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x\(^2\) + 5\(^2\) = 13\(^2\) x\(^2\) + 25 = 169 x\(^2\) = 144 x = \sqrt 144 = 12 Length of the chord AB = 2x 2\(\times\)12= 24cm

Surface Area of a sphere = 4\(\pi\)r\(^2\) S.A of a sphere = 154 = 4 x \(\frac{22}{7}\) x r\(^2\) r\(^2\) = \(\frac{154 \times 7}{4 \times 22}\) r\(^2\) = \(\frac{49}{4}\) r = \(\sqrt{\frac{49}{4}}\) r = \(\frac{7}{2}\) r = 3.5cm.

5 \(\times\) x\(^1\) + 5 \(\times\) x\(^0\) + 5 \(\times\) x\(^1\) + 2 \(\times\) x\(^0\) = 77 (change all to base 10) 5x + 5 + 5x + 2 = 77 10x + 7 = 77 10x = 77-7 10x = 70 x = 70/10 x = 7

Convert each mixed number to an improper fraction: 7 \(\frac{1}{12}\) = 7 + \(\frac{1}{12}\) = \(\frac{84}{12} + \frac{1}{12}\) = \(\frac{85}{12}\) 4 \(\frac{3}{4}\) = 4 + \(\frac{3}{4}\) = \(\frac{16}{4}\) + \(\frac{3}{4}\) = \(\frac{19}{4}\) 2 \(\frac{1}{2}\) = 2 + \(\frac{1}{2}\) = \(\frac{4}{2}\) + \(\frac{1}{2}\) = \(\frac{5}{2}\) Now calculate: \(\frac{85}{12} − \frac{19}{4} + \frac{5}{2}\) Common denominator = 12 \(\frac{85}{12}\) − (\(\frac{19}{4}\) \(\times\) \(\frac{3}{3}\)) + (\(\frac{5}{2}\) \(\times\) \(\frac{6}{6}\)) = \(\frac{85}{12}\) − \(\frac{57}{12}\) + \(\frac{30}{12}\) = (\(\frac{85 − 57 + 30)}{12}\) = \(\frac{(58)}{12}\) = \(\frac{29}{6}\) = 4 \(\frac{5}{6}\).

\(\frac{81.81 + 99.44}{20.09 + 36.16}\\ =\frac{181.25}{56.25}\\ =\frac{18125}{5625}\\ \frac{29}{9}\\ =3\frac{2}{9}\\ ≅ 3.22\)

C.P = N34000 + N2000 = N36000 Gain = 100 + 15 = 115% S.P = \(\frac{115}{100}\times \frac{N3600}{1}\) = N41400

Let his allowance = y. \(Books = \frac{y}{5}\) \(Remainder: y - \frac{y}{5} = \frac{4y}{5}\) \(Food: \frac{1}{2} \times \frac{4y}{5} = \frac{2y}{5}\) \(Contingencies: \frac{4y}{5} - \frac{2y}{5} = \frac{2y}{5}\) Therefore, he kept \(\frac{2}{5}\) of his allowance for contingencies.

Log 2 = 0.3010, Log 7 = 0.8451 ∴Log 280 = Log 28 x 10 = Log 7x4x10 =Log7 + Log4 + Log10 = Log7 + Log2\(^2\) + Log10 = Log7 + 2Log2 + Log10 = 0.8451 + 2(0.3010) + 1 = 0.8451 + 0.6020 + 1 = 2.4471

\(\frac{5+\sqrt{7}}{3+\sqrt{7}}=\frac{5+\sqrt{7}}{3+\sqrt{7}}\times \frac{3-\sqrt{7}}{3-\sqrt{7}}\) =\(\frac{(5+\sqrt{7})(3-\sqrt{7})}{3^2 - \sqrt{7}^2}\) =\(\frac{15-5\sqrt{7}+3\sqrt{7}-7}{9-7}\) = \(\frac{8-2\sqrt{7}}{2}\) Factorize, then divide by 2 \(=\frac{2(4-\sqrt{7}}{2})\) =4-\(\sqrt{7}\)

X = {n(^2\)+1:n is a positive integer and 1 \(\leq\) n \(\leq\) 5} Implies X = {2, 5, 10, 17, 26} i.e. put n= 1, 2, 3, 4 and 5 Y = {5n:n is a positive integer and 1 \(\leq\) n \(\leq\) 5} Put X = 1, 2, 3, 4, and 5 Y = {5, 10, 15, 20, 25} X \(\cap\) Y = {2, 5, 10, 17, 26} \(\cap\) {5, 10, 15, 20, 25} = {5, 10}

If S \(\subset\) T \(\subset\) W, S \(\cap\) T \(\cap\) W = S is true since S \(\cap\) T = S and S \(\cap\) W = S. S \(\cup\) T \(\cup\) W = W is also true. S \(\cup\) T = T and T \(\cup\) W = W. However, to say that T \(\cap\) W = S is not very true mathematically. Instead, it is safe to say S \(\subset\) (T \(\cap\) W).

\(p =\sqrt{\frac{rs^3}{t}}\) = p\(^2 =\frac{rs^3}{t}\) tp\(^2\) = rs\(^3\) r = \(\frac{p^2 t}{s^3}\)

If 4/3 and -3/5 are roots of a polynomial Imply x = 4/3 and - 3/5 3x = 4 and 5x = -3 ∴3x-4 = 0 and 5x+3 = 0 are factors (3x-4)(5x+3) = 0 product of the factors 15x2 + 9x – 20x – 12 = 0 By expansion 15x2 - 11x – 12 = 0

x = -2 and x = 1/4 x = -2 and 4x = 1 x+2 and 4x-1 (x+2)(4x-1) = 0 4x2 - x + 8x -2 = 0 4x2 + 7x – 2 = 0 but y intercept is positive. Multiply the equation by -1 -4x2 - 7x + 2 = 0 ∴y = 2 – 7x – 4x2

W ∝ U W = KU K = \(\frac{W}{U}\) K = \(\frac{5}{3}\) W = \(\frac{5}{3}U\) \(\frac{2}{7} = \frac{5}{3}\)U U = \(\frac{2}{7} \times \frac{3}{5}\) U = \(\frac{6}{35}\)