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6ax - 12by - 9ay + 8bx = 6ax - 9ay + 8bx - 12by = 3a(2x - 3y) + 4b(2x - 3y) = (3a + 4b)(2x - 3y)

Let the given positive number be x Then x + 42 = \(x^2\) 0 = \(x^2\) - x - 42 or \(x^2\) - x - 42 = 0 \(x^2\) - 7x + 6x - 42 = 0 x(x - 7) + 6(x - 7) = 0 = (x + 6)(x - 7) = 0 x = -6 or x = 7 Hence, x = 7

Using V = \(\frac{3}{1} \pi r^2h\), so, 38\(\frac{1}{2} = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 3\) \(\frac{77}{2} = \frac{22}{7} \times r^2\) r2 = \(\frac{77 \times 7}{2 \times 22}\) r2 = \(\frac{49}{4}\) Hence, r = \(\sqrt{\frac{49}{4}}\) = 3\(\frac{1}{2}\)

Volume of rectangular tank = L x B x H = 2 x 7 x 11 = 154cm3 volume of cylindrical tank = \(\pi r^2h\) 154 = \(\frac{22}{7} \times r^2 \times 4\) r2 = \(\frac{154 \times 7}{22 \times 4}\) = \(\frac{49}{4}\) r = \(\sqrt{\frac{49}{4}} = \frac{7}{2}\) = 3\(\frac{1}{2}\)m

Third quartile Q3 = \(\frac{3}{4}\)Nth Alternatively: Split the score into equal quarters 44, 47, 48, 49, 50, 51, 52, 53, 53, 54, 58, 59, 60, 61, 63, 65, 67, 70, 73, 75. Third quartile Q3 = [63 + 65] / 2 = 64

In the diagram, \(\frac{16}{12} = \frac{VX}{6}\) (similar \(\Delta\)s) VX = \(\frac{16 \times 6}{12}\) = 8cm

Let Tom's present age be x. Then x = 25 - n If Tom is 5 years younger than Bade, then Bade's present age is x + 5 = 25 - n + 5 = (30 - n)

\(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) = \(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) = \(\frac{\sqrt{2} \times \sqrt{3} + \sqrt{3} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) = \(\frac{\sqrt{6} + 3}{3}\) = \(\frac{3 + \sqrt{6}}{3}\) = Hence, (m + n) = 1 + \(\frac{1}{3}\) = 1\(\frac{1}{3}\)

T = \(\sqrt{\frac{TK - H}{K - H}}\) Taking the square of both sides, give T2 = \(\frac{TK - H}{K - H}\) T2(K - H) = TK - H T2K - T2H = TK - H T2K - TK = T2H - H K(T2 - T) = H(T2 - 1) K = \(\frac{H(T^2 - 1)}{T^2 - T}\)

\(\frac{3x - y}{xy} - \frac{2x + 3y}{2xy} + \frac{1}{2}\) = \(\frac{2(3x - y) - 1(2x + 3y) + xy}{2xy}\) = \(\frac{6x - 2y - 2x - 3y + xy}{2xy}\) = \(\frac{4x - 5y + xy}{2xy}\)

Let x represent the entire farmland then, \(\frac{2}{5}\)x + \(\frac{1}{3}\)[x - \(\frac{2}{3}x\)] + M = x Where M represents the part of the farmland used for growing maize, continuing \(\frac{2}{5}\)x + \(\frac{1}{3}\)x [1 - \(\frac{2}{3}x\)] + M = x \(\frac{2}{5}x + \frac{1}{3}\)x [\(\frac{3}{5}\)] + M = x \(\frac{2}{5}\)x + \(\frac{1x}{5}\) + M = x \(\frac{3x}{5} + M = x\) M = x - \(\frac{2}{5}\)x = x[1 - \(\frac{3}{5}\)] = x[\(\frac{2}{5}\)] = \(\frac{2x}{5}\) Hence the part of the land used for growing maize is \(\frac{2}{5}\)

Cost price CP of the 100 oranges = \(\frac{100}{5}\) x N40.00 selling price SP of the 100 oranges = \(\frac{100}{20}\) x N120 = N600.00 so, profit or loss per cent = \(\frac{SP - CP}{CP}\) x 100% = \(\frac{600 - 800}{800}\) x 100% = \(\frac{-200}{800}\) x 100% Hence, loss per cent = 25%

The shaded portion represents the elements that are in Q only, but not in P or R. In set notation, this is: Q ∩ P' ∩ R'

\(\frac{1}{4}\)p + 3p = 10...(1) 2p - \(\frac{1}{3}\)q = 7...(2) Multiply equation (2) by 3 to clear fraction 3 x 2p - 3 x \(\frac{1}{3}\)q = 3 x 7 6p - q = 21 6p - 21 = q....(3) substituting 6p - 21 for q in (1) \(\frac{1}{4}\)p + 3(6p - 21) = 10...(4) Multiply equation (4) by 4 to clear fraction 4 x \(\frac{1}{4}\)p + 4 x 3(6p - 21) = 4 x 10 p + 12(6p - 21) = 40 p + 72p - 252 73p = 292 p = \(\frac{292}{73}\) = 4

Area of floor = 3.5m x 2.2m = 7.7m2 Area of carpet = (3.5 - 0.5 - 0.5)m x (2.2 - 0.5 - 0.5)m 2.5m x 1.2m = 3m2 hence, area of margine = area of floor - area of carpet = (7.7 - 3)m2 = 4.7m2