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L\(\begin{pmatrix} x_1 & y_1 \\ -1 & -6 \end{pmatrix}\) m L\(\begin{pmatrix} x_2 & y_2 \\ -3 & -5 \end{pmatrix}\) D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) D = \(\sqrt{(-3 - (-1)^2 + (-5 -(-6)^2}\) D = \(\sqrt{(-3 + 1)^2 + (-5 + 6)^2}\) D = \(\sqrt{(-2)^2 + 1^2}\) D = \(\sqrt{4 + 1}\) D = \(\sqrt{5}\)

Roots; x and -c sum of roots = c + (-c) = 0 product of roots = c x -c = -c2 Equation; x2 - (sum of roots) x = product of roots = 0 x2 - (0)x + (-c2) = 0 x2 - c2 = 0

1 - 2x -\(\frac{4}{-6}\) = x > \(\frac{2}{3}\)

Given the 3rd and 5th terms of an A.P. are 6 and 10 respectively, we have: The formula for the \( n \)-th term of an A.P. is given by: \(T_n = a + (n-1)d\) where \( a \) is the first term and \( d \) is the common difference. For the 3rd term: \(T_3 = a + 2d = 6 \quad \text{(Equation 1)}\) For the 5th term: \(T_5 = a + 4d = 10 \quad \text{(Equation 2)}\) From these equations, we have: \(a + 2d = 6 \tag{1}\) \(a + 4d = 10 \tag{2}\) Subtract Equation 1 from Equation 2: \((a + 4d) - (a + 2d) = 10 - 6\) This simplifies to: \(2d = 4\) Thus, \(d = 2\) Substituting \( d = 2 \) into Equation 1: \(a + 2(2) = 6\) This simplifies to: \(a + 4 = 6\) Thus, \(a = 2\) The first term \( a \) is \( 2 \) and the common difference \( d \) is \( 2 \).

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

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

(a) \(\frac{2}{5}(x - 2) - \frac{1}{6}(x + 5) \leq 0\) Multiplying through by the LCM of 5 and 6 (i.e 30) \(12(x - 2) - 5(x + 5) \leq 0\) \(12x - 24 - 5x - 25 \leq 0\) \(7x - 49 \leq 0 \implies 7x \leq 49\) \(x \leq 7\). (b) (i) \(\frac{x^{2} - y^{2}}{x^{2} + xy}\) P = \(\frac{(x - y)(x + y)}{x (x + y)}\) P = \(\frac{x - y}{x}\) (ii) When x = -4, y = -6 \(P = \frac{-4 - (-6)}{-4}\) \(P = \frac{2}{-4}\) \(P = - \frac{1}{2}\)

(a) \(4x^{2} = 11x + 21 \implies 4x^{2} - 11x - 21 = 0\) \(a = 4, b = - 11 , c = -21\) \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) \(x = \frac{-(-11) \pm \sqrt{(-11)^{2} - 4(4)(-21)}}{2(4)}\) \(x = \frac{11 \pm \sqrt{121 + 336}}{8}\) \(x = \frac{11 \pm \sqrt{457}}{8}\) \(x = \frac{11 \pm 21.378}{8}\) \(x = \frac{11 + 21.378}{8}\) or \(x = \frac{11 - 21.378}{8}\) \(x = \frac{32.378}{8}\) or \(x = \frac{-10.378}{8}\)\ \(x = \text{4.047 or -1.297}\) \(x \approxeq \text{4.05 or -1.30}\).(2 decimal place). (b) (i) \(I = \frac{PRT}{100}\) \(I = £(1560 - 1500) = £60\) \(60 = \frac{1500 \times R \times 1}{100}\) \(60 = 15R \implies R = 4%\) (ii) Interest for the second year = \(\frac{1560 \times 4 \times 1}{100}\) = \(£62.40\) (c) Initial price of the car \(N300,000.00\) Depreciation after first year (25%) \(\frac{25}{100} \times 300,000 = N75,000\) Value after 1st year N225,000.00 Depreciation after 2nd year(20%) \(\frac{20}{100} \times 225,000 = N45,000\) Value after 2nd year N180,000.00

(a) x 0^\circ 60^\circ 120^\circ 180^\circ 240^\circ 300^\circ 360^\circ \(\sin x\) 0 0.8660 0.8660 0 -0.8660 -0.8660 0 \(\cos x\) 1 0.5 -0.5 -1 -0.5 -0.5 1 \(3\sin x\) 0 2.598 2.598 0 -2.598 -2.598 0 \(2\cos x\) 2 1 -1 -2 -1 -1 2 y 2 3.598 1.598 -2 -3.598 -3.598 2 y 2.00 3.60 1.60 -2.00 -3.60 -3.60 2.00 (b) (c) From graph, \(x_{1} = 120 + \frac{1}{5} \times 15 = 120 + 3 = 123^\circ\) \(x_{2} = 360^\circ - \frac{2}{3} \times 15 = 360 - 10 = 350^\circ\) (d) From graph, \(180^\circ - 15 < x < 300 + \frac{3}{5} \times 15\) \(165^\circ < x < 300 + 9 = 165^\circ < x < 309^\circ\)

d = \(\sqrt{\frac{42w}{5l}}\) square both side of the equation \(d^2 = \left(\sqrt{\frac{42W}{5l}}\right)^2\\ d^2 = \frac{42W}{5l}\\ 5ld^2=42W\\ l = \frac{42W}{5d^2}\)

(x\(^2\) + x - 12) \ge 0 (x + 4)(x - 3) \ge 0 Either x + 4 \ge 0 implies x \ge - 4 Or x - 3 \ge 0 implies x \ge 3 ∴ x \ge 3 or x \(\leq\) -4

-3(x-3) -12/-1 x > 12

3/2, 3, 7, 16, 35, 74, .... Using the method of substitution When n = 1, 5 . 2n-2 - 1 = 5 . 21-2 - 1 = 5 x 2-1 - 1 = 5 x 1/2 - 1 = 5/2 - 1 = 3/2 When n = 1, 5 . 2n-2 - n = 5 . 22-2 - 2 = 5 x 20 - 2 = 5 x 1 – 2 = 3 When n = 3, 5 . 2n-2 - n = 5 . 23-2 - 3 = 5 x 21 - 3 = 5 x 2 – 3 = 10 -3 = 7 When n = 4, 5 . 2n-2 - n = 5 . 24-2 - 4 = 5 x 22 - 4 = 5 x 4 – 4 = 20 – 4 = 16 When n = 5, 5 . 2n-2 - n = 5 . 25-2 - 5 = 5 x 23 - 5 = 5 x 8 – 5 = 40 – 5 = 35 When n = 6, 5 . 2n-2 - n = 5 . 26-2 - 6 = 5 x 24 - 6 = 5 x 16 – 6 = 80 – 6 = 74 ∴ 5 . 2n-2 - 1 = 3/2, 3, 7, 16, 35, 74

a = 2 r = \(\frac{3}{4}\) S = \(\frac{a}{1-r}\) S = \(\frac{2}{1-\frac{3}{4}}\) = \(\frac{2}{\frac{1}{4}}\) S = \(\frac{2}{1}\times \frac{4}{1}\) = 8.

If the point of intersection is (1, -2), it implies that x = 1 and y = -2 when the two equation are solved simultaneously. ∴ substitute x = 1 and y = -2 in any of the equations 2y - k x + 2 = 0 2(-2) - k(1) + 2 = 0 -4 - k + 2 = 0 -4 + 2 = k -2 = k