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2\(^7\) = 8\(^{5-x}\) 2\(^7\) = 2\(^{3(5 - x)}\) 7 = 15 - 3x 7 - 15 = - 3x -8 = -3x x = \(\frac{8}{3}\)

Using Pythagoras' theorem \(x^2+(x-1)^2 = (\sqrt{13})^2\) x\(^2\) + x\(^2\) - 2x + 1 = 13 2x\(^2 - 2x\) - 12 = 0 dividing through by 2 x\(^2\) - x- 6 = 0; (x-3)(x-2) = 0 x = 3 or -2

\(A=\pi r^2 \Rightarrow 77 = \frac{22}{7} \times \frac{r^2}{1}\Rightarrow r^2 = \frac{77\times 7}{22} \Rightarrow r^2 = \frac{49}{2}\\ \Rightarrow r = \sqrt{\frac{49}{2}}=\frac{7}{\sqrt{2}}. Since D = 2r ∴ D = 2 \times \frac{7}{\sqrt{2}} = \frac{14}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 7\sqrt{2}cm\)

\(log_9x= 1.5\Rightarrow x = 9^{1.5} = 9^{\frac{3}{2}}=(3^2)^{\frac{3}{2}}=3^3=27\)

To evaluate the expression \( \log 18 + \log 6 - \log 16^{\frac{1}{2}} \), we can use the properties of logarithms. Simplify \( \log 16^{\frac{1}{2}} \) Using the power rule of logarithms: \(\log 16^{\frac{1}{2}} = \frac{1}{2} \log 16\) Substitute back into the expression: Now rewrite the expression: \(\log 18 + \log 6 - \frac{1}{2} \log 16\) Combine the logarithms: Using the product property of logarithms: \(\log 18 + \log 6 = \log (18 \cdot 6) = \log 108\) So we can rewrite the expression as: \(\log 108 - \frac{1}{2} \log 16\) Rewrite \( \frac{1}{2} \log 16\) Using the power rule again: \(\frac{1}{2} \log 16 = \log 16^{\frac{1}{2}} = \log 4\) Substitute back; Now we have: \(\log 108 - \log 4\) Using the quotient property of logarithms: \(\log \left( \frac{108}{4} \right)\) Simplify \(\frac{108}{4}\) Calculating: \(\frac{108}{4}\) = 27 Thus, the expression simplifies to: log 27 = 3log3 NOTE: all is in base 10.

Given that \(x + \frac{1}{x} = 8\) We want to find \(x^2 + \frac{1}{x^2}\) We can use the identity: \(x^2 + \frac{1}{x^2} = \left( x + \frac{1}{x} \right)^2 - 2\) Substituting the given value: \(x^2 + \frac{1}{x^2} = 8^2 - 2\) Calculating \( 8^2 \): \(8^2 = 64\) Now substituting back: \(x^2 + \frac{1}{x^2} = 64 - 2 = 62\) Thus, the final answer is \(x^2 + \frac{1}{x^2} = 62\)

0.0025 \times 2.5 \times 10\(^2\) 0.0025 \times 2.5 \times 10\(^2\) = 0.0025 x 2.5 x 100 = 0.625 ≈ 0.63 to 2 significant figures. 6.3 \times 10\(^{-1}\)

No expanded working has been attached for this question yet. The verified answer is Option C.

(a) \(\log_{10} \sqrt{30} - \log_{10} \sqrt{6} + \log_{10} \sqrt{2}\) = \(\log_{10} (\sqrt{\frac{30}{6}} \times \sqrt{2})\) = \(\log_{10} (\sqrt{5} \times \sqrt{2})\) = \(\log_{10} \sqrt{10}\) = \(\log_{10} (10^{\frac{1}{2}})\) = \(\frac{1}{2} \log_{10} 10\) = \(\frac{1}{2}\). (b) (ii) \(A \cap C = { } \) (iii) \(A \cap B' = {1, 4}\).

(a) \(\sqrt{\frac{0.18 \times 12.5}{0.05 \times 0.2}}\) \(\frac{0.18 \times 12.5}{0.05 \times 0.2} = \frac{18 \times 10^{-2} \times 125 \times 10^{-1}}{5 \times 10^{-2} \times 2 \times 10^{-1}}\) = \(9 \times 25 \times 10^{-3 - (-3)}\) = \(9 \times 25\) = \(225\) \(\therefore \sqrt{\frac{0.18 \times 12.5}{0.05 \times 0.2}} = \sqrt{225}\) = \(15\). (b) \(\frac{8 - 4\sqrt{18}}{\sqrt{50}}\) \(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\) \(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\) \(\frac{8 - 4(3\sqrt{2})}{5\sqrt{2}} = \frac{8 - 12\sqrt{2}}{5\sqrt{2}}\) Rationalising, we have = \(\frac{8 - 12\sqrt{2}}{5\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\) = \(\frac{8\sqrt{2} - 24}{10}\) = \(0.8\sqrt{2} - 2.4\) = \(-2.4 + 0.8\sqrt{2}\) (c) \(x \propto \frac{y^{3}}{z^{2}}\) \(\implies x = \frac{ky^{3}}{z^{2}}\) \(108 = \frac{k \times 3^{3}}{4^{2}}\) \(k = \frac{108 \times 16}{27}\) \(k = 64\) \(\therefore x = \frac{64y^{3}}{z^{2}}\) \(\therefore 4000 = \frac{64 \times 10^{3}}{z^{2}}\) \(4 = \frac{64}{z^{2}}\) \(z^{2} = \frac{64}{4} = 16\) \(z = \sqrt{16} = \pm 4\)

\(\frac{X}{\sin 45^\circ} = \frac{15}{\sin 60^\circ}\) \(X = \frac{15 \sin 45^\circ}{\sin 60^\circ}\) \(X = \frac{15 \times \left( \frac{1}{\sqrt{2}} \right)}{\sqrt{\frac{3}{2}}}\) \(X = \frac{15}{\sqrt{2}} \times \frac{2}{\sqrt{3}}\) \(X = \frac{30}{\sqrt{6}}\) \(X = \frac{30}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{30\sqrt{6}}{6} = 5\sqrt{6}\)

X = 1, 2, 3, 4; ∑X = 10 x = ∑X/n = 10/4 = 2.5 X - x = -1.5, -0.5, 0.5, 1.5 lX - xl = 1.5, 0.5, 0.5, 1.5; ∑lX - xl = 4.0 mean deviation = (∑lX - xl)/n = 4.0/4 = 1.0

\(P\times Q =\sqrt{PQ}\) 4\(\times(8\times32) = 4\times\sqrt{8\times32}\) = 4\(\times\sqrt{256}\) =4\(\times16\) = \(\sqrt{4\times16}\) =\(\sqrt{64}\) = 8

6log\(_x2\) - 3log\(_x3\) = 3log\(_50.2\) = log\(_x2^6\) - 3log\(_x3^3\) = log\(_5(0.2)^3\) = logx(64/27) = log5(1/5)3 logx(64/27) = log5(1/125) let logx(64/27) = y ∴xy = 64/27 and log5(1/125) = y ∴ 5y = 1/125 5y = 125-1 5y = 5-3 ∴ y = -3 substitute y = -3 in xy = 64/27 implies x-3 = 64/27 1/x3 = 64/27 64x3 = 27 x3 = 27/64 x3 = 3\sqrt27/64 x = 3/4

\(\frac{1}{\sqrt{3}+2}\) = \(\frac{1}{\sqrt{3}+2}\times \frac{\sqrt{3} - 2}{\sqrt{3} - 2}\) = \(\frac{\sqrt{3}-2}{(\sqrt{3})^{2} -2^{2}}\) = \(\frac{\sqrt{3}-2}{3-4}\) = \(\frac{\sqrt{3}-2}{1}\) = -\(\sqrt{3}+2\) = 2- \(\sqrt{3}\)