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Let x = \(\frac{3}{4}\) or x = -4 i.e. 4x = 3 or x = -4 (4x - 3)(x + 4) = 0 therefore, 4x2 + 13x - 12 = 0

Given; y = x2 - x - 2, y = 2x - 1 Using y = y, gives x2 - x - 2 = 2x - 1 x2 - 3x - 2 + 1 = 0 therefore, x2 - 3x - 1 = 0

R \(\alpha\) D2 R = D2K R = 4 Litres when D = 15cm thus; 4 = 152k 4 = 225k k = \(\frac{4}{225}\) This gives R = \(\frac{4D^2}{225}\) Where R = 9litres equation gives 9 = \(\frac{4D^2}{225}\) 9 x 225 = 4d2 D2 = \(\frac{9 \times 225}{4}\) D = \(\sqrt{9 \times 225}{4}\) = \(\frac{3 \times 15}{2}\) = 22\(\frac{1}{2}\)km

(a) Let the sum of \(3\frac{2}{3}\) and \(2\frac{1}{5}\) be less than 7 by z. Then z = \(7 - (3\frac{2}{3} + 2\frac{1}{5})\) = \(7 - (\frac{11}{3} + \frac{11}{5})\) = \(7 - \frac{88}{15}\) = \(\frac{17}{15} = 1\frac{2}{15}\) Hence, the sum of \(3\frac{2}{3}\) and \(2\frac{1}{5}\) is less than 7 by \(1\frac{2}{15}\). (b) \(h = 6 + 4\cos (15p)^\circ\) (Given) (i) When p = 4, \(h = 6 + 4\cos (15 \times 4)^\circ\) \(h = 6 + 4 \cos 60^\circ\) = \(6 + 4 \times 0.5\) = \(6 + 2 = 8 m\) (ii) When h = 9 m, the given equation becomes \(9 = 6 + 4\cos (15p)^\circ\) \(9 - 6 = 4 \cos (15p)^\circ\) \(\cos (15p)^\circ = \frac{3}{4}\) \((15p)^\circ = \cos^{-1} (0.75)\) \((15p)^\circ = 41.4^\circ\) \(p^\circ = \frac{41.4}{15}\) \(p^\circ = 2.76^\circ\) \(p \approxeq 2.8^\circ\) (2 significant figures).

(a) (i) \(y = px^{2} - 5x + q\) x -3 -2 -1 0 1 2 3 4 5 y 21 6 -12 0 13 From the table, y = -12 when x = 0, i.e. \(-12 = p(0^{2}) - 5(0) + q\) \(q = -12\). y = 0 when x = 4, i.e. \(0 = p(4^{2}) - 5(4) + q\) \(0 = 16p - 20 + q \implies 20 = 16p + q .... (1) \) Substitute q = -12 in (1), \(20 = 16p - 12\) \(20 + 12 = 16p \implies p = \frac{32}{16} = 2\) Equation : \(y = 2x^{2} - 5x - 12 \). (ii) Given the equation \(y = 2x^{2} - 5x - 12\). The table becomes x -3 -2 -1 0 1 2 3 4 5 y 21 6 -5 -12 -15 -14 -9 0 13 (b) (c) From the graph, (i) When x = 1.8, y = -14.5 ; (ii) When y = -8, x = -0.65 or 3.15 or 3.2.

To find \( 110111_2 + 10100_2 \): Convert to decimal: \(110111_2 = 55_{10}, \quad 10100_2 = 20_{10}\) Add: \(55 + 20 = 75\) Convert back to binary: \(75_{10} = 1001011_2\) Final result: \(110111_2 + 10100_2 = 1001011_2\) OR \(\begin{array}{r} 110111_2 \\ + 010100_2 \\ \hline 1001011_2 \end{array}\) \(\therefore \; 110111_2 + 10100_2 = 1001011_2\)

\(P \propto \frac{Q}{R}\) \(P = K \frac{Q}{R}\) When Q = 36, R = 16, P = 27 Then substitute into the equation \(27 = K \frac{36}{16}\) \(K = \frac{27 \times 16}{36}\) \(K = 12\) So the equation connecting P, Q and R is \(P = \frac{12Q}{R}\)

\(\frac{x}{2} + \frac{3}{4} \leq \frac{5x}{6} - \frac{7}{12}\) \(12\frac{x}{2} + 12\frac{3}{4} \leq 12\frac{5x}{6} - 12\frac{7}{12}\) 6x + 9 \(\leq\) 10x - 7 6x - 10x \(\leq\) - 7 - 9 -4x \(\leq\) -16 -4x/-4 \(\geq\) -16/-4 x \(\geq\) 4

a + 3d = 13 .......... (1) a + 9d = 31 .......... (2) (2) - (1): 6d = 18 d = 18/6 = 3 From (1), a + 3(3) = 13 a + 9 = 13 a = 13 - 9 = 4 Hence, T24 = a + 23d T24 = 4 + 23(3) T24 = 4 + 69 T24 = 73

\(\begin{pmatrix}5 & -6 \\2 & -7\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix}7 \\-11\end{pmatrix}\) By matrices multiplication; 5x - 6y = 7 ........(1) 2x - 7y = -11 ......(2) 2 x (1): 10x - 12y = 14 .......(3) 5 x (2): 10x - 35y = -55 ......(4) (3) - (4): 23y = 69 y = 69/23 = 3

\(\begin{vmatrix}-x & 12 \\-1 & 4\end{vmatrix} = - 12\) -4x - (-1)12 = -12 -4x + 12 = -12 -4x = -12 - 12 -4x = - 24 x = 6

\(0 \begin{vmatrix}7 & 8 \\5 & 4\end{vmatrix} -3 \begin{vmatrix}1 & 8 \\0 & 4\end{vmatrix} +2 \begin{vmatrix}1 & 7 \\0 & 5\end{vmatrix}\) = 0(28 - 40) - 3(4 - 0) + 2(5 - 0) = 0(-12) - 3(4) + 2(5) = 0 - 12 + 10 = -2

(x1, y1) = (0,5) (x2, y2) = (5, 0) Using \(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\) \(\frac{y - 5}{0 - 5} = \frac{x - 0}{5 - 0}\) \(\frac{y - 5}{-5} = \frac{x}{5}\) 5(y - 5) = -5x y - 5 = -x x + y = 5 y = -x + 5

x2 = 3(2x + 9) x2 = 6x + 27 x2 - 6x - 27 = 0 x2 - 9x + 3x - 27 = 0 x(x - 9) + 3(x - 9) = (x + 3)(x - 9) = 0 x + 3 = 0 or x - 9 = 0 x = -3 or x = 9 x = -3, x = 9

let the number of N50 notes = x let the number of N20 notes = y from the first statement x + y = 13.....(1) from the second statement 50x + 20y = 530 ......(2) dividing through by 10, we have 5x + 2y = 53 ....(20 solving simultaneously by eliminating x + y = 12....(1) 5x + 2y = 53.....(2) multiply eqn (1) by 2 multiply eqn (2) by 1 2x + 2y = 26 -5x + 2y = 53 -3x = -27 x = 9 substitute x = 9 into equation (1) x + y = 13 y = 13 - x = 13 - 9 = 4 y = 4 the number of N50 notes = 9