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(a) r =\(\frac{9}{7-2x}\) r = \(\frac{5x+1}{7-2x}\) \(\frac{9}{7-2x}\) = \(\frac{5x+1}{7-2x}\) 9x9=(7-2x) (5x +17) 81 = 35x + 119 - 10x\(^2\) - 34x 10x\(^2\) - x - 38 (x -2)(10x+19) = 0 x = 2, x = \(\frac{19}{10}\) x = 2 (b) Let the integers be x and y x:y= 3:4 \(\frac{x}{y}\) = \(\frac{3}{4}\) x = \(\frac{3y}{4}\) 3x + 2y=68 3( \(\frac{3}{4}\)y)+ 2y = 68 \(\frac{9}{4}\)y + 2y = 68 9y + 8y = 272 17y = 272 y= 16 x = \(\frac{3}{4}\) * 16 →12

(a) y =(\(\frac{pr}{m} - p^2r\))\(^{\frac{-3}{2}}\) y = (\(\frac{pr - mrp^2}{m})^\frac{-3}{2}\) y = [\(\frac{m}{pr - mrp^2}]^\frac{3}{2}\) \(y^\frac{2}{3} = \frac{m}{r(p - mp^2)}\) r = \(\frac{m}{y^\frac{2}{3}[p - mp^2]}\) b) Given y = -8, m = 1 p = 3 r = \(\frac{m}{y^\frac{2}{3}[p - mp^2]}\) r = \(\frac{1}{(-8)^\frac{2}{3}[ 3 - (1)3^2]}\) r = \(\frac{1}{(-2)^2[3 - 9]}\) r = \(\frac{1}{[4][-6]}\) = \(\frac{-1}{24}\)

(a) 2x + 40 + 2y + 10 = 180^\circ...(1) 2x + 2y = 130 x + y = 65º 5x - 35 = 2(2y+ 10) ... (2) 5x - 35 = 4y +20 5x - 4y = 55 5x - 4(5-x) = 55 5x - 260 + 4x =55 9x = 315 x = 35^\circ y= 65 - 35 y=30^\circ ∠ABC + [2 x º35 + 40] + [5(35)-350)] + 40^\circ =360º ∠ABC + 110 + 140^\circ+ 40º = 360 ∠ABC + 290^\circ = 360^\circ ∠ABC = 70^\circ

a) m = tan 30º = \(\frac{\sqrt{3}, n = tan 45º = 1 \(\frac{ m - n}{mn}\), = \(\frac{\frac{\sqrt{3}}{3} - 1}{\frac{\sqrt{3}}{3}(1)}\) = \(\frac{\frac{\sqrt{3} - 3}{3}}{\frac{\sqrt{3}}{3}}\) = \(\frac{\sqrt{3} - 3}{3}\) X \(\frac{3}{\sqrt{3}}\) = \(\frac{\sqrt{3} - 3}{\sqrt{3}}\) = \(\frac{\sqrt{3} - 3}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) = \(\frac{3 - 3\sqrt{3}}{3}\) = 1 - \(\sqrt{3}\) b) see the diagram above.. Let the number of women wearing both wristwatches and glasses be x 15 - x + x + 10 - x = 20 25 - x = 20 x = 5 Therefore, the number of women wearing wrist watches only = 10 - x = 10 - 5 = 5 and number of women wearing glasses only = 15 - x = 15 - 5 = 10 Pr(both) = \(\frac{5}{20}\) = \(\frac{1}{4}\) = 0.25

(a) x-axis: 2 cm to 2 units or 1 cm to 1 unit y-axis: 2 cm to 10 units or 1 cm to 5 units (b) x = -2, x = 4 (x+2)(x-4) = 0 x\(^2\) -4x + 2x -8 = 0 x\(^2\) -2x - 8 = 0 y = 8 + 2x - x(curve has a max. point) m = -1,n = 2, r =8 (c) P(-5,-27),Q(3,5) Gradient = \(\frac{5+27}{3+5}\) = \(\frac{32}{8}\) = 4 (d) Range: {x:-2 < x < 4)

(a) Total cost 180 x 250 =N45,000.00 Total selling price = 300y + (180-y) x 250 x \(\frac{95}{100}\) 300y + 42750 -237.5y = 62.5y + 42750 Profit = 62.5y + 42750 -4500 7125 = 62.5y - 2250 9375 = 62.5y y = 150 (b)) Profit = \(\frac{x^2}{20} - 33x\) (24x+ 103) 9x - \(\frac{x^2}{20}\) - 103 (ii) Cost = 24(20) +103 = 583.00 Selling Price =33(20) - \(\frac{20^2}{20}\) = 640 Percentage profit = \(\frac{640-583}{583}\) x 100% = 9.78%

(a) 35 + 7.5 + 10 + 15 + 17.5 + x = 100 85 + x = 100. x = 100 - 85 x = 15 items % degree food & drinks 35 \(\frac{35}{100} * 360\) = 126º fuel 7.5 \(\frac{7.5}{100} * 360\) = 27º rent 10 \(\frac{10}{100} * 360\) = 36º building project 15 \(\frac{15}{100} * 360\) = 54º education 17.5 \(\frac{17.5}{100} * 360\) = 63º savings 15 \(\frac{15}{100} * 360\) = 54º (c) income tax = \(\frac{12}{100} * 28,800\) = $3,456.00 Monthly tax = \(\frac{3456}{12}\) = $288.00 (ii) Monthly net salary = \(\frac{1}{12} (288 - 3456)\) = $2,112.00 Amount saved each month = \(\frac{15}{100} * 2112\) = $316.80

xº 0 20 40 60 80 100 120 140 160 180 y 7.0 7.6 7.3 6.1 4.2 1.7 -0.9 -3.4 -5.6 7.0 (b) (C)(i) From the graph, When x is 150^\circ y = -4.6 \pm 0.2 (ii) Range of value of x for which y > is x: 0º \le x \le 113

Mean = \(\frac{\sum fx}{ \sum f}\) = (3x2) +(4x6) + (5x5) + (6x) + (7x6) + (8x9) + (9x8) + (10x5) \div (2 + 6 + 5 + x + 6 + 9 + 8 + 5) 7 = (291 + 6x) \div (41+x) 7(41+x) = 291 + 6x 287 + 7x = 291 + 6x x = 4 (b) Age(x in yrs) No.of child Fx\(^2\) 3 2 18 4 6 96 5 5 125 6 4 144 7 6 294 8 9 576 9 8 648 10 5 500 Σf = 45 Σfx\(^2\) = 2401 Standard deviation = \(\sqrt \frac{196}{45} = \sqrt 4.3555\) = 2.087

(a) ∠XYZ = \(\frac{1}{2}\) (146) = 73º ∠OZX = 34º ∠OZY = 73 - 34 ∠OYZ = ∠OZY = 39 (b) x = y - 7 42 + 38 + 57 + x + x + y + 2x - 15 + 3x - y = 360 122^\circ+7x = 360^\circ 7x = 360º - 122 = 238^\circ 7x = 238º x = 34 : 34 = y - 7 34 + 7 = y 41 = y

9(a)(i) p = \(\frac{1}{4}\), p = 1 - \(\frac{1}{4}\) p = \(\frac{3}{4}\) = \(\frac{3}{64}\) = 0.04687 ≈ 0.05 (ii) p(winning all) = \(\frac{3}{4}\) * \(\frac{3}{4}\) * \(\frac{3}{4}\) = \(\frac{27}{64}\) = 0.422 (iii) p(winning two) = (\(\frac{3}{4}\) * \(\frac{3}{4}\) * \(\frac{1}{4}\)) + (\(\frac{3}{4}\) * \(\frac{1}{4}\) * \(\frac{3}{4}\)) + (\(\frac{1}{4}\) * \(\frac{3}{4}\) * \(\frac{3}{4}\)) = \(\frac{9}{64}\) + \(\frac{9}{64}\) + \(\frac{9}{64}\) = \(\frac{27}{64}\) = 0.4218 = 0.42 (b) Volume of a cone = 1/3\pir\(^2\)h 1200 = \(\frac{1}{3}\) * \(\frac{22}{7}\) * r\(^2\) * 24 r\(^2\) = \(\frac{1200 * 3 * 7}{22 * 24}\) r\(^2\) = \(\frac{25200}{528}\) r\(^2\) = 47.727 r = \(\sqrt 47.727\) r = 6.9085cm Volume of a cone with same radius and height 84cm: v = \(\frac{1}{3}\) * \(\frac{22}{7}\) * (6.9085)\(^2\) * 84 = \(\frac{1}{3}\) * \(\frac{22}{7}\) * 47.727 * 84 volume = 4200cm\(^3\)

209 = 2 * \(\frac{22}{7}\) * \((\frac{7}{2})^2\) + 2 * \(\frac{22}{7}\) * \(\frac{7}{2}\) * h 209 = 77 + 22h 132 = 22h h = 6cm (b) tan 43º = \(\frac{h}{y}\) 0.9125y = h ...[i] tan 38º = \(\frac{h}{y+19}\) 0.7813(y+19) = h ...[ii] 0.7813(y+19) = 0.9125y 0.7813y + 14.8447 = 0.93257 14.8447 = 0.1512y y= 98.1792 = 98.2m h = 0.9325 x 98.1792 = 91.55 =91.6m

Higher and more complex animals like birds and mammals can reproduce only by sexual means.

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

The beaks of nectar sucking birds are long slender and pointed. This helps in sucking nectar and also feeding insects and food to the young ones. Birds like Hummingbirds have this kind of beaks.