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Line: 2y+8x-17=0 recall y = mx + c 2y = -8x + 17 y = -4x + \(\frac{17}{2}\) Slope m\(_1\) = 4 parallel lines: m\(_1\). m\(_2\) = -4 where Slope ( -4) = \(\frac{y_2 - y_1}{x_2 - x_1}\) at points (-1, -p) and (-2,2) -4( \(x_2 - x_1\) ) = \(y_2 - y_1\) -4 ( -2 - -1) = 2 - -p p = 4 - 2 = 2

The required locus is angle bisector of the two lines

The formula for midpoint = \(\frac{x_1 + x_2}{2}\), \(\frac{y_1 + y_2}{2}\) (-4,-2) = (x,y) x = \(\frac{x_1 + x_2}{2}\) -4 = \(\frac{-8 + p}{2}\) -4 * 2 = -8 + p -8 + 8 = p : p = 0 y = \(\frac{y_1 + y_2}{2}\) -2 = \(\frac{-12 + q}{2}\) -2 * 2 = -12 + q -4 + 12 = q : q = 8

diagram above(i) (i) Distance between Manu and Dogbe's farm =\sqrt(3\(^2\)+5\(^2\)) = 5.83km or 5.8 km (i) tanØ = 0.6 Ø= tan\(^{-1}\) (O.6) = 30.96^\circ Bearing = 30.96^\circ+200^\circ =230.96 = 231^\circ (nearest degree) (b) x\(^2\) + (x+2)\(^2\) = 10\(^2\) 2x\(^2\) + 4x - 96 = 0 (x-6) (x+8) = 0 x = 6,x = -8 : x = 6cm

Sin\theta = \(\frac{6}{12}\) Sin\theta = \(\frac{1}{2}\) \theta = Sin\(^0.5\) \theta = 30^\circ Bearing of Z from X, (270 - 30)^\circ = 240^\circ

Circumference of the circular track = 9km Distance covered = 302km Number of complete circles or revolutions from the starting point = 302/9 =33 circles and additional 5km. So, the distance of the cyclist from the starting point would be 5km

No expanded working has been attached for this question yet. The verified answer is Option C.

Speed = \(\frac{\text{Distance}}{\text{Time}}\) From the first statement, speed = xkm/h, t = 2hrs Therefore, distance = 2xkm From the second statement, speed = x + 2 ( since speed increased by 2km/h) and t = 3hrs distance = 3 X (x + 2) = 3x + 6 But total distance = 36km(given) Total distance = 2x + 3x + 6 = 36 i.e. (dist. from the first statement + dist. from the second statement) 5x + 6 = 36 5x = 36 - 6 = 30 x = (initial speed) = \(\frac{30}{5}\) = 6Km/h check = total distance = 36 = 2x + 3x + 6 = 12 + 18 + 6 = 36.

distance d more than 18 KM = d > 18 0r 18 < d distance d but not more than 23 KM = d \le 23 Mathematically; 18 < d \le 23

y = 2x\(^2\)(2x - 1) y = 4x\(^3\) - 2x\(^2\) dy/dx = 12x\(^2\) - 4x at x = -1 dy/dx = 12(-1)\(^2\) - 4(-1) = 12 + 4 = 16

Mean of 10 numbers = 16 The total sum of numbers = 16 x 10 = 160 Mean of 11 numbers = 18 Total sum of numbers = 11 x 18 = 198 The 11th no. = 198 - 160 = 38

Rearrange data in increasing order: 6,7,7,7,8,8,8,9,9 and 11 First quartile (ungrouped data) = \(\frac{n + 1}{4}\)th value = \(\frac{11}{4}\) = 2.75th value = \(\frac{7 + 7}{2}\) = 7.0

∑fx = (160 * 4) + (161 * 6) + (162 * 3) + (163 * 7) + (164 * 8) + (165 * 9) =640 + 966 + 486 + 1,141 + 1,312 + 1,485 = 6,030 ∑f = 4 + 6 + 3 + 7 + 8 + 9 = 37 = \(\frac{∑fx}{∑f}\) = \(\frac{6030}{37}\) = 162.97 or 163

Total angles = 360^\circ 100 + 60 + 80 + M = 360 M = 360 - 240 = 120 Orange (M) = 120^\circ and F ( fruits in total) For Apples: \(\frac{80}{360} \times F\) = 60 F = \(\frac{60 \times 360}{80}\) F = 270 fruits in total For Orange = \(\frac{120}{360} \times 270\) = 90 oranges

Given that 4, 16, 30, 20, 10, 14 and 26 Adding up = 120 nos \(\geq\) 16 are 16 + 30 + 20 + 26 = 92 The requires sum of angles = \(\frac{92}{120}\) x \(\frac{360^o}{1}\) = 276o