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The locus of a point which is 5cm from the line LM is a pair of lines on opposite sides of LM and parallel to it, each distance 5cm from LM.

y = 3cos(\(\frac{x}{3}\)) \(\frac{dy}{dx}\) = 3x(\(\frac{1}{3}\))x - sin (\(\frac{x}{3}\)) = - sin \(\frac{x}{3}\) But x = \(\frac{3\pi}{2}\) ∴ -sin(\(\frac{x}{3}\)) = -sin(\(\frac{3\pi}{6}\)) = -sin \(\frac{(3 \times 180)}{6}\) = - sin 90 = -1

y = (2 + 3x)(1 - x) y = 2 - 2x + 3x - 3x2 y = 2 + x - 3x2 dy/dx = 1 - 6x

\(V = \frac{2}{3} \pi r^{3}\) \(\frac{\mathrm d V}{\mathrm d r} = 2\pi r^{2}\) \(\frac{\mathrm d V}{\mathrm d r} (r = 2) = 2\pi (2)^{2}\) = \(8\pi\)

y = 2x\(^2\)(2x - 1) expand the bracket y = 4x\(^3\) - 2x\(^2\) \(\frac{\text{dy}}{\text{dx}}\) = 12x\(^2\) - 4x at x = -1 \(\frac{\text{dy}}{\text{dx}}\) = 12(-1)\(^2\) - 4(-1) = 12 + 4 = 16

\(\int_{1}^{3}(x^2\) - 1)dx = \(\left[\frac{1}{3}x^3 - x\right] ^{3}_{1}\) = (9-3) - \(\frac{1}{3}-1)\) = 6 - \(\left(-\frac{2}{3}\right)\) = 6+\(\frac{2}{3}\) =6\(\frac{2}{3}\)

Number of white balls = x Number of red balls = 12 Number of black balls = 16 Total number of balls = 28 + x P(white balls) = 3/7 But P(white balls) \(= \frac{x}{28+x}\) = \(\frac{3}{7} = \frac{x}{28+x}\) 3(28 + x) = 7x 84 + 3x = 7x 7x - 3x = 84 4x = 84 x = 21

X = 1, 2, 3, 4; ∑X = 10 x = ∑X/n = 10/4 = 2.5 X - x = -1.5, -0.5, 0.5, 1.5 lX - xl = 1.5, 0.5, 0.5, 1.5; ∑lX - xl = 4.0 mean deviation = (∑lX - xl)/n = 4.0/4 = 1.0

Number of times 5 was obtained = 20 P(obtaining 5) = 20/100 = 1/5

Sectorial angle of beans = 360 - (150 + 90 + 60) = 360 - 300 = 60º If 150º represents 3000 tonnes 1º = \(\frac{3000}{150}\) 60º will represents (\(\frac{3000}{150}\)) x (\(\frac{60}{1}\)) = 1200 tonnes

The number of ways of selecting the debtors = \(^5C_2\) ways \(^5C_2\) = \(\frac{5!}{(5 - 2)!2!}\) = \(\frac{5 . 4 . 3!}{(3!)2!}\) = \(\frac{5. 4}{2. 1}\) = 10ways (3!s cancels ou)

Let the number of students = x Total age of the students = 15x Total age including age of the teacher = 15x + 45 Mean of their ages; (15x + 45)/(x + 1) = 18 15x + 45 = 18(x + 1) 15x + 45 = 18x + 18 18x - 15x = 45 - 18 3x = 27 x = 9

Gold medals = 30 silver medals = 22 Bronze medals = 18/70 P(Gold medals) = 30/70 = 3/7 ∴ P(not Gold medal) = 1 - 3/7 = 4/7

The number of ways to choose 1 member from the house of assembly is: \(\binom{3}{1} = 3\) The number of ways to choose 5 members from the state commissioners is: \(\binom{9}{5}\) = 126 Thus, the total number of ways to form the committee is: \(\text{Total ways} = \binom{3}{1} \times \binom{9}{5}\) = 3 \(\times\) 126 = 378 The total number of ways to form the committee, including one member from the house of assembly, is 378 Ways.

Range = 18 - 15 = 3