This section is straight-forward: you substitute the unknown values for the values that the question gives you. We are practising this so that you can do it accurately and quickly.

\begin{examplebox}
    \begin{questions}
        \Question[2] Find the value of the subject when the pronumerals in the formula have the values indicated.
        \begin{parts}
            \part \(F=m a\), where \(a=10, m=3.5\)
            \begin{solutionordottedlines}[1in]
                \begin{align*}
                    F &= ma\\
                    &= 10\times 3.5\\
                    &= 35\\
                \end{align*}
            \end{solutionordottedlines}
            \part \(m=\frac{a+b}{2}\), where \(a=12, b=26\)
            \begin{solutionordottedlines}[1in]
                \begin{align*}
                    m &= \frac{a+b}{2}\\
                    &=\frac{12+26}{2}\\
                    &=19
                \end{align*}
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] The formula for the circumference \(C\) of a circle of radius \(r\) is \(C=2 \pi r\). Find the value of \(C\) when \(r=20\) :
        \begin{parts}
            \part in terms of \(\pi\) (that is, exactly) \(\quad\)
            \begin{solutionordottedlines}[2cm]
                \(C=2 \pi r\)
                \(=40 \pi\)
            \end{solutionordottedlines}
            \part correct to 2 decimal places
            \begin{solutionordottedlines}[2cm]
                \(C=40 \pi\)
                \(=125.663 \ldots\) (using a calculator)
                \(\approx 125.66\) (correct to 2 decimal places)
            \end{solutionordottedlines}
        \end{parts}
        \question[]
        \begin{parts}
            \Part[2] 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 \(11 \mathrm{~cm}\).
            \begin{solutionordottedlines}[2cm]
                \[
                \begin{aligned}
                A & =\frac{1}{2} b h \\
                & =\frac{1}{2} \times 16 \times 11 \\
                & =88
                \end{aligned}
                \]
                The area of the triangle is \(88 \mathrm{~cm}^{2}\).
            \end{solutionordottedlines}
            \Part[2] 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.
            \begin{solutionordottedlines}[2cm]
                \[
                \begin{aligned}
                I & =\frac{P r t}{100} \\
                & =\frac{1000 \times 3.5 \times 6}{100} \\
                & =210
                \end{aligned}
                \]

                The interest payable is \(\$ 210\).
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] 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 \(v=u+a t\).
        \begin{parts}
            \part Find \(u\) if \(a=2, v=15\) and \(t=7\).
                \begin{solutionordottedlines}[2cm]
                    \(v=u+a t\)
                    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 \(1 \mathrm{~m} / \mathrm{s}\).
                \end{solutionordottedlines}
            \part Find \(a\) if \(v=10, u=6\) and \(t=3\).
                \begin{solutionordottedlines}[2cm]
                    \(v=u+a t\)
                    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 \(\frac{4}{3} \mathrm{~m} / \mathrm{s}^{2}\).
                \end{solutionordottedlines}
        \end{parts}
        \Question[2] 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.
        \begin{parts}
            \part Find \(f\) if \(u=2\) and \(v=5\).
            \begin{solutionordottedlines}[3cm]
                \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)

                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 \(f=\frac{10}{7}\). 
            \end{solutionordottedlines}

            \part Find \(u\) if \(f=2\) and \(v=6\).
            \begin{solutionordottedlines}[3cm]
                \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
                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 \(u=3\).
            \end{solutionordottedlines}
        \end{parts}
        \Question[2] 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 \(r\) :
        \begin{parts}
            \part exactly
            \begin{solutionordottedlines}[3cm]
                \(A=\pi r^{2}\)

                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. \()\)
            \end{solutionordottedlines}
            \part correct to 2 decimal places
            \begin{solutionordottedlines}
                \(r \approx 2.52\)
                (correct to 2 decimal places)
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{examplebox}

 \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}