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(a) \((\frac{64}{125})^{-\frac{2}{3}} = (\frac{125}{64})^{\frac{2}{3}}\) = \((\frac{5}{4})^{3})^{\frac{2}{3}} \) = \((\frac{5}{4})^{2}\) = \(\frac{25}{16}\) \(\therefore 2 \div (\frac{64}{125})^{-\frac{2}{3}} = 2 \div \frac{25}{16}\) = \(2 \times \frac{16}{25}\) = \(\frac{32}{25}\) (b) \(3x + 5 = - 4x - 1\) \(3x + 4x = - 1 - 5\) \(7x = - 6\) \(x = -\frac{6}{7}\) \(y = 3x + 5\) ( You can use any of the given equations to get y) \(y = 3(-\frac{6}{7}) + 5 = -\frac{18}{7} + \frac{35}{7}\) = \(\frac{17}{7}\) \(k = (x, y) = (-\frac{6}{7}, \frac{17}{7})\)

(a) \(\frac{5}{8} of 2\frac{1}{2} - \frac{3}{4} \div \frac{3}{5}\) = \((\frac{5}{8} \times \frac{5}{2}) - (\frac{3}{4} \div \frac{3}{5})\) = \((\frac{25}{16}) - (\frac{5}{4})\) = \(\frac{25 - 20}{16}\) = \(\frac{5}{16}\) (b) \(V = \frac{1}{3} \times A \times h = \frac{1}{3} \pi r^{2} h\) \(\therefore \frac{1}{3} \times 154 \times h = \frac{1}{3} \times \pi \times r^{2} \times h\) Comparing the two equations, \(154 = \frac{22}{7} \times r^{2}\) \(r^{2} = \frac{154 \times 7}{22}\) \(r^{2} = 49\) \(\therefore r = 7 cm\).

(a) Sample - space 1 2 3 4 5 6 1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6 2 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6 4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6 5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6 6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6 (i) Same digit = {(1,1), (2,2),(3,3), (4,4),(5,5), (6,6)} P(same digit) = \(\frac{6}{36}\) (ii) P(total score greater than 5) = \(\frac{26}{36}\) = \(\frac{13}{18}\) (b) \(\frac{x^{2} + y^{2}}{y^{2} - x^{2}} = \frac{(\cos 30)^{2} + (\sin 30)^{2}}{(\sin 30)^{2} - (\cos 30)^{2}}\) = \(\frac{(\frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2}}{(\frac{1}{2})^{2} - (\frac{\sqrt{3}}{2})^{2}}\) = \(\frac{\frac{3}{4} + \frac{1}{4}}{\frac{1}{4} - \frac{3}{4}}\) = \(\frac{1}{-\frac{1}{2}}\) = \(-2\).

(a) \(< ADB = < ACB = 20^\circ\) (Angle in the same segment). In \(\Delta ACY\), \(\hat{C} = < ACB\) (i) \(\hat{C} + \hat{A} + \hat{Y} = 180^\circ\) \(20^\circ + \hat{A} + 60^\circ = 180^\circ\) \(\hat{A} = 180^\circ - 80^\circ = 100^\circ\) \(< CAY = \hat{A} = 100^\circ\) \(< CAX = 180^\circ - 100^\circ = 80^\circ\) (ii) In \(\Delta ACB\), \(\hat{B} = \hat{C} = \frac{180^\circ - 20^\circ}{2} \) = \(\frac{160^\circ}{2}\) = \(80^\circ\) In \(\Delta ABY\), \(< ABY = \hat{B} = 180^\circ - 80^\circ\) = \(100^\circ\) (b) \((m + 1)(m - 3) \equiv m^{2} - km + c\) \(m^{2} - 3m + m - 3 = m^{2} - 2m - 3\) \(m^{2} - 2m - 3 \equiv m^{2} - km + c\) \(\implies k = 2 ; c = -3\)

(a) \(\sqrt{\frac{0.18 \times 12.5}{0.05 \times 0.2}}\) \(\frac{0.18 \times 12.5}{0.05 \times 0.2} = \frac{18 \times 10^{-2} \times 125 \times 10^{-1}}{5 \times 10^{-2} \times 2 \times 10^{-1}}\) = \(9 \times 25 \times 10^{-3 - (-3)}\) = \(9 \times 25\) = \(225\) \(\therefore \sqrt{\frac{0.18 \times 12.5}{0.05 \times 0.2}} = \sqrt{225}\) = \(15\). (b) \(\frac{8 - 4\sqrt{18}}{\sqrt{50}}\) \(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\) \(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\) \(\frac{8 - 4(3\sqrt{2})}{5\sqrt{2}} = \frac{8 - 12\sqrt{2}}{5\sqrt{2}}\) Rationalising, we have = \(\frac{8 - 12\sqrt{2}}{5\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\) = \(\frac{8\sqrt{2} - 24}{10}\) = \(0.8\sqrt{2} - 2.4\) = \(-2.4 + 0.8\sqrt{2}\) (c) \(x \propto \frac{y^{3}}{z^{2}}\) \(\implies x = \frac{ky^{3}}{z^{2}}\) \(108 = \frac{k \times 3^{3}}{4^{2}}\) \(k = \frac{108 \times 16}{27}\) \(k = 64\) \(\therefore x = \frac{64y^{3}}{z^{2}}\) \(\therefore 4000 = \frac{64 \times 10^{3}}{z^{2}}\) \(4 = \frac{64}{z^{2}}\) \(z^{2} = \frac{64}{4} = 16\) \(z = \sqrt{16} = \pm 4\)

(a) (i) Let the exterior angle be x Let the interior angle be x + 120^\circ. Exterior angle + Interior angle = 180^\circ \(\therefore x + x + 120 = 180 \implies 2x + 120 = 180\) \(2x = 60 \implies x = 30^\circ\) Number of sides = \(\frac{360}{\text{exterior angle}}\) = \(\frac{360}{30} = 12 sides\) (ii) Sum of interior angle = \((2n - 4) \times 90^\circ\) = \((2(12) - 4) \times 90^\circ\) = \(20 \times 90\) = \(1800^\circ\) (b) Note : < PQR = 65^\circ + 34^\circ = 99^\circ\) \(|PR|^{2} = |PQ|^{2} + |QR|^{2} - 2(|QR|)(|PQ|) \cos 99\) \(|PR|^{2} = 6^{2} + 13^{2} - 2(6)(13) \cos 99\) \(|PR|^{2} = 205 - 156 \cos 99\) \(|PR|^{2} = 205 + 24.40 = 229.40\) \(|PR| = \sqrt{229.40}\) = \(15.146 km\) \(\approxeq 15 km\)

(a) Longitude difference = 85^\circ - 25^\circ = 60^\circ. Distance AB along the parallel of latitude = \(\frac{\theta}{360^\circ} \times 2\pi R \cos \alpha\) \(AB = \frac{60}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 53\) = \(\frac{1}{6} \times \frac{44}{7} \times 3851.62\) = \(4,035.03 km\) \(\approxeq 4035 km\) (b) \(Speed = \frac{Distance}{Time}\) \(\therefore Time = \frac{Distance}{Speed}\) = \(\frac{4035}{400}\) \(\approxeq 10 hours\). (c) Distance BC measured along the meridian \(BC = \frac{\theta}{360} \times 2 \pi R\) \(2000 = \frac{\theta}{360} \times 2 \times \frac{22}{7} \times 6400\) \(\theta = 17.898^\circ \approxeq 18^\circ\) \(\theta\) = Latitude difference Let the latitude of C = x. \(\theta = 53 - x\) \(18 = 53 - x\) \(x = 35^\circ\)

(a) \(x + 8 < 4x - 15 \implies 4x - 15 > x + 8\) \(4x - x > 8 + 15 \implies 3x > 23\) \(x > \frac{23}{3}\) \(x > 7\frac{2}{3}\). (b) Extra income = 5k of N200,000 = \(\frac{5}{100} \times N200,000\) = \(N10,000\) Total income = Monthly income + commission = N(2,500 + 10,000) = N12,500. (c) Area of semi- circle = \(\frac{\pi r^{2}}{2}\) = \(\frac{1}{2} \times \frac{22}{7} \times 35 \times 35\) = \(1925 cm^{2}\) Length of rectangular part = 50 cm Breadth = Diameter of semi- circle = 2(35 cm) = 70 cm Area of rectangular part = \(length \times \breadth\) = \(50 \times 70\) = \(3500 cm^{2}\) Total area of window = \((3500 + 1925) cm^{2}\) = \(5425 cm^{2}\)

(a) \(|AB| = 85m = \frac{85}{10}\) = \(8.5cm\) \(|TC| = 6cm \equiv 60 m\).

(a) Class Interval Freq (f) Classmark (x) \(fx\) 10 - 14 7 12 84 15 - 19 18 17 306 20 - 24 25 22 550 25 - 29 17 27 459 30 - 34 9 32 288 35 - 39 4 37 148 80 1835 \(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{1835}{80}\) \(\approxeq 22.9yrs\) (b)(i) Class Interval Freq (f) Class boundaries 10 - 14 7 9.5 - 14.5 15 - 19 18 14.5 - 19.5 20 - 24 25 19.5 - 24.5 25 - 29 17 24.5 - 29.5 30 - 34 9 29.5 - 34.5 35 - 39 4 34.5 - 39.5 (ii) Modal age = \(\frac{20 + 24}{2} = 22\) (c) P(picking from the modal class) = \(\frac{25}{80}\) = \(\frac{5}{16}\).

(a) In \(\Delta ACB, < CAB = 180^\circ - (90^\circ + 61^\circ)\) = \(180^\circ - 151^\circ\) = \(29^\circ\) In \(\Delta ADB, < DAB = (29 + x)^\circ\) \(\cos (29 + x) = \frac{3}{5} = 0.6\) \((29 + x)^\circ = \cos^{-1} (0.6)\) \((29 + x)^\circ = 53.13^\circ\) \(x = 53.13^\circ - 29^\circ\) = \(24.13^\circ\) \(\approxeq 24.1^\circ\) (1 decimal place). (b) (i) \(DB = 2DN = 2NB\) \(OA = OB = OC = OD = 4cm\) \(AB = BC = CD = DA = 2cm\) \(DB^{2} = BC^{2} + DC^{2}\) = \(2^{2} + 2^{2}\) = \(\sqrt{8}\) = \(2\sqrt{2} cm\) \(2DN = DB \implies DN = \frac{2\sqrt{2}}{2} = \sqrt{2} cm\) In \(\Delta DON\), \(OD^{2} = ON^{2} + DN^{2}\) \(4^{2} = h^{2} + (\sqrt{2})^{2} \implies h^{2} = 16 - 2\) \(h^{2} = 14 \implies h = \sqrt{14}\) (ii) \(V = \frac{1}{3} \times \base area \times \perp distance\) \(V = \frac{1}{3} \times 2^{2} \times \sqrt{14}\) = \(\frac{4\sqrt{14}}{3}\) = \(4.99 cm^{3}\)

(a) \(y = x^{2} - 2x - 2\) x -4 -3 -2 -1 0 1 2 3 4 \(x^{2}\) 16 9 4 1 0 1 4 9 16 \(-2x\) 8 6 4 2 0 -2 -4 -6 -8 \(-2\) -2 -2 -2 -2 -2 -2 -2 -2 -2 \(y\) 22 13 6 1 -2 -3 -2 1 6 (b) (c)(i) \(x^{2} - 2x - 2 = 0\) \(\therefore y = 0 ; x = \text{-0.7 or 2.7}\) (ii) \(x^{2} - 2x - 4\frac{1}{2} = 0\) \(x^{2} - 2x - 4\frac{1}{2} + 2\frac{1}{2} = 0 + 2\frac{1}{2}\) \(x^{2} - 2x - 2 = 2.5\) When y = 2.5, \(x = \text{-1.3 or 3.3}\). (iii) Line of symmetry at x = 1.

The sentence is using a conditional statement in the past perfect tense to express a hypothetical situation and its result. "If the rain hadn't fallen" implies that the rain did fall, and as a result, "we wouldn't have missed the match." This means that because the rain fell, they did not watch the match. Therefore, option B correctly explains the information conveyed in the sentence.

A. He bit off people's jaws: This option is not the answer because it takes the expression "jawing away" too literally. In this context, "jawing away" is an informal phrase that means talking or chatting continuously, not physically biting off people's jaws. B. He was always chewing in the class: This option is incorrect because it misinterprets the meaning of "jawing away." The phrase refers to talking or chatting, not chewing. C. He was always making trouble for the class: This option is not the answer because while "jawing away" might imply being talkative, it doesn't necessarily mean causing trouble. It's more about continuous talking or chatter. D. He talked continuously in the class. - This option is correct because it accurately captures the meaning of "jawing away" in this context. The sentence indicates that the new boy talked a lot in class, causing the class to become tired of it. So, option D is the best explanation for the information conveyed in the sentence.

A. He expected that the men would cooperate with him: This option is the best explanation because "pull together" is a common idiom that means to work together in cooperation and harmony. The sentence suggests that the new Headmaster hoped for unity and collaboration among his team.