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(x + 15)º + (2x - 45)º + (x + 10)º + (x - 30)º = (2n - 4)90º when n = 4 x + 15º + 2x - 45º + x - 30º + x + 10º = (2 x 4 - 4) 90º 5x - 50º = (8 - 4)90º 5x - 50º = 4 x 90º = 360º 5x = 360º + 50º 5x = 410º x = \(\frac{410^o}{5}\) = 82º Hence, the value of the least interior angle is (x - 30)º = (82 - 30) º = 52º

By comparing y = mx + c with y = -4x + 2, the gradient of y = -4x + 2 is m1 = -4 let the gradient of the line parallel to the given line be m2, then, m2 = m1 = -4 (condition for parallelism) using, y - y1 = m2(x - x1) Hence the equation of the parallel line is y - 3 = -4(x-2) y - 3 = -4 x + 8 y + 4x = 8 + 3 y + 4x = 11 y + 4x - 11 = 0

Let D denote the distance between (\(\frac{1}{2}\), -\(\frac{1}{2}\)) then using D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) = \(\sqrt{(-{\frac{1}{2} - \frac{1}{2}})^2 + (-{\frac{1}{2} - \frac{1}{2}})^2}\) = \(\sqrt{(-1)^2 + (-1)^2}\) = \(\sqrt{1 + 1}\) = \sqrt2

Let (x1, y1) = (1, 1) and (x2, y2)= (2, 5) then gradient m of \(\bar{PQ}\) is m = \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{5 - 1}{2 - 1}\) = \(\frac{4}{1}\) = 4

(a) \(\frac{FM}{FB} = \tan 35\) \(\therefore \frac{FM}{120} = \tan 35\) \(FM = 120 \tan 35 \) = \(120 \times 0.7002\) = \(84.024 m\) Therefore, the height of the cliff = \(2 \times 84.024 = 168.048m\) (b) Distance between P and Q = x km Speed for 1st motorist = 60 km/h \(\therefore \text{1st motorist's time} = \frac{x}{60} hrs\) Speed for 2nd motorist = 80 km/h \(\therefore \text{2nd motorist's time} = \frac{x}{80} hrs\) \(\frac{x}{60} - \frac{x}{80} = \frac{30}{60}\) \(\frac{4x - 3x}{240} = \frac{1}{2}\) \(\frac{x}{240} = \frac{1}{2} \implies 2x = 240\) \(x = 120 km\).

Given that y = -3 - 2x + X2 then, \(\frac{dy}{dx}\) = -2 + 2x At maximum value, \(\frac{dy}{dx}\) = O therefore, -2 + 2x 2x = 2 x = 2/2 = 1

If y = x sinx, then Let u = x and v = sinx \(\frac{du}{dx}\) = 1 and \(\frac{dv}{dx}\) = cosx Hence by the product rule, \(\frac{dy}{dx}\) = v \(\frac{du}{dx}\) + u\(\frac{dv}{dx}\) = (sin x) x 1 + x cosx = sinx + x cosx

\(\int^{2}_{0}(x^3 + x^2)\)dx = \(\int^{2}_{0}\)(\(\frac{x^4}{4} + {\frac {x^3}{3}}\)) = (\(\frac{2^4}{4} + {\frac {2^3}{3}}\)) - (\(\frac{0^4}{4} + {\frac {0^3}{3}}\)) = (\(\frac{16}{4} + {\frac {8}{3}}\)) - 0 = \(\frac{80}{12} = {\frac {20}{3}}\) or 6\(\frac{2}{3}\)

\(\begin{array}& x & x - \bar{x} & (x - \bar{x})^2 \\2 & -2 & 4 \\ 3 & -1 & 1 \\ 5 & 1 & 1 \\ 6 & 2 & 4\\ \hline \sum x = 16 & & \sum (x - \bar{x}^2) = 10 \end{array}\) ___________________________________ \(\bar{x}\) = \(\frac{\sum x }{N}\) = \(\frac{16}{4}\) = 4 S = \(\sqrt{\frac {(x - \bar{x})^2}{N}}\) = \(\sqrt{\frac {(10)}{4}}\) = \(\sqrt{\frac {(5)}{2}}\)

A committee of 2 women and 3 men can be chosen from 6 men and 5 women, in \(^{5}C_{2}\) x \(^{6}C_{3}\) ways = \(\frac{5!}{(5 - 2)!2!} \times {\frac{6!}{(6 - 3)!3!}}\) = \(\frac{5!}{3!2!} \times {\frac{6!}{3 \times 3!}}\) = \(\frac{5 \times 4 \times 3!}{3! \times 2!} \times {\frac{6 \times 5 \times 4 \times 3!}{3! \times 3!}}\) = \(\frac{5 \times 4}{1 \times 2} \times {\frac{6 \times 5 \times 4}{1 \times 2 \times 3}}\) = 10 x \(\frac{6 \times 20}{6}\) = 200

P(H) = \(\frac{1}{2}\) and P(T) = \(\frac{1}{2}\) Using the binomial prob. distribution, (H + T)3 = H3 + 3H2T1 + 3HT2 + T3 Hence the probability that three heads show in a toss of the three coins is H3 = (\(\frac{1}{2}\))3 = \(\frac{1}{8}\)

x - y = 2 ...........(1) x\(^2\) - y\(^2\) = 8 ........... (2) x - 2 = y ............ (3) Put y = x - 2 in (2) x\(^2\) - (x - 2)\(^2\) = 8 x\(^2\) - (x\(^2\) - 4x + 4) = 8 x\(^2\) - x\(^2\) + 4x - 4 = 8 4x = 8 + 4 = 12 x = \(\frac{12}{4}\) = 3 from (3), y = 3 - 2 = 1 therefore, x = 3, y = 1

0.000215215 x 0.000028 = 215 x 10-4 x 28 x 10-4 = 215 x 28 x 10-6-6 = 6020 x 10-12 = 6.020 x 103 x 10-12 = 6.02 x 103 - 12 = 6.02 x 10-9

x + y - ax = ay = (x + y) -a(x + y) = (x + y)(1 - a)

1 litre = N63.00 xlitres = N900.00 x x N63 = N900 x 1 litre x = \(\frac{900}{63} \times litre\) x = \(\frac{100}{7}\) litres Also, 1 litre = 14km \(\frac{100}{7}\) = y y x 1 litre = \(\frac{100litre}{7}\) x 14km y = 20km