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\(\frac{5+\sqrt{7}}{3+\sqrt{7}}=\frac{5+\sqrt{7}}{3+\sqrt{7}}\times \frac{3-\sqrt{7}}{3-\sqrt{7}}\) =\(\frac{(5+\sqrt{7})(3-\sqrt{7})}{3^2 - \sqrt{7}^2}\) =\(\frac{15-5\sqrt{7}+3\sqrt{7}-7}{9-7}\) = \(\frac{8-2\sqrt{7}}{2}\) Factorize, then divide by 2 \(=\frac{2(4-\sqrt{7}}{2})\) =4-\(\sqrt{7}\)

S = t\(^3\) - t\(^2\) - t + 5 ds/dt = 3t\(^2\) - 2t - 1 As ds = 0 3t\(^2\) - 2t - 1 = 0 (3t+1)(t-1) = 0 ∴ t = 1 or -1/3 At min pt t = 1 S = t\(^3\) - t\(^2\) - t + 5 put t = 1 = 1\(^3\) - \1(^2\) - 1 + 5 = 1 - 1 - 1 + 5 = 4

\(5^{2(x - 1)} \times 5^{x + 1} = 0.04\) \(5^{2x - 2} \times 5^{x + 1} = 5^{-2}\) \(2x - 2 + x + 1 = -2\) \(3x - 1 = -2 \implies 3x = -2 + 1 = -1\) \(x = -\frac{1}{3}\)

log (\(\frac{28}{3}) = \log (\frac{7 \times 4}{3})\) = log 7 + log 4 - log 3 log 7 + log \(2^2\) - log 3 = log 7 + 2log2 - log 3 = z + 2x - y = 2x + z - y

(2x + 3 )2 = 125 2n + 3 = 3\(\sqrt{125}\) 2x + 3 = 5 2x = 5 - 3 x = \(\frac{2}{2}\) = 1

92x = \(\frac{1}{3}\)(27x) (22)2x = \(\frac{(3^{3x})}{3}\) (34x) = 33x - 1 4x = 3x - 1 4x - 3x = -1 x = -1

\(\frac{1}{2}\sqrt{32} - \sqrt{18} + \sqrt{2}\) = \(\frac{1}{2} (4\sqrt{2}) - 3\sqrt{2} + \sqrt{2}\) = 2\(\sqrt{2} - 3\sqrt{2} + \sqrt{2}\) = 3\(\sqrt{2} - 3\sqrt{2} - 3\sqrt{2}\) = 0

(a) \((\sqrt{3} - 5\sqrt{2})(\sqrt{3} + \sqrt{2}) = \sqrt{3 \times 3} + \sqrt{3 \times 2} - 5\sqrt{2 \times 3} - 5\sqrt{2 \times 2}\) = \(3 + \sqrt{6} - 5\sqrt{6} - 5(2)\) = \(-7 - 4\sqrt{6}\) \(\therefore\) a = -7 and b = -4. (b) \(\frac{2^{1 - y} \times 2^{y - 1}}{2^{y + 2}} = 8^{2 - 3y}\) \(\frac{2^{1 - y + y - 1}}{2^{y + 2}} = (2^{3})^{2 - 3y}\) \(2^{0 - (y + 2)} = 2^{6 - 9y} \implies 2^{- y - 2} = 2^{6 - 9y}\) \(- y - 2 = 6 - 9y \implies - y + 9y = 6 + 2\) \(8y = 8 \implies y = 1\).

(a) \(\log (\frac{35 \times 49}{40 \div 56})\) Given \(\log 5 = 0.6990 , \log 7 = 0.8451 , \log 8 = 0.9031\) \(\log (\frac{35 \times 49}{40 \div 56} = \log (\frac{(7 \times 5) \times 7^{2}}{(8 \times 5) \div (8 \times 7)}\) = \(\log (\frac{7^{3} \times 5}{5 \div 7}\) = \(\log 7^{3} + \log 5 - (\log 5 - \log 7)\) = \(3 \log 7 + \log 5 - \log 5 + \log 7\) = \(4 \log 7\) = \(4 \times 0.8471\) = \(3.3804\) (b) (i) Since there are 60 more adults than children, then the number of adults = x + 60 \(\therefore D 5(x + 60) + D x(2) = D 1280\) \(5x + 300 + 2x = 1280\) \(7x + 300 = 1280 \implies 7x = 1280 - 300 = 980\) \(x = \frac{980}{7} = 140\) There were 140 children. (ii) Ratio of children to adults = \(x : x + 60\) = \(140 : (140 + 60)\) = \(140 : 200\) = \(7 : 10\) (iii) Total number of persons at the show : 140 + 200 = 340 persons. Total amount gotten = D 1280 Average paid per person = \( D \frac{1280}{340}\) = \(D \frac{64}{17}\) = D 3.765 (iv) Percentage profit Profit : D (1280 - 720) = D 560 % profit : \(\frac{560}{720} \times 100% = \frac{700}{9} %\) = \(77.78%\)

\(16^{\frac{-1}{2}}\times 4^{\frac{-1}{2}} \times 27^{\frac{1}{3}}\) =\(\frac{1}{16^{\frac{1}{2}}}\times \frac{1}{4^{\frac{1}{2}}}\times 27^{\frac{1}{3}}\) =\(\frac{1}{\sqrt{16}} \times \frac{1}{\sqrt{4}} \times \sqrt[3]{27}\) =\(\frac{1}{4} \times \frac{1}{2} \times \frac{3}{1}\) = \(\frac{3}{8}\)

If log\(_{x^{\frac{1}{2}}}\)64 = 3 (X\(^{\frac{1}{2}})^3\) = 64 (X\(^{\frac{1}{2}})^3\) = 4\(^3\) X\(^{\frac{1}{2}}\) = 4 X = 4\(^2\) X = 16

\(\frac{1+\sqrt{2}}{1-\sqrt{2}} \times \frac{1+\sqrt{2}}{1+\sqrt{2}}\) = \(\frac{1+(1+\sqrt{2})+\sqrt{2}(1+\sqrt{2})}{1^2 - (\sqrt{2})^2}\) = \(\frac{(1+\sqrt{2}+\sqrt{2}+2)}{1-2}\) = \(\frac{3+2\sqrt{2}}{-1}\) = - 3 - 2\(\sqrt{2}\) ∴ X and Y = -3 and -2

\(\sqrt{50} + \frac{10}}{\sqrt{2}} = \(\frac{\sqrt{50}}{1} + \sqrt{10}{\sqrt{2}}\) = \(\frac{\sqrt{50 \times 2} + 10}{\sqrt{2}}\) = \(\frac{\sqrt{100} + 10}{\sqrt{2}}\) = \(\frac{10 + 10}{\sqrt{2} = \frac{20}{\sqrt{2}}\) = \(\frac{20}{\sqrt{2}}\) \times \frac{\sqrt{2}}{\sqrt{2}}\) = \(\frac{20\sqrt{2}}{2}\) = 10\(\sqrt{2}\)

3\(\sqrt{27x^3y^9}\) = 3\(\sqrt{27} \times 3\sqrt{3^3} \times 3\sqrt{y^9}\) = 3 \(\times x \times y^3\) = 3xy3

Given; x = 2; y = \(\frac{-1}{4}\) = \(\frac{x^2y - 2xy}{5}\) = \(\frac{2^2(\frac{-1}{4}) - 2(2)(\frac{-1}{4})}{5}\) = \(\frac{4(\frac{-1}{4}) + 4(\frac{-1}{4})}{5}\) = \(\frac{1 + 1}{5}\) = \(\frac{0}{5}\) = 0