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\(\frac{54k^2 - 6}{3k + 1} = \frac{6(9k^2 - 1)}{3k + 1}\) = \(\frac{6(3k + 1) - (3k - 1)}{3k + 1}\) = 6(3k - 1)

q = \(\frac{3p}{r} + \frac{s}{2}\) q = \(\frac{6p + rs}{2r}\) 6p + rs = 2qr 6p = 2qr - rs p = \(\frac{2qr - rs}{6}\)

Curved surface area = 2\(\pi h\) 704 = 2 x \(\frac{22}{7} \times 8 \times h\) 704 = \(\frac{352}{7}\)h 352h = 704 x 7 h = \(\frac{704 \times 7}{352}\) = \(\frac{4928}{352}\) h = 14cm

y = \(\sqrt{8m}\); log y = log \(\sqrt{8m}\) log y = log \(\sqrt{8}\) + log \(\sqrt{m}\) log y = \(\frac{1}{2}\)log 2\(^3\) + log m\(\frac{1}{2}\) log y = \(\frac{3}{2}\) log 2 + \(\frac{1}{2}\) log m

(\(\frac{x -y}{y}\)); \(\frac{3}{1} - \frac{x y}{y}\) = \(\frac{3y - (x - y)}{y}\) = \(\frac{3y - x + y}{y}\) = \(\frac{4y - x}{y}\)

prime factor of 210 = 2, 3, 5, 7 prime numbers less than 10 = 2, 3, 5 , 7

let the total amount be Nx i.e (\(\frac{1}{4}\))x + (\(\frac{1}{3}\))x + 72,000 = x \(\frac{x}{4} + \frac{x}{4} + 72,000 = x\) \(\frac{3x + 4x + 86,400}{12} = x\) cross multiply to clear fraction 12x = 3x + 4x + 86,400 12x - 7x = 86,400 5x = 86,400 x - \(\frac{86,400}{5}\) = 172,800 amount spent on food = \(\frac{1}{4} \times 172,800\) = N43,200

(\(\frac{27}{125}\))-\(\frac{1}{3}\) x (\(\frac{4}{9}\))\(\frac{1}{2}\) = (\(\frac{3^3}{5^3}\))-\(\frac{1}{3}\)-\(\frac{1}{3}\) x (\(\frac{3^2}{3^2}\))\(\frac{1}{2}\) -\(\frac{1}{2}\) = \(\frac{3^{-1}}{3^{-1}} \times \frac{2}{3}\) = \(\frac{\frac{1}{3}}{\frac{1}{5}} \times \frac{2}{3}\) \(\frac{1}{3} \times \frac{5}{1} \times {2}{3} = \frac{10}{9}\)

(\(\frac{m}{2} - 1\frac{1}{4}\))(m + \(\frac{2}{3}\)) (\(\frac{m}{2} - \frac{3}{2}\))(\(\frac{m}{1} + \frac{2}{3}\)) = \(\frac{m^2}{3} + \frac{3m}{6} - \frac{6}{6}\) = \(\frac{2m - 9m}{6}\) = \(\frac{-7m}{6}\) = 1\(\frac{1}{6}\)

V = L x B x H V = \(2x \times x \times x = 2x^3\) = 54cm3 i.e. 2x3 = 54 x3 = \(\frac{54}{2}\) = 27 x = 3\(\sqrt{27}\) x = 3 L = 2x i.e. 2 x 3 = 6cm B = x i.e. x = 3cm H = x i.e. x = 3cm but A = 2(LB + LH + BH) = 2(6 x 3) + (6 x 3) + (3 x 3) = 2(18 + 18 + 9) = 2(45) = 90cm2

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

L + 30o - 180o(Sum of o - 30o = 150o L = q = 150o(opposite o(alt. o(sum of o + c 180 c = 180 - 30 c = 150o b = a = 30o (opp o (opp o = 180o (sum of o + k + 70o = 180 k + 100o = 180 k = 180 - 100 k = 80o x + 80o = 180(sum of o - 80o x = 100o

2x2 - 5x - 3 = 0 2x2 - 6x + x - 3 = 0 2x(x - 3) + 1(x - 3) = 0 (2x + 1)(x - 3) = 0 2x + 1 = 0 2x = -1 x = -\(\frac{1}{2}\) x - 3 = 0 -\(\frac{1}{2}\) < x < 3

(a) \(\frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}}\) \(1\frac{1}{4} + \frac{7}{9} = \frac{5}{4} + \frac{7}{9}\) = \(\frac{45 + 28}{36}\) = \(\frac{73}{36}\) \(1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64} = \frac{13}{9} - (\frac{8}{3} \times \frac{9}{64})\) = \(\frac{13}{9} - \frac{3}{8}\) = \(\frac{104 - 27}{72}\) = \(\frac{77}{72}\) \(\therefore \frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}} = \frac{73}{36} \div \frac{77}{72}\) = \(\frac{73}{36} \times \frac{72}{77}\) = \(\frac{146}{77}\) = \(1\frac{69}{77}\) (b) \(\sin x = \frac{2}{3}\) In \(\Delta ABC, 3^{2} = 2^{2} + |BC|^{2}\) \(\therefore |BC|^{2} = 9 - 4 = 5\) \(|BC| = \sqrt{5}\) \(\tan x = \frac{2}{\sqrt{5}} ; \cos x = \frac{\sqrt{5}}{3}\) \(\therefore \tan x - \cos x = \frac{2}{\sqrt{5}} - \frac{\sqrt{5}}{3}\) = \(\frac{6 - 5}{3\sqrt{5}}\) = \(\frac{1}{3\sqrt{5}}\) Rationalizing, we have \(\frac{1}{3\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{15}\)

Let Wale's present age = x. Then Sonny's age = 2x Four years ago, Wale = x - 4 Sonny = 4(x - 4) = 2x - 4 \(4x - 16 = 2x - 4 \implies 4x - 2x = -4 + 16\) \(2x = 12 \implies x = 6\) Sonny = 2x = 2(6) = 12 years In z years' time, the sum of their ages = 66 Wale = 6 + z Sonny = 12 + z \(6 + z + 12 + z = 66 \implies 18 + 2z = 66\) \(2z = 66 - 18 = 48 \implies z = 24\) Therefore, in 24 years time, the sum of their ages will be 66.