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No expanded working has been attached for this question yet. The verified answer is Option C.

T4 = 5 + (4 - 1)2; where n = 4 = 5 + (3)2 = 5 + 9 = 14 T = 5 + (6 - 1)2 where n = 6 = 5 + (5)2 = 5 + 25 = 30 T4 + T6 = 14 - 30 = -16

\(\frac{720 \times 1000}{8 \times 60 \times 60}\) = \(\frac{20 \times 10}{8}\) = \(\frac{200}{8}\) = 25 m/s

A = 3,500, P = N2,500 A = P + I But I = N3,500 - N2,500 I = N1,000 I = \(\frac{PRT}{100}\) N1,000 = \(\frac{2.500 \times R \times 4}{100}\) 1000 = 100R R = \(\frac{1000}{100}\) = 10%

\(\frac{1}{8} + \frac{2}{3x} = \frac{1}{3}\) = \(\frac{1}{2}\) \(\frac{5}{3x} = \frac{1}{3}\) 3x = 15 x = \(\frac{15}{3}\) = 5

\(\frac{54k^2 - 6}{3k + 1} = \frac{6(9k^2 - 1)}{3k + 1}\) = \(\frac{6(3k + 1) - (3k - 1)}{3k + 1}\) = 6(3k - 1)

q = \(\frac{3p}{r} + \frac{s}{2}\) q = \(\frac{6p + rs}{2r}\) 6p + rs = 2qr 6p = 2qr - rs p = \(\frac{2qr - rs}{6}\)

x + y = 2y - x + 1 = 5 x + y = 2y - x + 1 x + x + y - 2y = 1 2x - y = 1....(i) 2y - x + 1 = 5 -x + 2y = 5 + 1 -x = 2y = 4 x - 2y = -4 .....(ii) solve simultaneously (i) x 2x - y = 1 (ii) x x - 2y = -4 2x - y = 1 =2x - 4y = -8 3y = 9 y = \(\frac{9}{3}\) y = 3 substitute y = 3 into equation (i) 2x - y = 1 2x - 3 = 1 2x = 1 + 3 2x = 4 x = \(\frac{4}{2}\) = 2

12 = \(\frac{n}{3} - 2n = 1\), multiply through by 3 36 + n - 6n = 3 -5n = 3 - 36 -5n = -33 -5n + 33 = 0 5n - 33 = 0

\(\frac{-m}{2} - \frac{5}{4} \geq \frac{5m}{12} - \frac{7}{6}\) = \(\frac{-2m - 5}{4} \geq \frac{5m - 14}{12}\) 12(-2m - 5) \(\geq\) 4(5m - 14) -24m - 60 \(\geq\) 20m - 56 -24m - 20m \(\geq\) -56 + 60 44m \(\geq\) 4 m \(\leq \frac{4}{-44}\) m \(\leq \frac{-1}{11}\)

Curved surface area = 2\(\pi h\) 704 = 2 x \(\frac{22}{7} \times 8 \times h\) 704 = \(\frac{352}{7}\)h 352h = 704 x 7 h = \(\frac{704 \times 7}{352}\) = \(\frac{4928}{352}\) h = 14cm

Let 0 = angle of the minor sector angle of the major sector = 360 - \(\theta\)(angle at a point) 2 \(\pi r\) = 54 + 126(i.e circumference of minor and major arc) 2\(\pi r = 180^o\) r = \(\frac{180}{2\pi}\) = \(\frac{90}{\pi}\) Lenght of ninor arc = \(\frac{\theta}{360} \times 2 \pi r\) 54 = \(\frac{\theta}{360} \times 3 \pi r\) \(\theta = \frac{360 \times 54}{2 \pi r}\) but r = \(\frac{90}{\pi}\) substituting \(\frac{90}{\pi}\) for r \(\theta = \frac{360 \times 54 \times \pi}{2 \times \pi \times 90}\) \(\theta = 2 \times 54 = 108^o\) angle of the major sector = 360 - 108o = 252o

/YZ/\(^2\) = /XZ/\(^2\) + /XY/\(^2\) (Verify using Pythagoras' theorem)

cos(x + 40)o = 0.0872 x + 40 = cos-10.0872 x + 40o = 84.99o x = 84.99o - 40o x = 44.99 x = 45o

From the diagram above, Sin 54º = \(\frac{\text{OPP}}{\text{HYP}}\) Sin 54º = \(\frac{\text{x}}{50}\) x = 50 Sin 54º