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3x + y = 21 ... (i); xy = 30 ... (ii) From (ii), \(y = \frac{30}{x}\). Putting the value of y in (i), we have 3x + \(\frac{30}{x}\) = 21 \(\implies\) 3x\(^2\) + 30 = 21x 3x\(^2\) - 21x + 30 = 0 3x\(^2\) - 15x - 6x + 30 = 0 3x(x - 5) - 6(x - 5) = 0 (3x - 6)(x - 5) = 0 3x - 6 = 0 \(\implies\) x = 2. x - 5 = 0 \(\implies\) x = 5. If x = 2, y = \(\frac{30}{2}\) = 15; If x = 5, y = \(\frac{30}{5}\) = 6.

\(\tan \theta = \frac{9}{20} = 0.45\) \(\theta = \tan^{-1} (0.45) \) = 24.23^\circ \(\therefore\) The bearing of Y from X = 180^\circ - 24.23^\circ = 155.77^\circ = 156^\circ (to the nearest degree)

\(P = (\frac{Q(R - T)}{15})^{\frac{1}{3}}\) \(P^3 = \frac{Q(R - T)}{15}\) \(Q(R - T) = 15P^3\) \(R - T = \frac{15P^3}{Q}\) \(T = R - \frac{15P^3}{Q}\)

< BAC = \(\frac{130}{2}\) (angle subtended at the centre) < BAC = 65^\circ Also, x = 26^\circ (theorem) y = 65^\circ - 26^\circ = 39^\circ < AOC = 180^\circ - (39^\circ + 39^\circ) = 102^\circ

Since each interior angle = 140^\circ; Each exterior angle = 180^\circ - 140^\circ = 40^\circ Number of sides of the polygon = \(\frac{360^\circ}{40^\circ}\) = 9 Sum of angles in the polygon = 140^\circ x 9 = 1260^\circ

Cost price of the car = N 1,250.00 Selling price = N 1,085.00 Loss = N (1250 - 1085) = N 165.00 % loss = \(\frac{165}{1250} \times 100%\) = 13.2% loss

\(V = \frac{\pi h}{3} (R^2 + Rr + r^2)\) \(V = \frac{\pi R^2 h}{3} + \frac{\pi Rr h}{3} + \frac{\pi r^2 h}{3}\) \(\frac{\mathrm d V}{\mathrm d R} = \frac{2 \pi R h}{3} + \frac{\pi r h}{3}\) = \(\frac{\pi}{3} (2R + r)\)

\((0.0439 \div 3.62)\) = 0.01213 \(\approxeq\) 0.012 = \(\frac{12}{1000}\)

\(25^{1 - x} \times 5^{x + 2} \div (\frac{1}{125})^{x} = 625^{-1}\) \((5^2)^{(1 - x)} \times 5^{(x + 2)} \div (5^{-3})^x = (5^4)^{-1}\) \(5^{2 - 2x} \times 5^{x + 2} \div 5^{-3x} = 5^{-4}\) \(5^{(2 - 2x) + (x + 2) - (-3x)} = 5^{-4}\) Equating bases, we have \(2 - 2x + x + 2 + 3x = -4\) \(4 + 2x = -4 \implies 2x = -4 - 4\) \(2x = -8\) \(x = -4\)

Let the number of days worked by the assistant = t \(\therefore\) The bricklayer worked (t + 10) days. 1500(t + 10) + 500(t) = N 95,000 1500t + 15,000 + 500t = N 95,000 2000t = N 95,000 - N 15,000 2000t = N 80,000 t = 40 days \(\therefore\) The assistant worked for 40 days and received N (500 x 40) = N 20,000

Since A(x, y) is the point of equidistance between B and C, then AB = AC (AB)\(^2\) = (AC)\(^2\) Using the distance formula, (x - 0)\(^2\) + (y - 2)\(^2\) = (x - 2)\(^2\) + (y - 1)\(^2\) x\(^2\) + y\(^2\) - 4y + 4 = x\(^2\) - 4x + 4 + y\(^2\) - 2y + 1 x\(^2\) - x\(^2\) + y\(^2\) - y\(^2\) + 4x - 4y + 2y = 5 - 4 4x - 2y = 1

Education = \(\frac{15}{100} \times N 50,000\) = N 7,500 Food = N 13,600 Electricity = \(\frac{3}{100} \times N 50,000\) = N 1,500 Leftover : N (50,000 - (7,500 + 13,600 + 1,500)) = N 27,400

a \Delta b = a + 3b + 2 (3 \Delta 2) \Delta 5 = (3 + 3(2) + 2) \Delta 5 = 11 \Delta 5 = 11 + 3(5) + 2 = 28

\(M \propto N \) ; \(M \propto \frac{1}{\sqrt{P}}\). \(\therefore M \propto \frac{N}{\sqrt{P}}\) \(M = \frac{k N}{\sqrt{P}}\) when M = 3, N = 5 and P = 25; \(3 = \frac{5k}{\sqrt{25}}\) \(k = 3\) \(M = \frac{3N}{\sqrt{P}}\) when M = 2 and N = 6, \(2 = \frac{3(6)}{\sqrt{P}} \implies \sqrt{P} = \frac{18}{2}\) \(\sqrt{P} = 9 \implies P = 9^2\) P = 81

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