Correct Answer: C. N545

Explanation

Husband contributions = \(\frac{7}{100}\) x N5,500 = N385

Wife contribution = \(\frac{4}{100}\) x N4,000 = N160

Sum of their Annual contribution = N385 + N160 = N545

Correct Answer: C. \(\frac{3}{2}\)

Explanation

Let: (x1, y1) = (1, 2)

(x2, y2) = (5, 8)

The gradient m of \(\bar{PQ}\) is given by

m = \(\frac{y_2 y_1}{x_2 - x_1}\)

= \(\frac{8 - 2}{5 - 1}\)

= \(\frac{6}{4}\)

= \(\frac{3}{2}\)

Correct Answer: B. \(\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\)

Explanation

Let A = \(\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}\)

|A| = (3 x 2 - 5 x 1)

= 6 - 5

= 1

A\(^{-1}\) = \(\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\)

Explanation

(a) \(\sqrt{4x - 3} - \sqrt{2x - 5} = 2\)

Squaring both sides, we have

\(4x - 3 - 2\sqrt{8x^{2} - 26x + 15} + 2x - 5 = 4\)

\(6x - 8 - 4 = 2\sqrt{8x^{2} - 26x + 15}\)

\(6x - 12 = 2\sqrt{8x^{2} - 26x + 15}\)

\(3x - 6 = \sqrt{8x^{2} - 26x + 15}\)

Squaring both sides, we have

\(9x^{2} - 36x + 36 = 8x^{2} - 26x + 15\)

\(9x^{2} - 8x^{2} - 36x + 26x + 36 - 15 = 0\)

\(x^{2} - 10x + 21 = 0\)

Factorising, we have

\((x - 7)(x - 3) = 0\)

\(\text{x = 7 or 3}\).

(b)

We integrate with respect to y.

\(y^{2} = 4x \implies x = \frac{y^{2}}{4}\)

\(x + y = 0 \implies x = -y\)

To get the boundary, equate the two values of x;

\(\frac{y^{2}}{4} = - y \implies y^{2} = -4y\)

\(y^{2} + 4y = 0 \implies y(y + 4) = 0\)

\(y = 0; y = -4\)

\(\int_{-4}^{0} (\frac{y^{2}}{4}) \mathrm {d} y = [\frac{y^{3}}{12}]|_{-4}^{0}\)

\(0 - \frac{-4^{3}}{12} = 0 - \frac{-64}{12} = \frac{64}{12} sq. units\)

\(\int_{-4}^{0} - y \mathrm {d} y = [-\frac{y^{2}}{2}]|_{-4}^{0}\)

\(0 - \frac{-(-4^{2})}{2} = \frac{16}{2} = 8 sq. units\)

Area enclosed = \(8 - \frac{64}{12} = \frac{32}{12} = 2\frac{2}{3} sq. units\)

Correct Answer: B. N40,000

Explanation

\(Abu = 2x\\

Kayode = x\\

Uche = \frac{x}{2}\\

2x+x+\frac{x}{2}=140,000\\

\frac{7x}{2}=140,000\\

7x = 280,000\\

x=\frac{280,000}{7}\\

x=40,000\)Kayode receives N40,000

Correct Answer: B. 40√3m /3

Explanation

\(\tan 30 = \frac{x}{40}\)

\(x = 40 \tan 30 = \frac{40}{\sqrt{3}}\)

= \(\frac{40\sqrt{3}}{3} m\)

Correct Answer: B. \(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\)

Explanation

\(y = 2(2x + \sqrt{x})^{2}\)

Let \(u = 2x + \sqrt{x}\)

\(y = 2u^{2}\)

\(\frac{\mathrm d y}{\mathrm d u} = 4u\)

\(\frac{\mathrm d u}{\mathrm d x} = 2 + \frac{1}{2\sqrt{x}}\)

\(\therefore \frac{\mathrm d y}{\mathrm d x} = (\frac{\mathrm d y}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)

= \(4u(2 + \frac{1}{2\sqrt{x}}) \)

= \(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\)

Correct Answer: C. -i + 12j

Explanation

→PQ = →PR + →RQ

→PR = →PQ - →RQ

→PR = -2i + 5j - (-i - 7j)

→PR = -2i + 5j + i + 7j

→PR = -i + 12j.

Explanation

\(\frac{\sqrt{75} - 3}{\sqrt{3} + 1}\)

Rationalizing, \((\frac{\sqrt{75} - 3}{\sqrt{3} + 1})(\frac{\sqrt{3} - 1}{\sqrt{3} - 1}\)

\(\frac{\sqrt{3 \times 75} - \sqrt{75} - 3\sqrt{3} + 3}{3 - \sqrt{3} + \sqrt{3} - 1}\)

= \(\frac{15 - 5\sqrt{3} - 3\sqrt{3} + 3}{2}\)

= \(\frac{18 - 8\sqrt{3}}{2}\)

= \(9 - 4\sqrt{3}\)

(b)(i) The general form of the circle equation is \(x^{2} + y^{2} + 2gx + 2fy + c = 0\) where (x, y) is a point on the circle.

\(\therefore A(7, 3) = 7^{2} + 3^{2} + 7(2g) + 3(2f) + c = 0\)

\(14g + 6f + c = -58 .... (1)\)

\(B(2, 8) = 2^{2} + 8^{2} + 2(2g) + 2(2f) + c = 0\)

\(4g + 4f + c = -68 ..... (2)\)

\(C(-3, 3) = -3^{2} + 3^{2} - 3(2g) + 3(2f) + c = 0\)

\(6f - 6g + c = -18 ....(3)\)

Solving the equations, we get \(g = -2; f = -3; c = -12\)

\(\text{The equation of the circle is} x^{2} + y^{2} - 4x - 6y - 12 = 0\).

(b)(ii) Radius = \(\sqrt{g^{2} + f^{2} - c}\)

= \(\sqrt{-2^{2} + (-3^{2}) - (-12)}\)

\(\sqrt{25} = 5 units\)

Correct Answer: B. 0.719

Explanation

tan y = 0.404; y = tan^-1 0.0404(tables);y = 22^o

cos 2y = cos 2(22^o); cos 44^o

= 0.719