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(0.006)3 + (0.004)3 = 2.16 x 10-7 (2.16 + 0.64) x 10-7 = 2.8 x 10-7

Diameter = 42cm Length of the arc = \(\frac{\theta}{360^\circ} \times \pi d\) = \(\frac{60}{360} \times \frac{22}{7} \times 42cm\) = \(22cm\)

Area of a sector = \(\frac{\theta}{360^\circ} \times \pi r^{2}\) r = 6cm; \(\theta\) = 70^\circ. Area of the sector = \(\frac{70}{360} \times \frac{22}{7} \times 6^{2}\) = \(22 cm^{2}\)

p from LM = \(\sqrt{10^2 - 8^2}\) = \(\sqrt{36}\) = 6cm

2x - y .....(i) x + y.....(ii) from (i) y = 2x - 4 from (ii) y = -x + 2 2x - 4 = -x + 2 x = 2 y = -x + 2 = -2 + 2 = 0 \(x_1\) = 2 \(y_1\) = 0 \(x_2\) = 4 \(y_2\) = 3 Hence, dist. = \(\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}\) = \(\sqrt{(3 - 0)^2 + (4 - 2)^2}\) = \(\sqrt{3^2 + 2^2}\) = \(\sqrt{13}\)

\(\frac{40}{x}\) = tan 30o x = \(\frac{40}{tan 30}\) = \(\frac{40}{1\sqrt{3}}\) = 40\sqrt3m

(XZ)\(^2\) = 3\(^2\) + 4\(^2\) - 2 x 3 x 4 cos60º = 25 - 24\(\frac{1}{2}\) XZ = \(\sqrt{13}\)cm

Let \(3x^{2} - 2x = u\) \(y = \cos u \implies \frac{\mathrm d y}{\mathrm d u} = - \sin u\) \(\frac{\mathrm d u}{\mathrm d x} = 6x - 2\) \(\therefore \frac{\mathrm d y}{\mathrm d x} = (6x - 2) . - \sin u\) = \(- (6x - 2) \sin (3x^{2} - 2x)\)

\(\int \frac{1}{x} + \cos x = ln x - \sin x + k\)

x = \(\frac{\sum x}{N}\) = \(\frac{0}{11}\) = 0 \(\begin{array}{c|c} x & (x - x) & (x - x)^2 \\\hline -5 & -5 & 25 \\ -4 & -4 & 16 \\-3 & -3 & 9 \\ -2 & -2 & 4 \\ -1 & -1 & 1\\ 0 & 0 & 0\\ 1 & 1 & 1\\ 2 & 2 & 4\\ 3 & 3 & 9\\ 4 & 4 & 16 \\5 & 5 & 25\\ \hline & & 110\end{array}\) S.D = \(\sqrt{\frac{\sum(x - x)^2}{\sum f}}\) = \(\sqrt{\frac{110}{11}}\) = \(\sqrt{10}\)

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

\(\frac{5}{\sin 30^o} = \frac{2}{\sin \theta} = \frac{5}{\frac{1}{2}} = \frac{2}{\sin \theta}\) 10 = \(\frac{2}{\sin \theta}\) sin\(\theta\) = \(\frac{2}{10}\) = \(\frac{1}{5}\)

(a) \(\sin x = \frac{5}{13}\) \(AB = \sqrt{13^{2} - 5^{2}} = 12\) \(\cos x = \frac{12}{13}\) \(\tan x = \frac{5}{12}\) \(\frac{\cos x - 2 \sin x}{2 \tan x} = \frac{\frac{12}{15} - 2(\frac{5}{13})}{2(\frac{5}{12})}\) = \(\frac{\frac{12}{13} - \frac{10}{13}}{\frac{5}{6}}\) = \(\frac{2}{13} \times \frac{6}{5} = \frac{12}{65}\) (b) (i) < QRP = 180^\circ - 70^\circ = 110^\circ (angles on a straight line) < RQP = 180^\circ - (110^\circ + 30^\circ) = 40^\circ \(\frac{PR}{\sin 40^\circ} = \frac{PQ}{\sin 110^\circ}\) \(\frac{500}{\sin 40^\circ} = \frac{PQ}{\sin 110^\circ}\) \(PQ = \frac{500 \sin 110^\circ}{\sin 40^\circ} = 730.94m\) \(\approxeq 731 m\) (ii) \(\sin 30^\circ = \frac{QN}{PQ} = \frac{QN}{730.94}\) \(QN = 730.94 \times 0.5 = 365.47 m\) \(\approxeq 365 m\)

Angle of sector: \(Food = \frac{30}{100} \times 360^\circ = 108^\circ\) \(Rent = \frac{18}{100} \times 360^\circ = 64.8^\circ\) \(\text{Car Maintenance} = \frac{25}{100} \times 360^\circ = 90^\circ\) \(Savings = \frac{12}{100} \times 360^\circ = 43.2^\circ\) \(Taxes = \frac{5}{100} \times 360^\circ = 18^\circ\) \(Others = \frac{10}{100} \times 360^\circ = 36^\circ\) (a) (b) Annual Salary = N60,000.00 \(\therefore Savings = N(\frac{12}{100} \times 60,000) = N7,200\)

|AC| = 8.35 cm, < BAD = 70^\circ.