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let the chord be AB = 7cm radius OA = 3.7cm distance of the = OM using Pythagoras theorem OA2 = AM2 + OM2 3.72 = 3.52 + OM2 13.69 = 12.25 + OM2 13.69 - 12.25 = OM2 1.44 = OM2 OM = \(\sqrt{1.44}\) OM = 1.2cm distance of chord = 1.2cm

The range is a measure of dispersion that indicates the difference between the highest and lowest values in a dataset.

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

let the number of N50 notes = x let the number of N20 notes = y from the first statement x + y = 13.....(1) from the second statement 50x + 20y = 530 ......(2) dividing through by 10, we have 5x + 2y = 53 ....(20 solving simultaneously by eliminating x + y = 12....(1) 5x + 2y = 53.....(2) multiply eqn (1) by 2 multiply eqn (2) by 1 2x + 2y = 26 -5x + 2y = 53 -3x = -27 x = 9 substitute x = 9 into equation (1) x + y = 13 y = 13 - x = 13 - 9 = 4 y = 4 the number of N50 notes = 9

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

cos x\(\frac{12}{13}\) 132 = 122 + a2 169 = 144 + a2 a2 = 169 - 144 a2 = 25 a \(\sqrt{25}\) a = 5 tan x = \(\frac{5}{12}\) \(\frac{1 - \tan x}{\tan x} = \frac{1 - \frac{5}{12}}{\frac{5}{12}}\) \(\frac{\frac{1 - \frac{5}{12}}{12 - 5}}{12} = \frac{\frac{7}{12}}{\frac{5}{12}}\) = \(\frac{7}{2} \div \frac{5}{12}\) = \(\frac{7}{12} \times \frac{12}{5} = \frac{7}{5}\)

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

\(\sqrt{\frac{8^2 \times 4^{n + 1}}{2^{2n} \times 16}}\) = \(\sqrt{\frac{2^{3(2)} \times 2^{2(n + 1)}}{2^{2n} \times 2^4}}\) = \(\sqrt{\frac{2^6 \times 2^{2n + 2)}}{2^{2n} + 4}}\) = \(\sqrt{\frac{2^6 + 2^{2n + 2)}}{2^{2n} + 4}}\) = \(\sqrt{\frac{2^{2n + 8}}{2^{2n} + 4}}\) = \(\sqrt{2^{2n + 8} \div 2^{2n} + 4}\) = \(\sqrt{2^{2n - 2n} + 8 - 4}\) = \(\sqrt{2^4}\) = \(\sqrt{16}\) = 4

\(\frac{2}{x - 3} - \frac{3}{x - 2}\) = \(\frac{p}{(x - 3)(x - 2)}\) \(\frac{2(x - 2) - 3(x - 3)}{(x - 3)(x - 2)}\) = \(\frac{p}{(x - 3)(x - 2)}\) = \(\frac{2x - 4 - 3x + 9}{(x - 3)(x - 2)}\) = \(\frac{p}{(x - 3)(x - 2)}\) \(\frac{-x + 5}{(x - 3)(x - 2)} = \frac{p}{(x - 3)(x - 2)}\) p(x - 3)(x - 2) = -x + 5(x - 3)(x - 2) p = -x + 5 or p = 5 - x

\(\frac{1}{2}\)(a - b + c) + \(\frac{1}{2}\)(a + b - c) - [\(\frac{1}{2}\) (a - b - c)] \(\frac{1}{2}a - \frac{1}{2}b + \frac{1}{2}c + \frac{1}{2}a + \frac{1}{2}b - \frac{1}{2}c - \frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c\) = \(\frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c\) = \(\frac{1}{2}(a + b + c)\)

tan \(\theta = \frac{opp}{adj} = \frac{6.6}{13}\) tan \(\theta = 0.5077\) \(\theta\) = tan-1 0.5077 \(\theta = 27^o\)

perimeter of minor sector 2r + \(\frac{\theta}{360} \times 2 \pi r\) = 2 x 3.5 + \(\frac{120^o}{360^o} \times 2 \frac{22}{7} \times 3.5\) = 7 + \(\frac{154}{21}\) = 7 + 7.33 = 14.33 = 14\(\frac{1}{3}\)cm

There are 3 triangles: \(\bigtriangleup\)MRS, \(\bigtriangleup\) and \(\bigtriangleup\) PRQ Area PRMS has 2 triangles: \(\bigtriangleup\) MRS and \(\bigtriangleup\) RSP the fraction = \(\frac{2}{3}\)

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

To find the probability that the sum of a randomly selected number from \( P = \{1, 2, 3\} \) and \( Q = \{2, 3, 5\} \) is prime: Possible sums: 1 + 2 = 3 \(\text{prime})\) 1 + 3 = 4 \(\text{not prime})\) 1 + 5 = 6 \(\text{not prime})\) 2 + 2 = 4\(\text{not prime})\) 2 + 3 = 5 \(\text{prime})\) 2 + 5 = 7 \(\text{prime})\) 3 + 2 = 5 \(\text{prime})\) 3 + 3 = 6 \(\text{not prime})\) 3 + 5 = 8\(\text{not prime})\) Prime sums: \( 3, 5, 5, 7 \) (total: 4 favorable outcomes). Total outcomes: \( 3 \times 3 = 9 \). Probability: \(P = \frac{4}{9}\) Thus, the probability that the sum is prime is \(\frac{4}{9}\)