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For y = ax2 - bx + c for minimum y \(\frac{dy}{dx}\) = 2x - 2 = \(\frac{dy}{dx}\) = 0 ∴ 2x - 2 = 0 x = 1

Given the first and last term of an A.P, the sum of the terms is given by \(S_{n} = \frac{n}{2} [a + l]\) where a = first term; l = last term and n = number of terms. \(\therefore S_{20} = \frac{20}{2} [7 + 117]\) = \(10 (124)\) = 1240

BD = \(\sqrt{x^2 + x^2}\) = \(\sqrt{12^2 + 12^2}\) = \(\sqrt{144 + 144}\) = 2(144) = 12\(\sqrt{2cm}\)

\(\frac{x}{4}\) = \(\frac{\tan 60}{1}\) x = 4 tan60 = 4\(\sqrt{3}\) BD = 2x = 8\(\sqrt{3}\)

The sum of the exterior angles of a regular polygon = 360^\circ. \(\therefore x + (x + 5) + (x + 10) + (x + 15) + (x + 20) = 360^\circ\) \(5x + 50 = 360^\circ \implies 5x = 360^\circ - 50^\circ = 310^\circ\) \(x = \frac{310^\circ}{5} = 62^\circ\)

\(\frac{BC}{60}\) = \(\frac{tan 62}{1}\) BC = 60 tan 62 \(\frac{AC}{60}\) = \(\frac{tan 62}{1}\) AC = 60 tan 64 AB = AC - BC = 60(tan 64o - tan 62o)

\(\cos^{2} x (\sec^{2} x + \sec^{2} x \tan^{2} x)\) = \(\cos^{2} x \sec^{2} x + \cos^{2} x \sec^{2} x \tan^{2} x\) = \(1 + \tan^{2} x\) = \(1 + \frac{\sin^{2} x}{\cos^{2} x}\) = \(\frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x}\) = \(\frac{1}{\cos^{2} x} = \sec^{2} x\)

cosx = \(\sqrt{\frac{a}{b}}\) y2 + \(\sqrt{(a)^2}\) = \(\sqrt{(b)^2}\) by pythagoras y2 = b - a ∴ y = b - a cosec x = \(\frac{1}{sin x}\) = \(\frac{1}{y}\) \(\frac{b}{y}\) = \(\frac{\sqrt{b}}{\sqrt{b - a}}\) = \(\sqrt{\frac{b}{b - a}}\)

tan \(\theta\) = \(\frac{60\sqrt{3}}{60}\) = \(\sqrt{3}\) \(\theta\) = tan \(\frac{1}{3}\) = 60o Bearing of y from x = 060o

The total surface area of a cylinder = \(2\pi r (r + h)\) = \(2 \pi (4) (4 + 5)\) = \(8 \pi (9) = 72 \pi cm^{2}\)

Education = 11% = \(\frac{11}{100} \times 360^\circ = 39.6^\circ\) Food = 59% = \(\frac{59}{100} \times 360^\circ = 212.4^\circ\)

No of defective(x) No of boxes (f) \(fx\) 4 2 8 5 7 35 6 17 102 7 10 70 8 8 64 9 6 54 \(\sum\) 50 333 \(Mean = \frac{\sum fx}{\sum f}\) = \(\frac{333}{50} = 6.66 \approxeq 6.7\) The median is the average of the 25th and 26th position = 6.

% of boxes containing at least 5 defective bolts each = \(\frac{7 + 17 + 10 + 8 + 6}{50}\) = \(\frac{48}{50}\) x \(\frac{100}{1}\) = 96%

Coca-cola = 10 bottles, Fanta = 8 bottles Sprite = 6 bottles, Total = 24 P(cola-cola) = \(\frac{10}{24}\) P(not coca-cola) = 1 - \(\frac{10}{24}\) \(\frac{24 - 10}{24}\) = \(\frac{14}{24}\) = \(\frac{7}{12}\)

From the diagram above x = 60 + 70 = 130º (sum of angles on a straight line was used to obtain 60º and 70º, and alternate angles are equal to obtain the value of x)