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T\(_4\) = 3 = ar\(^3\) a = 81 3 = 81 r\(^3\) divide through by 81 r\(^3\) = \(\frac{1}{27}\) take cube root of both sides r = \(\sqrt[3]{\frac{1}{27}}\) = \(\frac{1}{3}\)

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

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

(a) \(\frac{1}{81^{(x - 2)}} = 27^{(1 - x)}\) \(81^{-(x - 2)} = 27^{(1 - x)}\) \(3^{4[-(x - 2)]} = 3^{3(1 - x)}\) \(3^{-4(x - 2)} = 3^{3(1 - x)}\) \(-4(x - 2) = 3(1 - x)\) \(-4x + 8 = 3 - 3x\) \(8 - 3 = -3x + 4x \implies x = 5\) (b) \(\frac{5}{\sqrt{7} - \sqrt{3}} + \frac{1}{\sqrt{7} + \sqrt{3}}\) = \(\frac{5(\sqrt{7} + \sqrt{3}) + (\sqrt{7} - \sqrt{3})}{(\sqrt{7} - \sqrt{3})(\sqrt{7} + \sqrt{3})}\) = \(\frac{5\sqrt{7} + 5\sqrt{3} + \sqrt{7} - \sqrt{3}}{7 - 3}\) = \(\frac{6\sqrt{7} + 4\sqrt{3}}{4}\) = \(\frac{3\sqrt{7} + 2\sqrt{3}}{2}\).

(a) \(5^{2} = 7^{2} + 7^{2} - 2(7)(7) \cos \theta\) \(25 = 98 - 98 \cos \theta\) \(\cos \theta = \frac{98 - 25}{98}\) \(\cos \theta = 0.7449\) \(\theta = \cos^{-1} (0.7449)\) \(\approxeq 42^\circ\). (b) (i) \((2n - 4) \times 90^\circ = 1440^\circ\) \(2n - 4 = \frac{1440}{90} = 16\) \(2n = 16 + 4 = 20\) \(n = 10\) (ii) Size of exterior angle = \(\frac{360}{\text{Number of sides}}\) = \(\frac{360}{10}\) = \(36^\circ\)

(a) ST = QR ; PS = height of tower. In \(\Delta\) PST, \(\frac{72}{ST} = \tan 62^\circ\) \(ST = \frac{72}{\tan 62}\) = 38.283m \(\approxeq\) 38.3m In \(\Delta\) PQR, \(\frac{h}{QR} = \tan 51^\circ\) \(h = QR \times \tan 51^\circ\) = \(38.3 \times 1.235\) = \(47.3m\) QS = RT (height of the building) QS = PS - PQ = 72 - 47.3 = 24.7m (b) (i) \(\Delta POR = \Delta QOR\) (isosceles triangle) \(\frac{PT}{30} = \sin 60\) \(PT = 30 \sin 60 = 30 \times \frac{\sqrt{3}}{2}\) = \(15\sqrt{3} cm\) PT = TR PR = 2PT = \(2 \times 15\sqrt{3} = 30\sqrt{3} cm\) = \(30 \times 1.732 = 51.96 cm\) \(\approxeq 52 cm\) (ii) Length of minor arc, PQR = \(\frac{\theta}{360} \times 2\pi r\) = \(\frac{120}{360} \times 2 \times 3.142 \times 30\) = \(62.84 cm\) (iii) Perimeter of shaded portion = chord PR + Arc PQR = 51.96 + 62.84 = 114.8 cm \(\approxeq\) 115cm.

(a) Marks Frequency Cum. freq 1 - 10 2 2 11 - 20 4 6 21 - 30 9 15 31 - 40 13 28 41 - 50 18 46 51 - 60 32 78 61 - 70 13 91 71 - 80 5 96 81 - 90 3 99 91 - 100 1 100 (b)(i) \(N = \sum f = 100\) \(Median = \frac{1}{2}N = \frac{1}{2}(100)\) = 50th percentile. \(\therefore Median = 52\). (ii) \(Q_{1} = \frac{1}{4} N = \frac{1}{4}(100)\) = 25th percentile. \(\therefore Q_{1} = 39\) (iii) 60th percentile = 55.

(a) \(1001_{2} = 1 \times 2^{3} + 0 \times 2^{2} + 0 \times 2^{1} + 1 \times 2^{0}\) = \(8 + 0 + 0 + 1\) = 9 \(\sqrt{9} = 3_{10}\) 2 3 2 1 r 1 0 r 1 \(\therefore \sqrt{1001_{2}} = 11_{2}\) (b) In \(\Delta QOP, OQ = OP\) \(\therefore QK = KP\) \(\Delta QOK = \Delta POK\) (right- angled triangle) In \(\Delta POK\), \(OP^{2} = PK^{2} + OK^{2}\) \(x^{2} = y^{2} + (\frac{z}{2})^{2}\) \(\frac{z^{2}}{4} = x^{2} - y^{2}\) \(z^{2} = 4x^{2} - 4y^{2}\) \(z = \sqrt{4(x^{2} - y^{2})}\) \(z = 2\sqrt{x^{2} - y^{2}}\) (c)(i) \(< PSQ = \frac{1}{2}(< POQ) = \frac{1}{2}(160^\circ)\) = 80^\circ In \(\Delta PSQ\), \(< PQS + < QPS + < PSQ = 180^\circ\) \(40^\circ + < QPS + 80^\circ = 180^\circ\) \(< QPS = 180^\circ - 120^\circ = 60^\circ\) (ii) \(< PSR = < PSQ + < QSR\) = \(80^\circ + 45^\circ\) = \(125^\circ\) \(< PQS = < PQO + < RQS \) = \(40^\circ + < RQS\) \(< PSR + < PQS = 180^\circ\) \(\therefore 125^\circ + 40^\circ + < RQS = 180^\circ\) \(< RQS = 180^\circ - 165^\circ\) = \(15^\circ\)

To find the values of \( x \) where the curve \(y = x^3 + 2x^2 - 5x - 6\) crosses the x-axis, we set \( y = 0 \): \(x^3 + 2x^2 - 5x - 6 = 0\) Testing for rational roots, we find that \( x = -1 \) is a root. Factoring the polynomial gives: \(x^3 + 2x^2 - 5x - 6 = (x + 1)(x^2 + x - 6)\) Solving the quadratic \( x^2 + x - 6 = 0 \) using the quadratic formula yields: \(x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot -6}}{2 \cdot 1} = \frac{-1 \pm 5}{2}\) Thus, the solutions are: \(x = -1, \quad x = 2, \quad x = -3\) Therefore, the values of \( x \) where the curve crosses the x-axis are \(x = -1, \quad x = 2, \quad x = -3\).

x = 4 and y = 2

Let the number of students that offer both subject be x hence, the number of students offering physics = 3x The number of students offering chemistry = 6x The number of students offering physics only = 3x - x = 2x The number of students offering chemistry only = 6x - x = 5x So, 2x + 5x + x = 40 8x = 40 x = \(\frac{40}{8}\) = 5. Therefore, the number of students offering physics only = 2x = 2 x 5 = 10.

You can either multiply directly in base 4 or you can decide to convert to base 10 and do the multiplication and re-convert.

Using a venn diagram, let x = number who offer both Maths and Physics. So that (32-x) offer Maths and another (24-x) offer Physics. (32-x) + (24-x) + (x) + (4 who offer neither) = 40 => 60 - x = 40 => -x = -20 Therefore x = 20.

\((\frac{1}{0.06} \div \frac{1}{0.042})^{-1}\) = \((\frac{100}{6} \div \frac{1000}{42})^{-1}\) = \((\frac{100}{6} \times \frac{42}{1000})^{-1}\) = \((\frac{7}{10})^{-1}\) = \(\frac{10}{7}\) = 1.4285 \(\approxeq\) 1.43.

C.P = N1800 S.P = (270/5) x 40 = N2160.00 Profit = S.P - C.P = N2160 - N1800 = N360.00