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In the figure above, \(\frac{x}{\sin 60^o} = \frac{10}{\sin 30^o}\) (Sine rule) x = \(\frac{10 \sin 60^o}{\sin 30^o}\) = 10 x \(\frac{\sqrt{3}}{2} \times \frac{1}{2}\) = 10 x \(\frac{\sqrt{3}}{2} \times \frac{2}{1}\) = 10\(\sqrt{3}\)cm

\(\sqrt{12} ( \sqrt{48} - \sqrt{3}\)) \(\sqrt{4 \times 3} (6 \times 3 - \sqrt{3}) = 2 \sqrt{3}(4 \sqrt{3} - \sqrt{3})\) = 2\(\sqrt{3} \times \sqrt{3} (4 - 1) 2\sqrt{9}(3) = 2 \times 3 \times 3 = 18\)

p(4, 3) Q(2 - 1) distance = \(\sqrt{(x_2 - x_1)^2 + (Y_2 - y_1)^2}\) = \(\sqrt{(2 - 4)^2 + (-1 - 3)^2}\) = \(\sqrt{(-2)^2 = (-4)^2}\) = \(\sqrt{4 + 16}\) = \(\sqrt{20}\) = \(\sqrt{4 \times 5}\) = 2\(\sqrt{5}\)

\(\sqrt{\frac{8^2 \times 4^{n + 1}}{2^{2n} \times 16}}\) = \(\sqrt{\frac{2^{3(2)} \times 2^{2(n + 1)}}{2^{2n} \times 2^4}}\) = \(\sqrt{\frac{2^6 \times 2^{2n + 2)}}{2^{2n} + 4}}\) = \(\sqrt{\frac{2^6 + 2^{2n + 2)}}{2^{2n} + 4}}\) = \(\sqrt{\frac{2^{2n + 8}}{2^{2n} + 4}}\) = \(\sqrt{2^{2n + 8} \div 2^{2n} + 4}\) = \(\sqrt{2^{2n - 2n} + 8 - 4}\) = \(\sqrt{2^4}\) = \(\sqrt{16}\) = 4

Log(0.0005957)\(^{\frac{1}{3}}\) = Log(\(\frac{5.957}{10000})^{\frac{1}{3}}\) = \(\frac{1}{3}\)Log(5.957) - \(\frac{1}{3}\)Log10\(^4\) = \(\frac{0.7750}{3}\) - \(\frac{4}{3}\) =\(\frac{\overline3.225}{3}\) = \(\overline{1}\).075

(a) \(3\sqrt{75} = 3 \times \sqrt{25 \times 3} = 3(5\sqrt{3}) = 15\sqrt{3}\) \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\) \(\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}\) \(\therefore 3\sqrt{75} - \sqrt{12} + \sqrt{108} \equiv 15\sqrt{3} - 2\sqrt{3} + 6\sqrt{3}\) = \((15 - 2 + 6)\sqrt{3}\) = \(19\sqrt{3}\). (b) \(124_{n} = 232_{five}\) \(124_{n} = (1 \times n^{2}) + (2 \times n^{1}) + (4 \times n^{0}) \) = \(n^{2} + 2n + 4\) \(232_{five} = (2 \times 5^{2}) + (3 \times 5^{1}) + (2 \times 5^{0})\) = \(50 + 15 + 2\) = \(67\) \(\implies n^{2} + 2n + 4 = 67\) \(n^{2} + 2n + 4 - 67 = 0\) \(n^{2} + 2n - 63 = 0\) \(n^{2} - 7n + 9n - 63 = 0\) \(n(n - 7) + 9(n - 7) = 0\) \((n + 9)(n - 7) = 0\) \(\text{n = -9 or 7}\) Hence, \(n = 7\) \(124_{seven} = 232_{five}\)

\(\frac{1.25 \times 0.025}{0.05}\) \( = \frac{125 \times 10^{-2} \times 25 \times 10^{-3}}{5 \times 10^{-2}}\) = 125 x 5 x 10-3 = 625 x 0.001 = 0.625 = 0.6 Approx to 1 d.p.

\(\frac{3^{-5n}}{9^{1-n}} \times 27^{n + 1}\) \(\frac{3^{-5n}}{3^{2(1-n)}} \times 3^{3(n + 1)}\) \(3^{-5n} \div 3^{2(1-n)} \times 3^{3(n + 1)}\) \(3^{-5n - 2(1-n) + 3(n + 1)}\) \(3^{-5n - 2 + 2n + 3n + 3}\) \(3^{-5n + 5n + 3 - 2}\) \(3^{1}\) = 3

\(\frac{\sqrt{5}(\sqrt{147} - \sqrt{12}}{\sqrt{15}}\) \(\frac{\sqrt{5}(\sqrt{49 \times 3} - \sqrt{4 \times 3}}{\sqrt{5 \times 3}}\) \(\frac{\sqrt{5}(7\sqrt{3} - 2\sqrt{3}}{\sqrt{5} \times \sqrt{3}}\) \(\frac{\sqrt{3} (7 - 2}{\sqrt{3}}\) = 5

\(\sqrt{50} - K\sqrt{8} = \frac{2}{\sqrt{2}}\) \(\sqrt{50} - \frac{2}{\sqrt{2}}\) = K\(\sqrt{8}\) = \(\sqrt{2} \times 25 - \frac{2}{\sqrt{2}}\) = K \(\sqrt{4 \times 2}\) \(\frac{5\sqrt{2}}{1} - \frac{2}{\sqrt{2}}\) = 2K\(\sqrt{2}\) \(\frac{5\sqrt{4} - 2}{\sqrt{2}} = 2K\sqrt{2}\) \(\frac{10 - 2}{\sqrt{2}} = 2K \sqrt{2}\) \(\frac{8}{\sqrt{2}} = \frac{2K\sqrt{2}}{1}\) = 2k\(\sqrt{2} \times \sqrt{2}\) = 8 2k \(\sqrt{4}\) = 8 2k x 2 = 8 4k = 8 k = \(\frac{8}{4}\) k = 2

\(\frac{x^{\frac{1}{2}} - y^{\frac{1}{3}}}{y - x^{\frac{2}{3}}}\) substitute x = 64 and y = 27 \(\frac{64^{\frac{1}{2}} - 27^{\frac{1}{3}}}{27 - 64^{\frac{2}{3}}} = \frac{\sqrt{64} - 3\sqrt{27}}{27 - (3\sqrt{64})^2}\) = \(\frac{8 - 3}{27 - 16}\) = \(\frac{5}{11}\)

From \(\bigtriangleup\) OMQ find /MQ/ by Pythagoras OQ2 = OM2 + MQ2 52 = 22 + MQ2 25 = 4 + MQ2 25 - 4 = MQ2 21 - MQ2 MQ2 = 21 MQ2 = \(\sqrt{21}\) Length of chord = 2 x \(\sqrt{21}\) = 2\(\sqrt{21}\)cm

2 log x (3\(\frac{3}{8}\)) = 6(divide both sides by 2) \(\frac{2 log_x(3 \frac{3}{8})}{2} = \frac{6}{2}\) \(\log_x \frac{27}{8} = 3\) \(\frac{27}{8} = x^3\) \(x^3 = \frac{27}{8}\) x = \(\sqrt{\frac{27}{8}}\) = (\(\frac{27}{8}\))\(\frac{1}{3}\) = \(\frac{(3^3)^{\frac{1}{3}}}{(2^3)^{\frac{1}{3}}}\) = \(\frac{3}{2}\)

(a) \(\frac{\frac{3}{4}(3\frac{3}{8} + 1\frac{5}{8})}{2\frac{1}{8} - 1\frac{1}{2}}\) \(\frac{3}{4}(3\frac{3}{8} + 1\frac{5}{8}) = \frac{3}{4}(\frac{27}{8} + \frac{13}{8})\) = \(\frac{3}{4}(\frac{40}{8})\) = \(\frac{15}{4}\) \(2\frac{1}{8} - 1\frac{1}{2} = \frac{17}{8} - \frac{3}{2}\) = \(\frac{17}{8} - \frac{12}{8}\) = \(\frac{5}{8}\) \(\therefore \frac{\frac{3}{4}(3\frac{3}{8} + 1\frac{5}{8})}{2\frac{1}{8} - 1\frac{1}{2}} = \frac{15}{4} \div \frac{5}{8}\) = \(\frac{15}{4} \times \frac{8}{5}\) = \(6\) (b) \(\log_{10} 2 = 0.3010 ; \log_{10} 3 = 0.4771\) \(\log_{10} 1.125 = \log_{10}(\frac{1125}{1000})\) (Dividing through with 125) = \(\log_{10} (\frac{9}{8})\) = \(\log_{10} 9 - \log_{10} 8\) \(\log_{10} 9 = \log_{10} 3^{2} = 2\log_{10} 3 = 2 \times 0.4771 = 0.9542\) \(\log_{10} 8 = \log_{10} 2^{3} = 3\log_{10} 2 = 3 \times 0.3010 = 0.9030\) \(\therefore \log_{10} (\frac{9}{8}) = 0.9542 - 0.9030 = 0.0512\) \(\approxeq 0.051\) ( 2 significant figures)

(a) Length of the chord = 14 cm Radius = \(\frac{3}{2} \times 14 cm = 21 cm\) The chord AB = \(2r \sin \frac{\theta}{2}\) \(14 = 2(21) \sin \frac{\theta}{2}\) \(\sin \frac{\theta}{2} = \frac{14}{42} = 0.333\) \(\frac{\theta}{2} = \sin^{-1} (0.333)\) \(\frac{\theta}{2} = 19.469^\circ\) \(\theta = 19.469^\circ \times 2 = 38.938^\circ\) Length of the arc = \(\frac{\theta}{360^\circ} \times 2\pi r\) \(\frac{38.938}{360} \times 2 \times \frac{22}{7} \times 21\) = \(14.277 cm\) Perimeter of the remaining portion = 22 + 12 + 12 + 5 + 3 + 14.277 = 68.277 cm \(\approxeq\) 68.28 cm (2 decimal places). (b) \(\frac{3}{\sqrt{3}}(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6}\) = \(\frac{3}{\sqrt{3}}(\frac{12 - \sqrt{36}}{6\sqrt{3}}\) = \(\frac{3}{\sqrt{3}}(\frac{12 - 6}{6\sqrt{3}}\) = \(\frac{3}{\sqrt{3}}(\frac{6}{6\sqrt{3}}\) = \(\frac{18}{6\sqrt{9}}\) = \(\frac{18}{18}\) = 1.