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Using Pythagoras theorem PR\(^2\) = 5\(^2\) + 12\(^2\) 25 + 144 = 169 PR = \sqrt(169)= 13km

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

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

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

\(\log_{10}\)(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

|MN| = \(\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\) = \(\sqrt{(-3 -5)^2 + (8 - 7)^2}\) = \(\sqrt{(-8)^2 + (8 + 7)^2}\) = \(\sqrt{64 + (15)^2}\) = \(\sqrt{64 + 225}\) = \(\sqrt{289}\) = 17 units

2 x 3, 4 x 6, 8 x 9, 16 x 12,... 2\(^1\) x 3 x 1, 2\(^2\) x 3 x 2, 2\(^3\) x 3 x 3, 2\(^4\) x 3 x 4,.... 2\(^n\) x 3n

From the diagram, h\(^2\) = 4\(^2\) + 3\(^2\) (pythagoras') h\(^2\) = 16 + 9 = 25 h = \(\sqrt{25}\) = 5 Hence, sin x - cos x = \(\frac{4}{5} - \frac{3}{5}\) = \(\frac{1}{5}\)

(a) (i) In the diagram, \(x^\circ + 90^\circ = (3x + 15)^\circ\) (Sum of opposite interior angles) Thus, \(90^\circ - 15^\circ = 3x - x \implies 75^\circ = 2x\) \(x = 37.5^\circ\) (ii) < RSQ = 90^\circ - 37.5^\circ = 52.5^\circ (b) \(2N4_{seven} = 15N_{nine}\) \(2N4_{seven} = (2 \times 7^{2}) + (N \times 7^{1}) + (4 \times 7^{0})\) = \(98 + 7N + 4\) = \(102 + 7N\) \(15N_{nine} = (1 \times 9^{2}) + (5 \times 9^{1}) + (N \times 9^{0})\) = \(81 + 45 + N\) = \(126 + N\) \(\implies 102 + 7N = 126 + N\) \(7N - N = 126 - 102 = 24\) \(6N = 24 \implies N = 4\) Therefore, \(244_{seven} = 154_{nine}\).

(a) In triangle SPQ, \(|SQ|^{2} = 5^{2} + 12^{2}\) (Pythagoras theorem) = \(25 + 144 = 169\) \(|SQ| = \sqrt{169} = 13 cm\) Angle b is common to triangles SPQ and PRS are similar. Using \(\sin b = \frac{12}{|SQ|} = \frac{|PR|}{5}\) \(\sin b = \frac{12}{13} = \frac{|PR|}{5}\) \(|PR| = \frac{12 \times 5}{13} \approxeq 4.62 cm\) (to 3 s.f) (b)(i) \(h^{2} = 10^{2} - 8^{2} = 36\) \(h = \sqrt{36} = 6 cm\) (ii) In the smaller triangle, \(\cos x = \frac{8}{10} = 0.8\) \(\cos^{-1} (0.8) = 36.87^\circ\) Since these are corresponding angles, x = x in the bigger triangle. \(\sin x = \frac{y}{12}\) \(y = 12 \sin x = 12 \sin 36.87\) = \(12 \times 0.6\) = 7.20 m The ladder M touches the wall at a height 7.2 m above the horizontal ground.

(a) Applying the laws of logarithm, log(y + 2) = log ax which amplies that y + 2 = ax and finding x to get x = \(\frac{y + 2}{a}\) (b)(i) The pie is shown (b)(i) The percentage of timber produced that year that was from Amenfi = \(\frac{900}{7200}\) x 100% = 12.5% (ii) Difference in revenue = $560.00(1800 - 600) = $672,000.00

Y = Px2 + q Y = 2x2 - 1 Px2 + q = 2x2 - 1 Px2 = 2x2 - 1 - q p = \(\frac{2x^2 - 1 - q}{x^2}\) at x = 2 P = \(\frac{2(2)^2 - 1 - q}{2^2}\) = \(\frac{2(4) - 1 -q}{4}\) = \(\frac{8 - 1 - q}{4}\) P = \(\frac{7 - q}{4}\)

3\(^{n - 1}\) \times \(\frac{27^{n + 1}}{81^n}\) = 3\(^{n - 1}\) \times \(\frac{3^{3(n + 1)}}{3^{4n}}\) = 3\(^{n - 1 + 3n + 3 − 4n}\) = 3\(^{4n − 4n − 1 + 3}\) = \(3^{2}\) = 9

when x=2, we have \(2 + 2(2^{2})\) = \(2 + 8 = 10\)

4\(\sqrt{27}\) + 5\(\sqrt{12}\) − 3\(\sqrt{75}\) = 4\(\sqrt{3}\) × 9 + 5\(\sqrt{3}\) × 4 − 3\(\sqrt{3}\) × 25 = 4 × 3\(\sqrt{3}\) + 5 × 2\(\sqrt{3}\) − 3 × 5\(\sqrt{3}\) = 12\(\sqrt{3}\) + 10\(\sqrt{3}\) − 15\(\sqrt{3}\) = (12 + 10 − 15)\(\sqrt{3}\) = 7\(\sqrt{3}\)