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Formula for an interior angle of a polygon is (2n - 4) x 90 (2n - 4) x 90 = 144n 180n - 360 = 144n 180n - 144n = 360 36n = 360 n = \(\frac{360}{36}\) = 10

\(\frac{3}{2x - 1}\) + \(\frac{2 - x}{x - 2}\) = \(\frac{3}{2x - 1}\) - \(\frac{x - 2}{x - 2}\) = \(\frac{3}{2x - 1}\) - 1 = \(\frac{3 - (2x - 1)}{2x - 1}\) = \(\frac{3 - 2x + 1}{2x - 1}\) = \(\frac{4 - 2x}{2x - 1}\)

5 + 3x - 2x2 = 0, (x + 1)(5 - 2x) (expand to check) (x + 1)(5 - 2x) = 0 when x = 1 = 0, x = -1 when 5 - 2x = 0, x = \(\frac{5}{2}\) - 1

3 kg of garri at N13.00 per kg = N39.00 half a cow = N160.00 24 oranges at 10k each = N2.40 balance = N20.60 Adding all together = N222.00

3x + 5y = 4, 4x + 3y = 5 3x + 5y = 4 x 4 4x + 3y = 5 x 3 12x + 20y = 16.....(i) 12x + 9y = 15.......(ii) subtract eqn.(ii) from eqn.(i) 11y = 1 y = \(\frac{1}{11}\) 12x + 20 x \(\frac{1}{11}\) = 16 12x = \(\frac{156}{11}\) x = \(\frac{13}{11}\) = \(\frac{13}{11}, \frac{1}{11}\)

Given: length of the arc AOB = 22cm ; L = 6cm Curved surface area of cone = \(\pi\)rl Length of an arc = \(\frac{\theta}{360}\) x 2 \(\pi\)L = 22cm but length of an arc = circumference of the cone = 2\(\pi\) where r is the radius of the cone circle 2\(\pi\)r = 22, r = \(\frac{22}{2\pi}\)r = 11 x \(\frac{7}{22}\) = \(\frac{7}{2}\) curved surface area = \(\pi\)rl = \(\frac{22}{7}\) x \(\frac{7}{2}\) x 6 = 66 sq.cm

Vol. of the 1st section is side x height vol. = 3 x 8 = 24cm3 vol. of the second section is 3 x 1 = 3 = 24cm3, 3cm3

f\(^{\frac{1}{2}}\)g\(^2\)h\(^{\frac{1}{3}}\) \(\div\) f\(^{\frac{5}{2}}\)g\(^o\)h\(^{\frac{7}{3}}\) = f\(^{\frac{1}{2} - \frac{5}{2}}\) g\(^{2 - 0}\) h\(^{\frac{1}{3} - \frac{7}{3}}\) = f\(^{-2}\) g\(^2\) h\(^{-2}\) = \(\frac{g^2}{f^2h^2}\) = (\(\frac{g}{fh}\))\(^2\)

\(\bar{x}\) = \(\frac{fx}{N}\) = \(\frac{6705}{50}\) = 1341 = N134.10

y = 5x, when x = 0 y = 0, when x = 1 y = 5 intercept on y axis is 0

\(\frac{x}{2}\) = sin60o = cos30o x = 2 sino = 2 x \(\frac{\sqrt{3}}{2}\) = \(\sqrt{3}\) length of the diagonal = 2 x \(\sqrt{2}\) = 2\(\sqrt{3}\)

y \(\alpha\) \(\frac{1}{x^2}\) y = \(\frac{k}{x^2}\) k = x\(^2\)y = (2)\(^2\) x 10 = 40 y = \(\frac{40}{x^2}\) = \(\frac{40}{(10)^2}\) = \(\frac{40}{100}\) = 0.4

The distinct sectors i the same circle substend 100o and 30o respectively at the centre of the circle. Their corresponding arcs are in ratio 100:30 = 10:3

1st 30 pupils at average age of 15.3 yrs. give total age of 15.3 x 30 = 459yrs 2nd group of 30 pupils at average age of 15.2 yrs give total age of 15.2 x 30 = 456yrs Difference in age = 459 - 456 = 3 yrs

From \(\bigtriangleup\)QPR, o - (25o + 90o) 180o - 115o = 65o From \(\bigtriangleup\)PSQ, Sin 65o = \(\frac{QPR}{hyp}\) = \(\frac{PS}{5}\) PS = 5 sin 65o From \(\bigtriangleup\)PSR, tan = \(\frac{OPP}{adj}\) = \(\frac{PS}{QS}\) but PS = 5 sin 65o QS tan 25o = PS QS tan 25o = 5 sin 65o QS = \(\frac{5 sin 65^o}{tan 25^o}\) = 5 tan 25o cos 65o