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\(\frac{av}{1 - v}\) = \(\sqrt[3]{\frac{2v + T}{a + 2T}}\) \(\frac{(av)^3}{(1 - v)^3}\) = \(\frac{2v + T}{a + 2T}\) \(\frac{a^3v^3}{(1^3 - v)^3}\) = \(\frac{2v + T}{a + 2T}\) = \(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\)

log10 a\(\frac{1}{3}\) + \(\frac{1}{4}\)log10 a - \(\frac{1}{12}\)log10a7 = log10 a\(\frac{1}{3}\) + log10\(\frac{1}{4}\) - log10 a\(\frac{7}{12}\) = log\(_{10}\) a\(\frac{7}{12}\) - log\(_{10}\) a\(\frac{7}{12}\) = log\(_{10}\) 1 = 0

(\(\frac{2}{3}\))m (\(\frac{3}{4}\))n = \(\frac{256}{729}\) \(\frac{2^m}{4^n}\) x \(\frac{3^n}{3^m}\) = \(\frac{2^{8}}{3^{6}}\) \(2^{m} \div 2^{2n} = 2^{8}; 3^{n} \div 3^{m} = 3^{-6}\) m - 2n = 8........(i) -m + n = -6........(ii) Solving the equations simultaneously m = 4, n = -2

log\(_7\)49 + log\(_7\)\(\frac{1}{7}\) = log\(_7\)7 = 1

By using cosine formula, p2 = Q2 + R2 - 2QR cos p Cos P = \(\frac{Q^2 + R^2 - p^2}{2 QR}\) = \(\frac{(3)^2 + 2(\sqrt{3})^2 - 3^2}{2\sqrt{3}}\) = \(\frac{3 + 12 - 9}{12}\) = \(\frac{6}{12}\) = \(\frac{1}{2}\) = 0.5 Cos P = 0.5 p = cos-1 0.5 = 60o = < P = 60o If < P = 60o and < Q = 30 < R = 180o - 90o angle P = 60o and angle R is 90o

\(\frac{1}{\sqrt{2} + \sqrt{5}}\) \(\frac{1}{\sqrt{2} + \sqrt{5}} \times \frac{(\sqrt{2} - \sqrt{5})}{(\sqrt{2} - \sqrt{5})}\) = \(\frac{\sqrt{2} - \sqrt{5}}{2 - 5}\) = \(\frac{\sqrt{2} - \sqrt{5}}{-3}\) = \(\frac{1}{3} (\sqrt{5} - \sqrt{2})\)

2log2 x - x log2 (1 + y) = 3 log2 \(\frac{x^2}{(1 + y)^x}\) = 3 = \(\frac{x^2}{(1 + y)^x}\) = 23 = 8 = x2 = 8(1 + y)x

The formula for calculating the diagonal of a cuboid is given by : \(\sqrt{l^2 + w^2 + h^2}\) where w = width, l = length and h = height 9 = \(\sqrt{l^2 + w^2 + h^2}\) 81 = 4\(^2\) + 4\(^2\) + h\(^2\) 81 - 16 - 16 = h\(^2\) h\(^2\) = 49 h = \(\sqrt{49}\) = 7cm NOTE:[l = w = 4 (square base)]

1 UM = 10\(^{-6}\)mm = 10\(^{-9}\)m 1.2 x 10\(^4\) x 10\(^{-9}\)m = 1.2 x 10\(^{-5}\)m = (0.000012)m

\(\frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}\) - \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\) \(\frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}\) = \(\frac{\sqrt{2}}{\sqrt{3}}\) - \(\frac{\sqrt{3}}{\sqrt{2}}\) \(\frac{\sqrt{3} + \sqrt{2}}{3 + \sqrt{2}}\) = \(\sqrt{6}\) + 2 \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\) = \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\) x \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}\) = 5 - 2\(\sqrt{6}\) \(\sqrt{6}\) + 2 - (5 - 2 \(\sqrt{6}\)) = \(\sqrt{6}\) + 2 - 5 + 2\(\sqrt{6}\) = 3\(\sqrt{6}\) - 3 = 3(\(\sqrt{6}\) - 1)

\(\log_{2} y = 3 - \log_{2} x^{\frac{3}{2}}\) When x = 4, \(\log_{2} y = 3 - \log_{2} 4^{\frac{3}{2}}\) \(\log_{2} y = 3 - \log_{2} 2^{3}\) \(\log_{2} y = 3 - 3 \log_{2} 2 = 3 - 3 = 0\) \(\log_{2} y = 0 \implies y = 2^{0} = 1\)

Given that log\(_{10}\)2 = 0.3010, If 10\(^x\) = 0.2 log\(_{10}\)10\(^x\) = log\(_{10}\) 0.2 = x = log\(_{10}\)0.2 But 0.2 = 2 x 10\(^{-1}\), so log\(_{10}\)0.2 = log\(_{10}(2 \times 10^{-1})\), x = log\(_{10}\)2 + log\(_{10}\)10\(^{-1}\) x = 0.3010 - 1 x = - 0.6990

\(\frac{6^{2n + 1} \times 9^n \times 4^{2n}}{18^n \times 2^n \times 12^{2n}}\) = \(\frac{(2 \times 3^{2n + 1} \times 3^{2n}) \times 2^{4n}}{(2 \times 9)^n \times 2^n \times (6 \times 2^{2n})}\) = \(\frac{(2^{2n + 1} \times 3^{2n + 1}) \times 3^{2n} \times 2^{4n}}{2^n \times 3^{2n} \times 2^n \times 2^{4n} \times 3^{2n}}\) = \(\frac{2^{2n} + 1 + 4^n \times 3^{2n} + 1 + 2^n}{2^{n + n + 4n} \times 3^{2n + 4n} \times 3^{2n + 2n}}\) = \(\frac{2^{6n + 1} \times 3^{4n + 1}}{2^{6n} x 3^{4n}}\) = 26n + 1 - 6n x 34n + 1 - 4n 2 x 3 = 6

\(\frac{6.3 \times 10^5}{8.1 \times 10^3}\) \(\frac{7}{9} \times 10^{2}\) = \(0.77778 \times 10^{2}\) = \(77.778 \approxeq 77.80\)

\(4t^{2} + 7t - 1 = 0\) Using a = 4, b = 7, c = -1. \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) = \(\frac{-7 \pm \sqrt{7^{2} - 4(4)(-1)}}{2(4)}\) = \(\frac{-7 \pm \sqrt{49 + 16}}{8}\) = \(\frac{-7 \pm \sqrt{65}}{8}\) = \(\frac{-7 \pm 8.0623}{8}\) The positive answer = \(\frac{-7 + 8.0623}{8} = \frac{1.0623}{8}\) \(\approxeq 0.1328\) (4 decimal place)