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Numbers divisible by 4 between 1 and 300 include 4, 8, 12, 16, 20 e.t.c.To get the number of figures divisible by 4, We solve by method of A.P Let x represent numbers divisible by 4, nth term = a + (n - 1)d a = 4, d = 4 Last term = 4 + (n - 1)4 296 = 4 + 4n - 4 = \(\frac{296}{4}\) = 74 rn(Note: 296 is the last Number divisible by 4 between 1 and 300) Prob. of x = \(\frac{74}{298}\) = \(\frac{37}{149}\)

\(\frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})}\) \(\log_{3} 27 = \log_{3} 3^{3} = 3\log_{3} 3 = 3\) \(\log_{\frac{1}{4}} 64 = \log_{\frac{1}{4}} (\frac{1}{4})^{-3} = -3\) \(\log_{3} (\frac{1}{81}) = \log_{3} 3^{-4} = -4\) \(\therefore \frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})} = \frac{3 - (-3)}{-4}\) = \(\frac{6}{-4} = \frac{-3}{2}\)

P(x, y), P(0, 3) If x increases by 1 unit and y by 4 units, then ratio of x : y = 1 : 4 \(\frac{x}{1}\) = \(\frac{y}{4}\) y = 4x Hence the sign of the graph is y + 3 = 4x

Total cost of 1035 oranges at N145 each = 1035 x 145 = 20025 Total selling price at N245 each = (103)5 x 245 = 30325 Hence his gain = 30325 - 20025 = 10305

\(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\) = \(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\) x \(\frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}}\) = \(\frac{(5 \times 7) + (5 \sqrt{7} \times 5) - (7 \times \sqrt{5} \times 7) (-7 \times 5)}{(\sqrt{7})^2}\) = \(\frac{5 \sqrt{35} - 7\sqrt{35}}{2}\) = \(\frac{-2\sqrt{35}}{2}\) = - \(\sqrt{35}\)

\(\frac{3^n - 3^{n - 1}}{3^3 \times 3^n - 27 \times 3^{n - 1}}\) = \(\frac{3^n - 3^{n - 1}}{3^3(3^n - 3^{n - 1})}\) = \(\frac{3^n - 3^{n - 1}}{27(3^n - 3^{n - 1})}\) = \(\frac{1}{27}\)

P \(\alpha\) \(\frac{q^2}{r}\) P = \(\frac{kq^2}{r}\) k = \(\frac{pr}{q^2}\) = \(\frac{36 x 4}{(3)^2}\) p = \(\frac{16q^2}{r}\) = \(\frac{16 \times 25}{2}\) = 200

6x2 - 14x - 12 = 6x2 - 18x + 4x - 12 (3x + 2)(2x - 6) = 3x(2x - 6) + 2(2x - 6) = (3x + 2) 2(x - 3)

When y = 5, y = x2 - 12x + 40, becomes x2 - 12x + 40 = 5 x2 - 12x + 40 - 5 = 0 x2 + 12x + 35 = 0 x2 - 7x - 5x + 35 = 0 x(x - 7) - 5(x - 7) = 0 = (x - 5)(x - 7) x = 5 or 7

If the graph of y = x3 - 3x + 1 is plotted,the graph crosses the x-axis in the ranges -2 < x < -1 and 0 < x < 1

Total distance covered by Musa in 2 hrs = x + 10 + 5x = 6x + 10 Ngozi = 118 km If they are equal, 6x + 10 = 118 but 6x + 10 < 118 6x < 108 = x < 18 0 < x < 18 = 0 \(\leq\) x < 18

2x + 3y = 1 x 2.......(i) x - 2y = 11 x 3.......(ii) 4x + 6y = 2........(iii) 3x - 6Y = 33........(4) 7x = 35 x = 5 Subt. for x 10 + 3y = 1 3y = -9 y = -3 x + y = 5 + -3 5 - 3 = 2

Tunde and Shola can do the work in 18 days. Both will do the work in \(\frac{1}{18}\) days. But Tunde can do the whole work in x days; Hence he does \(\frac{1}{x}\) of the work in 1 day. Shola does the work in (x + 15) days; hence, he does \(\frac{1}{x + 15}\) of the work in 1 day. \(\frac{1}{x} + \frac{1}{x + 15} = \frac{1}{18}\) \(\frac{2x + 15}{x^{2} + 15x} = \frac{1}{18}\) \(x^{2} + 15x = 36x + 270\) \(x^{2} + 15x - 36x - 270 = 0\) \(x^{2} - 21x - 270 = 0\)

f(x) = 2(x - 3)\(^2\) + 3(x - 3) + 4 = (2 + 3) (x - 3) + 4 = 5(x - 3) + 4 = 5x - 15 + 4 = 5x - 11 f(3) = 5 x 3 - 11 = 4 g(f(3)) = g(4) = \(\sqrt{5 + 4}\) = \(\sqrt{9}\) = 3 g(4) = 3 f(g(4)) = f(3) = 4 g[f(3)] and f[g(4)] = 3 and 4 respectively.

4a2 - 9b2, 8a3 + 27b3, (4a + 6b)2 = (2a + 3b)(2a - 3b) 8a3 + 27b3 = (2a)3 + (3b)3 = (2a + 3b)(4a - 6ab = 9a2) (4a + 6b)2 = 2(2a + 3b)2