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Note that \(\frac{10\sqrt{3}}{\sqrt{5}} = \frac{10\sqrt{3}}{\sqrt{5}} \times - \frac{\sqrt{5}}{\sqrt{5}}\) = \(\frac{10\sqrt{15}}{\sqrt{5}} = 2\sqrt{15}\) hence, (\(\frac{10\sqrt{3}}{\sqrt{5}} - \sqrt{15}\))2 = (\(2\sqrt{15} - \sqrt{15}\))2 = (\(2\sqrt{15} - \sqrt{15}\))(\(2\sqrt{15} - \sqrt{15}\)) = 4\(\sqrt{15 \times 15} - 2\sqrt{15 \times 15} - 2\sqrt{15 x 15} + \sqrt{15 \times 15}\) = 4 x 15 - 2 x 15 - 2 x 15 + 15 = 60 - 30 - 30 + 15 = 15

d \(\alpha\) t2 d = t2 k where k is a constant. d = 45cm, when t = 3s; thus 45 = 32 x t k = \(\frac{45}{9}\) = 5 thus equation connecting d and t is d = 5t2 when t = 6s, d = 5 x 62 = 5 x 36 = 180cm

From the venn diagram, Nigeria footballers from a subset of good footballers.

On a map, 1cm represents 5km. Then it follows that 1cm2 represents 25km2. Acm2 represents 100km2. By apparent cross-multiplication, 1cm2 x 100km2 = Acm2x 25km2 therefore A = \(\frac{100}{25}\) = 4cm2

\(\frac{3^{n - 1} \times 27^{n + 1}}{81^{n}}\) = \(\frac{3^{n - 1} \times 3^{3(n + 1)}}{3^{4n}}\) = 3\(^{n - 1 + 3n + 3 - 4n}\) = 3\(^{4n - 4n - 1 + 3}\) = 32 = 9

A = P + 1 I = A - P = 10,400 - P Now using I = \(\frac{P \times T \times R}{100}\) i.e. 10,400 - P = \(\frac{P \times 5 \times 6}{100}\) = 100(10,400 - P) = 30P 10(10,400 - P) = 3P 104,000 - 10P = 3P 104,000 - 10P = 3P 104,000 = 3P + 10P = 104,000 = 13P P = \(\frac{104,000}{100}\) P = D8,000

Let x = \(\frac{4}{3}\), x = -\(\frac{3}{7}\) Then 3x = 4, 7x = -3 3x - 4 = 0, 7x + 3 = 0 (3x - 4)(7x + 3) = 0 21x2 + 9x - 28x - 12 = 0 21x2 - 19x - 12 = 0

\(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\) Factorize the denominator; Y2 + 7y - 3y - 21 = y(y + 7) -3 (y + 7) = (y - 3)(y + 7) Hence the expression \(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\) is undefined when y2 + 4y - 21 = 0 ie. y = 3 or -7

2x + y = 7...(1) 3x - 2y = 3...(2) From (1), y = 7 - 2x for y in (2) 3x - 2(7 - 2x) = 3 3x - 14 + 4x = 3 7x + 3 + 14 = 17 x = \(\frac{17}{7}\) Hence, 7 x \(\frac{17}{7}\) = 17 - 10 = 7

\(\frac{2}{1 - x} - \frac{1}{x}\) = \(\frac{2x - 1(1 - x)}{x(1 - x)}\) = \(\frac{2x - 1 + x}{x(1 - x)}\) = \(\frac{3x - 1}{x(1 - x)}\)

P = S + \(\frac{sm^2}{nr}\) P = S(1 + \(\frac{m^2}{nr}\)) P = S(1 + \(\frac{nr + m^2}{nr}\)) nrp = S(nr + m2) S = \(\frac{nrp}{nr + m^2}\)

(2x + 3y)2 - (x - 4y)2 = (2x + 3y)(2x + 3y) - (x - 4y)(x - 4y) = 4x2 + 12xy + 9y2 - (x2 - 8xy+ 16y2) = 4x2 + 12xy + 19y2 - x2 + 8xy - 16y2 = 3x2 + 20xy - 7y2 = 3x2 + 21xy - xy - 7y2 = 3x(x + 7y) - y(x + 7y) = (3x - y)(x + 7y)

Curved surface area of cylinder = 2\(\pi\)rh 110 = 2 x \(\frac{22}{7}\) x r x 5 r = \(\frac{110 \times 7}{44 \times 5}\) = 3.5cm

Volume of pyramid = \(\frac{1}{3}\)lbh 90 = \(\frac{1}{3} \times x \times 6 \times 15\) x = \(\frac{90 \times 33}{6 \times 15}\) = 3

Let: (x1, y1) = (1, 2) (x2, y2) = (5, 8) The gradient m of \(\bar{PQ}\) is given by m = \(\frac{y_2 y_1}{x_2 - x_1}\) = \(\frac{8 - 2}{5 - 1}\) = \(\frac{6}{4}\) = \(\frac{3}{2}\)