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Using the cosine rule, we have \(p^{2} = q^{2} + r^{2} - 2qr \cos P\) \(p^{2} = 8^{2} + 6^{2} - 2(8)(6)(\frac{1}{12})\) = \(64 + 36 - 8\) \(p^{2} = 92 \therefore p = \sqrt{92} cm\)

\(\tan \theta = \frac{opp}{adj} = \frac{3}{4}\) \(hyp^{2} = opp^{2} + adj^{2}\) \(hyp = \sqrt{3^{2} + 4^{2}}\) = 5 \(\sin \theta = \frac{3}{5}; \cos \theta = \frac{4}{5}\) \(\sin \theta + \cos \theta = \frac{3}{5} + \frac{4}{5}\) = \(\frac{7}{5} = 1\frac{2}{5}\)

\(\int \frac{1 + x}{x^{3}} \mathrm d x\) = \(\int (\frac{1}{x^{3}} + \frac{x}{x^{3}}) \mathrm d x\) = \(\int (x^{-3} + x^{-2}) \mathrm d x\) = \(\frac{-1}{2x^{2}} - \frac{1}{x} + k\)

\(\int_{0} ^{\frac{\pi}{2}} \sin x \mathrm d x\) = \(- \cos x |_{0} ^{\frac{\pi}{2}\) = \(- \cos (\frac{\pi}{2}) - (- \cos 0)\) = \(0 + 1\) = 1

The expression(P n Q) u R is shaded by identifying two specific areas: 1. (P n Q): the overlapping region between circles P and R 2. u R: the entire area of the circle R

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

F = 50Cos30o = 43.3

This is because volume and the area have only magnitude but no direction

T2 = \(\sqrt{\frac{L^2 \times T_1^2}{L_1}}\) = \(\sqrt{\frac{0.8 \times 2^2}{0.4}}\) = 2\(\sqrt{2}s\)

t = \(\frac{v - u}{a}\) = \(\frac{0 - 20}{-2}\) = 10s

Wt loss = mg - m (g - a) = 70 x 10 - 70 (10 - 1.5) = 105N

\(\frac{0.25 \times 10^3} {1.25 \times 10^3}\) = 0.2

P.E = K.E Therefore, h = \(\frac{v^2}{2g} = \frac{5^2}{2 \times 10}\) = 1.25 Approx 1.3 (C)

P = \(\frac{mgh}{t}\) P = \(\frac{5000 \times 10 \times 60}{(15 \times 60)s}\) P = 3333.3333 P = 3.3 x 103js-1

W = \(\frac{1}{2}Fe\) W = \(\frac{1}{2}\times 20 \times 0.05\) W = 0.5J