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x + \(\frac{1}{x}\) = 4, find x2 + \(\frac{1}{x^2}\) = (x + \(\frac{1}{x}\))2 = x2 + \(\frac{1}{x^2}\) + 2 x2 + \(\frac{1}{x^2}\) = ( x + \(\frac{1}{x^2}\))2 - 2 = (4)2 - 2 = 16 - 2 = 14

4x\(^2\) - 4 = (2x)\(^2\) - 2\(^2\) (2x)\(^2\) - 2\(^2\) = a\(^2\) - b\(^2\) = (a - b)\(^2\) - 2ab - 2b\(^2\) (a - b)\(^2\) = a - b\(^2\) + (2ab + 2b)\(^2\) a = 2x and b = 2 2ab + 2b\(^2\) = 2(2x)(2) + 2(2)\(^2\) = 8x + 8 = 8(x + 1)

\(\frac{x}{x + y}\) + \(\frac{y}{x - y}\) - \(\frac{x^2}{x^2 - y^2}\) \(\frac{x}{x + y}\) + \(\frac{y}{x - y}\) - \(\frac{x^2}{(x + y)(x - y}\) = \(\frac{x(x - y) + y(x + y) - x^2}{(x + y)(x - y}\) = \(\frac{x^2 + xy + xy + y^2 - x^2}{(x + y)(x - y}\) = \(\frac{y^2}{(x + y)(x - y)}\) = \(\frac{y^2}{(x^2 - y^2)}\)

Given that x2 + y2 + z2 = 194, calculate z if x = 7 and \(\sqrt{y}\) = 3 x = 7 ∴ x2 = 49 \(\sqrt{y}\) = 3 ∴ y2 = 81 = x2 + y2 + z2 = 194 49 + 81 + z2 = 194 130 + z2 = 194 z2 = 194 - 130 = 64 z = \(\sqrt{64}\) = 8

3 + 6 + 12 + .....18thy term 1st term = 3, common ratio \(\frac{6}{3}\) = 2 n = 18, sum of GP is given by Sn = a\(\frac{(r^n - 1)}{r - 1}\) s18 = 3\(\frac{(2^{18} - 1)}{2 - 1}\) = 3(2^18 - 1)

x\(^2\) + x + 1 \(\frac{dy}{dx}\) = 2x + 1 At the turning point, \(\frac{dy}{dx}\) = 0 2x + 1 = 0 x = -\(\frac{1}{2}\)

\(\frac{h_1}{h_2}\) = \(\frac{2}{3}\) h2 = \(\frac{2h_1}{3}\) \(\frac{r_1}{r_2}\) = \(\frac{9}{8}\) r2 = \(\frac{9r_1}{8}\) v1 = \(\pi\)(\(\frac{9r_1}{8}\))2(\(\frac{2h_1}{3}\)) = \(\pi\)r1 2h1 x \(\frac{27}{32}\) v = \(\frac{\pi r_1 2h_1 \times \frac{27}{32}}{\pi r_1 2h_1}\) = \(\frac{27}{32}\) v2 : v1 = 27 : 32

\(\begin{array}{c|c} Scores & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{No. of students} & 1 & 4 & 5 & 6 & x & 2\end{array}\) Average = 3.5 3.5 = \(\frac{(1 \times 1) + (2 \times 4) + (3 \times 5) + (4 \times 6) + 5x + (6 \times 2)}{1 + 4 + 5 + 6 + x + 2}\) \(\frac{3.5}{1}\) = \(\frac{1 + 8 + 15 + 24 + 5x + 12}{18 + x}\) \(\frac{3.5}{1}\) = \(\frac{60 + 5x}{18 + x}\) 60 + 5x = 3.5(18 \(\div\) x) 60 + 5x = = 63 + 1.5x 5x - 1.5x = 63 - 60 1.5x = 3 x = \(\frac{3}{1.5}\) \(\frac{30}{15}\) = 2

\(\frac{x}{4} = \frac{6 + x}{6}\) 6x = 4(6 + x) = 24 + 4x x = 12cm CSA = \(\pi RL - \pi rl\) = \(\pi (6) \sqrt{{18^2} + 6^2} - \pi \times 4 \times \sqrt{{12^2} + 4^2}\) = 6\(\pi \sqrt{360} - 4 \pi \sqrt{160}\) = 36\(\pi \sqrt{10} - 16 \pi \sqrt{10}\) = 20\(\pi \sqrt{10}\)cm2

Error in length = 60cm - 50cm = 10cm Error in breadth = 40cm - 35cm = 5cm (i) % error in length = \(\frac{10}{50} \times 100% = 20%\) % error in breadth = \(\frac{5}{40} \times 100% = 12.5%\) (ii) Area of the actual table = \(50cm \times 40cm = 2000cm^{2}\) Area of the table in error = \(60cm \times 35cm = 2100cm^{2}\) Error in area = \(2100cm^{2} - 2000cm^{2} = 100cm^{2}\) % error in area = \(\frac{100}{2000} \times 100% = 5%\) (b) Product of the error in length and in breadth = \(20 \times 12.5 = 250%\)

(a) Formula : Volume = \(\frac{1}{3} \pi r^{2} h\) \(\pi = 3.142; r = 4cm; h = 6cm\) \(Volume = \frac{1}{3} \times 3.142 \times 4^{2} \times 6 = 100.544 cm^{3}\) (b) Volume of hemisphere = \(\frac{1}{2} \times \frac{4}{3} \pi r^{3}\) = \(\frac{2}{3} \times \frac{22}{7} \times 5^{3}\) = \(\frac{5500}{21} cm^{3} = 261.905 cm^{3}\) Volume of water discharged per second = \(\pi r^{2} h\) = \(\frac{22}{7} \times 0.2m \times 0.2m \times 2m = \frac{1.76}{7}\) = \(0.2514 m^{3} / sec\) \(\therefore \text{The time to fill the tank = } \frac{261.905}{0.2514 \times 60}\) = \(\frac{261.905}{15.084} = 17.363 mins\) \(\approxeq 17.4 mins\)

(a) The number that speak Hausa = n(H) = 25 The number that speak Igbo = n(I) = 16 The number that speak Yoruba = n(Y) = 21 Total no of students = U = 40 \(n(H \cap I) = 8 ; n(H \cap Y) = 11 ; n(Y \cap I) = 6\) (b) \(U = n(H) + n(I) + n(Y) - n(H \cap Y) - n(H \cap I) - n(I \cap Y) + n(H \cap Y \cap I)\) \(40 = 6 + x + 2 + x + 4 + x + 8 - x + 11 - x + 6 - x + x\) \(40 = 37 + x \implies x = 40 - 37 = 3\)

Degree for food items = 360^\circ - (120^\circ + 80^\circ + 43^\circ + 30^\circ + 18^\circ) = 360^\circ - 291^\circ = 69^\circ ∴∴ Amount spent on food items = \(\frac{69}{360}\)\times12,000.00 = N2300.00

This histogram is transferred into this frequency table \(\begin{array}{c|c} Marks & 20 & 40 & 60 & 80 & 100 \\ \hline students & 9 & 7 & 6 & 6 & 2\end{array}\) Students who scored more than 40 = 6 + 6 + 2 = 14 i.e. \(\frac{14}{30}\) x 100% = 46\(\frac{3}{4}\)%

loga3 + loga5 - loga6 log(a3 * a5/a6) = loga64 a2 = 64 a = 8