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\(\frac{(1.25 \times 10^{-4}) \times (2.0 \times 10^{-1})}{(6.25 \times 10^5)}\) = \(\frac{1.25 \times 2}{6.25}\) x 10\(^{-4 - 1 - 5}\) \(\frac{2.50}{6.25}\) x 10\(^{-10}\) = \(\frac{250}{625}\) x 10\(^{-10}\) 0.4 x 10\(^{-10}\) = 4.0 x 10\(^{-11}\)

5\(\sqrt{18}\) - 3\(\sqrt{72}\) + 4\(\sqrt{50}\) = 5(3\(\sqrt{2}\)) - 3(6\(\sqrt{2}\)) + 4(5\(\sqrt{2}\)) 15\(\sqrt{2}\) - 18\(\sqrt{2}\) + 20\(\sqrt{2}\) = 35\(\sqrt{2}\) - 18\(\sqrt{2}\) = 17\(\sqrt{2}\)

x = 3 - \(\sqrt{3}\) x2 = (3 - \(\sqrt{3}\))2 = 9 + 3 - 6\(\sqrt{34}\) = 12 - 6\(\sqrt{3}\) = 6(2 - \(\sqrt{3}\)) ∴ x2 + \(\frac{36}{x^2}\) = 6(2 - \(\sqrt{3}\)) + \(\frac{36}{6(2 - \sqrt{3})}\) 6(2 - \(\sqrt{3}\)) + \(\frac{6}{2 - \sqrt{3}}\) = 6(- \(\sqrt{3}\)) + \(\frac{6(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}\) = 6(2 - \(\sqrt{3}\)) + \(\frac{6(2 + \sqrt{3})}{4 - 3}\) 6(2 - \(\sqrt{3}\)) + 6(2 + \(\sqrt{3}\)) = 12 + 12 = 24

2x - 4y + 3 = 0 Required distance = \(\frac{(2 \times 2) + 3(-4) + 3}{\sqrt{2^2} + (-4)^2}\) = \(\frac{4 - 12 + 3}{\sqrt{20}}\) = \(\frac{-5}{-2\sqrt{5}}\) = \(\frac{\sqrt{5}}{2}\)

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

RY = YZ - RZ = 10 tan 45 - 10 tan 30 = 10(1 - \(\frac{1}{\sqrt{3}}\))

(a) \(9^{2x - 1} = \frac{81^{x - 2}}{3^{x}}\) \((3^{2})^{2x - 1} = \frac{(3^{4})^{x - 2}}{3^{x}}\) \(3^{4x - 2} = \frac{3^{4x - 8}}{3^{x}}\) \(3^{4x - 2} = 3^{4x - 8 - x}\) \(3^{4x - 2} = 3^{3x - 8}\) \(\therefore 4x - 2 = 3x - 8 \implies 4x - 3x = -8 + 2\) \(x = -6\) (b) \(\sqrt{\frac{0.81 \times 10^{-5}}{2.25 \times 10^{7}}}\) = \(\sqrt{\frac{0.81}{2.25}} \times \sqrt{10^{-5 - 7}}\) = \(\frac{0.9}{1.5} \times \sqrt{10^{-12}}\) = \(0.6 \times 10^{-6}\) = \(6 \times 10^{-7}\)

(a) \(2x + 5y = \frac{13}{2} \implies 4x + 10y = 13 ... (1)\) \(5x - 2y = 9 ... (2)\) \((2) \times 5 : 25x - 10y = 45 ... (2a)\) \((1) + (2a) : 29x = 58 \implies x = 2\) \(4(2) + 10y = 13 \implies 10y = 13 - 8 = 5\) \(y = \frac{5}{10} = 0.5\) (b) \(\log_{10} (2x + 1) - \log_{10} (3x - 2) = 1\) \(\log_{10} (\frac{2x + 1}{3x - 2}) = 1\) \(\frac{2x + 1}{3x - 2} = 10^{1} = 10\) \(2x + 1 = 10(3x - 2) \implies 2x + 1 = 30x - 20\) \(21 = 28x \implies x = \frac{3}{4}\)

(a) \(17x = 375^{2} - 356^{2}\) \(17x = (375 + 356)(375 - 356)\) \(17x = (731)(19)\) \(x = \frac{731 \times 19}{17} = 817\) (b) \(4^{x} = 2^{\frac{1}{2}} \times 8\) \(2^{2x} = 2^{\frac{1}{2}} \times 2^{3}\) \(2^{2x} = 2^{3\frac{1}{2}}\) \(\implies 2x = 3\frac{1}{2} \implies x = \frac{7}{4}\) (c) \(S_{n} = \frac{n}{2} (2a + (n - 1)d)\) (sum of terms of an A.P) \(S_{9} = \frac{9}{2} [2a + (9 - 1) d] = \frac{9}{2} [2a + 8d]\) \(72 = 9(a + 4d) \implies 8 = a + 4d ... (1)\) \(S_{9 + 4} = S_{13} = \frac{13}{2} [2a + (13 - 1)d] = \frac{13}{2} [2a + 12d]\) \(72 + 71 = 143 = 13(a + 6d) \implies 11 = a + 6d ... (2)\) \((2) - (1) : 2d = 3 \implies d = \frac{3}{2}\) \(a + 4(1\frac{1}{2}) = 8 \implies a + 6 = 8\) \(\implies a = 8 - 6 = 2\) \(\therefore \text{The A.P is } 2, 3\frac{1}{2}, 5, 6\frac{1}{2}, 8,...\)

log 6 + log 2 - log 12 = \(\log (\frac{6 \times 2}{12})\) = \(\log 1\) = 0

For the log to be 2.6025, there must be three digits before the decimal point.

\((\frac{1}{4})^{-1\frac{1}{2}}\) = \((\frac{1}{4})^{-\frac{3}{2}}\) = \((\sqrt{\frac{1}{4}})^{-3}\) = \((\frac{1}{2})^{-3}\) = \(2^3\) = 8

\(\frac{1}{4}(2^n - 2^{n + 2}\)) = \(\frac{1}{4}(2^n - 2^n \times2^2\)) \(\frac{1}{4} \times2^n(1 - 2^2\)) = \(2^{-2}\times2^n(1 - 2^2\)) \(2^{n - 2}(2^0 - 2^2\)) = \(2^{n - 2} - 2^n\)

2log \(\frac{2}{5}\) - log\(\frac{72}{125}\) + log 9 [\(\frac{2}{5}\))2 x 9] = log \(\frac{4}{25}\) x \(\frac{9}{1}\) x \(\frac{125}{72}\) = log \(\frac{72}{125}\) = log \(\frac{5}{2}\) = log \(\frac{10}{4}\) = log 10 - log 4 = log10 10 - log10 22 = 1 - 2 log2

\(\frac{1}{1 + \sqrt{5}}\) - \(\frac{1}{1 - \sqrt{5}}\) = \(\frac{1 - \sqrt{5} - 1 - \sqrt{5}}{(1 + \sqrt{5}) (1 - \sqrt{5}}\) = \(\frac{-2\sqrt{5}}{1 - 5}\) = \(\frac{-2\sqrt{5}}{- 4}\) = \(\frac{1}{2}\sqrt{5}\)