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\(Surface area (A) = 4\pi r^{2} = 616m^{2}\) solving, we have \(r = 7m\) \(Volume = \frac{4}{3}\pi r^{3}\) \(Volume = \frac{4\times 3.142 \times 7^{3}}{3}\) \(Volume = \frac{4310.824}{3}\) = \(1436.941m^{3} \approxeq 1400m^{3}(2 sig. fig)\)

C.S.A = 5,544 Volume = 38,808 S : V = 1 : 07

SP = #4600, profit % = 15 % profit = \(\frac{SP - CP}{CP}\) X 100 \(\frac{15}{100} = \frac{4600 - CP}{CP}\) 15CP = 460000 - 100CP 15CP + 100CP = 460000 CP = \(\frac{460000}{115}\) = #4000 Cost price of article = #4000 OR Profit % = 100 + 15 = 115 CP = \(\frac{4600 \times 100}{115}\) = #4000

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

To express 0.03785 in standard form to 3 significant figures: * Identify significant figures: 3.785 * Round to 3 figures: 3.79. * Write in standard form: 3.79 \times 10\(^{-2}\) Final answer: 3.79\times10\(^{−2}\)​

Convert (\(101.011)_2\) to base 10 \(1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 + 0 \times 2^{-1} + 1 \times 2^{-2} + 1 \times 2^{-3}\) 1 x 4 + 0 x 2 + 1 x 1 + 0 x \(\frac{1}{2}\) + 1 x \(\frac{1}{4}\) + 1 x \(\frac{1}{8}\) 4 + 1 + \(\frac{1}{4}\) + \(\frac{1}{8}\) = \(\frac{43}{8}\)

x\(^2\) - 3x - 10 = 0 Using factorisation method: x\(^2\) - 3x - 10 = 0 becomes x\(^2\) - 5x + 2x - 10 = 0 (x\(^2\) - 5x) + (2x - 10) = 0 x(x - 5) + 2(x -5) = 0 x - 5 = 0 or x + 2 = 0 \(\therefore\) x = 5 or -2

0^\circ ∠ \theta ∠ 90º, From the diagram above, the adjacent by Pythagoras theorem = 3 Cos \theta = \(\frac{\text{adj}}{\text{hyp}}\) = \(\frac{3}{5}\) Tan \theta = \(\frac{\text{opp}}{\text{adj}}\) = \(\frac{4}{3}\) cos \theta + tan \theta = \(\frac{3}{5}\) + \(\frac{4}{3}\) = \(\frac{29}{15}\)

using cosine rule, \(cos\theta\) = \(\frac{28^2 + 15^2 - 21^2}{2\times28\times15}\) = \(\frac{568}{840}\) \(\theta = cos^{-1}\) = 47.5^\circ

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

T H T T, T T, H H H , T H ,H Pro(one head) = \(\frac{3}{4}\) = \(\frac{3}{4}\)

(a) \(\log_{10} 54 \) \(54 = 2 \times 3^{3}\) \(\log_{10} 54 = \log_{10} (2 \times 3^{3})\) = \(\log_{10} 2 + \log_{10} 3^{3}\) = \(\log_{10} 2 + 3\log_{10} 3\) = \(0.3010 + (3 \times 0.4771)\) = \(0.3010 + 1.4313\) = \(1.7323\) (b) \(\log_{10} 0.24\) \(0.24 = \frac{24}{100} = \frac{2^{3} \times 3}{100}\) \(\log_{10} 0.24 = \log_{10} 24 - \log_{10} 100\) \(\log_{10} (2^{3} \times 3) - \log_{10} (10^{2})\) \(\log_{10} 2^{3} + \log_{10} 3 - 2\log_{10} 10\) = \(3\log_{10} 2 + \log_{10} 3 - 2\) = \((3 \times 0.3010) + 0.4771 - 2\) = \(0.9030 + 0.4771 - 2\) = \(-0.6199\)

(a) \(\frac{1}{3^{5n}} \times 9^{n - 1} \times 27^{n + 1}\) \(3^{-5n} \times (3^{2})^{n - 1} \times (3^{3})^{n + 1}\) \(3^{-5n} \times 3^{2n - 2} \times 3^{3n + 3}\) \(3^{-5n + 2n - 2 + 3n + 3}\) = \(3^{1} = 3\) (b) Let the daughter's age be c and the woman's age be d. \(c + d = 46 .... (1)\) In 4 years time, the daughter's age = c + 4 The woman's age = d + 4 \(\frac{d + 4}{c + 4} = \frac{7}{2}\) \(2(d + 4) = 7(c + 4) \implies 2d + 8 = 7c + 28\) \(2d - 7c = 28 - 8 \implies 2d - 7c = 20 ... (2)\) \(c + d = 46 \implies d = 46 - c\) \(\therefore 2(46 - c) - 7c = 20\) \(92 - 2c - 7c = 20 \implies 92 - 20 = 9c\) \(72 = 9c \implies c = 8\) \(d = 46 - c \implies d = 46 - 8 = 38\) Therefore, the daughter is 8 years old and the woman is 38 years.

(a) \((x + 8)^{2} = x^{2} + (x + 7)^{2}\) \(x^{2} + 16x + 64 = x^{2} + x^{2} + 14x + 49\) \(x^{2} + 16x + 64 - 2x^{2} - 14x - 49 = 0\) \(15 + 2x - x^{2} = 0\) \(x^{2} - 2x - 15 = 0\) \((x + 3)(x - 5) = 0\) x = 5m since measurements cannot be negative. (b) \(Area = length \times breadth\) Length = (x + 7) = (5 + 7) = 12 m Breadth = x m = 5 m \(Area = 12 \times 5 = 60 m^{2}\)

VEHICLES Jan Feb Mar CAR 10 40 5 LORRY 15 25 35 BUS 20 30 50 TOTAL 45 95 90 (a) Vehicles produced in February = 40 + 25 + 30 = 95 (b) Fraction of the vehicles in February that were cars = \(\frac{40}{95} = \frac{8}{19}\) (c) Total number of buses from January to March = 20 + 30 + 50 = 100 (d) Ratio of lorries in February to that in March = \(25 : 35 \equiv 5 : 7\).