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

Find the equation of the line through (5,7) parallel to the line 7x + 5y = 12 m\(_1\) = m\(_2\) condition for parallelism From 7x + 5y = 12 5y = -7x + 12 y = \(\frac{-7x}{5}\) + \(\frac{12}{5}\) m\(_1\) = \(\frac{-7}{5}\) = m\(_2\) equation of the line = y - y\(_1\) = m\(_2\)[x - x\(_1\)] y - 7 = \(\frac{-7}{5}\)[x - 5] 5(y - 7) = -7(x - 5) 5y - 35 = -7x + 35 5y + 7x = 35 + 35 5y + 7x = 70

Tan 74 = 150/x x = 150/tan 74 = 43.01cm

The word LEADER has 1L 2E 1A 1D and 1R making total of 6! \(\frac{6}{1!2!1!1!1!}\) = \(\frac{6!}{2!}\) = \(\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}\) = 360

From the figure, ∠ XMN = ∠ XZY Angle X is common So, ∠ XNM = ∠ XYZ Then from the angle relationship \(\frac{XM}{XZ}\) = \(\frac{XN}{XY}\) = \(\frac{MN}{ZY}\) XM = 8, XZ = 12 + 4 = 16, XN = 12, XY = 8 + YM \(\frac{8}{16}\) = \(\frac{12}{(8 + YM) }\) Cross multiply 8(8 + YM) = 192 64 + 8YM = 192 8YM = 128 YM = \(\frac{128}{8}\) = 16cm

The two numbers must be expressed in the same unit. To convert 495g to kg, it will be divided by 1000 495g = \(\frac{495}{1000}\) = 0.495kg To express in percentage, 0.495 will be divided by 16.5 and then multiplied by 100 % will be added to the answer \(\frac{0.4950}{16.5}\) x 100 = 3%

2\sqrt3 - 4) ( 2\sqrt3 + 4) = 12 + 8\sqrt3 - 8\sqrt3 – 16 = 12 – 16 = -4 The two expressions in the bracket are conjugate of each other

The gradient to the curve is found by differentiating the curve equation with respect to x So \(\frac{dy}{dx}\) 2x - 2 The gradient of the curve is the same with that of the tangent. At point (2, 0) \(\frac{dy}{dx}\) = 2(2) - 2 = 4 – 2 = 2 The equation of the tangent is given by (y - y1) \(\frac{dy}{dx}\) (x – x1) At point (x1, y1) = (2, 0) y - 0 = 2(x - 2) y = 2x - 4

The minimum value is the lowest value of the curve on y axis which gives a value of -5.3

log\(_2\) 8 – log\(_3\) \(\frac{1}{9}\) = log \(_2\) 2\(^3\) – log\(_3\) 9\(^{-1}\) = log\(_2\) 2\(^3\) – log\(_3\) 3\(^{-2}\) Based on law of logarithm = 3 log\(_2\) 2 – (-2 log\(_3\) 3) But log\(_2\) 2 = 1, log\(_3\) 3 = 1 So, = 3 + 2 = 5

First quadrant: Sin, Cos and Tan are all positive Second quadrant: Sin is positive, Cos is negative and Tan is negative Third quadrant: Tan is positive, Sin is negative and Cos is negative Fourth quadrant: Cos is positive, Sin is negative and Tan is negative The correct option is the third quadrant only where Tan\theta is positive and Sin\theta is negative

The total number of students is ∑ f = 3 + 4 + 5 + 6 + 7 + 6 + 5 + 4 = 40

A = (p, q, r, t, u, v) B = (r, s, t, u) A U B = Elements in both A and B = (p, q, r, s, t, u, v) (A U B)1 = elements in the universal set E but not in (A U B)= (w, x, y, z) n(A U B) 1 = number of the elements in (A U B)1 = 4

Locus is the path traced at by a point which moves in accordance with a certain law. It is also the set of all possible position occupied by an object The path traced from all possible location of 4cm from a given point P form a circle of radius 4cm with centre P.

Sin(5x - 28) = Cos(3x - 50)………..i But Sin\alpha = Cos(90 - \alpha) So Sin(5x - 28) = Cos(90 - [5x - 28]) Sin(5x - 28) = Cos(90 - 5x + 28) Sin(5x - 28) = Cos(118 - 5x)………ii Combining i and ii Cos(3x - 50) = Cos(118 - 5x) 3x - 50 = 118 - 5x Collecting the like terms 3x + 5x = 118 + 50 8x = 168 x = \(\frac{168}{8}\) x = 21\(^o\) Answer is B