\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} \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} \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}