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Using law of indices 25\(^{(\frac{1}{2}}\) \times 8\(\frac{-2}{3}\) = \(\sqrt{25}\) x (\(\sqrt[3]{8}\))\(^{ -2}\) = 5 x 2\(^{-2}\) = 5 x \(\frac{1}{2^2}\) = \(\frac{5}{4}\) = \(\frac{11}{4}\) The answer is A

Cos \theta = \(\frac{adj}{hyp}\) = \(\frac{300}{600}\) = 0.5 \theta = Cos - 10.5 = 60 ∠ RPQ = ∠ PQs So the bearing of P from Q is 180 + 60 = 240\(^o\) Answer is D

54 \(_{eight}\) and 67\(_{eight}\) = 1438 Starting with normal addition, 4 + 7 gives 11 (it is more than the base, 8) 8 goes in 11 just 1 time, remaining 3, the remainder will be written, and the 1 will be added to the sum of 5 and 6 which gives 12 altogether, 8 goes in 12 one time remaining 4, the remainder 4 was written and then the 1 that was the quotient was then written since nothing to add the 1 to. So answer is 143 in base eight

\(\sqrt{108} + \sqrt{125} - \sqrt{75}\) = \(\sqrt{3 \times 36} + \sqrt{5 \times 25} - \sqrt{3 \times 25}\) = \(6 \sqrt{3} + 5 \sqrt{5} - 5 \sqrt{3}\) = \(\sqrt{3} + 5\sqrt{5}\)

\([64^{\frac{1}{2}} + 125^{\frac{1}{3}}]^2\) = \([\sqrt{64} + \sqrt[3] {125}]^2\) \([8 + 5]^2\) = \([13]^2\) = 169

Y \(\alpha \frac{1}{x^2} \rightarrow y = \frac{k}{x^2}\) If x = 3 and y = 100, then, \(\frac{100}{1} = \frac{k}{3^2}\) \(\frac{100}{1} = \frac{k}{9}\) k = 100 x 9 = 900 Substitute 900 for k in y = \(\frac{k}{x^2}\); y = \(\frac{900}{x^2}\) = \(yx^2 = 900\)

\(32_4 = 22_x\) \(3 \times 4^1 + 2 \times 4^o\) = \(2 \times x^1 + 2 \times x^o\) 12 + 2 x 1 = 2x + 2 x 1 14 = 2x + 2 14 - 2 = 2x 12 = 2x x = \(\frac{12}{2}\) x = 6

Population of school = 250 + 150 = 400 60% of 250 = \(\frac{\text{60%}}{\text{100%}}\) x 250 = 150 40% of 150 = \(\frac{\text{40%}}{\text{100%}}\) x 150 = 60 Total number of students who plays football; 150 + 60 = 210 Percentage of school that play football; \(\frac{210}{400}\) x 100% = 52.5%

\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1 \(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = \(\log_{10}\)10 \(\log_{10}\)\(\frac{6x - 4}{2}\) - \(\log_{10}\)10 \(\frac{6x - 4}{2}\) = 10 6x - 4 = 2 x 10 = 20 6x = 20 + 4 6x = 20 x = \(\frac{24}{6}\) x = 4

F = \(\frac{9}{5}\)C + 32 When F = 98.6 98.6 = \(\frac{9}{5}\)C + 32 98.6 - 32 = \(\frac{9}{5}\)C 66.6= \(\frac{9}{5}\)C 66.6 x 5 = 9C C = \(\frac{66.6 \times 5}{9}\) = 37

y + 2x = 4 .....(1) y - 3x = -1 ......(2) Subtract (2) from (1) 2x - (-3x) = 4 - (-1) 2x + 3x = 4 + 1 5x = 5 X = \(\frac{5}{5}\) = 1 Substitute 1 for x in (1); y + 2(1) = 4 y + 2 = 4 y = 4 - 2 = 2 Hence, (x + y) = (1 + 2) = 3

If x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6x - 2y}\) \(\frac{x}{y}\) = \(\frac{2}{3}\) and \(\frac{y}{z}\) = \(\frac{3}{4}\) Thus; x = \(\frac{2}{3}T_1\) and z = \(\frac{3}{5}T_1\) y = \(\frac{3}{7}T_2\) and z = \(\frac{4}{7}T_2\) Using y = y \(\frac{3}{5}T_1\) = \(\frac{3}{7}T_2\); \(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\) \(\frac{T_1}{T_2}\) = \(\frac{15}{21}\) \(T_1\) = 15 and \(T_2\) = 21 Therefore; x = \(\frac{2}{5}\) x 15 = 6 y = \(\frac{3}{5}\) x 15 = 9 y = \(\frac{3}{7}\) x 21 = 9 (again) z = \(\frac{4}{7}\) x 21 = 12 Hence; \(\frac{9x + 3y}{6z - 2y}\) = \(\frac{9(6) + 3(9)}{6(12) - 2(9)}\) \(\frac{54 + 27}{72 - 18}\) = \(\frac{81}{54}\) = \(\frac{3}{2}\) = 1\(\frac{1}{2}\)

\(\frac{2 - 18m^2}{1 + 3m}\) = \(\frac{2(1 - 9)m^2}{1 + 3m}\) = \(\frac{2(1 + 3m)(1 - 3m)}{1 + 3m}\) = \(2(1 - 3m)\)

Since the curve cuts the x-axis at x = -2 and x = 3, (x + 2)(x - 3) = 0 \(x^2 - 3x + 2x - 6\) = 0 \(x^2 - x - 6\) = 0 Hence, the equation of the curve is y = \(x^2 - x - 6\)

Volume of a cylinder = \( \pi r^2\)h 385 = \(\frac{22}{7}\) x \(r^2\) x 10 385 x 7 = 22 x \(r^2\) x 10 \(r^2\) = \(\frac{385 \times 7}{22 \times 10}\) = 12.25 r = \(\sqrt{12.25}\) = 3.5m Hence, diameter of tank = 2r = 2 x 3.5 = 7m