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Let value of store = X Ratio of Taiwo's share to kehine's is 11:9 Keyinde sells \(\frac{2}{3}\) of her share to Taiwo for N720 \(\frac{2}{3}\) of 9 = 6 ∴ Sum of the ratio = 11 + 9 = 20 \{\frac{6}{20}\) of x = N720 \(\frac{6x}{20}\) = 720 ∴ x = \(\frac{720 \times 20}{6}\) x = N24,000

Let x represent total number of balls in the basket. If there are 10 blue balls, \(\frac{1}{8}\) of x = 10 x = 10 x 8 = 80 balls Green balls will be \(\frac{5}{8}\) x 80 = 50 and black balls = \(\frac{2}{8}\) x 80 = 20 Ratio = Green : black : blue 50 : 20 : 10 -10 : -10 : - ------------------ New Ratio 40 : 10 : 10 4 : 1 : 1

He pays tax on 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\)th of his income 20% is 490, 100% is \(\frac{100}{20}\) x 490, N2,450.00 = \(\frac{7}{8}\) of his income = N2,450.00 \(\frac{1}{\frac{7}{8}}\) x 2450 = \(\frac{8 \times 2450}{7}\) = \(\frac{19600}{7}\) = N2800.00

\(\frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}}\) = \(\frac{(2^3)^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{(2 \times 5)^{\frac{2}{3}}}\) = \(\frac{2 \times 5^{\frac{2}{3}}}{2^{\frac{2}{3}} \times 5^{\frac{2}{3}}}\) = \(2^{1 - \frac{2}{3}}\) x \(5^{\frac{2}{3} - \frac{2}{3}}\) = \(2^{\frac{1}{3}}\) x \(5^0\) = \(\sqrt[3]{2}\)

If \(log_{10} 2 = 0.3010\) and \(log_{10} 3 = 0.4771\), \(\log_{10} 4.5 = \log_{10} (\frac{3 \times 3}{2})\) \(log_{10} 3 + log_{10} 3 - log_{10} 2 = 0.4771 + 0.4771 - 0.3010\) = 0.6532

(m + \(\sqrt{3}\))(1 - \(\sqrt{3}\))2 = 6 - 2\(\sqrt{2}\) (m + \(\sqrt{3}\))(4 - 2\(\sqrt{3}\)) = 6 - 2\(\sqrt{2}\) = 6 - 2\(\sqrt{3}\) 4m - 6 + 4 - 2m\(\sqrt{3}\) = 6 - 2\(\sqrt{3}\) comparing co-efficients, 4m - 6 = 6.......(i) 4 - 2m = -2.......(ii) in both equations, m = 3

\(\frac{x + 2}{x + 1}\) - \(\frac{x - 2}{x + 2}\) = \(\frac{(x + 2)(x + 2) - (x - 2)(x + 1)}{(x + 1)(x + 2)}\) = \(\frac{(x^2 + 4x + 4) - (x^2 - x - 2)}{(x + 1)(x + 2)}\) = \(\frac{x^2 + 4x + 4 - x^2 + x + 2}{(x + 1)(x + 2)}\) = \(\frac{5x + 6}{(x + 1)(x + 2)}\)

\(x^2 + \frac{2x}{r^2}\) + \(\frac{1}{r^4}\) = 0 (x + \(\frac{1}{r^2}\)) = 0 x + \(\frac{1}{r^2}\) = 0 x = \(\frac{-1}{r^2}\)

-1 < 5 - 2x \(\geq\) 7 = -1 < 5 -2x and 5 - 2x \(\leq\) 7 = 2x < 5 + 1 and 5 - 7 \(\leq\) 2x = x < 3 and -1 \(\leq\) x Integral value of x are -1, 0, 1, 2

3x - 5y = 3, 2y - 6x = -5 -5y + 3x = 3........{i} x 2 2y - 6x = -5.........{ii} x 5 Substituting for x in equation (i) -5y + 3(\(\frac{19}{24}\)) = 3 -5y + 3 x \(\frac{19}{24}\) = 3 -5y = \(\frac{3 - 19}{8}\) -5 = \(\frac{24 - 19}{8}\) = \(\frac{5}{8}\) y = \(\frac{5}{8 \times 5}\) y = \(\frac{-1}{8}\) (x, y) = (\(\frac{19}{24}, \frac{-1}{8}\)

px2 + qx + b = 0 Using almighty formula \(\frac{-b \pm \sqrt{ b^2 - 4ac}}{2a}\).........(i) Where a = p, b = q and c = b substitute for this value in equation (i) = \(\frac{-q \pm \sqrt{ q^2 - 4bp}}{2p}\)

\(\frac{1}{x^2 + 5x + 6}\) + \(\frac{1}{x^2 + 3x + 2}\) = \(\frac{1}{(x + 2)(x + 3)}\) + \(\frac{1}{(x + 1)(x + 2)}\) \(\frac{(x + 1)+ (x + 3)}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{x + 1 + x + 3}{(x + 1)(x + 2)(x + 3)}\) \(\frac{2x + 4}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{2(x + 2)}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{2}{(x + 1)(x + 3)}\)

\(\cot \theta = \frac{x}{y}\) \(\implies \tan \theta = \frac{y}{x}\) \(opp = y; adj = x\) Using Pythagoras theorem, \(Hyp^{2} = Opp^{2} + Adj^{2}\) \(Hyp^{2} = y^{2} + x^{2}\) \(Hyp = \sqrt{y^{2} + x^{2}}\) \(\sin \theta = \frac{y}{\sqrt{y^{2} + x^{2}}}\) \(\therefore \csc \theta = \frac{\sqrt{y^{2} + x^{2}}}{y}\) = \(\frac{1}{y}(\sqrt{y^{2} + x^{2}})\)

PR2 = PQ2 + QR2 - 2(QR)(PQ) COS 120o PR2 = 12 + 22 - 2(1)(2) x - cos 60o = 5 - 2(1)(2) x -\(\frac{1}{2}\) = 5 + 2 = 7 PR = \(\sqrt{7}\)cm

The volume of the pipe is equal to the area of the cross section and length. let outer and inner radii be R and r respectively. Area of the cross section = (R2 - r2) where R = 6 and r = 6 - 1 = 5cm Area of the cross section = (62 - 52)\(\pi\) = (36 - 25)\(\pi\)cm sq vol. of the pipe = \(\pi\) (R2 - r2)L where length (L) = 10 volume = 11\(\pi\) x 10 = 110\(\pi\)cm3