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\(t^2\) + kt + \(\frac{25}{4}\) The general form of quadratic expression = ax\(^2\) + bx + c comparing \(t^2\) + kt + \(\frac{25}{4}\) and the general form a = 1, b = k and c = \(\frac{25}{4}\) using: b\(^2\) = 4ac we can find the value of k. ( note: there is/are another way(s) to find the value of k) k\(^2\) = 4 x 1 x \(\frac{25}{4}\) k\(^2\) = 25 taking the square root of both sides k = 5

to calculate for the high of the pole, we use sohcahtoa Let the height of the pole be x. Tan32º = \(\frac{opp}{adj}\) Tan32º = \(\frac{x}{68}\) x = Tan32º x 68 = 42.5m ( to 1 d.p)

VOLUME of CUBE = \(L^3\) 8000 = \(L^3\) L = \(\sqrt[3]{8000}\) = 20 S. A. CUBE = 6\(L^2\) S. A. CUBE = 6 x 20 x 20 = 2400\(cm^2\)

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

let one bottle of mirinda cost = #x, a bottle of malt will cost #3x 4 bottles of mirinda will cost = #4x, then 5 bottles of malt cost = #15x total cost = 4x + 15x = 380 x = 380/19 = 20, therefore a bottle of mirinda cost = #20 then a bottle of mirinda will cost 3 x 20 = #60.00

To find the probability that two beads drawn (with replacement) are of the same color, we consider two cases: both red and both white. Beads = 5, White beads = 7, Total beads = 12 Both Red: \(P(\text{Red}) = \frac{5}{12}, \quad P(\text{Both Red}) = \frac{5}{12} \times \frac{5}{12} = \frac{25}{144}\) Both White: \(P(\text{White}) = \frac{7}{12}, \quad P(\text{Both White}) = \frac{7}{12} \times \frac{7}{12} = \frac{49}{144}\) Total Probability of Same Color: \(P(\text{Same Color}) = \frac{25}{144} + \frac{49}{144} = \frac{74}{144} = \frac{37}{72}\) The probability that the two beads are the same color is \( \frac{37}{72}\).

\(\sqrt{6a^{-2}} \times \sqrt{24a^6}\) = \(\sqrt{6a^{-2} \times 24a^6}\) = \(\sqrt{144a^4}\) = \pm12a\(^2\).

9\(^x\) = 729 3\(^2\)\(^x\) = 3\(^6\) 2x = 6 x = 3

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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 + \(\frac{1}{4}\) + \(\frac{1}{8}\) = \(\frac{43}{8}\)