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Since ST = the radius r, => SOT is equilateral. Each angle = 60° Length of arc ST = θ/360 x 2πr = 60/360 x 2 x π x r = πr/3

y = 2kx2 + x + 1 Gradient function dy/dx = 4kx + 1 at dy/dx = 9, => 9 = 4kx + 1 => 8 = 4kx, and k = 8/4x. At x = 1, k = 8/4 = 2

\(\int 2(2x - 3)^{\frac{2}{3}} \mathrm d x\) Let \(u = 2x - 3\) \(\mathrm d u = 2 \mathrm d x\) = \(\int u^{\frac{2}{3}} \mathrm d u\) = \(\frac{u^{\frac{5}{3}}}{\frac{5}{3}} + k\) = \(\frac{3}{5} (2x - 3)^{\frac{5}{3}} + k\)

\(y = (2x + 5)^{2} (x - 4)\) \(\frac{\mathrm d y}{\mathrm d x} = (2x + 5)^{2} (1) + (x - 4)(2)(2)(2x + 5)\) = \((2x + 5)(2x + 5 + 4x - 16)\) = \((2x + 5)(6x - 11)\)

Hint: y = 4 - x\(^2\) and y = 2x + 1 => 4 - x2 = 2x + 1 => x2 + 2x - 3 = 0 (x+3)(x-1) = 0 thus x = 1 or x = -3. Integrating x\(^2\) + 2x - 3 = 3x - x\(^2\) - \(\frac{x^3}{3}\) from (1, to -3) : 3 (1) - 1\(^2\) - \(\frac{1^3}{3}\) - 3 (-3) - -3\(^2\) - \(\frac{-3^3}{3}\) = \(\frac{5}{3}\) + 9 will give 32/3 = 10 \(\frac{2}{3}\)

Volume of sphere, V = 4/3 x πr3 Rate of change of V = dv/dr Thus if V = 4/3 x πr3, => dv/dr = 4πr2 At r = 1, Rate = 4 x π x 1 = 4π

y = x sin x dy/dx = 1 x sinx + x cosx = sinx + x cosx At x = π/2, = sin (π/2) + (π/2) cos (π/2) = 1 + (π/2) x (0) = 1

Let the rectangle be a square of sides p/4. So that perimeter of square = 4p 4 x (p/4) = p.

Mode = score with highest frequency = 11. Square of 11 = 121

Mean = ∑fx/∑f Mean = (4x3) + (7x5) + ... + (8x1) all divided by 20 Mean = 174/20 = 8.7

In a football/soccer game, there are 3 possible outcomes: home win, draw, and away win. So, in this case, Pr(that the game ends in a draw) = \(\frac{1}{3}\)

6Pr = 6 Thus r = 1 N.B 6P1 = \(\frac{\text{6!}}{\text{(6 - 1)!}}\) = \(\frac{\text{6!}}{\text{5!}}\) = \(\frac{6 \times \text{5!}}{\text{5!}}\) = 6 6Pr + 1 = 6P2 = \(\frac{\text{6!}}{(6 - 2)\text{!}} = \frac{6\text{!}}{\text{4!}}\) = \(\frac{6 \times 5 \times \text{4!}}{\text{4!}}\) = 30

mean of X = x = (2+6+8+6+2+6)/6 = 30/6 = 5 X → 2,6,8,6,2,6 X - x = -3, 1, 3, 1, 3, 1 (X - x)2 = 9, 1, 9, 1, 9, 1 ∑(X-x) = 9+1+9+1+9+1 = 30 Variance = ∑(X-x)2/n = 30/6 = 5

To calculate the binomial coefficient \(^{12}C_{8}\): \(^{12}C_{8} = \frac{12!}{8!4!} = \frac{12 \times 11 \times 10 \times 9}{4!}\) Calculating \(4!\): \(4! = 24\) Thus, \(^{12}C_{8} = \frac{12 \times 11 \times 10 \times 9}{24} = \frac{11880}{24}\) = 495 = \(^{12}C_{8}\) = 495

\(\frac{1}{6}\), \(\frac{1}{3}\), \(\frac{3}{2}\), \(\frac{2}{3}\), \(\frac{8}{9}\) and \(\frac{4}{3}\). L.C.M of the denominators = 36 The fraction now becomes \(\frac{6}{36}\), \(\frac{12}{36}\), \(\frac{54}{36}\), \(\frac{24}{36}\), \(\frac{32}{36}\), and \(\frac{48}{36}\). Highest fraction = \(\frac{54}{36}\) Lowest fraction = \(\frac{6}{36}\) Range = Highest - Lowest ( \(\frac{54}{36}\)) - ( \(\frac{6}{36}\)) = \(\frac{48}{36}\) = \(\frac{4}{3}\).