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\((216^{\frac{1}{3}})^{-2}\) = \(216^{\frac{-2}{3}}\) = (3\sqrt216)\(^{-2}\) = (6)\(^{-2}\) = \(\frac{1}{6^2}\) =\(\frac{1}{36}\)

LCM of 15abc and 20a\(^2\)bc = 60a\(^2\)bc Factors of 15abc = 1, 3, 5,15, a, b, c and of 20a\(^2\)bc = 1, 2, 4, 5, 10, 20, a, b, c HCF of 15abc and 20a\(^2\)bc = 5abc → LCM \div HCF = 60a\(^2\)bc \div 5abc = 12a

|AB| = 4cm |BC| = 8cm In right-angled BAC; 8\(^2\) = 4\(^2\) + |AC|\(^2\) |AC|\(^2\) = 8\(^2\) - 4\(^2\) |AC\\(^2\) = 64 - 16 → 48 |AC| = \(\sqrt{48}\)cm → \(4\sqrt{3}cm\) The area of rectangle = L x B = |AB| x |AC| = \((4 \times 4\sqrt{3}cm^2\) =\(16\sqrt{3}cm^2\)

\(202^2_{three}\)when converted to base ten \(=(202_3)^2\) 202_3 = 2 \(\times 3^2 + 0 \times 3^1 + 2\times 3^0 = 18 + 0 + 2\) =20\(_{ten}; (202_3)^2 = (20)^2_{ten} = 400\) 112\(^2_{three}\)when converted to base ten \(= (112_3)^2\) 112_3 = 1 \(\times 3^2 + 1 \times 3^1 + 2\times 3^0 = 9+3+2=14_{ten}\) (112_3)\(^2 = (14)^2_{ten} = 196_{ten}\) Evaluate \(\Longrightarrow 400-196 = 204\) Reconvert to base three 204 \div 3 = 68 remainder 0 68 \div 3 = 22 remainder 2 22 \div 3 = 7 remainder 1 7 \div 3 = 2 remainder 1 2 \div 3 = 0 remainder 2 = 21120\(_3\)

\(2^{-2(2-y)}-x=2^{0}; -4 + 2y = 0\\ 2y = 4; y = 2\)

If \(log_q p\) = r convert from logarithm to indices P = q\(^r\)

\(1 + 2^2 = 5 + 3^2 = 14 + 4^2 = 30 + 5^2 = 55\) \(55 + 6^2 = 91 + 7^2 = 140\). 91, 140 are the next two terms.

\(\sqrt{128} +\sqrt{18}-\sqrt{k}=7\sqrt{2}\\ \sqrt{2\times 64}+\sqrt{9\times 2} - \sqrt{k} = 7\sqrt{2}\\ 8\sqrt{2} + 3\sqrt{2} - \sqrt{k} = 7\sqrt{2}; 11\sqrt{2} - \sqrt{k} = 7\sqrt{2}\\ -\sqrt{k}=7\sqrt{2}-11\sqrt{2}; -\sqrt{k} = -4\sqrt{2}; \sqrt{k}=4\sqrt{2}\\ =\sqrt{4^2\times 2} = \sqrt{16\times 2}; \sqrt{k}=\sqrt{32}; k = 32\)

Two angles of a triangle are 45º each and its longest side is 12cm, this simply describes an isosceles triangle two equal angles and two equal sides using Pythagoras theorem, \(x^2 + x^2 = 12^2\) \(2x^2\) = 144 \(x^2\) = 72( after dividing b/sides by 2) x = \(\sqrt{72} = 6\sqrt{2}\)cm.

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

\(\frac{\log 8 - \log 2}{\log 9 - \log 2} = \frac{2\log 2}{2\log 3 - \log 2}\) Numerator: \(2 \times 0.3010 = 0.6020\) Denominator: \(2 \times 0.4771 - 0.3010 = 0.6532\) \(\frac{0.6020}{0.6532} = \mathbf{0.92}\)

= \(\frac{0.422}{0.004}\) = \(\frac{4.22 \times 10^{-1}}{4 \times 10^{-3}}\) = \(1.055 \times 10^{-1 - (-3)}\) = \(1.055 \times 10^2\)

(a) \(\frac{1}{81^{(x - 2)}} = 27^{(1 - x)}\) \(81^{-(x - 2)} = 27^{(1 - x)}\) \(3^{4[-(x - 2)]} = 3^{3(1 - x)}\) \(3^{-4(x - 2)} = 3^{3(1 - x)}\) \(-4(x - 2) = 3(1 - x)\) \(-4x + 8 = 3 - 3x\) \(8 - 3 = -3x + 4x \implies x = 5\) (b) \(\frac{5}{\sqrt{7} - \sqrt{3}} + \frac{1}{\sqrt{7} + \sqrt{3}}\) = \(\frac{5(\sqrt{7} + \sqrt{3}) + (\sqrt{7} - \sqrt{3})}{(\sqrt{7} - \sqrt{3})(\sqrt{7} + \sqrt{3})}\) = \(\frac{5\sqrt{7} + 5\sqrt{3} + \sqrt{7} - \sqrt{3}}{7 - 3}\) = \(\frac{6\sqrt{7} + 4\sqrt{3}}{4}\) = \(\frac{3\sqrt{7} + 2\sqrt{3}}{2}\).

(a) \(1001_{2} = 1 \times 2^{3} + 0 \times 2^{2} + 0 \times 2^{1} + 1 \times 2^{0}\) = \(8 + 0 + 0 + 1\) = 9 \(\sqrt{9} = 3_{10}\) 2 3 2 1 r 1 0 r 1 \(\therefore \sqrt{1001_{2}} = 11_{2}\) (b) In \(\Delta QOP, OQ = OP\) \(\therefore QK = KP\) \(\Delta QOK = \Delta POK\) (right- angled triangle) In \(\Delta POK\), \(OP^{2} = PK^{2} + OK^{2}\) \(x^{2} = y^{2} + (\frac{z}{2})^{2}\) \(\frac{z^{2}}{4} = x^{2} - y^{2}\) \(z^{2} = 4x^{2} - 4y^{2}\) \(z = \sqrt{4(x^{2} - y^{2})}\) \(z = 2\sqrt{x^{2} - y^{2}}\) (c)(i) \(< PSQ = \frac{1}{2}(< POQ) = \frac{1}{2}(160^\circ)\) = 80^\circ In \(\Delta PSQ\), \(< PQS + < QPS + < PSQ = 180^\circ\) \(40^\circ + < QPS + 80^\circ = 180^\circ\) \(< QPS = 180^\circ - 120^\circ = 60^\circ\) (ii) \(< PSR = < PSQ + < QSR\) = \(80^\circ + 45^\circ\) = \(125^\circ\) \(< PQS = < PQO + < RQS \) = \(40^\circ + < RQS\) \(< PSR + < PQS = 180^\circ\) \(\therefore 125^\circ + 40^\circ + < RQS = 180^\circ\) \(< RQS = 180^\circ - 165^\circ\) = \(15^\circ\)

To evaluate the expression \( \log_{\sqrt{2}} 4 + \log_{\frac{1}{2}} 16 - \log_{4} 32 \), we can simplify each term separately. 1. Calculate \( \log_{\sqrt{2}} 4 \) \(\log_{\sqrt{2}} 4 = \frac{\log_2 4}{\log_2 \sqrt{2}}\) We know that \( \log_2 4 = 2 \) (since \( 2^2 = 4 \)) and \( \log_2 \sqrt{2} = \log_2 2^{1/2} = \frac{1}{2} \). Thus, \(log_{\sqrt{2}} 4 = \frac{2}{\frac{1}{2}} = 4\) 2. Calculate \( \log_{\frac{1}{2}} 16 \) \(\log_{\frac{1}{2}} 16 = \frac{\log_2 16}{\log_2 \frac{1}{2}}\) We know that \( \log_2 16 = 4 \) (since \( 2^4 = 16 \)) and \( \log_2 \frac{1}{2} = \log_2 2^{-1} = -1 \). Thus, \(\log_{\frac{1}{2}} 16 = \frac{4}{-1} = -4\) 3. Calculate \( \log_{4} 32 \) \(\log_{4} 32 = \frac{\log_2 32}{\log_2 4}\) We know that \( \log_2 32 = 5 \) (since \( 2^5 = 32 \)) and \( \log_2 4 = 2 \). Thus, \(\log_{4} 32 = \frac{5}{2}\) Now, we can substitute these results back into the original expression: \(\log_{\sqrt{2}} 4 + \log_{\frac{1}{2}} 16 - \log_{4} 32 = 4 - 4 - \frac{5}{2}\) Calculating this gives: \(4 - 4 - \frac{5}{2} = 0 - \frac{5}{2} = -\frac{5}{2}\) Thus, the final result is: \(-\frac{5}{2}\) = - 2.5