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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 ( \(\frac{54}{36}\)) - ( \(\frac{6}{36}\)) = \(\frac{48}{36}\) = \(\frac{4}{3}\).

cos\(\alpha\) = \(\frac{10}{16}\) = 0.625 \(\alpha\) = cos-1(0.625) = 51.3º ∠PQR = 2\(\alpha\) = 2(51.3º) = 102.6 = 103º

Given \(\cos x = -\frac{5}{13}\) where \(180^\circ < x < 270^\circ\): Using: \(\sin^2 x + \cos^2 x = 1\) \(\sin^2 x = 1 - \left(-\frac{5}{13}\right)^2 = \frac{144}{169} \implies \sin x = -\frac{12}{13}\) \(\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{12}{13}}{-\frac{5}{13}} = \frac{12}{5}\) Calculating \(\tan x - \sin x\): \(\tan x - \sin x = \frac{12}{5} - \left(-\frac{12}{13}\right) = \frac{12}{5} + \frac{12}{13}\) Finding a common denominator (65): \(\tan x - \sin x = \frac{12 \times 13}{65} + \frac{12 \times 5}{65} = \frac{156}{65} + \frac{60}{65} = \frac{216}{65}\) Thus, the value is: \(\tan x - \sin x = \frac{216}{65}\)

(\(\frac{1}{343}\))\(^{\frac{1}{3}}\) + (64)\(^{-\frac{1}{3}}\) - \(\frac{4}{9}\)\(^{-\frac{1}{2}}\) where \(343 = 7^3, 64 = 2^6, 4 = 2^2, 9 = 3^2\) \(7^{-3}\) \(^{\frac{1}{3}}\) + \(2^6\) \(^{\frac{-1}{3}}\) - \(\frac{2^2}{3^2}\) \(^{\frac{-1}{2}}\) \(7^{-1} + 2^{-2} - (\frac{2}{3})^{-1}\) \(\frac{1}{7} + \frac{1}{4} - \frac{3}{2}\) \(\frac{4+7-42}{28}\) -\(\frac{31}{28}\)

Given: \(\frac{3 - y}{2x} = \frac{6y + 7}{4x - 5}\) Substitute y = 2 \(\frac{3 - 2}{2x} = \frac{6(2) + 7}{4x - 5} \implies \frac{1}{2x} = \frac{19}{4x - 5}\) Cross-multiply: 1 \(\cdot\) (4x - 5) = 19 \(\cdot\) 2x 4x - 5 = 38x -5 = 38x - 4x -5 = 34x x = -\(\frac{5}{34}\)

first term (a) = \(\frac{5}{6}\) common difference = \(\frac{8}{6}\) - \(\frac{5}{6}\) → \(\frac{3}{6}\) or \(\frac{1}{2}\) A.P formula → T\(_n\) = a + (n - 1)d T\(_n\) = \(\frac{5}{6}\) + (19 - 1)\(\frac{1}{2}\) T\(_n\) = \(\frac{5}{6}\) + 9 → 9\(\frac{5}{6}\)