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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 \(\therefore \text{2nd motorist's time} = \frac{x}{80} hrs\) \(\frac{x}{60} - \frac{x}{80} = \frac{30}{60}\) \(\frac{4x - 3x}{240} = \frac{1}{2}\) \(\frac{x}{240} = \frac{1}{2} \implies 2x = 240\) \(x = 120 km\).

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

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

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 = \(\frac{1}{8}\)

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

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

Let the no. be x and x + 1 x + (x + 1) = \(\frac{5}{6}\) of x(x + 1) 2x + 1 = \(\frac{5}{6}\) x(x + 1) 6(2x + 1) = 5x2 + 5x 12x + 6 = 5x2 + 5x 5x2 + 5x - 12x - 6 = 0 5x2 - 7x - 6 = 0 5x2 - 10x + 3x - 6 = 0 5x(x - 2) + 3(x - 2) = 0 (5x + 3)(x - 2) = 0 (5x + 3)(x - 2) = 0 (5x + 3) = 0 x - 2 = 0 for (5x + 3) = 0 5x = -3 x = \(\frac{-3}{5}\) (Imposible since x is a whole number) x - 2 = 0 x = 2 x = \(\frac{-3}{5}\)(Impossible since x is a whole number) x - 2 = 0 x = 2 The numbers are x = 2 x + 1 = 2 + 1 = 3

Volume of water in tank = l x b x h = 180 litres but 1 litre = 1000cm2 250 x 120 x hw = 180 x 1000 hw = \(\frac{180 \times 1000}{250 \times 120}\) = 6cm

sin 3y = cos 2y, but sin \(\theta\) = cos(90 - \(\theta\)) sin 3y = cos(90 - 3y) cos(90 - 3y) = cos 2y 90 - 3y = 2y 5y = 90 y = \(\frac{90}{5}\) y = 18o

l = x; b = x - 1 perimeter = 2(l + b) = 16 l + b = \(\frac{16}{2}\) = 8 l + b = 8 x + x - 1 = 8 2x = 8 + 1 2x = 9 x = \(\frac{9}{2}\)cm x = 4\(\frac{1}{2}\)

Given tan x = 1 x = tan-1(1) x = 45o Now, \(\frac{1 - ( \frac{1}{\sqrt{2} )^2}}{\frac{1}{\sqrt{2}}}\) = \(\frac{1 - \frac{1}{2}}{\frac{1}{2}}\) = \(\frac{1}{2} + \frac{1}{\sqrt{2}}\) = \(\frac{1}{2} \times \frac{1}{\sqrt{2}}\) = \(\frac{\sqrt{2}}{2}\)

perimeter = 2(l + b) = 32 l + b = \(\frac{32}{2}\) l + b = 16 b = 16 - 1.......(1) Area = l + b = 39 lb = 39 .....(2) put (1) into (2) l(16 - 1) = 39 16l - l2 = 39 l2 = 13l - 3l + 39 = 0 l(l - 13) - 3(1 - 13) = 0 (l - 3)(l + 130 = 0 l - 3 = 0 or l - 13 = 0 l = 3cm or l = 13cm; The length in 13cm

sum of exterior angles = 360o Sum of interior angle = (n - 2) x 180 360 = \(\frac{1}{2}\) x(n - 2) x 180(90o) 360 = \(\frac{1}{2}\) x(n - 2) x 90o \(\frac{360}{90}\) = a - 2 4 = n - 2 n = 4 + 2 = 6