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(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.

To find the father's monthly income, we follow these steps: Let \( I \) = Father's monthly income, \( A \) = Total allowance. The total allowance is given by: \(A = 0.2 I\) The second child receives 30% of the allowance: \(0.3 A = 3000\) Solving for \( A \): \(A = \frac{3000}{0.3} = 10000\) Father's Income: Now, substituting \( A \) into the equation for income: \(0.2 I = 10000 \implies I = \frac{10000}{0.2} = 50000\) The father's monthly income is: ₦50,000

To divide \(6x^2 - 13x + 5\) by \(2x - 1\): \(\begin{array}{r|l} 3x - 5 & \\ \hline 2x - 1 & 6x^2 - 13x + 5 \\ & -(6x^2 - 3x) \\ \hline & -10x + 5 \\ & -(-10x + 5) \\ \hline & 0 \\ \end{array}\) Thus, the quotient is: 3x - 5

To find the determinant of the matrix \(P = \begin{pmatrix} 1 & 3 & 2 \\ 4 & 5 & -1 \\ -3 & 2 & 0 \end{pmatrix}\) We use the formula: \(\text{det}(P) = a(ei - fh) - b(di - fg) + c(dh - eg)\) Substituting the values: - \(a = 1\), \(b = 3\), \(c = 2\) - \(d = 4\), \(e = 5\), \(f = -1\) - \(g = -3\), \(h = 2\), \(i = 0\) Calculating: \(ei - fh = 5 \cdot 0 - (-1) \cdot 2 = 2,\) \(di - fg = 4 \cdot 0 - (-1) \cdot -3 = -3,\) \(dh - eg = 4 \cdot 2 - 5 \cdot -3 = 8 + 15 = 23.\) Thus, \(\text{det}(P) = 1 \cdot 2 - 3 \cdot (-3) + 2 \cdot 23 = 2 + 9 + 46 = 57\) The determinant of matrix \( P \) is 57

Let the universal set be \(U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}\) and the set of prime factors of 30 be \(P = \{2, 3, 5\}.\) The complement of \( P \) is \(P' = U - P = \{1, 4, 6, 7, 8, 9\}\) Thus, the complement of \( P \) is {1, 4, 6, 7, 8, 9}.

To find the coordinates of the midpoint \( M \) of the line segment joining the points \( A(3, -4) \) and \( B(-1, 10) \), we use the midpoint formula: \(M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\) Substituting the coordinates of points \( A \) and \( B \): - \( x_1 = 3 \) - \( y_1 = -4 \) - \( x_2 = -1 \) - \( y_2 = 10 \) Calculating the coordinates of the midpoint: \(M = \left( \frac{3 + (-1)}{2}, \frac{-4 + 10}{2} \right) = \left( \frac{3 - 1}{2}, \frac{6}{2} \right) = \left( \frac{2}{2}, 3 \right) = (1, 3)\) Thus, the coordinates of the midpoint are (1, 3).

To evaluate \( 2M - 3N \) where \(M = \begin{pmatrix} 1 & 3 & 2 \\ 4 & 5 & -1 \\ -3 & 2 & 0 \end{pmatrix}\) and \(N = \begin{pmatrix} 1 & -2 & 3 \\ 4 & -1 & 5 \\ 2 & -3 & -1 \end{pmatrix}\) Calculate \( 2M \) \(2M = 2 \times \begin{pmatrix} 1 & 3 & 2 \\ 4 & 5 & -1 \\ -3 & 2 & 0 \end{pmatrix} = \begin{pmatrix} 2 \cdot 1 & 2 \cdot 3 & 2 \cdot 2 \\ 2 \cdot 4 & 2 \cdot 5 & 2 \cdot -1 \\ 2 \cdot -3 & 2 \cdot 2 & 2 \cdot 0 \end{pmatrix} = \begin{pmatrix} 2 & 6 & 4 \\ 8 & 10 & -2 \\ -6 & 4 & 0 \end{pmatrix}\) Calculate \( 3N \) \(3N = 3 \times \begin{pmatrix} 1 & -2 & 3 \\ 4 & -1 & 5 \\ 2 & -3 & -1 \end{pmatrix} = \begin{pmatrix} 3 \cdot 1 & 3 \cdot -2 & 3 \cdot 3 \\ 3 \cdot 4 & 3 \cdot -1 & 3 \cdot 5 \\ 3 \cdot 2 & 3 \cdot -3 & 3 \cdot -1 \end{pmatrix} = \begin{pmatrix} 3 & -6 & 9 \\ 12 & -3 & 15 \\ 6 & -9 & -3 \end{pmatrix}\) Calculate \( 2M - 3N \) \(2M - 3N = \begin{pmatrix} 2 & 6 & 4 \\ 8 & 10 & -2 \\ -6 & 4 & 0 \end{pmatrix} - \begin{pmatrix} 3 & -6 & 9 \\ 12 & -3 & 15 \\ 6 & -9 & -3 \end{pmatrix}\) Subtracting the matrices element-wise: \(= \begin{pmatrix} 2 - 3 & 6 - (-6) & 4 - 9 \\ 8 - 12 & 10 - (-3) & -2 - 15 \\ -6 - 6 & 4 - (-9) & 0 - (-3) \end{pmatrix}\) Calculating each element: \(= \begin{pmatrix} -1 & 12 & -5 \\ -4 & 13 & -17 \\ -12 & 13 & 3 \end{pmatrix}\) Final Result Thus, the result of \( 2M - 3N \) is \(\begin{pmatrix} -1 & 12 & -5 \\ -4 & 13 & -17 \\ -12 & 13 & 3 \end{pmatrix}\)

Length of an arc = \(\frac{\theta}{360}\) x 2\(\pi\)r 22 = \(\frac{3x}{360}\) x \(\frac{22}{7}\) x 7 (since D = 2r) 22 x 7 x 360 = 3x X 22 X 7 x = \(\frac{22 \times 7 \times 360}{3 \times 22 \times 7\times 2}\) = 60º

PQ = \(\sqrt{(\beta - 1)^{2} + (1 - \alpha)^{2}}\) 3 = \(\sqrt{(\beta^{2} -2\beta^{2} + 1 + 1 - 2\alpha + \alpha^{2})}\) 3 = \(\sqrt{(\alpha^{2} + \beta^{2} - 2\alpha + 2\beta + 2)}\) 3 = \(\sqrt{(\alpha^{2} + \beta^{2} - 2(\alpha + \beta) + 2)}\) 3 = \(\sqrt{(\alpha^{2} + \beta^{2} - 2 * 2 + 2)}\) 3 = \(\sqrt{(\alpha^{2} + \beta^{2} - 2)}\) 9 = \(\alpha^{2} + \beta^{2} - 2)\) \(\alpha^{2} + \beta^{2}\) = 9 + 2 \(\alpha^{2} + \beta^{2}\) = 11

6x + 6y = (n - 2) 180 6x + 6y= (5 - 2) 180 6(x + y) = 3 * 180 x + y = (3 * 180)/6 x + y = 90o y = 90 - x

Area of parallelogram PQTU = base * height 32 = 4 * h w = 32/4 w = 8 ∴ = Area of Trapezium PQRU = (1/2)(4 + 14) *8 = 1/2 * (18 * 8) = 72 cm\(^2\)

Mid point = \(\frac{(x_1 + x_2)}{2}\) ; \(\frac{(y_1 + y_2)}{2}\) = \(\frac{(-3 + 5)}{2} ; \frac{(5 - 3)}{2}\) = \(\frac{2}{2}\) ; \(\frac{2}{2}\) = (1, 1)

The diagonal QS bisects the angle formed by PQ and QR ∴ [A]

\(\frac{X}{\sin 45^\circ} = \frac{15}{\sin 60^\circ}\) \(X = \frac{15 \sin 45^\circ}{\sin 60^\circ}\) \(X = \frac{15 \times \left( \frac{1}{\sqrt{2}} \right)}{\sqrt{\frac{3}{2}}}\) \(X = \frac{15}{\sqrt{2}} \times \frac{2}{\sqrt{3}}\) \(X = \frac{30}{\sqrt{6}}\) \(X = \frac{30}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{30\sqrt{6}}{6} = 5\sqrt{6}\)

xo = 35 + 20 (Exterior angle = sum of two interior opposite angles) x = 55o (∠ at the center twice ∠ at circumference) y = 110o ∠PSO = ∠SPO (base ∠S of 1sc \Delta b/c PO = SO) ∴ ∠PSO = (180 - 110)/2 = 35o