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Log2(3x - 1) = 5 Log2(3x - 1) = Log225 Log2(3x - 1) = Log232 3x - 1 = 32 3x = 32 + 1 = 33 x = \(\frac{33}{3}\) = 11

Volume of sphere = Volume of cylinder i.e. \(\frac{4}{3} \pi r^3 = \pi r^2 h\) \(\frac{4}{3} \pi r^3 = \pi \times 3^2 \times 4\) r3 = \(\frac{\pi \times 9 \times 4 \times 3}{4 \pi}\) r = 3\(\sqrt{27}\) = 3

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

From the venn diagram given, M = (a, b, c), N = (c, f, g) U = (a, b, c, d, e, f, g) Thus M' \(\cap\) N = (e, f, g) \(\cap\) (c, f, g) = (f, g)

First, reduce 20(mod 9) to its simplest form in mod 9; 9 x 2 + 2 = 2(mod 9) If 2(mod 9) y(mod 6), then y = 2 by comparism

\(\frac{(p - r)^2 - r^2}{2p^2 - 4pr}\) = \(\frac{(p - r)(p - r) - r^2}{2p^2 - 4pr}\)\ = \(\frac{p^2 - 2pr + r^2 - r^2}{2p(p - 2r}\) = \(\frac{p^2 - 2pr}{2p(p - 2r)}\) = \(\frac{p(p - 2r)}{2p(p - 2r)}\) = \(\frac{1}{2}\)

Y = x2 + 2x + k (given) y = o when x = 2 thus 0 = 22 + 2 x 2 + k 0 = 4 + 4 + k given k = -8

Let a represent an interior angle; e represent an exterior angle. A section of the polygon is down in the diagram. \(\frac{e}{a}\) = \(\frac{l}{11}\) given a = 11e a + e = 180o(angles on a straight line) 11e + e = 180o 12e = 180o e = \(\frac{180^o}{12}\) = 15o Hence, number of sides = \(\frac{360^o}{\tect{size of one exterior angle}\) = \(\frac{360^o}{14^o}\) = 24

Halima's age = n years old Her brother's age = \(\frac{n}{2}\) + 5

In the diagram give, \(\alpha\) 50o(alternative angles) tan\(\alpha\) = \(\frac{h}{6}\) h = 6tan\(\alpha\) = 6tan 50o = 6 x 1.1918 = 7.1508 = 7.15m (2d.p)

(a) \(7(x + 4) - \frac{2}{3}(x - 6) \leq 2[x - 3(x + 5)]\) Multiply through by 3 to clear the fraction : \(21(x + 4) - 2(x - 6) \leq 6[x - 3(x + 5)]\) \(21x + 84 - 2x + 12 \leq 6x - 18x - 90\) \(19x + 96 \leq -12x - 90\) \(19x + 12x \leq -90 - 96\) \(31x \leq - 186\) Hence, \(x \leq -6\). (b) Let : t represent number of tricycles; and b represent number of taxicabs Then \(t + b = 20 ... (1)\) \(2t + 4b = 66 .... (2)\) From (1), b = 20 - t. Put that into eqtn (2), we have: \(2t + 4(20 - t) = 2t + 80 - 4t = 66\) \(80 - 66 = 14 = 2t \) \(t = 7\) The transport company has 7 tricycles.

(a) From the table, \(\sum f = x + 155\) P(obtaining a 3) = \(\frac{\text{no of times a 3 was obtained}}{\sum f}\) \(\frac{x}{x + 155} = 0.225\) \(x = 0.255x + (155 \times 0.225)\) \(x - 0.225x = 34.875 \) \(0.775x = 34.875\) \(x = \frac{39.875}{0.775}\) \(x = 45\) \(\sum f = 155 + x = 155 + 45 = 200\) Hence, the die was thrown 200 times. (b) Set of even numbers = {2, 4, 6} Set of prime numbers = {2, 4, 5} Therefore, \(N \cup M = (2, 3, 4, 5, 6)\) \(n(N \cup M) = 30 + 45 + 28 + 40 + 32 = 175\) \(P(N \cup M) = \frac{175}{200} = 0.875\)

(a) (b)(i) From the Venn diagram, x + x + 7 + 8 = n(B) 2x + 15 = 47 2x = 47 - 15 = 32 x = 16 (ii) Number of people who travelled by at least 2 means = 16 + 7 + 8 + 3 = 34

(a) \(\frac{x - 2}{4} = \frac{x + 2}{2x}\) \(2x(x - 2) = 4(x + 2)\) \(2x^{2} - 4x = 4x + 8\) \(2x^{2} - 4x - 4x - 8 = 2x^{2} - 8x - 8 = 0\) Divide through by 2, we have \(x^{2} - 4x - 4 = 0\) \(x^{2} - 4x = 4\) Taking the square of \(\frac{b}{2}\) and add to both sides, \(x^{2} - 4x + (-2)^{2} = 4 + (-2)^{2}\) \((x - 2)^{2} = 8\) \(x - 2 = \pm {\sqrt{8}}\) \(x = 2 \pm \sqrt{8}\) \(x = 2 \pm 2.828\) \(x = 4.828\) or \(x = -0.828\). Hence, x = 4.83 or -0.83 (2 decimal place). (b)(i) In the diagram above, < QRS = 180^\circ - 52^\circ = 128^\circ (opp. angles of a cyclic quadrilateral). \(x_{1} = x_{2}\) (base angles of isosceles triangle) \(x_{1} + x_{2} + 128^\circ = 180^\circ\) (sum of angles of a triangle) \(2x_{2} = 180^\circ - 128^\circ = 52^\circ\) \(x_{2} = 26^\circ\) Hence, < SQT = 26^\circ. (ii) < PQS = 90^\circ (angles in a semi-circle) Hence, < PQT = 90^\circ - 26^\circ = 64^\circ.

(a) First, change the given linear equation to the gradient- intercept form: \(2x + y = 6 \implies y = 6 - 2x\) Thus, \(m_{1} = -2, c = 6\). Thus, for a line to be parallel, the gradient \(m_{2} = -2\) (condition for parallelism). Using \(y - y_{1} = m(x - x_{1})\), we have \(y - (-3) = -2(x - 2)\) \(y + 3 = - 2x + 4\) \(2x + y = 4 - 3 = 1\) Thus, the line passing through (2, -3) parallel to 2x + y = 6 is 2x + y = 1. (b) (i) \(\Delta\) 2 3 5 7 2 0 3 1 7 3 3 7 7 7 5 1 7 3 7 7 7 7 7 7 (ii) (a) \(2 \Delta (5 \Delta 7) = 2 \Delta 7 = 7\) (b) \(2 \Delta n = 5 \Delta 7\) \(2 \Delta n = 7\) By comparison, n = 7.