{solutionordottedlines}                                                                    
                 \part \(\sqrt[5]{7}\)                                                                          
                 \begin{solutionordottedlines}[2cm]                                                             
                     7^{\frac{1}{5}} \approx 1.4758                                                             
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
         \Question[6] Evaluate:                                                                                 
         \begin{parts}                                                                                          
             \begin{multicols}{3}                                                                               
                 \part \(4^{\frac{1}{2}}\)                                                                      
                 \begin{solutionordottedlines}[2cm]                                                             
                     2                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(27^{\frac{1}{3}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     3                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(64^{\frac{1}{2}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     8                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(25^{\frac{1}{2}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     5                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(32^{\frac{1}{5}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     2                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(625^{\frac{1}{4}}\)                                                                    
                 \begin{solutionordottedlines}[2cm]                                                             
                     5                                                                                          
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
         \Question[6] Evaluate:                                                                                 
         \begin{parts}                                                                                          
             \begin{multicols}{3}                                                                               
                 \part \(4^{\frac{5}{2}}\)                                                                      
                 \begin{solutionordottedlines}[2cm]                                                             
                     32                                                                                         
                 \end{solutionordottedlines}                                                                    
                 \part \(25^{\frac{3}{2}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     125                                                                                        
                 \end{solutionordottedlines}                                                                    
                 \part \(32^{\frac{2}{5}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     4                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(81^{\frac{3}{4}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     27                                                                                         
                 \end{solutionordottedlines}                                                                    
                 \part \((\sqrt[4]{16})^{3}\)                                                                   
                 \begin{solutionordottedlines}[2cm]                                                             
                     8                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \((\sqrt[3]{27})^{2}\)                                                                   
                 \begin{solutionordottedlines}[2cm]                                                             
                     9                                                                                          
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
         \Question[8] Simplify:                                                                                 
         \begin{parts}                                                                                          
             \begin{multicols}{2}                                                                               
                 \part \(\left(a^{\frac{1}{2}}\right)^{2}\)                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     a                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(\left(b^{\frac{1}{3}}\right)^{6}\)                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     b^2                                                                                        
                 \end{solutionordottedlines}                                                                    
                 \part \(x^{\frac{1}{2}} \times x^{\frac{3}{2}}\)                                               
                 \begin{solutionordottedlines}[2cm]                                                             
                     x^2                                                                                        
                 \end{solutionordottedlines}                                                                    
                 \part \(y^{\frac{1}{3}} \times y^{\frac{2}{3}}\)                                               
                 \begin{solutionordottedlines}[2cm]                                                             
                     y                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(x^{\frac{3}{2}} \div x^{\frac{1}{2}}\)                                                 
                 \begin{solutionordottedlines}[2cm]                                                             
                     x                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(y^{\frac{2}{3}} \div y^{\frac{1}{3}}\)                                                 
                 \begin{solutionordottedlines}[2cm]                                                             
                     y^{\frac{1}{3}}                                                                            
                 \end{solutionordottedlines}                                                                    
                 \part \(\left(4 m^{6}\right)^{\frac{1}{2}}\)                                                   
                 \begin{solutionordottedlines}[2cm]                                                             
                     2m^3                                                                                       
                 \end{solutionordottedlines}                                                                    
                 \part \(\left(27 n^{12}\right)^{\frac{1}{3}}\)                                                 
                 \begin{solutionordottedlines}[2cm]                                                             
                     3n^4                                                                                       
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
         \Question[4] Evaluate:                                                                                 
         \begin{parts}                                                                                          
             \begin{multicols}{2}                                                                               
                 \part \(4^{-\frac{1}{2}}\)                                                                     
                 \begin{solutionordottedlines}[2cm]                                                             
                     \frac{1}{2}                                                                                
                 \end{solutionordottedlines}                                                                    
                 \part \(25^{-\frac{1}{2}}\)                                                                    
                 \begin{solutionordottedlines}[2cm]                                                             
                     \frac{1}{5}                                                                                
                 \end{solutionordottedlines}                                                                    
                 \part \(\left(\frac{1}{81}\right)^{-\frac{1}{4}}\)                                             
                 \begin{solutionordottedlines}[2cm]                                                             
                     3                                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(81^{-\frac{1}{4}}\)                                                                    
                 \begin{solutionordottedlines}[2cm]                                                             
                     \frac{1}{3}                                                                                
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
         \Question[6] Simplify, expressing the answer with positive indices.                                    
         \begin{parts}                                                                                          
             \begin{multicols}{3}                                                                               
                 \part \(\left(a^{\frac{1}{2}}\right)^{-2}\)                                                    
                 \begin{solutionordottedlines}[2cm]                                                             
                     a^{-1}                                                                                     
                 \end{solutionordottedlines}                                                                    
                 \part \(\left(b^{-\frac{2}{3}}\right)^{6}\)                                                    
                 \begin{solutionordottedlines}[2cm]                                                             
                     b^{-4}                                                                                     
                 \end{solutionordottedlines}                                                                    
                 \part \(y^{\frac{1}{3}} \times y^{-\frac{2}{3}}\)                                              
                 \begin{solutionordottedlines}[2cm]                                                             
                     y^{-\frac{1}{3}}                                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \(p^{\frac{3}{4}} \times p^{-\frac{2}{5}}\)                                              
                 \begin{solutionordottedlines}[2cm]                                                             
                     p^{\frac{7}{20}}                                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \(p^{\frac{3}{4}} \div p^{-\frac{2}{5}}\)                                                
                 \begin{solutionordottedlines}[2cm]                                                             
                     p^{\frac{19}{20}}                                                                          
                 \end{solutionordottedlines}                                                                    
                 \part \(q^{\frac{3}{2}} \div q^{-\frac{2}{3}}\)                                                
                 \begin{solutionordottedlines}[2cm]                                                             
                     q^{\frac{13}{6}}                                                                           
                 \end{solutionordottedlines}                                                                    
             \end{multicols}                                                                                    
         \end{parts}                                                                                            
     \end{questions}                                                                                            
 \end{exercisebox}                                                                                              

|728984434|kitty|/Applications/kitty.app|0|
                                                                                                                                
 \begin{problembox}                                                                                                             
     \subsection{Formulae}                                                                                                      
     \begin{questions}                                                                                                          
         \Question[3] Temperatures can be measured in either degrees Fahrenheit or degrees Celsius. To convert from one scale t 
 the other, the following formula is used: \(F=\frac{9}{5} C+32\).                                                              
         \begin{parts}                                                                                                          
             \part Rearrange the formula to make \(C\) the subject.                                                             
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(C = \frac{5}{9}(F - 32)\)                                                                                    
             \end{solutionordottedlines}                                                                                        
             \part On a particular day in Melbourne, the temperature was \(28^{\circ} \mathrm{C}\). What is this temperature    
 measured in Fahrenheit?                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(F = \frac{9}{5} \times 28 + 32 = 82.4^{\circ} \mathrm{F}\)                                                   
             \end{solutionordottedlines}                                                                                        
             \part In Boston, USA, the minimum overnight temperature was \(4^{\circ} \mathrm{F}\). What is this temperature     
 measured in Celsius?                                                                                                           
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(C = \frac{5}{9}(4 - 32) = -15.56^{\circ} \mathrm{C}\)                                                        
             \end{solutionordottedlines}                                                                                        
             \part What number represents the same temperature in \({ }^{\circ} \mathrm{C}\) and \({ }^{\circ} \mathrm{F}\) ?   
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(C = F\) \\                                                                                                   
                 \(\frac{5}{9}(F - 32) = F\) \\                                                                                 
                 \(5F - 160 = 9F\) \\                                                                                           
                 \(4F = 160\) \\                                                                                                
                 \(F = 40^{\circ}\)                                                                                             
             \end{solutionordottedlines}                                                                                        
             \part An approximate conversion formula, used frequently when converting oven temperatures, is \(F=2 C+30\). Use   
 this to convert these temperatures:                                                                                            
             \begin{subparts}                                                                                                   
                 \subpart an oven temperature of \(180^{\circ} \mathrm{C}\)                                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(F = 2 \times 180 + 30 = 390^{\circ} \mathrm{F}\)                                                             
             \end{solutionordottedlines}                                                                                        
                 \subpart an oven temperature of \(530^{\circ} \mathrm{F}\)                                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(C = \frac{530 - 30}{2} = 250^{\circ} \mathrm{C}\)                                                            
             \end{solutionordottedlines}                                                                                        
             \end{subparts}                                                                                                     
         \end{parts}                                                                                                            
         \Question[2] Gareth the gardener has a large rectangular vegetable patch and he wishes to put in a path around it usin 
 concrete pavers that measure \(50 \mathrm{~cm} \times 50 \mathrm{~cm}\). The path is to be 1 paver wide. Let \(n\) be the numb 
 of pavers required. If the vegetable patch measures \(x\) metres by \(y\) metres, find a formula for \(n\) in terms of \(x\) a 
 \(y\).                                                                                                                         
         \begin{solutionordottedlines}[2cm]                                                                                     
             \(n = 2(x + y) \times 2\) because each side needs \(x \times 2\) and \(y \times 2\) pavers, and there are two      
 lengths and two widths.                                                                                                        
         \end{solutionordottedlines}                                                                                            
         \Question[3] If \(s=\frac{n}{2}(2 a+(n-1) d)\) :                                                                       
         \begin{parts}                                                                                                          
             \part find the value of \(s\) when \(n=10, a=6\) and \(d=3\)                                                       
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(s = \frac{10}{2}(2 \times 6 + (10 - 1) \times 3) = 5(12 + 27) = 5 \times 39 = 195\)                          
             \end{solutionordottedlines}                                                                                        
             \part find the value of \(a\) when \(s=350, n=20\) and \(d=4\)                                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(350 = \frac{20}{2}(2a + (20 - 1) \times 4)\) \\                                                              
                 \(350 = 10(2a + 76)\) \\                                                                                       
                 \(35 = 2a + 76\) \\                                                                                            
                 \(a = \frac{35 - 76}{2} = -20.5\)                                                                              
             \end{solutionordottedlines}                                                                                        
             \part find the value of \(d\) when \(s=460, n=10\) and \(a=10\)                                                    
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(460 = \frac{10}{2}(2 \times 10 + (10 - 1)d)\) \\                                                             
                 \(460 = 5(20 + 9d)\) \\                                                                                        
                 \(92 = 20 + 9d\) \\                                                                                            
                 \(d = \frac{92 - 20}{9} = 8\)                                                                                  
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] The formula for the geometric mean \(m\) of two positive numbers \(a\) and \(b\) is \(m=\sqrt{a b}\).     
         \begin{parts}                                                                                                          
             \part Find \(m\) if \(a=16\) and \(b=25\).                                                                         
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(m = \sqrt{16 \times 25} = \sqrt{400} = 20\)                                                                  
             \end{solutionordottedlines}                                                                                        
             \part Find \(a\) if \(m=7\) and \(b=16\).                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(7 = \sqrt{a \times 16}\) \\                                                                                  
                 \(49 = a \times 16\) \\                                                                                        
                 \(a = \frac{49}{16} = 3.0625\)                                                                                 
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] If \(x=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\) :                                                              
         \begin{parts}                                                                                                          
             \part find \(x\) if \(b=4, a=1\) and \(c=-24\)                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(x = \frac{-4 + \sqrt{4^2 - 4 \times 1 \times (-24)}}{2 \times 1}\) \\                                        
                 \(x = \frac{-4 + \sqrt{16 + 96}}{2}\) \\                                                                       
                 \(x = \frac{-4 + \sqrt{112}}{2}\) \\                                                                           
                 \(x = \frac{-4 + 10.58}{2}\) \\                                                                                
                 \(x = 3.29\)                                                                                                   
             \end{solutionordottedlines}                                                                                        
             \part find \(c\) if \(a=1, x=6\) and \(b=2\)                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(6 = \frac{-2 + \sqrt{2^2 - 4 \times 1 \times c}}{2 \times 1}\) \\                                            
                 \(12 = -2 + \sqrt{4 - 4c}\) \\                                                                                 
                 \(14 = \sqrt{4 - 4c}\) \\                                                                                      
                 \(196 = 4 - 4c\) \\                                                                                            
                 \(c = \frac{4 - 196}{4} = -48\)                                                                                
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] A pillar is in the shape of a cylinder with a hemispherical top. If \(r\) metres is the radius of the bas 
 and \(h\) metres is the total height, the volume \(V\) cubic metres is given by the formula \(V=\frac{1}{3} \pi r^{2}(3 h-r)\) 
         \begin{parts}                                                                                                          
             \part Rearrange the formula to make \(h\) the subject.                                                             
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(V = \frac{1}{3} \pi r^2 (3h - r)\) \\                                                                        
                 \(3V = \pi r^2 (3h - r)\) \\                                                                                   
                 \(3V = 3\pi r^2 h - \pi r^3\) \\                                                                               
                 \(3V + \pi r^3 = 3\pi r^2 h\) \\                                                                               
                 \(h = \frac{3V + \pi r^3}{3\pi r^2}\)                                                                          
             \end{solutionordottedlines}                                                                                        
             \part Find the height of the pillar, correct to the nearest centimetre, if the radius of the pillar is \(0.5       
 \mathrm{~m}\) and the volume is \(10 \mathrm{~m}^{3}\).                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(h = \frac{3 \times 10 + \pi \times 0.5^3}{3\pi \times 0.5^2}\) \\                                            
                 \(h = \frac{30 + \frac{1}{8}\pi}{\frac{3}{4}\pi}\) \\                                                          
                 \(h = \frac{30 + 0.3927}{2.3562}\) \\                                                                          
                 \(h = \frac{30.3927}{2.3562}\) \\                                                                              
                 \(h = 12.90 \mathrm{~m}\) \\                                                                                   
                 \(h \approx 1290 \mathrm{~cm}\)                                                                                
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[3] Rearrange each of these formulas to make the pronumeral in brackets the subject. (All of the pronumerals  
 represent positive numbers.)                                                                                                   
         \begin{parts}                                                                                                          
             \part \(A=\ell \times w\hfill{}(\ell)\)                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(\ell = \frac{A}{w}\)                                                                                         
             \end{solutionordottedlines}                                                                                        
             \part \(V=\pi r^{2} h\hfill{}(r)\)                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(r = \sqrt{\frac{V}{\pi h}}\)                                                                                 
             \end{solutionordottedlines}                                                                                        
             \part \(\frac{1}{x}+\frac{1}{y}=\frac{2}{z}\hfill{}(z)\)                                                           
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(\frac{1}{x}+\frac{1}{y}=\frac{2}{z}\) \\                                                                     
                 \(\frac{y+x}{xy}=\frac{2}{z}\) \\                                                                              
                 \(z = \frac{2xy}{x+y}\)                                                                                        
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] If a stone is dropped off a cliff, the number of metres it has fallen after a certain number of seconds i 
 found by multiplying the square of the number of seconds by 4.9.                                                               
         \begin{parts}                                                                                                          
             \part Find the formula for the distance \(d\) metres fallen by the stone in \(t\) seconds.                         
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(d = 4.9t^2\)                                                                                                 
             \end{solutionordottedlines}                                                                                        
             \part Find the distance fallen in 1.5 seconds.                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(d = 4.9 \times 1.5^2 = 4.9 \times 2.25 = 11.025 \mathrm{~m}\)                                                
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[3] If \(t=\sqrt{\frac{M}{M-m}}\) :                                                                           
         \begin{parts}                                                                                                          
             \part express the formula with \(m\) as the subject                                                                
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(t^2 = \frac{M}{M-m}\) \\                                                                                     
                 \(t^2(M-m) = M\) \\                                                                                            
                 \(t^2M - t^2m = M\) \\                                                                                         
                 \(t^2m = t^2M - M\) \\                                                                                         
                 \(m = \frac{t^2M - M}{t^2}\)                                                                                   
             \end{solutionordottedlines}                                                                                        
             \part express the formula with \(M\) as the subject                                                                
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(t^2 = \frac{M}{M-m}\) \\                                                                                     
                 \(t^2(M-m) = M\) \\                                                                                            
                 \(M - t^2m = M/t^2\) \\                                                                                        
                 \(M(1 - 1/t^2) = t^2m\) \\                                                                                     
                 \(M = \frac{t^2m}{1 - 1/t^2}\)                                                                                 
             \end{solutionordottedlines}                                                                                        
             \part find the value of \(M\) if \(m=3\) and \(t=\sqrt{2}\).                                                       
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(M = \frac{(\sqrt{2})^2 \times 3}{1 - 1/(\sqrt{2})^2}\) \\                                                    
                 \(M = \frac{2 \times 3}{1 - 1/2}\) \\                                                                          
                 \(M = \frac{6}{1/2}\) \\                                                                                       
                 \(M = 12\)                                                                                                     
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] The total surface area \(S \mathrm{~cm}^{2}\) of a cylinder is given in terms of its radius \(r           
 \mathrm{~cm}\) and height \(h \mathrm{~cm}\) by the formula \(S=2 \pi r(r+h)\).                                                
         \begin{parts}                                                                                                          
             \part Express this formula with \(h\) as the subject.                                                              
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(S = 2 \pi r^2 + 2 \pi r h\) \\                                                                               
                 \(h = \frac{S - 2 \pi r^2}{2 \pi r}\)                                                                          
             \end{solutionordottedlines}                                                                                        
             \part What is the height of such a cylinder if the radius is \(7 \mathrm{~cm}\) and the total surface area is \(50 
 \mathrm{~cm}^{2}\) ? Calculate your answer in centimetres, correct to 2 decimal places.                                        
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(h = \frac{500 - 2 \pi \times 7^2}{2 \pi \times 7}\) \\                                                       
                 \(h = \frac{500 - 2 \pi \times 49}{2 \pi \times 7}\) \\                                                        
                 \(h = \frac{500 - 307.88}{43.98}\) \\                                                                          
                 \(h = \frac{192.12}{43.98}\) \\                                                                                
                 \(h = 4.37 \mathrm{~cm}\)                                                                                      
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] The sum \(S\) of the squares of the first \(n\) whole numbers is given by the formula \(S=\frac{n(n+1)(2  
 n+1)}{6}\). Find the sum of the squares of:                                                                                    
         \begin{parts}                                                                                                          
             \part the first 20 whole numbers                                                                                   
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(S = \frac{20(20+1)(2 \times 20+1)}{6}\) \\                                                                   
                 \(S = \frac{20 \times 21 \times 41}{6}\) \\                                                                    
                 \(S = \frac{17220}{6}\) \\                                                                                     
                 \(S = 2870\)                                                                                                   
             \end{solutionordottedlines}                                                                                        
             \part all the numbers from 5 to 21 inclusive                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(S_{21} = \frac{21(21+1)(2 \times 21+1)}{6}\) \\                                                              
                 \(S_{21} = \frac{21 \times 22 \times 43}{6}\) \\                                                               
                 \(S_{21} = 3311\) \\                                                                                           
                 \(S_{4} = \frac{4(4+1)(2 \times 4+1)}{6}\) \\                                                                  
                 \(S_{4} = \frac{4 \times 5 \times 9}{6}\) \\                                                                   
                 \(S_{4} = 30\) \\                                                                                              
                 \(S = S_{21} - S_{4} = 3311 - 30 = 3281\)                                                                      
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] For the formula \(D=\sqrt{\frac{f+x}{f-x}}\), make \(x\) the subject.                                     
         \begin{solutionordottedlines}[2cm]                                                                                     
             \(D^2 = \frac{f+x}{f-x}\) \\                                                                                       
             \(D^2(f-x) = f+x\) \\                                                                                              
             \(D^2f - D^2x = f+x\) \\                                                                                           
             \(D^2x + x = D^2f - f\) \\                                                                                         
             \(x(D^2 + 1) = D^2f - f\) \\                                                                                       
             \(x = \frac{D^2f - f}{D^2 + 1}\)                                                                                   
         \end{solutionordottedlines}                                                                                            
|728984984|kitty|/Applications/kitty.app|0|
 \begin{problembox}                                                                                                             
     \subsection{Formulae}                                                                                                      
     \begin{questions}                                                                                                          
         \Question[3] Temperatures can be measured in either degrees Fahrenheit or degrees Celsius. To convert from one scale t 
 the other, the following formula is used: \(F=\frac{9}{5} C+32\).                                                              
         \begin{parts}                                                                                                          
             \part Rearrange the formula to make \(C\) the subject.                                                             
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(C = \frac{5}{9}(F - 32)\)                                                                                    
             \end{solutionordottedlines}                                                                                        
             \part On a particular day in Melbourne, the temperature was \(28^{\circ} \mathrm{C}\). What is this temperature    
 measured in Fahrenheit?                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(F = \frac{9}{5} \times 28 + 32 = 82.4^{\circ} \mathrm{F}\)                                                   
             \end{solutionordottedlines}                                                                                        
             \part In Boston, USA, the minimum overnight temperature was \(4^{\circ} \mathrm{F}\). What is this temperature     
 measured in Celsius?                                                                                                           
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(C = \frac{5}{9}(4 - 32) = -15.56^{\circ} \mathrm{C}\)                                                        
             \end{solutionordottedlines}                                                                                        
             \part What number represents the same temperature in \({ }^{\circ} \mathrm{C}\) and \({ }^{\circ} \mathrm{F}\) ?   
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(C = F\) \\                                                                                                   
                 \(\frac{5}{9}(F - 32) = F\) \\                                                                                 
                 \(5F - 160 = 9F\) \\                                                                                           
                 \(4F = 160\) \\                                                                                                
                 \(F = 40^{\circ}\)                                                                                             
             \end{solutionordottedlines}                                                                                        
             \part An approximate conversion formula, used frequently when converting oven temperatures, is \(F=2 C+30\). Use   
 this to convert these temperatures:                                                                                            
             \begin{subparts}                                                                                                   
                 \subpart an oven temperature of \(180^{\circ} \mathrm{C}\)                                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(F = 2 \times 180 + 30 = 390^{\circ} \mathrm{F}\)                                                             
             \end{solutionordottedlines}                                                                                        
                 \subpart an oven temperature of \(530^{\circ} \mathrm{F}\)                                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(C = \frac{530 - 30}{2} = 250^{\circ} \mathrm{C}\)                                                            
             \end{solutionordottedlines}                                                                                        
             \end{subparts}                                                                                                     
         \end{parts}                                                                                                            
         \Question[2] Gareth the gardener has a large rectangular vegetable patch and he wishes to put in a path around it usin 
 concrete pavers that measure \(50 \mathrm{~cm} \times 50 \mathrm{~cm}\). The path is to be 1 paver wide. Let \(n\) be the numb 
 of pavers required. If the vegetable patch measures \(x\) metres by \(y\) metres, find a formula for \(n\) in terms of \(x\) a 
 \(y\).                                                                                                                         
         \begin{solutionordottedlines}[2cm]                                                                                     
             \(n = 2(x + y) \times 2\) because each side needs \(x \times 2\) and \(y \times 2\) pavers, and there are two      
 lengths and two widths.                                                                                                        
         \end{solutionordottedlines}                                                                                            
         \Question[3] If \(s=\frac{n}{2}(2 a+(n-1) d)\) :                                                                       
         \begin{parts}                                                                                                          
             \part find the value of \(s\) when \(n=10, a=6\) and \(d=3\)                                                       
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(s = \frac{10}{2}(2 \times 6 + (10 - 1) \times 3) = 5(12 + 27) = 5 \times 39 = 195\)                          
             \end{solutionordottedlines}                                                                                        
             \part find the value of \(a\) when \(s=350, n=20\) and \(d=4\)                                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(350 = \frac{20}{2}(2a + (20 - 1) \times 4)\) \\                                                              
                 \(350 = 10(2a + 76)\) \\                                                                                       
                 \(35 = 2a + 76\) \\                                                                                            
                 \(a = \frac{35 - 76}{2} = -20.5\)                                                                              
             \end{solutionordottedlines}                                                                                        
             \part find the value of \(d\) when \(s=460, n=10\) and \(a=10\)                                                    
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(460 = \frac{10}{2}(2 \times 10 + (10 - 1)d)\) \\                                                             
                 \(460 = 5(20 + 9d)\) \\                                                                                        
                 \(92 = 20 + 9d\) \\                                                                                            
                 \(d = \frac{92 - 20}{9} = 8\)                                                                                  
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] The formula for the geometric mean \(m\) of two positive numbers \(a\) and \(b\) is \(m=\sqrt{a b}\).     
         \begin{parts}                                                                                                          
             \part Find \(m\) if \(a=16\) and \(b=25\).                                                                         
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(m = \sqrt{16 \times 25} = \sqrt{400} = 20\)                                                                  
             \end{solutionordottedlines}                                                                                        
             \part Find \(a\) if \(m=7\) and \(b=16\).                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(7 = \sqrt{a \times 16}\) \\                                                                                  
                 \(49 = a \times 16\) \\                                                                                        
                 \(a = \frac{49}{16} = 3.0625\)                                                                                 
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] If \(x=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\) :                                                              
         \begin{parts}                                                                                                          
             \part find \(x\) if \(b=4, a=1\) and \(c=-24\)                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(x = \frac{-4 + \sqrt{4^2 - 4 \times 1 \times (-24)}}{2 \times 1}\) \\                                        
                 \(x = \frac{-4 + \sqrt{16 + 96}}{2}\) \\                                                                       
                 \(x = \frac{-4 + \sqrt{112}}{2}\) \\                                                                           
                 \(x = \frac{-4 + 10.58}{2}\) \\                                                                                
                 \(x = 3.29\)                                                                                                   
             \end{solutionordottedlines}                                                                                        
             \part find \(c\) if \(a=1, x=6\) and \(b=2\)                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(6 = \frac{-2 + \sqrt{2^2 - 4 \times 1 \times c}}{2 \times 1}\) \\                                            
                 \(12 = -2 + \sqrt{4 - 4c}\) \\                                                                                 
                 \(14 = \sqrt{4 - 4c}\) \\                                                                                      
                 \(196 = 4 - 4c\) \\                                                                                            
                 \(c = \frac{4 - 196}{4} = -48\)                                                                                
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] A pillar is in the shape of a cylinder with a hemispherical top. If \(r\) metres is the radius of the bas 
 and \(h\) metres is the total height, the volume \(V\) cubic metres is given by the formula \(V=\frac{1}{3} \pi r^{2}(3 h-r)\) 
         \begin{parts}                                                                                                          
             \part Rearrange the formula to make \(h\) the subject.                                                             
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(V = \frac{1}{3} \pi r^2 (3h - r)\) \\                                                                        
                 \(3V = \pi r^2 (3h - r)\) \\                                                                                   
                 \(3V = 3\pi r^2 h - \pi r^3\) \\                                                                               
                 \(3V + \pi r^3 = 3\pi r^2 h\) \\                                                                               
                 \(h = \frac{3V + \pi r^3}{3\pi r^2}\)                                                                          
             \end{solutionordottedlines}                                                                                        
             \part Find the height of the pillar, correct to the nearest centimetre, if the radius of the pillar is \(0.5       
 \mathrm{~m}\) and the volume is \(10 \mathrm{~m}^{3}\).                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(h = \frac{3 \times 10 + \pi \times 0.5^3}{3\pi \times 0.5^2}\) \\                                            
                 \(h = \frac{30 + \frac{1}{8}\pi}{\frac{3}{4}\pi}\) \\                                                          
                 \(h = \frac{30 + 0.3927}{2.3562}\) \\                                                                          
                 \(h = \frac{30.3927}{2.3562}\) \\                                                                              
                 \(h = 12.90 \mathrm{~m}\) \\                                                                                   
                 \(h \approx 1290 \mathrm{~cm}\)                                                                                
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[3] Rearrange each of these formulas to make the pronumeral in brackets the subject. (All of the pronumerals  
 represent positive numbers.)                                                                                                   
         \begin{parts}                                                                                                          
             \part \(A=\ell \times w\hfill{}(\ell)\)                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(\ell = \frac{A}{w}\)                                                                                         
             \end{solutionordottedlines}                                                                                        
             \part \(V=\pi r^{2} h\hfill{}(r)\)                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(r = \sqrt{\frac{V}{\pi h}}\)                                                                                 
             \end{solutionordottedlines}                                                                                        
             \part \(\frac{1}{x}+\frac{1}{y}=\frac{2}{z}\hfill{}(z)\)                                                           
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(\frac{1}{x}+\frac{1}{y}=\frac{2}{z}\) \\                                                                     
                 \(\frac{y+x}{xy}=\frac{2}{z}\) \\                                                                              
                 \(z = \frac{2xy}{x+y}\)                                                                                        
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] If a stone is dropped off a cliff, the number of metres it has fallen after a certain number of seconds i 
 found by multiplying the square of the number of seconds by 4.9.                                                               
         \begin{parts}                                                                                                          
             \part Find the formula for the distance \(d\) metres fallen by the stone in \(t\) seconds.                         
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(d = 4.9t^2\)                                                                                                 
             \end{solutionordottedlines}                                                                                        
             \part Find the distance fallen in 1.5 seconds.                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(d = 4.9 \times 1.5^2 = 4.9 \times 2.25 = 11.025 \mathrm{~m}\)                                                
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[3] If \(t=\sqrt{\frac{M}{M-m}}\) :                                                                           
         \begin{parts}                                                                                                          
             \part express the formula with \(m\) as the subject                                                                
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(t^2 = \frac{M}{M-m}\) \\                                                                                     
                 \(t^2(M-m) = M\) \\                                                                                            
                 \(t^2M - t^2m = M\) \\                                                                                         
                 \(t^2m = t^2M - M\) \\                                                                                         
                 \(m = \frac{t^2M - M}{t^2}\)                                                                                   
             \end{solutionordottedlines}                                                                                        
             \part express the formula with \(M\) as the subject                                                                
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(t^2 = \frac{M}{M-m}\) \\                                                                                     
                 \(t^2(M-m) = M\) \\                                                                                            
                 \(M - t^2m = M/t^2\) \\                                                                                        
                 \(M(1 - 1/t^2) = t^2m\) \\                                                                                     
                 \(M = \frac{t^2m}{1 - 1/t^2}\)                                                                                 
             \end{solutionordottedlines}                                                                                        
             \part find the value of \(M\) if \(m=3\) and \(t=\sqrt{2}\).                                                       
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(M = \frac{(\sqrt{2})^2 \times 3}{1 - 1/(\sqrt{2})^2}\) \\                                                    
                 \(M = \frac{2 \times 3}{1 - 1/2}\) \\                                                                          
                 \(M = \frac{6}{1/2}\) \\                                                                                       
                 \(M = 12\)                                                                                                     
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] The total surface area \(S \mathrm{~cm}^{2}\) of a cylinder is given in terms of its radius \(r           
 \mathrm{~cm}\) and height \(h \mathrm{~cm}\) by the formula \(S=2 \pi r(r+h)\).                                                
         \begin{parts}                                                                                                          
             \part Express this formula with \(h\) as the subject.                                                              
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(S = 2 \pi r^2 + 2 \pi r h\) \\                                                                               
                 \(h = \frac{S - 2 \pi r^2}{2 \pi r}\)                                                                          
             \end{solutionordottedlines}                                                                                        
             \part What is the height of such a cylinder if the radius is \(7 \mathrm{~cm}\) and the total surface area is \(50 
 \mathrm{~cm}^{2}\) ? Calculate your answer in centimetres, correct to 2 decimal places.                                        
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(h = \frac{500 - 2 \pi \times 7^2}{2 \pi \times 7}\) \\                                                       
                 \(h = \frac{500 - 2 \pi \times 49}{2 \pi \times 7}\) \\                                                        
                 \(h = \frac{500 - 307.88}{43.98}\) \\                                                                          
                 \(h = \frac{192.12}{43.98}\) \\                                                                                
                 \(h = 4.37 \mathrm{~cm}\)                                                                                      
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] The sum \(S\) of the squares of the first \(n\) whole numbers is given by the formula \(S=\frac{n(n+1)(2  
 n+1)}{6}\). Find the sum of the squares of:                                                                                    
         \begin{parts}                                                                                                          
             \part the first 20 whole numbers                                                                                   
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(S = \frac{20(20+1)(2 \times 20+1)}{6}\) \\                                                                   
                 \(S = \frac{20 \times 21 \times 41}{6}\) \\                                                                    
                 \(S = \frac{17220}{6}\) \\                                                                                     
                 \(S = 2870\)                                                                                                   
             \end{solutionordottedlines}                                                                                        
             \part all the numbers from 5 to 21 inclusive                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(S_{21} = \frac{21(21+1)(2 \times 21+1)}{6}\) \\                                                              
                 \(S_{21} = \frac{21 \times 22 \times 43}{6}\) \\                                                               
                 \(S_{21} = 3311\) \\                                                                                           
                 \(S_{4} = \frac{4(4+1)(2 \times 4+1)}{6}\) \\                                                                  
                 \(S_{4} = \frac{4 \times 5 \times 9}{6}\) \\                                                                   
                 \(S_{4} = 30\) \\                                                                                              
                 \(S = S_{21} - S_{4} = 3311 - 30 = 3281\)                                                                      
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] For the formula \(D=\sqrt{\frac{f+x}{f-x}}\), make \(x\) the subject.                                     
         \begin{solutionordottedlines}[2cm]                                                                                     
             \(D^2 = \frac{f+x}{f-x}\) \\                                                                                       
             \(D^2(f-x) = f+x\) \\                                                                                              
             \(D^2f - D^2x = f+x\) \\                                                                                           
             \(D^2x + x = D^2f - f\) \\                                                                                         
             \(x(D^2 + 1) = D^2f - f\) \\                                                                                       
             \(x = \frac{D^2f - f}{D^2 + 1}\)                                                                                   
         \end{solutionordottedlines}                                                                                            
|728985018|kitty|/Applications/kitty.app|0|
\begin{problembox}
    \subsection{Formulae}
    \begin{questions}
        \Question[3] Temperatures can be measured in either degrees Fahrenheit or degrees Celsius. To convert from one scale to the other, the following formula is used: \(F=\frac{9}{5} C+32\).
        \begin{parts}
            \part Rearrange the formula to make \(C\) the subject.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part On a particular day in Melbourne, the temperature was \(28^{\circ} \mathrm{C}\). What is this temperature measured in Fahrenheit?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part In Boston, USA, the minimum overnight temperature was \(4^{\circ} \mathrm{F}\). What is this temperature measured in Celsius?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part What number represents the same temperature in \({ }^{\circ} \mathrm{C}\) and \({ }^{\circ} \mathrm{F}\) ?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part An approximate conversion formula, used frequently when converting oven temperatures, is \(F=2 C+30\). Use this to convert these temperatures:
            \begin{subparts}
                \subpart an oven temperature of \(180^{\circ} \mathrm{C}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
                \subpart an oven temperature of \(530^{\circ} \mathrm{F}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \end{subparts}
        \end{parts}
        \Question[2] Gareth the gardener has a large rectangular vegetable patch and he wishes to put in a path around it using concrete pavers that measure \(50 \mathrm{~cm} \times 50 \mathrm{~cm}\). The path is to be 1 paver wide. Let \(n\) be the number of pavers required. If the vegetable patch measures \(x\) metres by \(y\) metres, find a formula for \(n\) in terms of \(x\) and \(y\).
        \Question[3] If \(s=\frac{n}{2}(2 a+(n-1) d)\) :
        \begin{parts}
            \part find the value of \(s\) when \(n=10, a=6\) and \(d=3\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part find the value of \(a\) when \(s=350, n=20\) and \(d=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part find the value of \(d\) when \(s=460, n=10\) and \(a=10\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] The formula for the geometric mean \(m\) of two positive numbers \(a\) and \(b\) is \(m=\sqrt{a b}\).
        \begin{parts}
            \part Find \(m\) if \(a=16\) and \(b=25\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part Find \(a\) if \(m=7\) and \(b=16\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] If \(x=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\) :
        \begin{parts}
            \part find \(x\) if \(b=4, a=1\) and \(c=-24\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part find \(c\) if \(a=1, x=6\) and \(b=2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] A pillar is in the shape of a cylinder with a hemispherical top. If \(r\) metres is the radius of the base and \(h\) metres is the total height, the volume \(V\) cubic metres is given by the formula \(V=\frac{1}{3} \pi r^{2}(3 h-r)\).
        \begin{parts}
            \part Rearrange the formula to make \(h\) the subject.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part Find the height of the pillar, correct to the nearest centimetre, if the radius of the pillar is \(0.5 \mathrm{~m}\) and the volume is \(10 \mathrm{~m}^{3}\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] Rearrange each of these formulas to make the pronumeral in brackets the subject. (All of the pronumerals represent positive numbers.)
        \begin{parts}
            \part \(A=\ell \times w\hfill{}(\ell)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(V=\pi r^{2} h\hfill{}(r)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{1}{x}+\frac{1}{y}=\frac{2}{z}\hfill{}(z)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] If a stone is dropped off a cliff, the number of metres it has fallen after a certain number of seconds is found by multiplying the square of the number of seconds by 4.9.
        \begin{parts}
            \part Find the formula for the distance \(d\) metres fallen by the stone in \(t\) seconds.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part Find the distance fallen in 1.5 seconds.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] If \(t=\sqrt{\frac{M}{M-m}}\) :
        \begin{parts}
            \part express the formula with \(m\) as the subject
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part express the formula with \(M\) as the subject
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part find the value of \(M\) if \(m=3\) and \(t=\sqrt{2}\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] The total surface area \(S \mathrm{~cm}^{2}\) of a cylinder is given in terms of its radius \(r \mathrm{~cm}\) and height \(h \mathrm{~cm}\) by the formula \(S=2 \pi r(r+h)\).
        \begin{parts}
            \part Express this formula with \(h\) as the subject.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part What is the height of such a cylinder if the radius is \(7 \mathrm{~cm}\) and the total surface area is \(500 \mathrm{~cm}^{2}\) ? Calculate your answer in centimetres, correct to 2 decimal places.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] The sum \(S\) of the squares of the first \(n\) whole numbers is given by the formula \(S=\frac{n(n+1)(2 n+1)}{6}\). Find the sum of the squares of:
        \begin{parts}
            \part the first 20 whole numbers
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part all the numbers from 5 to 21 inclusive
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] For the formula \(D=\sqrt{\frac{f+x}{f-x}}\), make \(x\) the subject.
|728985023|kitty|/Applications/kitty.app|0|
        \Question[2] Cans in a supermarket are displayed in a triangular stack with one can at the top, two cans in the second row from the top, three cans in the third row from the top, and so on. What is the number of cans in the display if the number of rows is:
        \begin{parts}
            \part \(n\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 35
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] Rearrange this formula to make the pronumeral in brackets the subject. \[v^{2}=u^{2}+2 a s\hfill{}(u)\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Question[2] For the formula \(I=\frac{180 n-360}{n}\) :
        \begin{parts}
            \part find \(I\) when \(n=6\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part make \(n\) the subject of the formula and find \(n\) when \(I=108\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[4] Find the formula connecting \(x\) and \(y\) for each of these statements making \(y\) the subject.
        \begin{parts}
            \part \(y\) is four more than twice the square of \(x\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(x\) and \(y\) are complementary angles.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part A car travelled \(x \mathrm{~km}\) in \(y\) hours at a speed of \(100 \mathrm{~km} / \mathrm{h}\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part A car travelled \(100 \mathrm{~km}\) in \(y\) hours at a speed of \(x \mathrm{~km} / \mathrm{h}\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \subsubsection{Challenge Problems}
        \Question[10] A builder wishes to place a circular cap of a given height above an existing window. To do this he needs to know the location of the centre of the circle (the cap is not necessarily a semicircle) and the radius of the circle. \(O\) is the centre of the required circle, the radius of the required circle is \(r \mathrm{~cm}\), the width of the window is \(2 d \mathrm{~cm}\) and the height of the circular cap is \(h \mathrm{~cm}\).
        \includegraphics[width=0.5\textwidth]{2023_12_09_5c1d855de2ec0db78c30g-19}
        \begin{parts}
            \part Express each of these in terms of \(r, d\) and \(h\).
            \begin{subparts}
                \subpart \(A B\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
                \subpart \(O A\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \end{subparts}
            \part Show that \(r=\frac{h^{2}+d^{2}}{2 h}\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part If the window is \(120 \mathrm{~cm}\) wide and the cap is \(40 \mathrm{~cm}\) high, find:
            \begin{subparts}
                \subpart the radius of the circle
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
                \subpart how far below the top of the window the centre of the circle must be placed
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \end{subparts}
            \part If the builder used a circle of radius \(50 \mathrm{~cm}\) and this produced a cap of height \(20 \mathrm{~cm}\), what was the width of the window?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[4] A group of \(n\) people attend a club meeting. Before the meeting begins, they all shake hands with each other. Write a formula to find \(H\), the number of handshakes exchanged.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{questions}
    \subsection{Indices}
    \begin{questions}
        \Question[2] Evaluate:
        \begin{parts}
            \part \(2^{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(10^{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[1] Express \(120^{2}\) as a product of powers of prime numbers.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Question[8] Simplify and evaluate where possible.
        \begin{multicols}{2}\begin{parts}
            \part \(a^{6} \times a^{7}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(2 x^{3} \times 5 x^{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(a^{7} \div a^{4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{12 b^{7}}{6 b^{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{18 p^{10}}{9 p}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(a^{4}\right)^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(2 a^{7}\right)^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(3 b^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}\end{multicols}
        \Question[7] Simplify and evaluate where possible.
        \begin{parts}
            \part \[4 a^{2} b^{3} \times 5 a b^{4}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\frac{20 a^{4} b^{2}}{5 a^{2} b}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\frac{24 m^{9} n^{4}}{18 m^{6} n^{2}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\left(3 a^{3} b\right)^{4}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\left(5 a^{2} b\right)^{2} \times 4 a^{4} b^{3}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\frac{8 m^{4} n^{2}}{7 m^{3} n} \div \frac{3 m^{3} n^{5}}{14 m^{9} n^{16}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\frac{\left(2 x^{2} y\right)^{3}}{5 x^{6} y^{2}} \times\left(\frac{x^{3}}{2 y^{2}}\right)^{3}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[6] Evaluate:
        \begin{multicols}{2}\begin{parts}
            \part \(6^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(8^{-3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(2^{-7}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(\frac{4}{5}\right)^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(\frac{2}{3}\right)^{-4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(16^{\frac{1}{4}}\right)^{-3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}\end{multicols}
        \Question[4] Simplify, expressing the answer with positive indices.
        \begin{parts}
            \part \(\frac{4 m^{2} n^{5} p^{-6}}{16 m^{-2} n^{5} p^{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(2^{2} y^{3}\right)^{-5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(5^{-2} x^{3}\right)^{-5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(3^{-3} a^{2} b^{-1}\right)^{-4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] Simplify, expressing the answer with positive indices.
        \begin{parts}
            \part \(4 a^{2} \times 5 a^{-3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(14 a^{-4} \div 7 a^{-5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{2 m^{3} n^{4}}{(5 m)^{2}} \times \frac{10 m}{3 n^{-4}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] Evaluate:
        \begin{parts}
            \part \(49^{\frac{1}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(125^{\frac{2}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(\frac{1}{8}\right)^{-\frac{2}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[4] Simplify, expressing the answer with positive indices.
        \begin{parts}
            \part \(3 b^{\frac{2}{3}} \times 4 b\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(p^{\frac{2}{3}} \div p^{\frac{1}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(2 x^{-\frac{1}{3}}\right)^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(8 p^{-2} q^{3}\right)^{\frac{1}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] Write in scientific notation.
        \begin{parts}
            \part 164000000
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 0.0047
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 0.0035
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] Write in decimal form.
        \begin{parts}
            \part \(6.8 \times 10^{4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(9.4 \times 10^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(3.2 \times 10^{-4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] Simplify, writing each answer in scientific notation.
        \begin{parts}
            \part \(\left(3.1 \times 10^{4}\right) \times\left(2 \times 10^{-2}\right)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{\left(3 \times 10^{4}\right)^{3}}{9 \times 10^{-2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[6] Write in scientific notation correct to the number of significant figures indicated in the brackets.
        \begin{multicols}{2}\begin{parts}
            \part 18.62\hfill{}(2)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 18.62\hfill{}(3)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 18.62\hfill{}(1)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 0.004276\hfill{}(2)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 5973.4\hfill{}(2)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 0.473952\hfill{}(3)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}\end{multicols}
    \end{questions}
\end{problembox}

|728985029|kitty|/Applications/kitty.app|0|
     \Question[2] Cans in a supermarket are displayed in a triangular stack with one can at the top, two cans in the second row from the top, three 
 cans in the third row from the top, and so on. What is the number of cans in the display if the number of rows is:                                 
         \begin{parts}                                                                                                                              
             \part \(n\)                                                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                                     
                 The number of cans is given by the \(n\)th triangular number, which is \(\frac{n(n+1)}{2}\).                                       
             \end{solutionordottedlines}                                                                                                            
             \part 35                                                                                                                               
             \begin{solutionordottedlines}[2cm]                                                                                                     
                 For \(n = 35\), the number of cans is \(\frac{35(35+1)}{2} = \frac{35 \times 36}{2} = 630\).                                       
             \end{solutionordottedlines}                                                                                                            
         \end{parts}                                                                                                                                
     \Question[2] Rearrange this formula to make the pronumeral in brackets the subject. \[v^{2}=u^{2}+2 a s\hfill{}(u)\]                           
         \begin{solutionordottedlines}[2cm]                                                                                                         
             To make \(u\) the subject, we rearrange the formula: \(u = \sqrt{v^{2} - 2 a s}\).                                                     
         \end{solutionordottedlines}                                                                                                                
     \Question[2] For the formula \(I=\frac{180 n-360}{n}\) :                                                                                       
         \begin{parts}                                                                                                                              
             \part find \(I\) when \(n=6\)                                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                                     
                 For \(n = 6\), \(I = \frac{180 \times 6 - 360}{6} = \frac{1080 - 360}{6} = \frac{720}{6} = 120\).                                  
             \end{solutionordottedlines}                                                                                                            
             \part make \(n\) the subject of the formula and find \(n\) when \(I=108\)                                                              
             \begin{solutionordottedlines}[2cm]                                                                                                     
                 To make \(n\) the subject, we rearrange the formula: \(n = \frac{180}{I + 2}\). For \(I = 108\), \(n = \frac{180}{108 + 2} =       
 \frac{180}{110} = \frac{18}{11}\).                                                                                                                 
             \end{solutionordottedlines}                                                                                                            
         \end{parts}                                                                                                                                
     \Question[4] Find the formula connecting \(x\) and \(y\) for each of these statements making \(y\) the subject.                                
         \begin{parts}                                                                                                                              
             \part \(y\) is four more than twice the square of \(x\).                                                                               
             \begin{solutionordottedlines}[2cm]                                                                                                     
                 \(y = 2x^{2} + 4\).                                                                                                                
             \end{solutionordottedlines}                                                                                                            
             \part \(x\) and \(y\) are complementary angles.                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                                     
                 \(y = 90 - x\).                                                                                                                    
             \end{solutionordottedlines}                                                                                                            
             \part A car travelled \(x \mathrm{~km}\) in \(y\) hours at a speed of \(100 \mathrm{~km} / \mathrm{h}\).                               
             \begin{solutionordottedlines}[2cm]                                                                                                     
                 \(y = \frac{x}{100}\).                                                                                                             
             \end{solutionordottedlines}                                                                                                            
             \part A car travelled \(100 \mathrm{~km}\) in \(y\) hours at a speed of \(x \mathrm{~km} / \mathrm{h}\).                               
             \begin{solutionordottedlines}[2cm]                                                                                                     
                 \(y = \frac{100}{x}\).                                                                                                             
             \end{solutionordottedlines}                                                                                                            
         \end{parts}                                                                                                                                
     \subsubsection{Challenge Problems}                                                                                                             
     \Question[10] A builder wishes to place a circular cap of a given height above an existing window. To do this he needs to know the location of 
 the centre of the circle (the cap is not necessarily a semicircle) and the radius of the circle. \(O\) is the centre of the required circle, the   
 radius of the required circle is \(r \mathrm{~cm}\), the width of the window is \(2 d \mathrm{~cm}\) and the height of the circular cap is \(h     
 \mathrm{~cm}\).                                                                                                                                    
     % Image is omitted in the solution                                                                                                             
     \begin{parts}                                                                                                                                  
         \part Express each of these in terms of \(r, d\) and \(h\).                                                                                
         \begin{subparts}                                                                                                                           
             \subpart \(A B\)                                                                                                                       
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(AB = 2d\).                                                                                                                           
         \end{solutionordottedlines}                                                                                                                
             \subpart \(O A\)                                                                                                                       
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(OA = r\).                                                                                                                            
         \end{solutionordottedlines}                                                                                                                
         \end{subparts}                                                                                                                             
         \part Show that \(r=\frac{h^{2}+d^{2}}{2 h}\).                                                                                             
         \begin{solutionordottedlines}[2cm]                                                                                                         
             By the intersecting chords theorem, \(OA^2 = OD \times OB\), where \(OD = r - h\) and \(OB = r + h\). So, \(r^2 = (r - h)(r + h) = r^2 
 h^2\). Adding \(h^2\) to both sides gives \(r^2 + h^2 = 2rh\). Dividing by \(2h\) gives \(r = \frac{h^2 + d^2}{2h}\).                              
         \end{solutionordottedlines}                                                                                                                
         \part If the window is \(120 \mathrm{~cm}\) wide and the cap is \(40 \mathrm{~cm}\) high, find:                                            
         \begin{subparts}                                                                                                                           
             \subpart the radius of the circle                                                                                                      
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(r = \frac{40^2 + 60^2}{2 \times 40} = \frac{1600 + 3600}{80} = \frac{5200}{80} = 65 \mathrm{~cm}\).                                  
         \end{solutionordottedlines}                                                                                                                
             \subpart how far below the top of the window the centre of the circle must be placed                                                   
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(OD = r - h = 65 - 40 = 25 \mathrm{~cm}\).                                                                                            
         \end{solutionordottedlines}                                                                                                                
         \end{subparts}                                                                                                                             
         \part If the builder used a circle of radius \(50 \mathrm{~cm}\) and this produced a cap of height \(20 \mathrm{~cm}\), what was the width 
 of the window?                                                                                                                                     
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(d = \sqrt{2rh - h^2} = \sqrt{2 \times 50 \times 20 - 20^2} = \sqrt{2000 - 400} = \sqrt{1600} = 40 \mathrm{~cm}\). So the width of th 
 window is \(2d = 80 \mathrm{~cm}\).                                                                                                                
         \end{solutionordottedlines}                                                                                                                
     \end{parts}                                                                                                                                    
     \Question[4] A group of \(n\) people attend a club meeting. Before the meeting begins, they all shake hands with each other. Write a formula t 
 find \(H\), the number of handshakes exchanged.                                                                                                    
         \begin{solutionordottedlines}[2cm]                                                                                                         
             Each person shakes hands with \(n - 1\) others, and there are \(n\) people, so there are \(n(n - 1)\) handshakes. However, this counts 
 each handshake twice, so \(H = \frac{n(n - 1)}{2}\).                                                                                               
         \end{solutionordottedlines}                                                                                                                
 \end{questions}                                                                                                                                    
 \subsection{Indices}                                                                                                                               
 \begin{questions}                                                                                                                                  
     \Question[2] Evaluate:                                                                                                                         
     \begin{parts}                                                                                                                                  
         \part \(2^{6}\)                                                                                                                            
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(2^{6} = 64\).                                                                                                                        
         \end{solutionordottedlines}                                                                                                                
         \part \(10^{6}\)                                                                                                                           
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(10^{6} = 1,000,000\).                                                                                                                
         \end{solutionordottedlines}                                                                                                                
     \end{parts}                                                                                                                                    
     \Question[1] Express \(120^{2}\) as a product of powers of prime numbers.                                                                      
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(120^{2} = (2^{3} \times 3 \times 5)^{2} = 2^{6} \times 3^{2} \times 5^{2}\).                                                         
         \end{solutionordottedlines}                                                                                                                
     \Question[8] Simplify and evaluate where possible.                                                                                             
     \begin{multicols}{2}\begin{parts}                                                                                                              
         \part \(a^{6} \times a^{7}\)                                                                                                               
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(a^{6} \times a^{7} = a^{6+7} = a^{13}\).                                                                                             
         \end{solutionordottedlines}                                                                                                                
         \part \(2 x^{3} \times 5 x^{6}\)                                                                                                           
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(2 x^{3} \times 5 x^{6} = 10 x^{3+6} = 10 x^{9}\).                                                                                    
         \end{solutionordottedlines}                                                                                                                
         \part \(a^{7} \div a^{4}\)                                                                                                                 
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(a^{7} \div a^{4} = a^{7-4} = a^{3}\).                                                                                                
         \end{solutionordottedlines}                                                                                                                
         \part \(\frac{12 b^{7}}{6 b^{2}}\)                                                                                                         
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(\frac{12 b^{7}}{6 b^{2}} = 2 b^{7-2} = 2 b^{5}\).                                                                                    
         \end{solutionordottedlines}                                                                                                                
         \part \(\frac{18 p^{10}}{9 p}\)                                                                                                            
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(\frac{18 p^{10}}{9 p} = 2 p^{10-1} = 2 p^{9}\).                                                                                      
         \end{solutionordottedlines}                                                                                                                
         \part \(\left(a^{4}\right)^{3}\)                                                                                                           
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(\left(a^{4}\right)^{3} = a^{4 \times 3} = a^{12}\).                                                                                  
         \end{solutionordottedlines}                                                                                                                
         \part \(\left(2 a^{7}\right)^{3}\)                                                                                                         
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(\left(2 a^{7}\right)^{3} = 2^{3} a^{7 \times 3} = 8 a^{21}\).                                                                        
         \end{solutionordottedlines}                                                                                                                
         \part \(3 b^{0}\)                                                                                                                          
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(3 b^{0} = 3 \times 1 = 3\).                                                                                                          
         \end{solutionordottedlines}                                                                                                                
     \end{parts}\end{multicols}                                                                                                                     
     \Question[7] Simplify and evaluate where possible.                                                                                             
     \begin{parts}                                                                                                                                  
         \part \[4 a^{2} b^{3} \times 5 a b^{4}\]                                                                                                   
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(4 a^{2} b^{3} \times 5 a b^{4} = 20 a^{2+1} b^{3+4} = 20 a^{3} b^{7}\).                                                              
         \end{solutionordottedlines}                                                                                                                
         \part \[\frac{20 a^{4} b^{2}}{5 a^{2} b}\]                                                                                                 
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(\frac{20 a^{4} b^{2}}{5 a^{2} b} = 4 a^{4-2} b^{2-1} = 4 a^{2} b\).                                                                  
         \end{solutionordottedlines}                                                                                                                
         \part \[\frac{24 m^{9} n^{4}}{18 m^{6} n^{2}}\]                                                                                            
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(\frac{24 m^{9} n^{4}}{18 m^{6} n^{2}} = \frac{4}{3} m^{9-6} n^{4-2} = \frac{4}{3} m^{3} n^{2}\).                                     
         \end{solutionordottedlines}                                                                                                                
         \part \[\left(3 a^{3} b\right)^{4}\]                                                                                                       
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(\left(3 a^{3} b\right)^{4} = 3^{4} a^{3 \times 4} b^{4} = 81 a^{12} b^{4}\).                                                         
         \end{solutionordottedlines}                                                                                                                
         \part \[\left(5 a^{2} b\right)^{2} \times 4 a^{4} b^{3}\]                                                                                  
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(\left(5 a^{2} b\right)^{2} \times 4 a^{4} b^{3} = 25 a^{4} b^{2} \times 4 a^{4} b^{3} = 100 a^{8} b^{5}\).                           
         \end{solutionordottedlines}                                                                                                                
         \part \[\frac{8 m^{4} n^{2}}{7 m^{3} n} \div \frac{3 m^{3} n^{5}}{14 m^{9} n^{16}}\]                                                       
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(\frac{8 m^{4} n^{2}}{7 m^{3} n} \div \frac{3 m^{3} n^{5}}{14 m^{9} n^{16}} = \frac{8 m^{4} n^{2}}{7 m^{3} n} \times \frac{14 m^{9}   
 n^{16}}{3 m^{3} n^{5}} = \frac{8 \times 14 m^{10} n^{17}}{7 \times 3 m^{6} n^{6}} = \frac{112 m^{4} n^{11}}{21} = \frac{16}{3} m^{4} n^{11}\).     
         \end{solutionordottedlines}                                                                                                                
         \part \[\frac{\left(2 x^{2} y\right)^{3}}{5 x^{6} y^{2}} \times\left(\frac{x^{3}}{2 y^{2}}\right)^{3}\]                                    
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(\frac{\left(2 x^{2} y\right)^{3}}{5 x^{6} y^{2}} \times\left(\frac{x^{3}}{2 y^{2}}\right)^{3} = \frac{8 x^{6} y^{3}}{5 x^{6} y^{2}}  
 \times \frac{x^{9}}{8 y^{6}} = \frac{8}{5} y \times \frac{x^{9}}{8 y^{6}} = \frac{x^{9}}{5 y^{5}}\).                                               
         \end{solutionordottedlines}                                                                                                                
     \end{parts}                                                                                                                                    
     \Question[6] Evaluate:                                                                                                                         
     \begin{multicols}{2}\begin{parts}                                                                                                              
         \part \(6^{-2}\)                                                                                                                           
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(6^{-2} = \frac{1}{6^{2}} = \frac{1}{36}\).                                                                                           
         \end{solutionordottedlines}                                                                                                                
         \part \(8^{-3}\)                                                                                                                           
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(8^{-3} = \frac{1}{8^{3}} = \frac{1}{512}\).                                                                                          
         \end{solutionordottedlines}                                                                                                                
         \part \(2^{-7}\)                                                                                                                           
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(2^{-7} = \frac{1}{2^{7}} = \frac{1}{128}\).                                                                                          
         \end{solutionordottedlines}                                                                                                                
         \part \(\left(\frac{4}{5}\right)^{-2}\)                                                                                                    
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(\left(\frac{4}{5}\right)^{-2} = \left(\frac{5}{4}\right)^{2} = \frac{25}{16}\).                                                      
         \end{solutionordottedlines}                                                                                                                
         \part \(\left(\frac{2}{3}\right)^{-4}\)                                                                                                    
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(\left(\frac{2}{3}\right)^{-4} = \left(\frac{3}{2}\right)^{4} = \frac{81}{16}\).                                                      
         \end{solutionordottedlines}                                                                                                                
         \part \(\left(16^{\frac{1}{4}}\right)^{-3}\)                                                                                               
         \begin{solutionordottedlines}[2cm]                                                                                                         
             \(\left(16^{\frac{1}{4}}\right)^{-3} = \left(2\right)^{-3} = \frac{1}{2^{3}} = \frac{1}{8}\).                                          
         \end{solutionordottedlines}                                                                                                                
     \end{parts}\end{multicols}                                                                                                                     
|728985366|kitty|/Applications/kitty.app|0|
\begin{problembox}
    \subsection{Formulae}
    \begin{questions}
        \Question[3] Temperatures can be measured in either degrees Fahrenheit or degrees Celsius. To convert from one scale to the other, the following formula is used: \(F=\frac{9}{5} C+32\).
        \begin{parts}
            \part Rearrange the formula to make \(C\) the subject.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part On a particular day in Melbourne, the temperature was \(28^{\circ} \mathrm{C}\). What is this temperature measured in Fahrenheit?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part In Boston, USA, the minimum overnight temperature was \(4^{\circ} \mathrm{F}\). What is this temperature measured in Celsius?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part What number represents the same temperature in \({ }^{\circ} \mathrm{C}\) and \({ }^{\circ} \mathrm{F}\) ?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part An approximate conversion formula, used frequently when converting oven temperatures, is \(F=2 C+30\). Use this to convert these temperatures:
            \begin{subparts}
                \subpart an oven temperature of \(180^{\circ} \mathrm{C}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
                \subpart an oven temperature of \(530^{\circ} \mathrm{F}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \end{subparts}
        \end{parts}
        \Question[2] Gareth the gardener has a large rectangular vegetable patch and he wishes to put in a path around it using concrete pavers that measure \(50 \mathrm{~cm} \times 50 \mathrm{~cm}\). The path is to be 1 paver wide. Let \(n\) be the number of pavers required. If the vegetable patch measures \(x\) metres by \(y\) metres, find a formula for \(n\) in terms of \(x\) and \(y\).
        \Question[3] If \(s=\frac{n}{2}(2 a+(n-1) d)\) :
        \begin{parts}
            \part find the value of \(s\) when \(n=10, a=6\) and \(d=3\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part find the value of \(a\) when \(s=350, n=20\) and \(d=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part find the value of \(d\) when \(s=460, n=10\) and \(a=10\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] The formula for the geometric mean \(m\) of two positive numbers \(a\) and \(b\) is \(m=\sqrt{a b}\).
        \begin{parts}
            \part Find \(m\) if \(a=16\) and \(b=25\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part Find \(a\) if \(m=7\) and \(b=16\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] If \(x=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\) :
        \begin{parts}
            \part find \(x\) if \(b=4, a=1\) and \(c=-24\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part find \(c\) if \(a=1, x=6\) and \(b=2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] A pillar is in the shape of a cylinder with a hemispherical top. If \(r\) metres is the radius of the base and \(h\) metres is the total height, the volume \(V\) cubic metres is given by the formula \(V=\frac{1}{3} \pi r^{2}(3 h-r)\).
        \begin{parts}
            \part Rearrange the formula to make \(h\) the subject.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part Find the height of the pillar, correct to the nearest centimetre, if the radius of the pillar is \(0.5 \mathrm{~m}\) and the volume is \(10 \mathrm{~m}^{3}\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] Rearrange each of these formulas to make the pronumeral in brackets the subject. (All of the pronumerals represent positive numbers.)
        \begin{parts}
            \part \(A=\ell \times w\hfill{}(\ell)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(V=\pi r^{2} h\hfill{}(r)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{1}{x}+\frac{1}{y}=\frac{2}{z}\hfill{}(z)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] If a stone is dropped off a cliff, the number of metres it has fallen after a certain number of seconds is found by multiplying the square of the number of seconds by 4.9.
        \begin{parts}
            \part Find the formula for the distance \(d\) metres fallen by the stone in \(t\) seconds.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part Find the distance fallen in 1.5 seconds.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] If \(t=\sqrt{\frac{M}{M-m}}\) :
        \begin{parts}
            \part express the formula with \(m\) as the subject
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part express the formula with \(M\) as the subject
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part find the value of \(M\) if \(m=3\) and \(t=\sqrt{2}\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] The total surface area \(S \mathrm{~cm}^{2}\) of a cylinder is given in terms of its radius \(r \mathrm{~cm}\) and height \(h \mathrm{~cm}\) by the formula \(S=2 \pi r(r+h)\).
        \begin{parts}
            \part Express this formula with \(h\) as the subject.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part What is the height of such a cylinder if the radius is \(7 \mathrm{~cm}\) and the total surface area is \(500 \mathrm{~cm}^{2}\) ? Calculate your answer in centimetres, correct to 2 decimal places.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] The sum \(S\) of the squares of the first \(n\) whole numbers is given by the formula \(S=\frac{n(n+1)(2 n+1)}{6}\). Find the sum of the squares of:
        \begin{parts}
            \part the first 20 whole numbers
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part all the numbers from 5 to 21 inclusive
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] For the formula \(D=\sqrt{\frac{f+x}{f-x}}\), make \(x\) the subject.
        \Question[2] Cans in a supermarket are displayed in a triangular stack with one can at the top, two cans in the second row from the top, three cans in the third row from the top, and so on. What is the number of cans in the display if the number of rows is:
        \begin{parts}
            \part \(n\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 35
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] Rearrange this formula to make the pronumeral in brackets the subject. \[v^{2}=u^{2}+2 a s\hfill{}(u)\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Question[2] For the formula \(I=\frac{180 n-360}{n}\) :
        \begin{parts}
            \part find \(I\) when \(n=6\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part make \(n\) the subject of the formula and find \(n\) when \(I=108\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[4] Find the formula connecting \(x\) and \(y\) for each of these statements making \(y\) the subject.
        \begin{parts}
            \part \(y\) is four more than twice the square of \(x\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(x\) and \(y\) are complementary angles.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part A car travelled \(x \mathrm{~km}\) in \(y\) hours at a speed of \(100 \mathrm{~km} / \mathrm{h}\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part A car travelled \(100 \mathrm{~km}\) in \(y\) hours at a speed of \(x \mathrm{~km} / \mathrm{h}\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \subsubsection{Challenge Problems}
        \Question[10] A builder wishes to place a circular cap of a given height above an existing window. To do this he needs to know the location of the centre of the circle (the cap is not necessarily a semicircle) and the radius of the circle. \(O\) is the centre of the required circle, the radius of the required circle is \(r \mathrm{~cm}\), the width of the window is \(2 d \mathrm{~cm}\) and the height of the circular cap is \(h \mathrm{~cm}\).
        \includegraphics[width=\textwidth]{2023_12_09_5c1d855de2ec0db78c30g-19}
        \begin{parts}
            \part Express each of these in terms of \(r, d\) and \(h\).
            \begin{subparts}
                \subpart \(A B\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
                \subpart \(O A\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \end{subparts}
            \part Show that \(r=\frac{h^{2}+d^{2}}{2 h}\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part If the window is \(120 \mathrm{~cm}\) wide and the cap is \(40 \mathrm{~cm}\) high, find:
            \begin{subparts}
                \subpart the radius of the circle
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
                \subpart how far below the top of the window the centre of the circle must be placed
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \end{subparts}
            \part If the builder used a circle of radius \(50 \mathrm{~cm}\) and this produced a cap of height \(20 \mathrm{~cm}\), what was the width of the window?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[4] A group of \(n\) people attend a club meeting. Before the meeting begins, they all shake hands with each other. Write a formula to find \(H\), the number of handshakes exchanged.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{questions}
    \subsection{Indices}
    \begin{questions}
        \Question[2] Evaluate:
        \begin{parts}
            \part \(2^{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(10^{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[1] Express \(120^{2}\) as a product of powers of prime numbers.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Question[8] Simplify and evaluate where possible.
        \begin{multicols}{2}\begin{parts}
            \part \(a^{6} \times a^{7}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(2 x^{3} \times 5 x^{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(a^{7} \div a^{4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{12 b^{7}}{6 b^{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{18 p^{10}}{9 p}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(a^{4}\right)^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(2 a^{7}\right)^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(3 b^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}\end{multicols}
        \Question[7] Simplify and evaluate where possible.
        \begin{parts}
            \part \[4 a^{2} b^{3} \times 5 a b^{4}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\frac{20 a^{4} b^{2}}{5 a^{2} b}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\frac{24 m^{9} n^{4}}{18 m^{6} n^{2}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\left(3 a^{3} b\right)^{4}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\left(5 a^{2} b\right)^{2} \times 4 a^{4} b^{3}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\frac{8 m^{4} n^{2}}{7 m^{3} n} \div \frac{3 m^{3} n^{5}}{14 m^{9} n^{16}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\frac{\left(2 x^{2} y\right)^{3}}{5 x^{6} y^{2}} \times\left(\frac{x^{3}}{2 y^{2}}\right)^{3}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[6] Evaluate:
        \begin{multicols}{2}\begin{parts}
            \part \(6^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(8^{-3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(2^{-7}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(\frac{4}{5}\right)^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(\frac{2}{3}\right)^{-4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(16^{\frac{1}{4}}\right)^{-3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}\end{multicols}
|728985459|kitty|/Applications/kitty.app|0|
\begin{problembox}
    \subsection{Formulae}
    \begin{questions}
        \Question[3] Temperatures can be measured in either degrees Fahrenheit or degrees Celsius. To convert from one scale to the other, the following formula is used: \(F=\frac{9}{5} C+32\).
        \begin{parts}
            \part Rearrange the formula to make \(C\) the subject.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part On a particular day in Melbourne, the temperature was \(28^{\circ} \mathrm{C}\). What is this temperature measured in Fahrenheit?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part In Boston, USA, the minimum overnight temperature was \(4^{\circ} \mathrm{F}\). What is this temperature measured in Celsius?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part What number represents the same temperature in \({ }^{\circ} \mathrm{C}\) and \({ }^{\circ} \mathrm{F}\) ?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part An approximate conversion formula, used frequently when converting oven temperatures, is \(F=2 C+30\). Use this to convert these temperatures:
            \begin{subparts}
                \subpart an oven temperature of \(180^{\circ} \mathrm{C}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
                \subpart an oven temperature of \(530^{\circ} \mathrm{F}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \end{subparts}
        \end{parts}
        \Question[2] Gareth the gardener has a large rectangular vegetable patch and he wishes to put in a path around it using concrete pavers that measure \(50 \mathrm{~cm} \times 50 \mathrm{~cm}\). The path is to be 1 paver wide. Let \(n\) be the number of pavers required. If the vegetable patch measures \(x\) metres by \(y\) metres, find a formula for \(n\) in terms of \(x\) and \(y\).
        \Question[3] If \(s=\frac{n}{2}(2 a+(n-1) d)\) :
        \begin{parts}
            \part find the value of \(s\) when \(n=10, a=6\) and \(d=3\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part find the value of \(a\) when \(s=350, n=20\) and \(d=4\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part find the value of \(d\) when \(s=460, n=10\) and \(a=10\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] The formula for the geometric mean \(m\) of two positive numbers \(a\) and \(b\) is \(m=\sqrt{a b}\).
        \begin{parts}
            \part Find \(m\) if \(a=16\) and \(b=25\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part Find \(a\) if \(m=7\) and \(b=16\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] If \(x=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\) :
        \begin{parts}
            \part find \(x\) if \(b=4, a=1\) and \(c=-24\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part find \(c\) if \(a=1, x=6\) and \(b=2\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] A pillar is in the shape of a cylinder with a hemispherical top. If \(r\) metres is the radius of the base and \(h\) metres is the total height, the volume \(V\) cubic metres is given by the formula \(V=\frac{1}{3} \pi r^{2}(3 h-r)\).
        \begin{parts}
            \part Rearrange the formula to make \(h\) the subject.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part Find the height of the pillar, correct to the nearest centimetre, if the radius of the pillar is \(0.5 \mathrm{~m}\) and the volume is \(10 \mathrm{~m}^{3}\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] Rearrange each of these formulas to make the pronumeral in brackets the subject. (All of the pronumerals represent positive numbers.)
        \begin{parts}
            \part \(A=\ell \times w\hfill{}(\ell)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(V=\pi r^{2} h\hfill{}(r)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{1}{x}+\frac{1}{y}=\frac{2}{z}\hfill{}(z)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] If a stone is dropped off a cliff, the number of metres it has fallen after a certain number of seconds is found by multiplying the square of the number of seconds by 4.9.
        \begin{parts}
            \part Find the formula for the distance \(d\) metres fallen by the stone in \(t\) seconds.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part Find the distance fallen in 1.5 seconds.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] If \(t=\sqrt{\frac{M}{M-m}}\) :
        \begin{parts}
            \part express the formula with \(m\) as the subject
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part express the formula with \(M\) as the subject
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part find the value of \(M\) if \(m=3\) and \(t=\sqrt{2}\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] The total surface area \(S \mathrm{~cm}^{2}\) of a cylinder is given in terms of its radius \(r \mathrm{~cm}\) and height \(h \mathrm{~cm}\) by the formula \(S=2 \pi r(r+h)\).
        \begin{parts}
            \part Express this formula with \(h\) as the subject.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part What is the height of such a cylinder if the radius is \(7 \mathrm{~cm}\) and the total surface area is \(500 \mathrm{~cm}^{2}\) ? Calculate your answer in centimetres, correct to 2 decimal places.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] The sum \(S\) of the squares of the first \(n\) whole numbers is given by the formula \(S=\frac{n(n+1)(2 n+1)}{6}\). Find the sum of the squares of:
        \begin{parts}
            \part the first 20 whole numbers
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part all the numbers from 5 to 21 inclusive
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] For the formula \(D=\sqrt{\frac{f+x}{f-x}}\), make \(x\) the subject.
        \Question[2] Cans in a supermarket are displayed in a triangular stack with one can at the top, two cans in the second row from the top, three cans in the third row from the top, and so on. What is the number of cans in the display if the number of rows is:
        \begin{parts}
            \part \(n\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 35
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] Rearrange this formula to make the pronumeral in brackets the subject. \[v^{2}=u^{2}+2 a s\hfill{}(u)\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Question[2] For the formula \(I=\frac{180 n-360}{n}\) :
        \begin{parts}
            \part find \(I\) when \(n=6\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part make \(n\) the subject of the formula and find \(n\) when \(I=108\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[4] Find the formula connecting \(x\) and \(y\) for each of these statements making \(y\) the subject.
        \begin{parts}
            \part \(y\) is four more than twice the square of \(x\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(x\) and \(y\) are complementary angles.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part A car travelled \(x \mathrm{~km}\) in \(y\) hours at a speed of \(100 \mathrm{~km} / \mathrm{h}\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part A car travelled \(100 \mathrm{~km}\) in \(y\) hours at a speed of \(x \mathrm{~km} / \mathrm{h}\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \subsubsection{Challenge Problems}
        \Question[10] A builder wishes to place a circular cap of a given height above an existing window. To do this he needs to know the location of the centre of the circle (the cap is not necessarily a semicircle) and the radius of the circle. \(O\) is the centre of the required circle, the radius of the required circle is \(r \mathrm{~cm}\), the width of the window is \(2 d \mathrm{~cm}\) and the height of the circular cap is \(h \mathrm{~cm}\).
        \includegraphics[width=\textwidth]{2023_12_09_5c1d855de2ec0db78c30g-19}
        \begin{parts}
            \part Express each of these in terms of \(r, d\) and \(h\).
            \begin{subparts}
                \subpart \(A B\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
                \subpart \(O A\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \end{subparts}
            \part Show that \(r=\frac{h^{2}+d^{2}}{2 h}\).
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part If the window is \(120 \mathrm{~cm}\) wide and the cap is \(40 \mathrm{~cm}\) high, find:
            \begin{subparts}
                \subpart the radius of the circle
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
                \subpart how far below the top of the window the centre of the circle must be placed
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \end{subparts}
            \part If the builder used a circle of radius \(50 \mathrm{~cm}\) and this produced a cap of height \(20 \mathrm{~cm}\), what was the width of the window?
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[4] A group of \(n\) people attend a club meeting. Before the meeting begins, they all shake hands with each other. Write a formula to find \(H\), the number of handshakes exchanged.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
    \end{questions}
    \subsection{Indices}
    \begin{questions}
        \Question[2] Evaluate:
        \begin{parts}
            \part \(2^{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(10^{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[1] Express \(120^{2}\) as a product of powers of prime numbers.
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \Question[8] Simplify and evaluate where possible.
        \begin{multicols}{2}\begin{parts}
            \part \(a^{6} \times a^{7}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(2 x^{3} \times 5 x^{6}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(a^{7} \div a^{4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{12 b^{7}}{6 b^{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{18 p^{10}}{9 p}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(a^{4}\right)^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(2 a^{7}\right)^{3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(3 b^{0}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}\end{multicols}
        \Question[7] Simplify and evaluate where possible.
        \begin{parts}
            \part \[4 a^{2} b^{3} \times 5 a b^{4}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\frac{20 a^{4} b^{2}}{5 a^{2} b}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\frac{24 m^{9} n^{4}}{18 m^{6} n^{2}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\left(3 a^{3} b\right)^{4}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\left(5 a^{2} b\right)^{2} \times 4 a^{4} b^{3}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\frac{8 m^{4} n^{2}}{7 m^{3} n} \div \frac{3 m^{3} n^{5}}{14 m^{9} n^{16}}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \[\frac{\left(2 x^{2} y\right)^{3}}{5 x^{6} y^{2}} \times\left(\frac{x^{3}}{2 y^{2}}\right)^{3}\]
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[6] Evaluate:
        \begin{multicols}{2}\begin{parts}
            \part \(6^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(8^{-3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(2^{-7}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(\frac{4}{5}\right)^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(\frac{2}{3}\right)^{-4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(16^{\frac{1}{4}}\right)^{-3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}\end{multicols}
        \Question[4] Simplify, expressing the answer with positive indices.
        \begin{parts}
            \part \(\frac{4 m^{2} n^{5} p^{-6}}{16 m^{-2} n^{5} p^{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(2^{2} y^{3}\right)^{-5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(5^{-2} x^{3}\right)^{-5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(3^{-3} a^{2} b^{-1}\right)^{-4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] Simplify, expressing the answer with positive indices.
        \begin{parts}
            \part \(4 a^{2} \times 5 a^{-3}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(14 a^{-4} \div 7 a^{-5}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{2 m^{3} n^{4}}{(5 m)^{2}} \times \frac{10 m}{3 n^{-4}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] Evaluate:
        \begin{parts}
            \part \(49^{\frac{1}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(125^{\frac{2}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(\frac{1}{8}\right)^{-\frac{2}{3}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[4] Simplify, expressing the answer with positive indices.
        \begin{parts}
            \part \(3 b^{\frac{2}{3}} \times 4 b\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(p^{\frac{2}{3}} \div p^{\frac{1}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(2 x^{-\frac{1}{3}}\right)^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\left(8 p^{-2} q^{3}\right)^{\frac{1}{2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] Write in scientific notation.
        \begin{parts}
            \part 164000000
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 0.0047
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 0.0035
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[3] Write in decimal form.
        \begin{parts}
            \part \(6.8 \times 10^{4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(9.4 \times 10^{-2}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(3.2 \times 10^{-4}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] Simplify, writing each answer in scientific notation.
        \begin{parts}
            \part \(\left(3.1 \times 10^{4}\right) \times\left(2 \times 10^{-2}\right)\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part \(\frac{\left(3 \times 10^{4}\right)^{3}}{9 \times 10^{-2}}\)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
        \end{parts}
        \Question[6] Write in scientific notation correct to the number of significant figures indicated in the brackets.
        \begin{multicols}{2}\begin{parts}
            \part 18.62\hfill{}(2)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 18.62\hfill{}(3)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 18.62\hfill{}(1)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 0.004276\hfill{}(2)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 5973.4\hfill{}(2)
            \begin{solutionordottedlines}[2cm]
            \end{solutionordottedlines}
            \part 0.473952\hfill{}(3)
        \end{parts}\end{multicols}
    \end{questions}
\end{problembox}

|728985486|kitty|/Applications/kitty.app|0|
         \Question[4] Simplify, expressing the answer with positive indices.                                                                                                    
         \begin{parts}                                                                                                                                                          
             \part \(\frac{4 m^{2} n^{5} p^{-6}}{16 m^{-2} n^{5} p^{3}}\)                                                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=\frac{4}{16} \cdot m^{2-(-2)} \cdot n^{5-5} \cdot p^{-6-3}\) \\                                                                                             
                 \(=\frac{1}{4} \cdot m^{4} \cdot p^{-9}\) \\                                                                                                                   
                 \(=\frac{m^{4}}{4p^{9}}\)                                                                                                                                      
             \end{solutionordottedlines}                                                                                                                                        
             \part \(\left(2^{2} y^{3}\right)^{-5}\)                                                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=(2^{2 \cdot -5}) \cdot (y^{3 \cdot -5})\) \\                                                                                                                
                 \(=2^{-10} \cdot y^{-15}\) \\                                                                                                                                  
                 \(=\frac{1}{2^{10} y^{15}}\)                                                                                                                                   
             \end{solutionordottedlines}                                                                                                                                        
             \part \(\left(5^{-2} x^{3}\right)^{-5}\)                                                                                                                           
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=(5^{-2 \cdot -5}) \cdot (x^{3 \cdot -5})\) \\                                                                                                               
                 \(=5^{10} \cdot x^{-15}\) \\                                                                                                                                   
                 \(=\frac{5^{10}}{x^{15}}\)                                                                                                                                     
             \end{solutionordottedlines}                                                                                                                                        
             \part \(\left(3^{-3} a^{2} b^{-1}\right)^{-4}\)                                                                                                                    
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=(3^{-3 \cdot -4}) \cdot (a^{2 \cdot -4}) \cdot (b^{-1 \cdot -4})\) \\                                                                                       
                 \(=3^{12} \cdot a^{-8} \cdot b^{4}\) \\                                                                                                                        
                 \(=\frac{3^{12} b^{4}}{a^{8}}\)                                                                                                                                
             \end{solutionordottedlines}                                                                                                                                        
         \end{parts}                                                                                                                                                            
         \Question[3] Simplify, expressing the answer with positive indices.                                                                                                    
         \begin{parts}                                                                                                                                                          
             \part \(4 a^{2} \times 5 a^{-3}\)                                                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=4 \cdot 5 \cdot a^{2-3}\) \\                                                                                                                                
                 \(=20 a^{-1}\) \\                                                                                                                                              
                 \(=\frac{20}{a}\)                                                                                                                                              
             \end{solutionordottedlines}                                                                                                                                        
             \part \(14 a^{-4} \div 7 a^{-5}\)                                                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=\frac{14}{7} \cdot a^{-4-(-5)}\) \\                                                                                                                         
                 \(=2 a^{1}\) \\                                                                                                                                                
                 \(=2a\)                                                                                                                                                        
             \end{solutionordottedlines}                                                                                                                                        
             \part \(\frac{2 m^{3} n^{4}}{(5 m)^{2}} \times \frac{10 m}{3 n^{-4}}\)                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=\frac{2 m^{3} n^{4}}{25 m^{2}} \cdot \frac{10 m}{3 n^{-4}}\) \\                                                                                             
                 \(=\frac{2}{25} \cdot \frac{10}{3} \cdot m^{3-2+1} \cdot n^{4+4}\) \\                                                                                          
                 \(=\frac{20}{75} \cdot m^{2} \cdot n^{8}\) \\                                                                                                                  
                 \(=\frac{4}{15} m^{2} n^{8}\)                                                                                                                                  
             \end{solutionordottedlines}                                                                                                                                        
         \end{parts}                                                                                                                                                            
         \Question[3] Evaluate:                                                                                                                                                 
         \begin{parts}                                                                                                                                                          
             \part \(49^{\frac{1}{2}}\)                                                                                                                                         
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=\sqrt{49}\) \\                                                                                                                                              
                 \(=7\)                                                                                                                                                         
             \end{solutionordottedlines}                                                                                                                                        
             \part \(125^{\frac{2}{3}}\)                                                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=\sqrt[3]{125^{2}}\) \\                                                                                                                                      
                 \(=\sqrt[3]{15625}\) \\                                                                                                                                        
                 \(=25\)                                                                                                                                                        
             \end{solutionordottedlines}                                                                                                                                        
             \part \(\left(\frac{1}{8}\right)^{-\frac{2}{3}}\)                                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=(8^{\frac{2}{3}})\) \\                                                                                                                                      
                 \(=\sqrt[3]{8^{2}}\) \\                                                                                                                                        
                 \(=\sqrt[3]{64}\) \\                                                                                                                                           
                 \(=4\)                                                                                                                                                         
             \end{solutionordottedlines}                                                                                                                                        
         \end{parts}                                                                                                                                                            
         \Question[4] Simplify, expressing the answer with positive indices.                                                                                                    
         \begin{parts}                                                                                                                                                          
             \part \(3 b^{\frac{2}{3}} \times 4 b\)                                                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=3 \cdot 4 \cdot b^{\frac{2}{3}+1}\) \\                                                                                                                      
                 \(=12 b^{\frac{5}{3}}\)                                                                                                                                        
             \end{solutionordottedlines}                                                                                                                                        
             \part \(p^{\frac{2}{3}} \div p^{\frac{1}{2}}\)                                                                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=p^{\frac{2}{3}-\frac{1}{2}}\) \\                                                                                                                            
                 \(=p^{\frac{4}{6}-\frac{3}{6}}\) \\                                                                                                                            
                 \(=p^{\frac{1}{6}}\)                                                                                                                                           
             \end{solutionordottedlines}                                                                                                                                        
             \part \(\left(2 x^{-\frac{1}{3}}\right)^{-2}\)                                                                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=2^{-2} \cdot x^{\frac{2}{3}}\) \\                                                                                                                           
                 \(=\frac{1}{4} x^{\frac{2}{3}}\)                                                                                                                               
             \end{solutionordottedlines}                                                                                                                                        
             \part \(\left(8 p^{-2} q^{3}\right)^{\frac{1}{2}}\)                                                                                                                
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=8^{\frac{1}{2}} \cdot p^{-1} \cdot q^{\frac{3}{2}}\) \\                                                                                                     
                 \(=2 \cdot \frac{q^{\frac{3}{2}}}{p}\)                                                                                                                         
             \end{solutionordottedlines}                                                                                                                                        
         \end{parts}                                                                                                                                                            
         \Question[3] Write in scientific notation.                                                                                                                             
         \begin{parts}                                                                                                                                                          
             \part 164000000                                                                                                                                                    
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=1.64 \times 10^{8}\)                                                                                                                                        
             \end{solutionordottedlines}                                                                                                                                        
             \part 0.0047                                                                                                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=4.7 \times 10^{-3}\)                                                                                                                                        
             \end{solutionordottedlines}                                                                                                                                        
             \part 0.0035                                                                                                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=3.5 \times 10^{-3}\)                                                                                                                                        
             \end{solutionordottedlines}                                                                                                                                        
         \end{parts}                                                                                                                                                            
         \Question[3] Write in decimal form.                                                                                                                                    
         \begin{parts}                                                                                                                                                          
             \part \(6.8 \times 10^{4}\)                                                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=68000\)                                                                                                                                                     
             \end{solutionordottedlines}                                                                                                                                        
             \part \(9.4 \times 10^{-2}\)                                                                                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=0.094\)                                                                                                                                                     
             \end{solutionordottedlines}                                                                                                                                        
             \part \(3.2 \times 10^{-4}\)                                                                                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=0.00032\)                                                                                                                                                   
             \end{solutionordottedlines}                                                                                                                                        
         \end{parts}                                                                                                                                                            
         \Question[2] Simplify, writing each answer in scientific notation.                                                                                                     
         \begin{parts}                                                                                                                                                          
             \part \(\left(3.1 \times 10^{4}\right) \times\left(2 \times 10^{-2}\right)\)                                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=3.1 \cdot 2 \times 10^{4-2}\) \\                                                                                                                            
                 \(=6.2 \times 10^{2}\)                                                                                                                                         
             \end{solutionordottedlines}                                                                                                                                        
             \part \(\frac{\left(3 \times 10^{4}\right)^{3}}{9 \times 10^{-2}}\)                                                                                                
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=\frac{3^{3} \times 10^{12}}{9 \times 10^{-2}}\) \\                                                                                                          
                 \(=\frac{27}{9} \times 10^{12+2}\) \\                                                                                                                          
                 \(=3 \times 10^{14}\)                                                                                                                                          
             \end{solutionordottedlines}                                                                                                                                        
         \end{parts}                                                                                                                                                            
         \Question[6] Write in scientific notation correct to the number of significant figures indicated in the brackets.                                                      
         \begin{multicols}{2}\begin{parts}                                                                                                                                      
             \part 18.62\hfill{}(2)                                                                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=1.9 \times 10^{1}\)                                                                                                                                         
             \end{solutionordottedlines}                                                                                                                                        
             \part 18.62\hfill{}(3)                                                                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=1.86 \times 10^{1}\)                                                                                                                                        
             \end{solutionordottedlines}                                                                                                                                        
             \part 18.62\hfill{}(1)                                                                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=2 \times 10^{1}\)                                                                                                                                           
             \end{solutionordottedlines}                                                                                                                                        
             \part 0.004276\hfill{}(2)                                                                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=4.3 \times 10^{-3}\)                                                                                                                                        
             \end{solutionordottedlines}                                                                                                                                        
             \part 5973.4\hfill{}(2)                                                                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=6.0 \times 10^{3}\)                                                                                                                                         
             \end{solutionordottedlines}                                                                                                                                        
             \part 0.473952\hfill{}(3)                                                                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(=4.74 \times 10^{-1}\)                                                                                                                                       
             \end{solutionordottedlines}                                                                                                                                        
         \end{parts}\end{multicols}                                                                                                                                             
     \end{questions}                                                                                                                                                            
 \end{problembox}                                                                                                                                                               
                                                                                                                                                                                
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n|728985668|kitty|/Applications/kitty.app|0|
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9781108404297|729063212|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|
sqlite3 foo.db "INSERT INTO CASHIER VALUES (\"foo\",\"bar\");"|729064142|Brave Browser Beta|/Applications/Brave Browser Beta.app|0|