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Tan \(x^o = \frac{100m}{80}\) Tan \(x^o = Tan^{-1} 1.25\) x = 51.34\(^o\)

\(60^2 + x^2 + x^2\) \(360^2 = 2x^2\) \(x^2\) = 1800 x = \(\sqrt{1800}\) x = 42.4264 x = 42.4264 perimeter = 4x = 4 x 42.4264 = 169.7056 = 120\(\sqrt{2}\) = 120\(\sqrt{2}\)

Find the minimum value of y is the point where the base of the curve corresponds to the value of y i.e. at -2.1

RSP = 90 < substance in semi a circle RPS = 180 - (90 + 24) = 180 - (114) = 66 TPS = 180 - 66 = 114 RST = 24 < STR = 180 - (114 + 24) = 180 - 138 = 42\(^o\)

(a) \((y - 1)\log_{10} 4 = y\log_{10} 16\) \((y - 1)\log_{10} 4 = y \log_{10} 4^{2}\) \((y - 1)\log_{10} 4 = 2y\log_{10} 4\) Equating both sides, we have \(y - 1 = 2y \implies -1 = 2y - y\) \(\therefore y = -1\) (b) Let the distance from my house to the office = c. At 4km/h, the time taken to get to the office from the house = \(\frac{c}{4} hr\) At 5km/h, the time taken to get to the office from the house = \(\frac{c}{5} hr\) \(\frac{c}{4} = \frac{c}{5} + \frac{30}{60}\) \(\frac{c}{4} - \frac{c}{5} = \frac{1}{2}\) \(\frac{c}{20} = \frac{1}{2} \implies c = 10km\)

(a) \(\frac{2}{3}(3x - 5) - \frac{3}{5}(2x - 3) = 3\) \(\frac{10(3x - 5) - 9(2x - 3)}{15} = 3\) \(30x - 50 - 18x + 27 = 3 \times 15 = 45\) \(12x = 45 + 23 = 68\) \(x = \frac{68}{12} = 5\frac{2}{3}\) (b) \(n + r = 80^\circ .... (1)\) \(180^\circ - m + r = 88^\circ .... (2)\) From (1), r = 80^\circ - n. Putting into (2), we have \(180 - m + 80 - n = 88 \implies - m - n = 88 - 260 = - 172^\circ\) \(-(m + n) = - 172 \implies m + n = 172^\circ\).

(a) \(< P = 23.6^\circ\) (Alternate angle) \(\tan 23.6 = \frac{x}{50}\) \(x = 50\tan 23.6\) = \(50 \times 0.4369\) = \(21.845 m \approxeq 22 m\) (to the nearest metre) (b) Given that \(\Delta TRU = 45cm^{2}\) \(\implies \frac{1}{2}(bh) = 45\) \(10h = 45 \times 2 = 90\) \(h = 9cm\) Area of trapezium = \(\frac{1}{2}(a + b)h\), where \(a = 6cm; b = 10 + 6 = 16cm; h = 9cm\) = \(\frac{1}{2}(6 + 16) \times 9\) = \(11 \times 9\) = \(99 cm^{2}\)

(a) \(T_{n} = a + (n - 1)d\) (For an AP series) \(T_{6} = 37 = a + (6 - 1) d = a + 5d\) \(a + 5d = 37 .... (1)\) \(S_{n} = \frac{n}{2}(2a + (n - 1) d)\) or \(S_{n} = \frac{n}{2}(a + l)\) where a = first term; and l = last term. Since, we are given the sum of the first 6 terms and also given the sixth term which is the last term in the sum, we can use the second formula where the sixth term is the last term in this case. \(S_{6} = 147 = \frac{6}{2} (a + 37)\) \(147 = 3(a + 37) \implies 147 = 3a + 111 \) \(3a = 147 - 111 = 36 \implies a = 12\) (b) From equation (1) above, \(a + 5d = 37\) \(12 + 5d = 37\) \(5d = 37 - 12 = 25\) \(d = 5\) \(S_{15} = \frac{15}{2} (2(12) + (15 - 1)(5))\) \(\frac{15}{2} (24 + 70) = \frac{15}{2} (94)\) = \(15 \times 47\) = \(705\).

(a) (b) \(x + 45 + x + 11 = 120 \implies 2x + 56 = 120\) \(2x = 120 - 56 = 64\) \(x = 32\) Therefore, the number of customers who bought bags = 32. (b) the number of customers that bought shoes = 11 + 32 + 45 = 88. (c) \(P(\text{a random customer bought bags}) = \frac{\text{no of customers that bought bags}}{\text{total no of customers}}\) No of customers that bought bags = 32 + 45 = 77 \(P(\text{customer bought bag}) = \frac{77}{120}\).

(a) Worth of raw materials = N30,450.00 Worth of raw material to get processed materials = N87.00 Amount of raw material to get processed material = \(\frac{30,450}{87}\) Time required for direct labour = 3 hours Amount that will be budgeted for direct labour at N18.25 per hour = \(3 \times 18.25 \times \frac{30,450}{87}\) = \(\frac{1667137.5}{87}\) = N19,162.50 (b) Total amount invested \(N(x + y) = N8,000,000 ....(1)\) Interest from bank M = \(\frac{x \times 4 \times 6}{100} = 0.24x\) Interest from bank N = \(\frac{y \times 4 \times 8}{100} = 0.32y\) \(0.24x + 0.32y = 2,320,000 .... (2)\) (i) Multiplying (1) by 0.24 in order to eliminate x, we have \(0.24(x + y) = 0.24(8,000,000.00)\) \(0.24x + 0.24y = 1,920,000.00 .... (3)\) (2) - (3) : \(0.32y - 0.24y = 2,320,000 - 1,920,000 = 400,000\) \(0.08y = 400,000 \implies y = \frac{400,000}{0.08} = N5,000,000\) \(x = 8,000,000 - y\) = \(8,000,000 - 5,000,000\) = \(N3,000,000\) Therefore, x = N3,000,000 and y = N5,000,000. (ii) Interest paid by bank N = 0.32y = \(0.32 \times N5,000,000\) = \(N1,600,000\)

(a) x -3 -2 -1 0 1 2 3 4 5 6 y 30 13 0 -9 -14 -15 -12 -5 6 21 (b) (c) (i) From the graph, the roots of the equation 2x\(^2\) - 7x = 26 can be obtained by drawing a line through y = 17 to meet the curve 2x\(^2\) - 7x - 9. The roots of the equation 2x\(^2\) - 7x = 26 : -2.2 and 5.8. (ii) The coordinates of the minimum point of y : 1.8, -15. (iii) The range of values for which 2x\(^{2}\) - 7x < 9 : - 0.9 < x < 4.5.

Mark (x) 1 2 3 4 5 Total Frequency (f) m + 2 m - 1 2m - 3 m + 5 3m - 4 8m - 1 fx m + 2 2m - 2 6m - 9 4m + 20 15m - 20 28m - 9 (a) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\) \(\frac{75}{23} = \frac{28m - 9}{8m - 1}\) \(75(8m - 1) = 23(28m - 9) \implies 600m - 75 = 644m - 207\) \(-75 + 207 = 644m - 600m\) \(132 = 44m \implies m = 3\) (b)(i) Interquartile range = Third quartile - First quartile Frequency = 8(3) - 1 = 24 - 1 = 23 \(Q_{3} = \frac{3}{4} \times 23 = 17.25th\) position = 4 \(Q_{1} = \frac{1}{4} \times 23 = 5.75th\) position = 2 Interquartile range : 4 - 2 = 2 (ii) P(at least 4 marks) = \(\frac{(m + 5 + 3m - 4)}{23} = \frac{4m + 1}{23}\) = \(\frac{4(3) + 1}{23} \) = \(\frac{13}{23}\)

(ii) (a) < RTS = 65^\circ - 34^\circ (exterior angles of triangle PST) = 31^\circ (ii) (b) < SRP = 65^\circ (angles in alternate segment are equal) \(\therefore < SRP = 180^\circ - 65^\circ = 115^\circ\). (b) If < XYZ = 128^\circ then < XZY = 128^\circ + 20^\circ = 148^\circ < XZV = 180^\circ - 148^\circ = 32^\circ < YVZ = < VYZ = 52^\circ (Isoceles triangle) < WYZ = 52^\circ - 32^\circ = 20^\circ

(a) \(\sin x = \frac{5}{13}\) Using SOHCAHTOA, then Opp = 5 and Hyp = 13 \(13^{2} = 5^{2} + Adj^{2}\) \(169 = 25 + Adj^{2}\) \(Adj = \sqrt{169 - 25} = \sqrt{144} = 12\) \(\cos x = \frac{12}{13}\) \(\tan x = \frac{5}{12}\) \(\frac{\cos x - 2\sin x}{2\tan x} = \frac{\frac{12}{13} - 2(\frac{5}{13})}{2(\frac{5}{12})}\) = \(\frac{\frac{2}{13}}{\frac{5}{6}}\) = \(\frac{12}{65}\) (b) Taking \(\Delta LPB\), Using Pythagoras theorem, \(/BP/^{2} = /BL/^{2} - /LP/^{2}\) \(i.e. /BP/ = \sqrt{12^{2} - 9.6^{2}}\) = \(\sqrt{144 - 92.16} = \sqrt{51.84} = 7.2m\) \(/PA/ = 10m - 7.2m = 2.8m\) \(/LA/^{2} = x^{2}\) \(x^{2} = /LP/^{2} + /PA/^{2}\) \(x^{2} = 9.6^{2} + 2.8^{2}\) = \(92.16 + 7.84\) \(x^{2} = 100 \implies x = 10m\) (ii) \(\tan \theta_{y} = \frac{9.6}{2.8} = 3.4286\) \(\theta_{y} = \tan^{-1} (3.4286) \approxeq 74^\circ\)