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(a)(i) < PQR = (180^\circ - 130^\circ) + 50^\circ = 100^\circ \(PR^{2} = PQ^{2} + QR^{2} - 2(PQ)(QR) \cos < PQR\) = \(8^{2} + 13^{2} - 2(8)(13) \cos 100\) = \(233 + 36.118\) \(PR^{2} = 269.118\) \(PR = 16.405 m \approxeq 16.4 m\) (ii) \(\frac{QR}{\sin < QPR} = \frac{PR}{\sin < PQR}\) \(\frac{13}{\sin < QPR} = \frac{16.4}{\sin 100}\) \(\sin < QPR = \frac{13 \times \sin 100}{16.4}\) \(\sin < QPR = 0.7806\) \(< QPR = 51.32^\circ \approxeq 51.3^\circ\) The bearing of R from P = 50^\circ + 51.3^\circ = 101.3^\circ \(\approxeq\) 101^\circ. (b) In \(\Delta PQD\), \(\frac{h}{8} = \sin 51.3\) \(h = 8 \sin 51.3\) = \(6.243 m \) \(\approxeq 6.24 m\) Area of \(\Delta PQR = \frac{1}{2} \times PR \times h\) \(\frac{1}{2} \times 6.24 \times 16.4 = 51.166 m^{2}\) \(\approxeq 51.2 m^{2}\)

(a) Class Interval Freq (f) Tally Mid-point (x) \(fx\) 45 - 49 1 | 47 47 50 - 54 3 ||| 52 156 55 - 59 6 |||| | 57 342 60 - 64 7 |||| || 62 434 65 - 69 6 |||| | 67 402 70 - 74 4 |||| 72 288 75 - 79 2 || 77 154 80 - 84 1 | 82 82 \(\sum\) 30 1905 (b) (c) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{1905}{30}\) = \(63.5\). (d) P(iron rod in modal class) = \(\frac{7}{30}\)

(a) The roots of the equation are -1.5 or 2. (b) C = 6 (intercept on the positive vertical axis) \(ax^{2} + bx + c = 0\) When x = 2, c = 6 \(4a + 2b + 6 = 0 \implies 4a + 2b = -6 ....(1)\) When x = -1.5, c = 6 \(a(-1.5)^{2} + b(-1.5) = -6\) \(\frac{9a}{4} - \frac{3b}{2} = -6 \implies 9a - 6b = -24 ... (2)\) Multiply (1) by 3 : \(12a + 6b = -18 ... (1a)\) (1a) + (2) : \(12a + 9a = -18 - 24 \implies 21a = -42\) \(a = -2\) \(4a + 2b = -6 \implies 4(-2) + 2b = -6\) \(2b = -6 + 8 = 2 \implies b = 1\) (a, b, c) = (-2, 1, 6) Hence the relation illustrated in the graph is \(y = -2x^{2} + x + 6\). (c) When \(x = 0.2, y_{max} = 6.2\) (d) When \(y = 2, x = \text{ -1.2 or 1.7}\)

Hint: Use BODMAS, in other words, do multiplication of the second and the last first before subtracting value obtained from the first.

\((\sqrt[3]{64a^{3}})^{-1} = (\sqrt[3]{(4a)^{3}})^{-1}\) = \((4a)^{-1} \) = \(\frac{1}{4a}\)

\(\frac{p^{2} - q^{2}}{2pq} = \frac{(p + q)(p - q)}{2pq}\) = \(\frac{(1 + \sqrt{2} - (1 - \sqrt{2}))(1 + \sqrt{2} + 1 - \sqrt{2})}{2(1 + \sqrt{2})(1 - \sqrt{2})}\) = \(\frac{(2\sqrt{2})(2)}{-2}\) = \(-2\sqrt{2}\)

Total cost = N(250,000 + 70,000) = N320,000 Selling price = N400,000 (given) Gain = SP - CP = N(400,000 - 320,000) = N80,000 Gain % = gain/CP x 100 = (80,000/320,000) x 100 Gain % = 25%

\(x = \frac{y}{2} \) \(\left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^{2}}{y^{2}}\right)\) \(\frac{x^3}{y^3} + \frac{1}{2} = (\frac{y}{2})^{3} \div y^{3} + \frac{1}{2}\) = \(\frac{y^{3}}{8} \times \frac{1}{y^3} + \frac{1}{2}\) = \(\frac{1}{8} + \frac{1}{2}\) = \(\frac{5}{8}\) \(\frac{1}{2} - \frac{x^2}{y^2} = \frac{1}{2} - (\frac{y}{2})^{2} \div y^2)\) = \(\frac{1}{2} - \frac{y^2}{4} \times \frac{1}{y^2}\) = \(\frac{1}{2} - \frac{1}{4}\) = \(\frac{1}{4}\) \(\therefore \left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^{2}}{y^{2}}\right) = \frac{5}{8} \div \frac{1}{4}\) = \(\frac{5}{2}\)

Principal, P = Amount, A - Interest, I. A = P + I I = (P.T.R)/100 = (P x 5 x 2)/100 = 10P/100 = P/10 But A = P + I, => 5500 = P + (P/10) => 55000 = 10P + P => 55000 = 11P Thus P = 55000/11 = N5,000

\(\frac{(0.14)^{2} \times 0.275}{7(0.02)} = \frac{(0.14)^{2} \times 0.275}{0.14}\) = \(0.14 \times 0.275\) = \(0.0385 \approxeq 0.039\) Note: Embedded zeroes are counted just the same as non-zero digits and For the number of decimal places stated, count that number of digits to the right of the decimal and underline it. The next number to its right is called the 'rounder decider'. If the 'rounder decider' is 5 or more, then round the previous digit up by 1

\(\frac{a^{3x} - 26a^{2x} + 156a^{x} - 216}{a^{2x} - 24a^{x} + 108}\) Let \(a^{x} = z\) \(\therefore = \frac{z^{3} - 26z^{2} + 156z - 216}{z^{2} - 24y + 108} ... (i)\) Dividing (i) above, we get \(z - 2\) = \(a^{x} - 2\)

\(y = px^{2} + q ... (i)\) \(y = 2x^{2} - 1 ... (ii)\) At x = 2, (i): \(y = p(2^{2}) + q = 4p + q\) (ii): \(y = 2(2^{2}) - 1 = 7\) \(\therefore \text{The coordinates of the point of intersection = (2, 7)}\) (i): \(7 = 4p + q \implies p = \frac{7 - q}{4}\)

m2 + n2 = 29 .......(1) m + n = 7 ............(2) From (2), m = 7 - n but m2 + n2 = 29, substituting; (7-n)2 + n2 = 29 49 - 14n + n2 + n2 = 29 => 2n2 -14n + 20 = 0 Thus n2 -7n + 10 = 0 Factorizing; (n-5)(n-2) = 0 n - 5 = 0, => n = 5 n - 2 = 0, => n = 2. When n = 5, m + n = 7, => m = 2, When n = 2, m + n = 7, => m = 5. Thus (m,n) = (5,2) and (2,5)

By definition a*b = a + b + 1. Let the inverse of the element 2 be x, Therefore 2*x = -1 i.e. 2 + x + 1 = -1 3 + x = -1 x = -1 - 3 x = -4

1st statement: U6 = 1/2(U12) a + (n -1)d = 1/2[a + (n-1)d] a + 5d = a + 11d 2(a + 5d) = a + 11d 2a + 10d = a + 11d Solving, => a = d Hence the first term is equal to the common difference