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= \(\sqrt{(5 - 2)^2 + (6 - 2)^2}\) = \(\sqrt{3^2 + 4^2}\) = \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5 units

Equation: 4y = 7x + 3 \(\implies y = \frac{7}{4} x + \frac{3}{4}\) Slope = coefficient of x = \(\frac{7}{4}\) Slope of perpendicular line = \(\frac{-1}{\frac{7}{4}}\) = \(\frac{-4}{7}\) The perpendicular line passes (-3, 1) \(\therefore\) Using the equation of line \(y = mx + b\) m = slope and b = intercept. \(y = \frac{-4}{7} x + b\) To find the intercept, substitute y = 1 and x = -3 in the equation. \(1 = \frac{-4}{7} (-3) + b\) \(1 = \frac{12}{7} + b\) \(b = \frac{-5}{7}\) \(\therefore y = \frac{-4}{7} x - \frac{5}{7}\) \(7y + 4x + 5 = 0\)

Marks (x) 1 2 3 4 5 Frequency (f) 2y - 2 y - 1 3y - 4 3 - y 6 - 2y 3y + 2 fx 2y - 2 2y - 2 9y - 12 12 - 4y 30 - 10y 26 - y Mean = \(\frac{\sum fx}{\sum f}\) \(3 = \frac{26 - y}{3y + 2}\) \(3(3y + 2) = 26 - y\) \(9y + 6 = 26 - y\) \(9y + y = 26 - 6\) \(10y = 20 \implies y = 2\)

Length of arc = \(\frac{\theta}{360^\circ} \times 2 \pi r\) = \(\frac{57}{360} \times 2 \times \frac{22}{7} \times 7\) = 6.97 cm

|AC| = |DF| = 13 cm Using Pythagoras theorem, |AC|\(^2\) = |AB|\(^2\) + |BC|\(^2\) 13\(^2\) = 12\(^2\) + |BC|\(^2\) |BC|\(^2\) = 169 - 144 = 25 |BC| = \(\sqrt{25}\) = 5 cm Volume of triangular prism = \(\frac{1}{2} \times base \times length \times height\) = \(\frac{1}{2} \times 5 \times 12 \times 18\) = 540 cm\(^3\)

\(\sin x = \frac{opp}{Hyp} = \frac{4}{5}\) 5\(^2\) = 4\(^2\) + adj\(^2\) adj\(^2\) = 25 - 16 = 9 adj = \(\sqrt{9}\) = 3 \(\tan x = \frac{4}{3}\) \(\cot x = \frac{1}{\frac{4}{3}} = \frac{3}{4}\) \(\cot^2 x = (\frac{3}{4})^2 = \frac{9}{16}\) \(\csc x = \frac{1}{\sin x}\) = \(\frac{1}{\frac{4}{5}} = \frac{5}{4}\) \(\csc^2 x = (\frac{5}{4})^2 = \frac{25}{16}\) \(\therefore \frac{1 + \cot^2 x}{\csc^2 x - 1} = \frac{1 + \frac{9}{16}}{\frac{25}{16} - 1}\) = \(\frac{25}{16} \div \frac{9}{16}\) = \(\frac{25}{9}\)

\(\frac{2\sin 30 + 5\tan 60}{\sin 60}\) \(\sin 30 = \frac{1}{2}; \tan 60 = \sqrt{3}; \sin 60 = \frac{\sqrt{3}}{2}\) \(\therefore \frac{2\sin 30 + 5\tan 60}{\sin 60} = \frac{2(\frac{1}{2}) + 5(\sqrt{3})}{\frac{\sqrt{3}}{2}}\)\) = \(\frac{1 + 5\sqrt{3}}{\frac{\sqrt{3}}{2}}\) = \(\frac{2(1 + 5\sqrt{3})}{\sqrt{3}}\) = \(\frac{2 + 10\sqrt{3}}{\sqrt{3}}\) Rationalizing, we get = \(\frac{2\sqrt{3} + 30}{3}\) = \(\frac{2}{3} \sqrt{3} + 10\)

(a) majority of the candidates multiplied through by the L.C.M which is 14 to arrive at 7(1+4x) - 2(5+2x) < 14(x - 2) and when simplified yielded 10x < -25 x -2\(\frac{1}{2}\) (b) it was expected that they express x = \(\frac{3y}{5}\) or y = \(\frac{5x}{3}\) and thereafter, substitute into the given expression to have \(\frac{2x^2 - y^2}{y^2 - x^2} = \frac{2x^2 - (\frac{5x}{3})^2}{(\frac{5x}{3})^2 - x^2} = \frac{x^(2 - \frac{25}{9})}{x^2(\frac{25}{9} - 1)} = \frac{-7}{16}\)

(a) The second term is a + d= x - 1 while the fourth term is a + 3d = x + 1. Subtracting the second term from the fourth term yielded 2d = 2 so that d = 1. (b). Substituting for d in the expression for the sixth term to get a + 5d = 7; a + 5(1) = 7 and solving for a to obtain a = 2. (c). Substituted for a and d in the expression for the second term( a + d = x - 1) and found x = 2 + 1 + 1 = 4.

(a) finding a = 10\(^{1.3010}\) and b = 10\(^{1.4771}\) Then, ab = 10\(^{1.3010}\) x 10\(^{1.4771}\) = 10\(^{2.4771}\) = 599.9 (b)(i) they were expected to find < ABC = 180^\circ - 62^\circ = 118^\circ Exterior < AOC a 2 x 118^\circ - 236^\circ (< at centre is twice angle at the circumference) Interior < AOC = 360^\circ - 236^\circ = 124^\circ(angle at a point) as required. (b)(ii) < BAC in 180^\circ - 39^\circ - 118^\circ = 23^\circ (sum of angles of a triangle)

(a) \(\frac{\log_28 + \log_216 - 4\log_22}{\log_416}\) = \(\frac{\log_22^3 + \log_2 2^4 - 4\log_22}{\log_44^2}\) = \(\frac{3 \log_2 2 + 4\log_22 - 4 \log_2^2}{2 \log 4^4}\) But log\(_2\) 2 = \(\log_44 = 1\) Then, \(\frac{\log_28+\log_216 - 4 \log_2^2}{\log_416}\) = \(\frac{3 + 4 - 4}{2}\) = 1\(\frac{1}{2}\) (b) Subtract 241\(_{five}\) to get 1101\(_{five}\). Thereafter, converting to base ten to find the value of x = (5\(^3\) x 1) + (5\(^2\) x 1 + (5\(^1\) x 0) + (5\(^o\) x 1) = 125 + 25 + 0 + 1 = 151 (b) They subtracted 241\(_{five}\) from 1342\(_{five}\) to get 1101\(_{five}\). Thereafter, converting to base ten to find the value of x = (5\(^3\) x 1) + (5\(^2\) x 1) + (5\(^1\) x 0) + (5^\circx 1) = 125 + 25 + + 1 = 151.

Let C be the cost and n the number of tourists. then C = x + ny. Where x and y are constants 740 = x + 20y...........(1) 960 = x + 30y.............(2) Solving (1) and (2) simultaneously will yield x = 300 and y = 22. Then, C = 300 + 22n. If we substitute 15 for n this implies that C = 300 + (22 x 15)= $630.00

(a) Selling price = \(\frac{90 \times 5600}{100}\) = $5,040.00. Thereafter, to obtain the amount invested = 5,040 - 1310 = $3,730.00 The interest amount invested is \(\frac{3730 \times 6 \times 3}{100}\) = $671.40 (b) 2\(^x\)(2\(^{-14}\)) = 2. This will be simplified and solved for x to get x = 15. Similarly, 3\(^{-x}\)(3\(^{4y}\)) = 3 and this will yield -x + 4y = 1. Substituting for x and solving for y yielded y = 4.

(a) Using length of chord = 2r sin (\(\frac{\theta}{2}\)) = 22.42 and simplifying sin(\(\frac{\theta}{2}\)) = \(\frac{22.42}{33.18}\) = 0.6757 Taking the sine inverse of 0.6757 to obtain \(\frac{\theta}{2}\) = 42.51\(^o\) and multiplying through to get \(\theta\) = 58.02\(^o\) = 885\(^o\) (b(i), the expected diagram is (b)(ii) since T is equidistant from P and Q, then ITPI = ITQI. Using sum of the angles in a triangle. 70^\circ + x + x = 180^\circ and solving for x. x = 55^\circ. Therefore. the bearing of Q from P = 270^\circ — 30^\circ — 55^\circ = 185^\circ.

(a) The range; 20 - 15 = 5. (b) They drew the frequency distribution table as \(\sum\) f = 11\(\sum\) fx = 191 \(\sum\)f\(^2\) = 3337 Age(x) Tally Freq (f) Com. Freq fx fx\(^2\) 15 16 17 18 19 20 I II III III I I 1 2 3 3 1 1 1 3 6 9 10 11 15 32 51 54 19 20 225 512 867 972 361 400 (c) Median age as 17. (d)(i) Mean = \(\frac{\sum fx}{\sum f}\) = \(\frac{191}{11}\) = 17.3636 (d)(ii). The standard deviation was \(\sqrt{\frac{2227}{11} - (\frac{191}{11})^2}\) = \(\sqrt{1.8677}\) = 1.37(correct to two decimal places)