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12 = \(\frac{n}{3} - 2n = 1\), multiply through by 3 36 + n - 6n = 3 -5n = 3 - 36 -5n = -33 -5n + 33 = 0 5n - 33 = 0

\(\frac{-m}{2} - \frac{5}{4} \geq \frac{5m}{12} - \frac{7}{6}\) = \(\frac{-2m - 5}{4} \geq \frac{5m - 14}{12}\) 12(-2m - 5) \(\geq\) 4(5m - 14) -24m - 60 \(\geq\) 20m - 56 -24m - 20m \(\geq\) -56 + 60 44m \(\geq\) 4 m \(\leq \frac{4}{-44}\) m \(\leq \frac{-1}{11}\)

Curved surface area = 2\(\pi h\) 704 = 2 x \(\frac{22}{7} \times 8 \times h\) 704 = \(\frac{352}{7}\)h 352h = 704 x 7 h = \(\frac{704 \times 7}{352}\) = \(\frac{4928}{352}\) h = 14cm

Let 0 = angle of the minor sector angle of the major sector = 360 - \(\theta\)(angle at a point) 2 \(\pi r\) = 54 + 126(i.e circumference of minor and major arc) 2\(\pi r = 180^o\) r = \(\frac{180}{2\pi}\) = \(\frac{90}{\pi}\) Lenght of ninor arc = \(\frac{\theta}{360} \times 2 \pi r\) 54 = \(\frac{\theta}{360} \times 3 \pi r\) \(\theta = \frac{360 \times 54}{2 \pi r}\) but r = \(\frac{90}{\pi}\) substituting \(\frac{90}{\pi}\) for r \(\theta = \frac{360 \times 54 \times \pi}{2 \times \pi \times 90}\) \(\theta = 2 \times 54 = 108^o\) angle of the major sector = 360 - 108o = 252o

/YZ/\(^2\) = /XZ/\(^2\) + /XY/\(^2\) (Verify using Pythagoras' theorem)

cos(x + 40)o = 0.0872 x + 40 = cos-10.0872 x + 40o = 84.99o x = 84.99o - 40o x = 44.99 x = 45o

From the diagram above, Sin 54º = \(\frac{\text{OPP}}{\text{HYP}}\) Sin 54º = \(\frac{\text{x}}{50}\) x = 50 Sin 54º

\(\begin{array}{c|c} x & x - x & (x - \bar{x})^2\\ \hline 15 & -6 & 36\\21 & 0 & 0\\17 & -4 & 16\\ 26 & 5 & 25 \\ 18 & -3 &9 \\ 29 & 8 & 64 \end{array}\) \(E(x - \bar{x})^2\) = 150 N = 6 S.D = \(\sqrt{\frac{(x - x)^2}{N}}\) S.D = \(\sqrt{\frac{150}{6}}\) = 5

Prob(RB + BR) = Total balls = 4 + 6 = 10 = prob(\(\frac{4}{10} \times \frac{6}{9}\)) + prob(\(\frac{6}{10} \times \frac{4}{9}) = \frac{24}{90} + \frac{24}{90}\) = \(\frac{48}{90} = \frac{16}{30} = \frac{8}{15}\)

y = \(\sqrt{8m}\); log y = log \(\sqrt{8m}\) log y = log \(\sqrt{8}\) + log \(\sqrt{m}\) log y = \(\frac{1}{2}\)log 2\(^3\) + log m\(\frac{1}{2}\) log y = \(\frac{3}{2}\) log 2 + \(\frac{1}{2}\) log m

x + 0.4y = 3...(i) y = \(\frac{1}{2}\)x x = 2y x - 2y = 0....(ii) solve simultaneously; x + 0.4y = 3 - x - 2y = 0 2.4 = 3 y = \(\frac{3 \times 10}{2.4 \times 10} \) = \(\frac{30}{24} = \frac{5}{4}\) x - 2(\(\frac{5}{4}\)) = 0 x - \(\frac{5}{2}\) = 0 x = \(\frac{5}{2}\) x + y = \(\frac{5}{2} + \frac{5}{4}\) \(\frac{10 + 5}{4} = \frac{15}{4}\) = 3\(\frac{3}{4}\)

(\(\frac{x -y}{y}\)); \(\frac{3}{1} - \frac{x y}{y}\) = \(\frac{3y - (x - y)}{y}\) = \(\frac{3y - x + y}{y}\) = \(\frac{4y - x}{y}\)

prime factor of 210 = 2, 3, 5, 7 prime numbers less than 10 = 2, 3, 5 , 7

let the total amount be Nx i.e (\(\frac{1}{4}\))x + (\(\frac{1}{3}\))x + 72,000 = x \(\frac{x}{4} + \frac{x}{4} + 72,000 = x\) \(\frac{3x + 4x + 86,400}{12} = x\) cross multiply to clear fraction 12x = 3x + 4x + 86,400 12x - 7x = 86,400 5x = 86,400 x - \(\frac{86,400}{5}\) = 172,800 amount spent on food = \(\frac{1}{4} \times 172,800\) = N43,200

(\(\frac{27}{125}\))-\(\frac{1}{3}\) x (\(\frac{4}{9}\))\(\frac{1}{2}\) = (\(\frac{3^3}{5^3}\))-\(\frac{1}{3}\)-\(\frac{1}{3}\) x (\(\frac{3^2}{3^2}\))\(\frac{1}{2}\) -\(\frac{1}{2}\) = \(\frac{3^{-1}}{3^{-1}} \times \frac{2}{3}\) = \(\frac{\frac{1}{3}}{\frac{1}{5}} \times \frac{2}{3}\) \(\frac{1}{3} \times \frac{5}{1} \times {2}{3} = \frac{10}{9}\)