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The angle of elevation = Tan \theta = \(\frac{opp}{adj}\) Tan \theta = \(\frac{12- 1.4}{10}\) Tan \theta = \(\frac{10.6}{10}\) \theta = Tan\(^{-1}\)(1.06) \theta ≈ 47º

\(\frac{x^3 + 6x - 7}{x-1}\): When numerator is differentiated → 3x\(^2\) + 6 When denominator is differentiated → 1 : \(\frac{3x^2 + 6}{1}\) substitute x for 1 \(\frac{3 * 1^2 + 6}{1}\) = \(\frac{3 + 6}{1}\) = \(\frac{9}{1}\) = 9

The initial velocity (v) is obtained by differentiating the displacement (S) with respect to time, which gives: v = \(\frac{dS}{dt}\) = \(\frac{d}{dt}\)(\(3t^2\) + 5t - 10) = 6t + 5 The initial velocity is the velocity of the particle at time t = 0. Substituting t = 0 into the velocity equation: v(0) = 6(0) + 5 = 5 Therefore, the initial velocity of the particle is 5 m/s.

Recall chain rule: u = 2x +1; du = 2dx → dx = \(\frac{du}{2}\) u\(^3\) = ∫ u\(^3\) \(\frac{du}{2}\) → \(\frac{1}{2}\) ∫ u\(^3\) = \(\frac{1*u^4}{2*4}\) = \(\frac{u^4}{8}\) → \(\frac{{2x+1}^4}{8}\) + C

Firstly; solving for x 6 = \(\frac{2 + 5 + x+1 + x+2 + 7 + 9}{6}\) cross multiply to have: 6 * 6 = 2 + 5 + x+1 + x+2 + 7 + 9 36 = 2x + 26 36 - 26 = 2x 10 = 2x x = 5 Median = \(\frac{7+6}{2}\) → 6.5

When rearranged: 10, 10, 13, 14, 17, 18 median = \(\frac{13+14}{2}\) = \(\frac{27}{2}\) or 13.5

mean (4) = \(\frac{2x + 4x + 2x - 13 + 6x}{4}\) 16 = 14x - 13 16 + 13 = 14x x = 29/14 or 2.07

10th number must be over 56 (as it increases the average) 55 x 9 = 495 = sum of first 9 numbers 56 x 10 = 560 = sum of all 10 numbers 560 - 495 = 65 = 10th number

Rearranged: 12,13,15, 36, 39, 41, 41, 43, 43, 47 and 49 Lower quartile or First quartile = \(\frac{total number of items}{4}\) If odd: items numbers + 1 → \(\frac{11 items + 1}{4}\) = \(\frac{12}{4}\) Lower quartile = 3rd iems →15

\(\frac{12 + 47 + 49 + 15 + 43 + 41 + 13 + 39 + 43 + 41 + 36}{11}\) = \(\frac{379}{11}\) = 34.45

Zoology has 7 letters in total, with O repeated thrice \(\frac{7!}{3!}\) → \(\frac{7*6*5*4*3*2*1}{3*2*1}\) = 840ways

\(^{3^2 + 1}\)C\(_{3+5}\) \(^{9 + 1}\)C\(_{3+5}\) \(^{10}\)\(C_8\) = \(\frac{10!}{8! 2!}\) \(\frac{10 * 9 * 8!}{8! 2!}\) = \(\frac{10 * 9}{2}\) = 45

OR When a coin is tossed twice, the sample space is given by S ={HH HT TH TT} and the event of obtaining at least a head is provided by A ={HH HT TH}. Therefore the probability of getting at least a head is given by: P = n(A) \div n(S) P = 3 \div 4 P = 0.75

Girl = \(\frac{1}{2}\) Blue eyes = \(\frac{1}{4}\) probability that she will give birth to a girl with blue eyes: \(\frac{1}{2}\) x \(\frac{1}{4}\) = \(\frac{1}{8}\)

Using pythagoras formula: Hyp\(^2\) = adj\(^2\) + opp\(^2\) 5\(^2\) = opp\(^2\) + 3\(^2\) 5\(^2\) - 3\(^2\) = adj\(^2\) 4 = adj length of the chord = 2 * 4 = 8cm