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PQ\(^2\) = (x\(_2\) - x\(_1\))\(^2\) + (y\(_2\) - y\(_1\))\(^2\) = 12\(^2\) + 5\(^2\) = 144 + 25 = 169 PQ = \(\sqrt{169}\) = 13 But PQ = diameter = 2r, r = \(\frac{\text{PQ}}{2}\) = 6.5 units

The line y = 4 is a horizontal line in the Cartesian plane. The distance from any point( x, y) to this line is given by d = | y - 4| and this distance to be equal to k | y - 4 | = k case1, y - 4 = k, y = k + 4 ⇒ 4 + k case2, y - 4 = - k, y = - k + 4 ⇒ 4 - k combining the two solutions above, we have y = 4 \(\pm\) k

Since ST = the radius r, => SOT is equilateral. Each angle = 60° Length of arc ST = θ/360 x 2πr = 60/360 x 2 x π x r = πr/3

y = 2kx2 + x + 1 Gradient function dy/dx = 4kx + 1 at dy/dx = 9, => 9 = 4kx + 1 => 8 = 4kx, and k = 8/4x. At x = 1, k = 8/4 = 2

y = x sin x dy/dx = 1 x sinx + x cosx = sinx + x cosx At x = π/2, = sin (π/2) + (π/2) cos (π/2) = 1 + (π/2) x (0) = 1

Let the rectangle be a square of sides p/4. So that perimeter of square = 4p 4 x (p/4) = p.

cos\(\alpha\) = \(\frac{10}{16}\) = 0.625 \(\alpha\) = cos-1(0.625) = 51.3º ∠PQR = 2\(\alpha\) = 2(51.3º) = 102.6 = 103º

Given \(\cos x = -\frac{5}{13}\) where \(180^\circ < x < 270^\circ\): Using: \(\sin^2 x + \cos^2 x = 1\) \(\sin^2 x = 1 - \left(-\frac{5}{13}\right)^2 = \frac{144}{169} \implies \sin x = -\frac{12}{13}\) \(\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{12}{13}}{-\frac{5}{13}} = \frac{12}{5}\) Calculating \(\tan x - \sin x\): \(\tan x - \sin x = \frac{12}{5} - \left(-\frac{12}{13}\right) = \frac{12}{5} + \frac{12}{13}\) Finding a common denominator (65): \(\tan x - \sin x = \frac{12 \times 13}{65} + \frac{12 \times 5}{65} = \frac{156}{65} + \frac{60}{65} = \frac{216}{65}\) Thus, the value is: \(\tan x - \sin x = \frac{216}{65}\)

N ∝ M ∴N = kM where k is a constant N = 8 when M = 20 ∴ 8 = k x 20 \(k = \frac{8}{20}\) = \(\frac{2}{5}\) ∴ N = \(\frac{2 M}{5}\) If N = 7 ∴ 7 = \(\frac{2 M}{5}\) M = \(\frac{35}{2}\) = 17\(\frac{1}{2}\)

0.0462 in standard form = 4.62 x \(10^{-2}\) Option D

= 180^\circ - 40^\circ = 140^\circ

If \(sin P = \frac{5}{13}\) from right angled triangle from Pythagoras theorem \(BC^2 = 13^2 - 5^2\) = 169 - 25 BC = \(\sqrt{144}\) = 12 ∴ cos P - tan P = \(\frac{12}{13} - \frac{5}{12}\) =\(\frac{79}{156}\)

The size of an exterior angle = \(\frac{360}{n}\) From the options, all options can perfectly divide 360^\circ, except 21^\circ.

< PQS = 50^\circ (PQ = PR) < PQS = < SPQ = 50^\circ (QS = PS) In triangle PSQ, < PQS + < QPS + < PSQ = 180^\circ 50^\circ + 50^\circ + < PSQ = 180^\circ < PSQ = 180^\circ - 100^\circ = 80^\circ

In \(\Delta\) PRQ, < SRQ = 90^\circ - 58^\circ < SRQ = < QTS = x (angles in the segment) \(\therefore\) x = 32^\circ