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Angle of major sector = 360^\circ - 120^\circ = 240^\circ Area of major sector : \(\frac{\theta}{360} \times \pi r^{2}\) r = \(\frac{4\sqrt{3}}{2} = 2\sqrt{3} cm\) Area : \(\frac{240}{360} \times \pi \times (2\sqrt{3})^{2}\) = \(8\pi cm^{2}\)

If you take any point that is equidistant from the line PQ, it lies at a constant distance above and below the line. Therefore, the correct answer is: a pair of parallel lines to PQ.

Midpoint of a line PQ where P has coordinates (\(x_1\), \(y_1\)) and Q has coordinates (\(x_2\), \(y_2\)) is given as \((\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})\). \(\therefore\) If Q has coordinates (r, s), then \(\frac{-2 + r}{2} = 2\) and \(\frac{1 + s}{2} = 3\) \(-2 + r = 4 \implies r = 6\) \(1 + s = 6 \implies s = 5\) Q = (6, 5)

Given P(2, -3) and Q(-5, 1) Midpoint = \((\frac{2 + (-5)}{2}, \frac{-3 + 1}{2})\) = \((\frac{-3}{2}, -1)\) Slope of the line PQ = \(\frac{1 - (-3)}{-5 - 2}\) = \(-\frac{4}{7}\) The slope of the perpendicular line to PQ = \(\frac{-1}{-\frac{4}{7}}\) = \(\frac{7}{4}\) The equation of the perpendicular line: \(y = \frac{7}{4}x + b\) Using a point on the line (in this case, the midpoint) to find the value of b (the intercept). \(-1 = (\frac{7}{4})(\frac{-3}{2}) + b\) \(-1 + \frac{21}{8} = \frac{13}{8} = b\) \(\therefore\) The equation of the perpendicular bisector of the line PQ is \(y = \frac{7}{4}x + \frac{13}{8}\) \(\equiv 8y = 14x + 13 \implies 8y - 14x - 13 = 0\)

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