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P(2, -8) and Q(1, y) Gradient of slope(m) = \(\frac{y_2 - y_1}{x_2 - x_1}\) - 4 = \(\frac{y - (-8))}{1 - 2}\) -4 = \(\frac{y + 8}{- 1}\) cross multiplying 4 = y + 8 y = 4 - 8 = -4

\(\angle\)XWY = \(\angle\)ZYW = 44º (alternate \(\angle\)s) Considering \(\triangle\)WTX \(\angle\)W + \(\angle\)T + \(\angle\)X = 180º 44 + \(\angle\)T + 50 = 180 \(\angle\)T = 180 - 44 - 50 = 86º

Let the length of the garden be x, then, the width = x - 7 perimeter of rectangle = 2(L + B) = 90 m = 2(x + x - 7) = 90 = 2x - 7 = 45 ( after dividing through by 2) = 2x = 45 + 7 = 52 x = \(\frac{52}{2}\) = 26 m Thus, the length of the rectangular garden = 26 m.

Hexagon = 6 sides Sum of interior angles of a regular hexagon = (n - 2)180 = (6 - 2)180 = 4 x 180 = 720º But 4 of these angles sum up to 420º and the remaining ones are equal. Let the remaining two that are equal = 2x SuM of the angles = 420 + 2x = 720º 2x = 720 - 420 = 300 x = \(\frac{300}{2}\) = 150º Thus, the value of each of the remaining angles = 150º

\(\angle\)N + \(\angle\)P + \(\angle\)M = 180 ( sum of angles in \(\triangle\) NPM) 65º + 55º + \(\angle\)M = 180 \(\angle\)M = 180 - 65 - 55 = 60º x + \(\angle\)M = 180(opp interior angles of parallel lines are supplementary) x = 180 - 60 = 120º

Mean(\(\overline{x}\)) = \(\frac{\sum{x}}{n}\) Mean(\(\overline{x}\)) = \(\frac{ 59 + 44 + 53 + 57 + 49 + 40 + 48 + 50}{8}\) Mean(\(\overline{x}\)) = \(\frac{397}{8}\) ≈ 50 kg

Variance(s\(^2\)) = \(\frac{\sum{(x - \overline{x})^2}}{n}\) But mean = 50 (from question 39 same information) x x - \(\overline{x}\) (x - \(\overline{x})^2\) 59 9 81 44 - 6 36 53 3 9 57 7 49 49 - 1 1 40 -10 100 48 - 2 4 50 0 0 \(\sum{(x - \overline{x})^2}\) = 280 Variance(s\(^2\)) = \(\frac{\sum{(x - \overline{x})^2}}{n}\) = \(\frac{280}{8}\) = 35

Opposite sides of a rectangle are equal, then (5x + 3) = (2x + 9) 5x - 2x = 9 - 3 3x = 6 x = \(\frac{6}{3}\) = 2. Put x = 2 in either (5x + 3) or (2x + 9) = 13m Adjacent side (6x - 7) (given) also put x = 2 into it, 6(2) - 7 = 5 Area of rectangle = length x breadth(width) = 13 x 5 = 65 m\(^2\)

Sample space = { 1, 2, 3, 4, 5, 6} Prime numbers = ( 2, 3, and 5) Pr(prime Nos) = \(\frac{\text{number of required outcome}}{\text{total possible outcome}} = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Area of a sector = \(\frac{\theta}{360}\) x \(\pi\) r\(^2\) 51.3 = \(\frac{\theta}{360}\) x \(\frac{22}{7}\) x 7 x 7 \(\theta\) = \(\frac{51.3 \times 360}{22 \times 7}\) = 119.9 ≈ 120º to the nearest whole number.

From the figure above, using SOH CAH TOA Tan\(\alpha\) = \(\frac{\text{opposite}}{\text{adjacent}}\) Tan\(\alpha\) = \(\frac{87}{22}\) = 3.9545 \(\alpha\) = Tan\(^{-1}\)(3.9545) = 76º

\(\angle\)SPT = 35º (alternate segment) From \(\triangle\)TSP 44 + \(\angle\)SPT + \(\angle\)PST = 180º (sum of angles in a triangle) \(\angle\)PST = 180 - 44 - 35 = 101º

Pr(Amaka passes) = \(\frac{3}{7}\), Pr(Bala passes) = \(\frac{4}{9}\). Pr(both Amaka and Bala pass the exam) = \(\frac{3}{7}\) x \(\frac{4}{9}\) = \(\frac{4}{21}\).

If a point lies on a particular line. It will satisfy the equation of the line. Test the points given in the options any that satisfy the equation of the line, is the point. Thus, the point(1, -1) lies on the line 3x - 8y = 11 That is, putting x = 1 and y = -1 into 3x - 8y to see if the result will equal 11 3(1) - 8(-1) = 11.

Range = max. value - min. value max. value = 39, min.value = 15 Range = max. value - min. value = 39 - 15 = 24.