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(a) \(\frac{0.6 \times 32 \times 0.004}{1.2 \times 0.008 \times 0.16}\) = \(\frac{6 \times 10^{-1} \times 32 \times 4 \times 10^{-3}}{12 \times 10^{-1} \times 8 \times 10^{-3} \times 16 \times 10^{-2}}\) = \(\frac{4}{2 \times 4 \times 10^{-2}}\) = \(\frac{1}{2} \times \frac{1}{10^{-2}\) = \(0.5 \times 10^{2} \) = \(5 \times 10^{1}\) (b) In the diagram above, \(< GHB = 180 - 7x^\circ\) (angles on a straight line) \(< HBN = 180 - 7x^\circ\) (alternate angles) \(< EBN = 3x^\circ\) (corresponding angles) Then \(3x^\circ + (180 - 7x)^\circ = 120^\circ\) \(\implies 3x^\circ - 7x^\circ + 180^\circ = 120^\circ\) \(\implies 180^\circ - 120^\circ = 7x^\circ - 3x^\circ\) \(60^\circ = 4x^\circ\) \(x = 15^\circ\) Hence, \(< GHB = 180^\circ - (7 \times 15^\circ) = 180^\circ - 105^\circ\) = \(75^\circ\).

(a) x 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ 210^\circ 240^\circ 270^\circ y 1.0 2.0 2.7 3.0 2.7 2.0 1.0 0.0 -0.7 -1.0 (b) (c) \(\sin x = \frac{1}{4}\) (Given) Multiply through by 2 ; \(2 \sin x = 2 \times \frac{1}{4} = \frac{1}{2}\) \(2 \sin x = \frac{1}{2} \implies 2 \sin x - \frac{1}{2} = 0\) Add \(1\frac{1}{2}\) to both sides ; \(2 \sin x - \frac{1}{2} + 1\frac{1}{2} = 0 + 1\frac{1}{2}\) \(\implies 2 \sin x + 1 = 1\frac{1}{2}\) Draw the line \(y = 1\frac{1}{2}\) on the same axis as \(y = 2 \sin x + 1\). The line \(y = 1\frac{1}{2}\) cuts the curve at points P and Q where x = 15^\circ and x = 168^\circ. Hence, the values of x for which \(\sin x = \frac{1}{4}\)are 15^\circ and 168^\circ.

(a) \((x - 2)(x - 3) = 12\) \(x^{2} - 3x - 2x + 6 = 12 \implies x^{2} - 5x + 6 - 12 = 0\) \(x^{2} - 5x - 6 = 0\) \(x^{2} - 6x + x - 6 = 0 \implies x(x - 6) + 1(x - 6) = 0\) \( (x - 6)(x + 1) = 0 \implies \text{x = 6 or -1}\) (b) The shaded portion comprises two shaded segments labelled A and B. Area of segment A = \(\frac{60^\circ}{360^\circ} \times \pi r^{2} - \frac{1}{2} r^{2} \sin 60^\circ\) \(\frac{1}{6} \times \frac{22}{7} \times 7 \times 7 - \frac{1}{2} \times 7 \times 7 \times 0.8660\) = \(25.667 - 21.217\) = \(4.45 cm^{2}\) Also, the area of the segment B = \(4.45 cm^{2}\). Hence, the area of the shaded portion = \(4.45 cm^{2} + 4.45 cm^{2}\) = \(8.9 cm^{2}\) \(\approxeq 9 cm^{2}\) (to the nearest whole number).

(a) \(5 \cos (x + 8.5)^\circ - 1 = 0\) \(5 \cos (x + 8.5)^\circ = 1\) \(\cos (x + 8.5)^\circ = \frac{1}{5} = 0.2\) \((x + 8.5)^\circ = \cos^{-1}(0.2) = 78.463^\circ\) \(x = 78.463^\circ - 8.5 = 69.963^\circ\) \(\approxeq 70^\circ\) (to the nearest degree). (b)(i) (ii) By the cosine rule, \(|QR|^{2} = 32^{2} + 24^{2} - 2 \times 32 \times 24 \times \cos 45\) \(|QR|^{2} = 1024 + 576 - 1536 \cos 45\) = \(1600 - 1086.1056\) \(|QR|^{2} = 513.8944\) \(|QR| = \sqrt{513.8944} = 22.669 km\) \(\approxeq 22.67 km\) ( 2 decimal place) (iii) By the sine rule, \(\frac{32}{\sin \alpha} = \frac{22.67}{\sin 45}\) \(\sin \alpha = \frac{32 \times \sin 45}{22.67}\) = \(0.9981\) \(\alpha = \sin^{-1} (0.9981) = 86.4787^\circ\) The diagram below shows all the angles at Q; The bearing of R from Q is given by the reflex angle NQR. Thus reflex < NQR = 360^\circ - (86.47^\circ + 30^\circ) = 360^\circ - 116.47^\circ = 243.53^\circ Hence, the bearing of R from Q = 244^\circ (to the nearest degree).

Perimeter of the sector = \(2r + \frac{\theta}{360^\circ} \times 2\pi r\) = \(2(10.5) + \frac{120}{360} \times 2 \times \frac{22}{7} \times 10.5\) = \(21 + 22\) = 43 cm

Angle in a semi-circle is 90º when OM is subtended : (90 - 25)º = 65º

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

Using \(S.I =\frac{P \times T \times R}{100}\) \(600 =\frac{3000 \times T \times 8}{100}\) \(T =\frac{600 \times 100}{3000 \times 8}\) \(\frac{20}{8}\) = 2\(\frac{1}{2}\) years

S = \(\sqrt{t^2 - 4t + 4}\) S2 = t2 - 4t + 4 t2 - 4t + 4 - S2 = 0 Using \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Substituting, we have; Using \(t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(4 - S^2)}}{2(1)}\) \(t = \frac{4 \pm \sqrt{16 - 4(4 - S^2)}}{2}\) \(t = \frac{4 \pm \sqrt{16 - 16 + 4S^2}}{2}\) \(t = \frac{4 \pm \sqrt{4S^2}}{2}\) \(t = \frac{2(2 \pm S)}{2}\) Hence t = 2 + S or t = 2 - S

y = x sin x Where u = x and v = sin x Then \(\frac{\delta u}{\delta x}\) = 1 and \(\frac{\delta v}{\delta x}\) = cos x By the chain rule, \(\frac{\delta y}{\delta x} = v\frac{\delta u}{\delta x} + u\frac{\delta v}{\delta x}\) = (sin x)1 + x cos x = sin x + x cos x

A = \(\pi\)r2, \(\frac{\delta A}{\delta r}\) = 2πr So, using \(\frac{\delta A}{\delta t}\) = \(\frac {\delta A}{\delta r}\) x \(\frac {\delta A}{\delta t}\) = 2\(\pi\)r x 0.02 = 2\(\pi\) x 7 x 0.02 = 2 x \(\frac{22}{7}\) x 0.02 = 0.88cm2s-1

Using x = \(\frac{\sum x}{N}\) in each case, we get; \(\sum_{6}^{i=1} x_i\) = 10 x 7 = 70 \(\sum_{7}^{i=1} x_i\) = 2 + 4 + 8 + 14 + 16 + 18 = 62 Hence the missing number can be obtained from \(\sum_{6}^{i=1} x_i - \sum_{7}^{i=1} x_i\) = 70 - 62 = 8 So, all the seven numbers are 2, 4, 8, 8, 14, 16, 18 Mode = 8

Let \(\delta^2\) and \(\delta\) denote the variance and standard deviation of the distribution respectively. But \(\delta^2\) = 4 (given) Hence \(\delta\) = \(\sqrt{4}\) = 2

Sum of the angles in a triangle = 180^\circ 180^\circ - x + 60 + 70 = 180 130^\circ - x = 180^\circ - 180^\circ = 0^\circ x = 130^\circ

5x^\circ + (16x - 24)^\circ + 5x^\circ + (4x + 12)^\circ + (6x + 12)^\circ = 360^\circ 36x^\circ - 24 + 12 + 12 = 360^\circ 36x^\circ = 360^\circ x^\circ = \(\frac{360^\circ}{36}\) = 10^\circ Thus, the angle of sector representing Mathematics is 5 x 10^\circ = 50^\circ. Hence the number of students who offer mathematics is \(\frac{50}{360} \times 80 \approx 11\)