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\(25^{1 - x} \times 5^{x + 2} \div (\frac{1}{125})^{x} = 625^{-1}\) \((5^2)^{(1 - x)} \times 5^{(x + 2)} \div (5^{-3})^x = (5^4)^{-1}\) \(5^{2 - 2x} \times 5^{x + 2} \div 5^{-3x} = 5^{-4}\) \(5^{(2 - 2x) + (x + 2) - (-3x)} = 5^{-4}\) Equating bases, we have \(2 - 2x + x + 2 + 3x = -4\) \(4 + 2x = -4 \implies 2x = -4 - 4\) \(2x = -8\) \(x = -4\)

Education = \(\frac{15}{100} \times N 50,000\) = N 7,500 Food = N 13,600 Electricity = \(\frac{3}{100} \times N 50,000\) = N 1,500 Leftover : N (50,000 - (7,500 + 13,600 + 1,500)) = N 27,400

a \Delta b = a + 3b + 2 (3 \Delta 2) \Delta 5 = (3 + 3(2) + 2) \Delta 5 = 11 \Delta 5 = 11 + 3(5) + 2 = 28

\(M \propto N \) ; \(M \propto \frac{1}{\sqrt{P}}\). \(\therefore M \propto \frac{N}{\sqrt{P}}\) \(M = \frac{k N}{\sqrt{P}}\) when M = 3, N = 5 and P = 25; \(3 = \frac{5k}{\sqrt{25}}\) \(k = 3\) \(M = \frac{3N}{\sqrt{P}}\) when M = 2 and N = 6, \(2 = \frac{3(6)}{\sqrt{P}} \implies \sqrt{P} = \frac{18}{2}\) \(\sqrt{P} = 9 \implies P = 9^2\) P = 81

\(T_n = ar^{n - 1}\) (terms of a G.P) \(T_3 = ar^2 = 9\) ... (i) \(T_7 = ar^6 = \frac{1}{9}\) ... (ii) Divide (i) by (ii); \(\frac{ar^6}{ar^2} = \frac{\frac{1}{9}}{9}\) \(r^4 = \frac{1}{81}\) \(r^4 = (\frac{1}{3})^4\) \(r = \frac{1}{3}\)

\mu = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} A = {3, 6, 9, 12, 15, 18} B = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} A ∩ B = {3, 9, 15}

For the committee to include 2 females, we must have 3 males, so that there should be 5 members. That is, \(^4C_3 \times ^3C_2\) = \(\frac{4!}{(4 - 3)! 3!} \times \frac{3!}{(3 - 2)! 2!}\) = 4 \times 3 = 12 ways

\(\frac{2\log_{3} 9 \times \log_{3} 81^{-2}}{\log_{5} 625}\) = \(\frac{2\log_3 3^2 \times \log_3 (3^4)^{-2}}{\log_5 (5^4)}\) = \(\frac{4\log_3 3 \times -8\log_3 3}{4\log_5 5}\) = -8.

\(\frac{2 + 2x}{3} - 2 \geq \frac{4x - 6}{5}\) \(\frac{2 + 2x - 6}{3} \geq \frac{4x - 6}{5}\) \(\frac{2x - 4}{3} \geq \frac{4x - 6}{5}\) \(5(2x - 4) \geq 3(4x - 6)\) \(10x - 20 \geq 12x - 18\) \(10x - 12x \geq -18 + 20\) \(-2x \geq 2\) \(x \leq -1\)

y = 8x\(^3\) - 3x\(^2\) + 7x - 1 \(\frac{\mathrm d^2 y}{\mathrm d x^2} = \frac{\mathrm d}{\mathrm d x} (\frac{\mathrm d y}{\mathrm d x})\) = \(\frac{\mathrm d}{\mathrm d x} (24x^2 - 6x + 7)\) \(\frac{\mathrm d^2 y}{\mathrm d x^2} = 48x - 6\)

|AC| = |DF| = 13 cm Using Pythagoras theorem, |AC|\(^2\) = |AB|\(^2\) + |BC|\(^2\) 13\(^2\) = 12\(^2\) + |BC|\(^2\) |BC|\(^2\) = 169 - 144 = 25 |BC| = \(\sqrt{25}\) = 5 cm Volume of triangular prism = \(\frac{1}{2} \times base \times length \times height\) = \(\frac{1}{2} \times 5 \times 12 \times 18\) = 540 cm\(^3\)

\(\frac{2\sin 30 + 5\tan 60}{\sin 60}\) \(\sin 30 = \frac{1}{2}; \tan 60 = \sqrt{3}; \sin 60 = \frac{\sqrt{3}}{2}\) \(\therefore \frac{2\sin 30 + 5\tan 60}{\sin 60} = \frac{2(\frac{1}{2}) + 5(\sqrt{3})}{\frac{\sqrt{3}}{2}}\)\) = \(\frac{1 + 5\sqrt{3}}{\frac{\sqrt{3}}{2}}\) = \(\frac{2(1 + 5\sqrt{3})}{\sqrt{3}}\) = \(\frac{2 + 10\sqrt{3}}{\sqrt{3}}\) Rationalizing, we get = \(\frac{2\sqrt{3} + 30}{3}\) = \(\frac{2}{3} \sqrt{3} + 10\)

(a) \(\frac{\log_28 + \log_216 - 4\log_22}{\log_416}\) = \(\frac{\log_22^3 + \log_2 2^4 - 4\log_22}{\log_44^2}\) = \(\frac{3 \log_2 2 + 4\log_22 - 4 \log_2^2}{2 \log 4^4}\) But log\(_2\) 2 = \(\log_44 = 1\) Then, \(\frac{\log_28+\log_216 - 4 \log_2^2}{\log_416}\) = \(\frac{3 + 4 - 4}{2}\) = 1\(\frac{1}{2}\) (b) Subtract 241\(_{five}\) to get 1101\(_{five}\). Thereafter, converting to base ten to find the value of x = (5\(^3\) x 1) + (5\(^2\) x 1 + (5\(^1\) x 0) + (5\(^o\) x 1) = 125 + 25 + 0 + 1 = 151 (b) They subtracted 241\(_{five}\) from 1342\(_{five}\) to get 1101\(_{five}\). Thereafter, converting to base ten to find the value of x = (5\(^3\) x 1) + (5\(^2\) x 1) + (5\(^1\) x 0) + (5^\circx 1) = 125 + 25 + + 1 = 151.

(a) since the curved surface areas were the same, it was expected that they have 5 x 12\(pi\) = 6 \(\pi\)l and solving for the slant height to get l = 10 cm. Then by the use of Pythagoras theorem, the height of the second cone is h\(^2\) = 10\(^2\) - 6\(^2\) and when simplified will result in h = 8 cm. (b(i), they were to find BA = \(\begin{pmatrix} 3 & -2 \\4 & 1\end{pmatrix}\) \(\begin{pmatrix} 2 & 5 \\ -1 & -3\end{pmatrix}\) = \(\begin{pmatrix} 8 & 21 \\ 7 & 17 \end{pmatrix}\) (b)(ii), the determinant of BA = (8 x 17) - (21 x 7) = 136 - 147 = -11

(a)110\(_x\) - 4\(0_5\) 1 x \(x^2}\) + 1 x \(x^1\) + 0 x \(x^o\) x = 4 x 5\(^1\) + 0 x 5\(^o\) x\(^2\) + x = 4 x 5 + 0 x\(^2\) + x - 20 = 0 (x - 4)(x + 5) = 0 x - 4 = 0 or x + 5 = 0 x = 4 or x = -5 x = 4 (b) Simplify \(\frac{15}{\sqrt{75}}\) + \(\sqrt{108}\) + \(\sqrt{432}\) = \(\frac{15}{\sqrt{25 \times 3}}\) + \(\sqrt{36 \times 3}\) = \(\frac{15}{5 \sqrt{3}\) + 6\(\sqrt{3}\) + 12\(\sqrt{3}\) = \(\frac{15 \sqrt{3}}{15 \times \sqrt{3} \times \sqrt{3}}\) + 18\(\sqrt{3}\) = \(\frac{15\sqrt{3}}{15}\) + 18\(\sqrt{3}\) \(\sqrt{3}\) + 18\(\sqrt{3}\) = 19\(\sqrt{3}\)