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If log\(_9\)x = 1.5, 9\(^{1.5}\) = x 9^\(\frac{3}{2}\) = x (\sqrt9)\(^3\) = 3 ∴ x = 27

log\(_{10}\) 24 = log\(_{10}\) 8 \(\times\) log\(_{10}\) 3 where log\(_{10}\) 8 = 3 log\(_{10}\) 2 = 3 \(\times\) m and log\(_{10}\) 3 = n : log\(_{10}\) 24 = 3m + n

\(\sqrt{27}\) + \(\frac{3}{\sqrt{3}}\) = \(\sqrt{9 \times 3}\) + \(\frac{3 \times {\sqrt{3}}}{{\sqrt{3}} \times {\sqrt{3}}}\) = 3\(\sqrt{3}\) + \(\sqrt{3}\) = 4\(\sqrt{3}\)

2\sqrt7- \(\frac{14}{\sqrt7} + \frac{7}{\sqrt21}\) L.C.M is \sqrt21 \(\frac{\sqrt21 * 2\sqrt7 - \sqrt3 *14 + 7}{\sqrt21}\) =\(\frac{2 * 7\sqrt3 - 14\sqrt3 + 7}{\sqrt21}\) = \(\frac{14\sqrt3 - 14\sqrt3 + 7}{\sqrt21}\) =\(\frac{7}{\sqrt21}\) or \(\frac{7 \times \sqrt21}{\sqrt21 \times \sqrt21}\) = \(\frac{7 \times \sqrt21}{21}\) or \(\frac{\sqrt21}{3}\)

\(\frac{2}{(2r - 1)}\) - \(\frac{5}{3}\) = \(\frac{1}{(r + 2)}\) \(\frac{2}{(2r - 1)}\) - \(\frac{1}{(r + 2)}\) = \(\frac{5}{3}\) The L.C.M.: (2r - 1) (r + 2) \(\frac{2(r + 2) - 1(2r - 1)}{(2r - 1) (r + 2)}\) = \(\frac{5}{3}\) \(\frac{2r + 4 - 2r + 1}{ (2r - 1) (r + 2)}\) = \(\frac{5}{3}\) cross multiply the solution 3 * 5 = (2r - 1) (r + 2) * 5 divide both sides 5 3 = 2r\(^2\) + 3r - 2 (when expanded) collect like terms 2r\(^2\) + 3r - 2 - 3 = 0 2r\(^2\) + 3r - 5 = 0 Factors are -2r and +5r 2r\(^2\) -2r + 5r - 5 = 0 [2r\(^2\) -2r] [+ 5r - 5] = 0 2r(r-1) + 5(r-1) = 0 (2r+5) (r-1) = 0 r = 1 or - \(\frac{5}{2}\)

use "T" to represent the total profit. The first receives \(\frac{1}{3}\) T remaining, 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)T The seconds receives the remaining, which is \(\frac{2}{3}\) also \(\frac{2}{3}\) x \(\frac{2}{3}\) x \(\frac{4}{9}\) The third receives the left over, which is \(\frac{2}{3}\)T - \(\frac{4}{9}\)T = (\(\frac{6 - 4}{9}\))T = \(\frac{2}{9}\)T The third receives \(\frac{2}{9}\)T which is equivalent to N12000 If \(\frac{2}{9}\)T = N12, 000 T = \(\frac{12 000}{\frac{2}{9}}\) = N54, 000

To round to three significant figures, look at the fourth significant figure. It's a 5 , so round up or round down if below 5. To round to two significant figures, look at the third significant figure. It's an 8 , so round up.

The sum of 2 \(\frac{1}{6}\) and 2\(\frac{7}{12}\) = \(\frac{13}{6}\) + \(\frac{31}{12}\) = \(\frac{13 \times 2 + 31}{12}\) = \(\frac{26 + 31}{12}\) = \(\frac{57}{12}\) What should be subtracted from \(\frac{57}{12}\) to give 3\(\frac{1}{4}\) \(\frac{57}{12}\) - y = 3\(\frac{1}{4}\) : y = \(\frac{57}{12}\) - 3\(\frac{1}{4}\) = \(\frac{57}{12}\) - \(\frac{13}{4}\) y = \(\frac{57 - 3 \times 13}{12}\) = \(\frac{57 - 39}{12}\) y = \(\frac{18}{12}\) y = \(\frac{3}{2}\) or 1\(\frac{1}{2}\)

\(\frac{2}{x-3}\) - \(\frac{3}{x-2}\) = \(\frac{2(x -2) -3(x - 3)}{(x-3)(x -2)}\) = \(\frac{2x -4 -3x + 9}{(x-3)(x -2)}\) = \(\frac{5 - x}{(x-3)(x -2)}\)

The mean = sum of 1\(\frac{1}{2}\), 2\(\frac{2}{3}\), 3\(\frac{3}{4}\), 4\(\frac{4}{5}\), and 5\(\frac{5}{6}\) divided by their quantity. Mean = ( 1\(\frac{1}{2}\) + 2\(\frac{2}{3}\) + 3\(\frac{3}{4}\) + 4\(\frac{4}{5}\) + and 5\(\frac{5}{6}\) ) / 5 = 3.708 or 3.71( 2 d.p )

Total number = 40 number of red = 10 Pr (R) = \(\frac{10}{40}\) Pr(blue) = \(\frac{12}{40}\) probability (neither red nor blue) = 1 - \(\frac{10}{40}\) - \(\frac{12}{40}\) = \(\frac{40 -12 -10}{40}\) = \(\frac{18}{40}\) or \(\frac{9}{20}\)

(a) interests = \(\frac{25000 *3 * 21}{100}\) = $ 15750.00 amount paid = 25000 + 15750 = $40,750.00 amounts paid by year = 40750.00 / 2 = $ 20375.00 (b) let's be consecutive numbers be X and X + 1 3X + 2(X + 1) = 17 3X + 2X + 2 = 17 5x = 15 x = 3 some of the numbers = 3 + 4 = 7 Percentage = \(\frac{3}{7}\) * 100 = 42.857 or 42.9%