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(3a - xy)(3a + xy); (3a)2 - (xy)2 difference of two sqs; 9a2 - x2y2

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

o (base of isosceles) RPS = 50o; o < PQR = 180 - 80 = 100

5 + 6 + 4 + 8 + 5 + 2 + x = 35 30 + x = 35 x = 35 - 30 x = 5

at least 2 kinds of food two or more foods i.e(R and p) + (p and Y) + (y and R) + R and y and p = 6 + 2 + 4 + 5 = 17

x : y = 3.5, y : z = 4 : 7 x = 3 x 4 = 12 y = 5 x 4 = 20 y = 4 x 5 = 20 z = 7 x 5 = 35 then x : y : z = 12 : 20 : 35

\(\frac{n}{4} = \frac{m}{4 + 12}\) \(\frac{n}{4} = \frac{m}{16}\) 4m = 16n \(\frac{4m}{4} = \frac{16n}{4}\) m = 4n

Convert \( (42_5) \) to base three: First, convert to decimal: \(42_5 = 4 \times 5 + 2 = 22_{10}\) Now convert \( (22_{10}) \) to base three: 22 \(\div 3 = 7 \quad ; \, r1 \) 7 \(\div 3 = 2 \quad ; \, r1 \) 2 \(\div 3 = 0 \quad ; \, r2\) Read the remainders from bottom to top: \(22_{10} = 211_3\).

(a) \(\frac{3\frac{1}{12} + \frac{7}{8}}{2\frac{1}{4} - \frac{1}{6}}\) \(3\frac{1}{12} + \frac{7}{8} = \frac{37}{12} + \frac{7}{8}\) = \(\frac{74 + 21}{24}\) = \(\frac{95}{24}\) \(2\frac{1}{4} - \frac{1}{6} = \frac{9}{4} - \frac{1}{6}\) = \(\frac{27 - 2}{12}\) = \(\frac{25}{12}\) \(\therefore \frac{3\frac{1}{12} + \frac{7}{8}}{2\frac{1}{4} - \frac{1}{6}} = \frac{95}{24} \div \frac{25}{12}\) \(\frac{95}{24} \times \frac{12}{25} = \frac{19}{10}\) = \(1.9\) (b)(i) \(p = \frac{m}{2} - \frac{n^{2}}{5m}\) \(\frac{n^{2}}{5m} = \frac{m}{2} - p\) \(n^{2} = 5m(\frac{m}{2} - p)\) \(n = \pm \sqrt{5m(\frac{m}{2} - p)}\) (ii) When p = 14 and m = -8, \(n = \sqrt{5(-8)(\frac{-8}{2} - 14)}\) \(n = \sqrt{-40(- 4 - 14)}\) \(n = \sqrt{720}\) \(n = 12\sqrt{5}\) = \(\pm 26.83\)

(a) No Log 0.004592 \(\bar{3}.6620\) \((0.004592)^{\frac{1}{3}}\) \(\frac{\bar{3}.6620}{3} = \bar{1}.2207\) Antilog of \(\bar{1}.2207 = 0.1663\) (b) \(\log_{10} y + 3 \log_{10} x = 2\) \(\log_{10} y + \log_{10} x^{3} = 2\) \(\log_{10} (yx^{3}) = 2\) \(yx^{3} = 10^{2}\) \(yx^{3} = 100\) \(y = \frac{100}{x^{3}}\) (c) \(3x - 2y = 21 .... (1)\) \(4x + 5y = 5 ..... (2)\) Multiply (1) by 4 and (2) by 3, \(12x - 8y = 84 .... (1a)\) \(12x + 15y = 15 ... (2a)\) (1a) - (2a), \(- 8y - 15y = 84 - 15\) \(- 23y = 69 \implies y = -3\) \(3x - 2y = 21 \implies 3x - 2(-3) = 21\) \(3x + 6 = 21 \implies 3x = 15\) \(x = 5\)

(a) \(3(x + y) + 45 = 180^\circ\) ( sum of angles on a straight line) \(3x + 3y + 45 = 180 \implies 3x + 3y = 135\) \(3x = 135 - 3y \implies x = 45 - y\) Also, \(5x + y + y = 180\) \(5x + 2y = 180..... (*)\) Put x = 45 - y into (*) above, \(5(45 - y) + 2y = 180 \implies 225 - 5y + 2y = 180\) \(225 - 3y = 180 \implies 3y = 45\) \(y = \frac{45}{3} = 15^\circ\) \(x = 45 - y\) \(x = 45 - 15 = 30^\circ\) (b) Let the distance BD = x and CD = y. In \(\Delta ABD\), \(\tan 53 = \frac{33}{x}\) \(x = \frac{33}{\tan 53}\) = \( 24.868 cm\) In \(\Delta ACD\), \(\tan 67 = \frac{33}{y}\) \(y = \frac{33}{\tan 67}\) \(y = 14.01 cm\) \(\therefore BC = x + y \) \(24.868 + 14.01 = 38.878 cm\) \(\approxeq 38.9 cm\)

The question will be answered in \(^6C_4 = \frac{6!}{(6-4)!4!}\) =\(\frac{6!}{2!4!}\) =\(\frac{6\times5\times4!}{2\times1\times4!}\) = 15ways.

\(^{n+1}C_3 = 4(^nC_3)\\\frac{(n+1)!}{(n+1-3)!3!} = 4\left(\frac{n!}{(n-3)!3!}\right)\\\frac{(n+1)n!}{(n-2)(n-3)!}=4\left(\frac{n!}{n-3!}\right)\\=\frac{n+1}{n-2}=\frac{4}{1}\\n+1 = 4(n-2)\\n+1 = 4n-8\\-3n = -9\\\frac{-9}{-3}\\n=3\)

y = 3x\(^3\) - 9x\(^2\) \(\frac{dy}{dx}\) = 9x\(^2\) - 18x As \(\frac{dy}{dx}\) = 0 9x\(^2\) - 18x = 0 9x(x-2) = 0 9x = 0 which implies x = 0 x - 2 = 0 implies x = 2 \(\frac{d^2y}{dx^2}\) = 18x - 18 when x = 0 \(\frac{d^2y}{dx^2}\) > 0 ∴ x is minimum when x = 2, \(\frac{d^2y}{dx^2}\) = 18 ∴ the value > 0 x is minimum

\(\int_{-4}^0(1 - 2x)\)dx = \([x - x^2]_{-4}^0\) = (0 - 0 + C) - (-4 - (-4)\(^2\) + C) = C - ( - 4 - 16 +C ) = C - ( - 20 + C) = C + 20 - C = 20