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Sum of all 4 angles of a quadrilateral = 360^\circ (5x-30) + (3x + 61) + (60-x) + (4x+ 60) = 360^\circ 11x + 151 = 360^\circ 11x = 360 - 151 = 209 x = 209/11 = 19^\circ Each angles is : 5x - 30 = 65^\circ 4x+ 60 = 136^\circ 60 - x =41^\circ 3x + 61 = 118^\circ Smallest of these angles is 41^\circ

In a polygon with n sides, the sum of the angles = (n - 2)180, where n = the number of sides. In our problem, n = k. So, we have: (3k - 10)90 = (k - 2)180 270k - 900 = 180k - 360 Simplifying: 90k = 540 k = 6. So, we have a regular hexagon. Now, each exterior angle = 360/n We have: 360/6 = 60 Conclusion: Each exterior angle is 60 degrees.

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

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

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