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The two points of contact of the curve with the x-axis are x = 1 and x = 4. The roots of the equation are x = 1 and x = 4

The factor for the curve are (x-1)(x-4). The product off the curves gives \(=x^2-5x+4=y\)

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

(a) \((\frac{64}{125})^{-\frac{2}{3}} = (\frac{125}{64})^{\frac{2}{3}}\) = \((\frac{5}{4})^{3})^{\frac{2}{3}} \) = \((\frac{5}{4})^{2}\) = \(\frac{25}{16}\) \(\therefore 2 \div (\frac{64}{125})^{-\frac{2}{3}} = 2 \div \frac{25}{16}\) = \(2 \times \frac{16}{25}\) = \(\frac{32}{25}\) (b) \(3x + 5 = - 4x - 1\) \(3x + 4x = - 1 - 5\) \(7x = - 6\) \(x = -\frac{6}{7}\) \(y = 3x + 5\) ( You can use any of the given equations to get y) \(y = 3(-\frac{6}{7}) + 5 = -\frac{18}{7} + \frac{35}{7}\) = \(\frac{17}{7}\) \(k = (x, y) = (-\frac{6}{7}, \frac{17}{7})\)

(a) \(x + 8 < 4x - 15 \implies 4x - 15 > x + 8\) \(4x - x > 8 + 15 \implies 3x > 23\) \(x > \frac{23}{3}\) \(x > 7\frac{2}{3}\). (b) Extra income = 5k of N200,000 = \(\frac{5}{100} \times N200,000\) = \(N10,000\) Total income = Monthly income + commission = N(2,500 + 10,000) = N12,500. (c) Area of semi- circle = \(\frac{\pi r^{2}}{2}\) = \(\frac{1}{2} \times \frac{22}{7} \times 35 \times 35\) = \(1925 cm^{2}\) Length of rectangular part = 50 cm Breadth = Diameter of semi- circle = 2(35 cm) = 70 cm Area of rectangular part = \(length \times \breadth\) = \(50 \times 70\) = \(3500 cm^{2}\) Total area of window = \((3500 + 1925) cm^{2}\) = \(5425 cm^{2}\)

(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 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

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 \(\begin{pmatrix} -1 & 12 & -5 \\ -4 & 13 & -17 \\ -12 & 13 & 3 \end{pmatrix}\)

-1 < 3 -2x and 3 - 2x \(\leq\) 5 2x < 3 + 1 and -2x \(\leq\) 5 - 3 2x < 4 and -2x \(\leq\) 2 x < 2 and x \(\geq\) 2/-2 = x \(\geq\) -1 If -1 \(\leq\) x < 2 it implies that x = -1, 0, 1

Qn = 3 * 2n - 2 Um = 3 * 22m - 3 Q2 * U2 = (3 * 22 - 2) * (3 * 22(2) - 3) = 3 * 20 * 3 * 21 = 3 * 1 * 3 * 2 = 18

\(a=\frac{1}{2}\\r=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac{1}{6}\times\frac{2}{1}=\frac{1}{3}\\S_\infty = \frac{a}{1-r}\\S_\infty = \frac{\frac{1}{2}}{1-\frac{1}{3}}\\=\frac{\frac{1}{2}}{\frac{3}{2}}\\=\frac{1}{2}\times\frac{3}{2}\\=\frac{3}{4}\)

Given that \( P \propto \frac{1}{\sqrt{r}} \), we can express this relationship as: \(P = k \cdot \frac{1}{\sqrt{r}}\) for some constant \( k \). From the information provided, when \( P = 3 \) and \( r = 16 \): \(3 = k \cdot \frac{1}{\sqrt{16}}\) Since \( \sqrt{16} = 4 \), we have: \(3 = k \cdot \frac{1}{4}\) Thus, \(k = 3 \cdot 4 = 12\) Now we can write the equation as: \(P = \frac{12}{\sqrt{r}}\) Next, we need to find \( r \) when \( P = \frac{3}{2} \): \(\frac{3}{2} = \frac{12}{\sqrt{r}}\) Cross-multiplying gives: \(3\sqrt{r} = 24\) Dividing both sides by 3: \(\sqrt{r} = 8\) Squaring both sides: \(r = 64\). Thus, r = 64

first term[a] = 3 * d → 3 * 8 = 24 sixth term of the A.P = a + 5d → 24 + [5 * 8] ⇒ 24 + 40 = 64

Coordinates of the line are (x1 y1)(0,2) respectively and (x2 y2)(2,0) respectively Gradient (m) of the line = (y2 - y1) / (x2 - x1) = (0-6) / (2-0) = -6/2 = -3 Equation of the line y-y1 = m(x-x1) y - 6 = -3(x - 0) y - 6 = -3x y + 3x = 6 inequality; from the y axis, the shaded region is below the line ∴ y + 3x ∠ 6 and line is a broken line