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\(\frac{\frac{7}{9}-\frac{2}{3}}{\frac{1}{3}+\frac{\frac{2}{5}}{\frac{4}{5}}}\\ =\frac{\frac{7-6}{9}}{\frac{1}{3}+\frac{\frac{2}{5}}{\frac{4}{5}}}\\ =\frac{\frac{1}{9}}{\frac{1}{3}+\left(\frac{2}{5}\times \frac{5}{4}\right)}\\ =\frac{\frac{1}{9}}{\frac{1}{3}+\frac{1}{2}}\ =\frac{\frac{1}{9}}{\frac{2+3}{6}}\\ =\frac{\frac{1}{9}}{\frac{5}{6}}\\ =\frac{1}{9}\times\frac{6}{5}\\ =\frac{2}{15}\)

m : n = 13 : 11 m\(^2\) - n\(^2\) : (m + n)2 = 13\(^2\) - 11\(^2\) : (13 + 11)\(^2\) = 169 - 121 : 24\(^2\) = 48 : 576 = 1 : 12

\(3^{-4}\times 9^2 \times 81^{-1}\) =\(log_9 (3^{-4}\times 9^2 \times 81^{-1})\) = \(log_9 \left(\frac{1}{3^4}\times 9^2 \times \frac{1}{81}\right)\) = \(log_9 \left(\frac{1}{81}\times \frac{81}{1}\times \frac{1}{81}\right)\) = \(log_9 \frac{1}{81}\) = \(log_9 \frac{1}{9^2}\) = \(log_9 9^{-2}\) = -2\(log_9 9\) - 2 \(\times 1\) = -2

(K2)\(_6\) * 3\(_6\) = 3\(_5\)(K4)\(_5\) (K*6\(^1\) + 2*6\(^0\)) 3*6\(^0\) = 3*5\(^0\)(K*5\(^1\) + 4*5\(^0\)) 3(6K + 2) = 3(5K + 4) 6K + 2 = 5K + 4 6K - 5K = 4 - 2 K = 2

Let n(L∩P) = x n(L) only = 350 - x n(P) only = 200 - x ∴350 - x + x + 200 - x = 500 550 - x = 500 x = 550 - 500 x = 50 speak both ∴x% = (50/500) * (100/1) = 10%

N20,000 - N10,000 = N10,000 1st tax on N5000 at N50 per N1000 = 5*50 = N250 2nd tax on the remaining N5000 at N55 per N1000 = 5*55 = N275 Total tax paid = N250 + N275 = N525

\(\frac{2}{6-5\sqrt{3}} = \frac{2}{6-5\sqrt{3}} \times \frac{6+5\sqrt{3}}{6+5\sqrt{3}}\\ =\frac{2(6+5\sqrt{3})}{(6-5\sqrt{3})(6+5\sqrt{3})}\\ =\frac{12+10\sqrt{3}}{36-25(3)}\\ =\frac{12+10\sqrt{3}}{36-75}\\ =\frac{12+10\sqrt{3}}{-39}\\ =-\left(\frac{12}{39}+\frac{10\sqrt{3}}{39}\right)\)

1100112 + 111112 = 10100102

(25)\(^{\frac{-1}{2}}\) x (27)\(^{\frac{1}{3}}\) + (121)\(^{\frac{-1}{2}}\) x (625)\(^{\frac{-1}{4}}\) 5\(^{2 \times \frac{-1}{2}}\) x 3\(^{3 \times \frac{1}{3}}\) + 11\(^{2 \times \frac{-1}{2}}\) x 5\(^{4 \times \frac{-1}{4}}\) 5\(^{-1}\) x 3\(^1\) x 11\(^{-1}\) x 5\(^{-1}\) \(\frac{1}{5} \times \frac{3}{1} + \frac{1}{11} \times \frac{1}{5}\) \(\frac{3}{5} + \frac{1}{55} = \frac{33+1}{55}\) = \(\frac{34}{55}\)

1st convert to base 10 2232\(_4\) = 2 x 4\(^3\) + 2 x 4\(^2\) + 3 x 4\(^1\) + 2 x 4\(^0\) = 2 x 64 + 2 x 16 + 3 x 4 + 2 x 1 = 128 + 32 + 12 + 2 = 174 convert to base 6 6/174 6/29 R 0 6/4 R 5 6/0 R 4 = 450\(_6\)

Area of rectangle PQRS = \(10 \times 7 = 70cm^2\) Area of semi-circle: \(\frac{\pi r^{2}}{2}\) r = \(\frac{7}{2} cm\) Area of semi-circle = \(\frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) = \(\frac{77}{4} = 19.25 cm^{2}\) Area of figure = \((70 + 19.25) cm^{2}\) = 89.25 cm\(^{2} \approxeq\) 89 cm\(^{2}\)

Tan θ = 5/4 x2 = 52 + 42 = 25 + 16 = 41 x = √41 sin2θ - cos2θ = (5/√41)2 - (4/√41)2 = 25/41 - 16/41 = 9/41

Gradient PQ, P(1,q) and Q(3,2) \(=\frac{(2-q)}{(3-1)} = \frac{(2-q)}{2}\) Gradient of RS : R(3,4) and S(5,2q) \(= \frac{(2q-4)}{(5-3)}= \frac{(2q-4)}{2} = \frac{2(q-2)}{2}\) = q-2 Since PQ and RS are parallel, their gradients are equal \(∴ \frac{(2-q)}{2} = q-2\) 2-q = 2(q-2) 2-q = 2q-4 2+4 = 2q+q 6 = 3q q = 2

∠yxz = 180 - (45 + 15) = 180 - 60 = 120o Using sine rule \(\frac{x}{sinx}=\frac{7}{sinz}\) \(\frac{x}{sin 120}=\frac{7}{sin 45}\) x = \(\frac{7 sin 120}{sin45}\) x = \(\frac{7sin(180-120)}{sin 45}\) x = \(\frac{7 sin 60}{sin 45}=\left(7\left(\frac{\sqrt{3}}{2}\right)\div \frac{1}{\sqrt{2}}\right)\) x = \(\left(7\left(\frac{\sqrt{3}}{2}\right)\div \frac{1}{\sqrt{2}}\right)\) x = 7\(\left(\frac{\sqrt{6}}{2}\right)\)

y = 55 (alternate angles) x = x (same reason) 260 + y + x = 360 (∠s at a point) 260 + 55 + x = 360 315 + x = 360 x = 360 - 315 x = 45o