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\(1011_2 + X_7 = 25_{10} = 1011_2 = 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0\) = 8 + 0 + 2 + 1 = \(11_{10}\) \(X_7 = 25_{10} - 11_{10}\) = \(14_{10}\) \(\begin{array}{c|c} 7 & 14 \\ 7 & 2 R 0 \\ & 0 R 2 \end{array}\) X = \(20_7\)

\(\frac{k}{\sqrt{3} + \sqrt{2}}\) = k\(\sqrt{3 - 2}\) \(\frac{k}{\sqrt{3} + \sqrt{2}}\) x \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}\) = k\(\sqrt{3 - 2}\) K(\(\sqrt{3}\) - K\(\sqrt{2}\) ) = K\(\sqrt{3}\) - 2 K\(\sqrt{3}\) - K\(\sqrt{3}\) - K\(\sqrt{2}\) = - 2 - K\(\sqrt{2}\) = - 2 K = \(\frac{- 2}{- \sqrt{2}}\) K = \(\frac{2}{\sqrt{2}}\) x \(\frac{\sqrt{2}}{\sqrt{2}}\) K = \(\frac{2\sqrt{2}}{2}\) = \(\sqrt{2}\)

d2 = (y - y)2 + (x - x)2 5 = 4x2 + 1 = 25= 4x2 + 1 = 4x2 = 25 - 1= 24 x2 = \(\frac{24}{4}\) x = \(\sqrt{6}\)

Length of chord = \(2r \times \sin(\frac{\theta}{2})\) = \( 2 \times \sqrt{3} \times \sin(\frac{60}{2})\) = \(2 \times \sqrt{3} \times \frac{1}{2}\) = \(\sqrt{3}\) cm.

(a)(i) \(\log_{10} 35 = \log_{10} (7 \times 5) = \log_{10} 7 + \log_{10} 5\) = \(0.8451 + 0.6990 = 1.5441\) (ii) \(\log_{10} 2.8 = \log_{10} (\frac{4 \times 7}{10} = \log_{10} 4 + \log_{10} 7 - \log_{10} 10\) = \(\log_{10} 2^{2} + \log_{10} 7 - 1\) = \(2\log_{10} 2 + \log_{10} 7 - 1 = 2(0.3010) + 0.8451 - 1\) = \(0.4471\) (b) \(N^{0.8942} = 2.8\) \(0.8942 \log N = \log 2.8\) \(\log N = \frac{\log 2.8}{0.8942} = \frac{0.4471}{0.8942}\) \(\log N = 0.5\)

(a) \(2Ax - Kx^{2}\) When x = 1, \(2A(1) - K(1^{2}) = 7\) \(2A - K = 7 ... (1)\) When x = 2, \(2A(2) - K(2^{2}) = 4\) \(4A - 4K = 4 \implies A - K = 1 ... (2)\) From (2), \(A = 1 + K\) (1) becomes : \(2(1 + K) - K = 7\) \(2 + 2K - K = 7 \implies 2 + K = 7\) \(\therefore K = 5\) \(A = 1 + K = 1 + 5 = 6\) \(A, K = 6, 5\) (b) \(x^{2} - 3x - 1 = 0\) \(a = 1, b = -3, c = -1\) Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) \(x = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4(1)(-3)}}{2(1)}\) \(x = \frac{3 \pm \sqrt{9 + 12}}{2}\) \(x = \frac{3 \pm \sqrt{21}}{2} = \frac{3 \pm 4.583}{2}\) \(x = \frac{3 + 4.583}{2} ; x = \frac{3 - 4.583}{2}\) \(x = \frac{7.583}{2} = 3.7915 \approxeq 3.8\) (to 1 d.p) \(x = \frac{-1.583}{2} = -0.7915 \approxeq -0.8\) (to 1 d.p)

(64.764 - 35.236)(64.764 + 35.236) = 29.528 x 100 = 2950 to 3 sig. fig.

r = log9/log3 = 2log3/log3 = 2

\(\frac{1}{2} \log_{10} P = 1\) \(\log_{10} P^{\frac{1}{2}} = 1\) \(P^{\frac{1}{2}} = 10^{1}\) \(\implies P = 10^{2}\)

8\({\frac{x}{2}}\) = \(2^{\frac{3}{8}})(4^{\frac{3}{4}}\)) applying the laws of indices. (\(2^{3\times (\frac{x}{2})}) = (2^{\frac{3}{8}})(2^{2\times (\frac{3}{4})}\)) \(2^{\frac{3x}{2}} = (2^{\frac{3}{8}})(2^{(\frac{3}{2})}\)) \(\frac{3x}{2} = \frac{3}{8} + \frac{3}{2}\) \(\frac{3x}{2} = \frac{3 + 12}{8}\) \(\frac{3x}{2} = \frac{15}{8}\) 24x = 30 x = \(\frac{30}{24} = \frac{5}{4}\)

\(\frac{(2\sqrt{3} + 3\sqrt{5})(3\sqrt{5} + 2\sqrt{3})}{(3\sqrt{5} - 2\sqrt{3})(3\sqrt{5} - 2\sqrt{3})}\) = \(\frac{57 + 12\sqrt{15}}{33}\) =\(\frac{19 + 4\sqrt{15}}{11}\)

p from LM = \(\sqrt{10^2 - 8^2}\) = \(\sqrt{36}\) = 6cm

2x - y .....(i) x + y.....(ii) from (i) y = 2x - 4 from (ii) y = -x + 2 2x - 4 = -x + 2 x = 2 y = -x + 2 = -2 + 2 = 0 \(x_1\) = 2 \(y_1\) = 0 \(x_2\) = 4 \(y_2\) = 3 Hence, dist. = \(\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}\) = \(\sqrt{(3 - 0)^2 + (4 - 2)^2}\) = \(\sqrt{3^2 + 2^2}\) = \(\sqrt{13}\)

(XZ)\(^2\) = 3\(^2\) + 4\(^2\) - 2 x 3 x 4 cos60º = 25 - 24\(\frac{1}{2}\) XZ = \(\sqrt{13}\)cm

x = \(\frac{\sum x}{N}\) = \(\frac{0}{11}\) = 0 \(\begin{array}{c|c} x & (x - x) & (x - x)^2 \\\hline -5 & -5 & 25 \\ -4 & -4 & 16 \\-3 & -3 & 9 \\ -2 & -2 & 4 \\ -1 & -1 & 1\\ 0 & 0 & 0\\ 1 & 1 & 1\\ 2 & 2 & 4\\ 3 & 3 & 9\\ 4 & 4 & 16 \\5 & 5 & 25\\ \hline & & 110\end{array}\) S.D = \(\sqrt{\frac{\sum(x - x)^2}{\sum f}}\) = \(\sqrt{\frac{110}{11}}\) = \(\sqrt{10}\)