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\(\frac{p^{2} - q^{2}}{2pq} = \frac{(p + q)(p - q)}{2pq}\) = \(\frac{(1 + \sqrt{2} - (1 - \sqrt{2}))(1 + \sqrt{2} + 1 - \sqrt{2})}{2(1 + \sqrt{2})(1 - \sqrt{2})}\) = \(\frac{(2\sqrt{2})(2)}{-2}\) = \(-2\sqrt{2}\)

Cos 30 = \(\frac{5}{\text{x}}\) x cos 30 = 5, => x = 5\(\sqrt{3}\) Coordinates of P = - 5\(\sqrt{3}\), 0 Coordinates of Q = 0, 5 Gradient of PQ = (y\(_2\) - y\(_1\)) (X\(_2\) - X\(_1\)) = \(\frac{(5 - 0)}{(0 - 5\sqrt{3}}\)) = \(\frac{5}{5\sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\) Equation of PQ = y - y\(_1\) = m (x -x\(_1\)) y - 0 = \(\frac{1}{\sqrt{3}}\) (x -( -5\(\sqrt{3}\))) Thus: \(\sqrt{3}\)y = x + 5\(\sqrt{3}\)

PQ\(^2\) = (x\(_2\) - x\(_1\))\(^2\) + (y\(_2\) - y\(_1\))\(^2\) = 12\(^2\) + 5\(^2\) = 144 + 25 = 169 PQ = \(\sqrt{169}\) = 13 But PQ = diameter = 2r, r = \(\frac{\text{PQ}}{2}\) = 6.5 units

\(log_{10}5 + log_{10}20\) \(log_{10}[5\times 20]\) \(log_{10}100\) \(log_{10}10^2\) \(2log_{10}10 = 2\)

\(y \propto \frac{1}{\sqrt{x}}\) \(y = \frac{k}{\sqrt{x}}\) When x = 16, y = 2. \(2 = \frac{k}{\sqrt{16}} \implies 2 = \frac{k}{4}\) \(k = 8\) \(y = \frac{8}{\sqrt{x}}\) When y = 24, \(24 = \frac{8}{\sqrt{x}}\) \(\sqrt{x} = \frac{8}{24} = \frac{1}{3}\) \(\therefore x = (\frac{1}{3})^2\) \(x = \frac{1}{9}\)

If \(sin P = \frac{5}{13}\) from right angled triangle from Pythagoras theorem \(BC^2 = 13^2 - 5^2\) = 169 - 25 BC = \(\sqrt{144}\) = 12 ∴ cos P - tan P = \(\frac{12}{13} - \frac{5}{12}\) =\(\frac{79}{156}\)

\((PR)^2 = (PQ)^2 + (QR)^2\) \(10^2 = 5^2 + (QR)^2\) \((QR)^2 = 100 - 25\) \(QR = \sqrt{75}\) = \(8.660\) \(\approxeq\) 8.7 km

Area of trapezium = \(\frac{1}{2}\)( a + b) x h a = 9cm, b = 15cm, c = (15 - 9) = 6cm since height was not given, so we calculate for it using trig . ratio h = 2\(\sqrt{3}\)cm Area of trapezium = \(\frac{1}{2}\)( a + b ) x h = \(\frac{1}{2}\)( 9 + 15 ) x 2\(\sqrt{3}\) = 24 x \(\sqrt{3}\) = 24\(\sqrt{3}\)cm\(^2\)

Using cosine formula (t\(\sqrt{3}\))2 = t2 + t2 - 2t2 cos\(\theta\) 3t2 = 2t2 - 2t2 cos\(\theta\) = 2t2(1 - cos\(\theta\)) 1 - cos\(\theta\) = \(\frac{3t^2}{2t^2}\) = \(\frac{3}{2}\) cos = 1 - \(\frac{3}{2} = -\frac{1}{2}\) \(\theta\) = cos-1(-\(\frac{1}{2}\)) = 120o and 240o N.B 0 \(\geq\) \(\theta\) 360

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

9\(^x\) = 729 3\(^2\)\(^x\) = 3\(^6\) 2x = 6 x = 3

(a) \(\log_{10} 54 \) \(54 = 2 \times 3^{3}\) \(\log_{10} 54 = \log_{10} (2 \times 3^{3})\) = \(\log_{10} 2 + \log_{10} 3^{3}\) = \(\log_{10} 2 + 3\log_{10} 3\) = \(0.3010 + (3 \times 0.4771)\) = \(0.3010 + 1.4313\) = \(1.7323\) (b) \(\log_{10} 0.24\) \(0.24 = \frac{24}{100} = \frac{2^{3} \times 3}{100}\) \(\log_{10} 0.24 = \log_{10} 24 - \log_{10} 100\) \(\log_{10} (2^{3} \times 3) - \log_{10} (10^{2})\) \(\log_{10} 2^{3} + \log_{10} 3 - 2\log_{10} 10\) = \(3\log_{10} 2 + \log_{10} 3 - 2\) = \((3 \times 0.3010) + 0.4771 - 2\) = \(0.9030 + 0.4771 - 2\) = \(-0.6199\)

(a) \(\frac{1}{3^{5n}} \times 9^{n - 1} \times 27^{n + 1}\) \(3^{-5n} \times (3^{2})^{n - 1} \times (3^{3})^{n + 1}\) \(3^{-5n} \times 3^{2n - 2} \times 3^{3n + 3}\) \(3^{-5n + 2n - 2 + 3n + 3}\) = \(3^{1} = 3\) (b) Let the daughter's age be c and the woman's age be d. \(c + d = 46 .... (1)\) In 4 years time, the daughter's age = c + 4 The woman's age = d + 4 \(\frac{d + 4}{c + 4} = \frac{7}{2}\) \(2(d + 4) = 7(c + 4) \implies 2d + 8 = 7c + 28\) \(2d - 7c = 28 - 8 \implies 2d - 7c = 20 ... (2)\) \(c + d = 46 \implies d = 46 - c\) \(\therefore 2(46 - c) - 7c = 20\) \(92 - 2c - 7c = 20 \implies 92 - 20 = 9c\) \(72 = 9c \implies c = 8\) \(d = 46 - c \implies d = 46 - 8 = 38\) Therefore, the daughter is 8 years old and the woman is 38 years.

Rationalize \(\frac{(2\sqrt{3}-\sqrt{2})}{(\sqrt{3}+2\sqrt{2})}\) and equate to \(m +n\sqrt{6}\). Such that m = -2, and n = 1. Use \(\sqrt{3}-2\sqrt{2}\) as the conjugate for Rationalization \(\frac{(2\sqrt{3}-\sqrt{2})}{(\sqrt{3}+2\sqrt{2})}\) X \(\frac{(\sqrt{3}-2\sqrt{2})}{(\sqrt{3}-2\sqrt{2})}\) \(\frac{6 - 4\sqrt{6} - \sqrt{6} + 4}{3 - 2\sqrt{6} + 2\sqrt{6} - 8}\) =\(\frac{10 - 5\sqrt{6}}{-5}\) = -2 + \(\sqrt{6}\) = \(m +n\sqrt{6}\) ∴ m = -2 and n = 1

\(\frac{2.813 \times 10^{-3} \times 1.063}{5.637 \times 10^{-2}}\) = \(\frac{0.002813 \times 1.063}{0.05637}\) \(\approxeq \frac{0.0028 \times 1.1}{0.056}\) = \(0.055\)