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P(Kodjo passing) = \(\frac{3}{4}\); P(Adoga passing) = \(\frac{3}{5}\) P(Kodjo failing) = \(\frac{1}{4}\); P(Adoga failing) = \(\frac{2}{5}\) P(both fail) = \(\frac{1}{4} \times \frac{2}{5}\) = \(\frac{1}{10}\)

< Q = < R (OQ = OR = radii) < QOR = 180^\circ - 2(40^\circ) = 100^\circ < QSR = < RPQ = \(\frac{1}{2}\) < QOR = \(\frac{100}{2} = 50^\circ\)

From the figure above In △BOC, 2x + 30º = 180º Therefore, x = \(\frac{180º - 30º}{2}\) = 75º Each interior angle = 2 times x = 2 x 75 = 150º OR To calculate the interior angle of a regular convex polygon where each side subtends an angle of \(30^\circ\) at the center, we follow these steps: Determine the number of sides (n) The angle subtended at the center by each side is given as \(30^\circ\). The total angle around a point is \(360^\circ\). \(n = \frac{360^\circ}{30^\circ}\) = 12 So, the polygon has 12 sides. Calculate the interior angle. The formula for the interior angle A of a regular polygon is: \(A = \frac{(n - 2) \times 180^\circ}{n}\) Substituting n = 12: \(A = \frac{(12 - 2) \times 180^\circ}{12} = \frac{10 \times 180^\circ}{12} = \frac{1800^\circ}{12} = 150^\circ\) Each interior angle of the regular convex polygon is \(150^\circ\).

\(27^{(1+x)}\) = 9 \(3^{3(1+x)} = 3^2\) 3(1+x) = 2 3 + 3x = 2 3x = -1 x = \(\frac{-1}{3}\)

\(2x^2 + 3x - 14\) \(2x^2 + 7x - 4x - 14\) \(x(2x + 7) - 2(2x + 7)\) = \((x - 2)(2x + 7)\) The other factor = (x - 2).

Let the three items be M, Y and P. n{M ∩ Y} only = 4-3 = 1 n{M ∩ P) only = 5-3 = 2 n{ Y ∩ P} only = 2 n{M} only = 12-(1+3+2) = 6 n{Y} only = 10-(1+2+3) = 4 n{P} only = 14-(2+3+2) = 7 n{M∩P∩Y} = 3 Number of women in the group = 6+4+7+(1+2+2+3) as above =25 women.

\(log_810\) = X = \(log_8{2 x 5}\) \(log_82\) + \(log_85\) = X Base 8 can be written as \(2^3\) \(log_82 = y\) therefore \(2 = 8^y\) \(y = \frac{1}{3}\) \(\frac{1}{3} = log_82\) taking \(\frac{1}{3}\) to the other side of the original equation \(log_85 = X-\frac{1}{3}\) explanation courtesy of Oluteyu and Ifechuks OR Given \(\log_8 10 = X\), we can express \(\log_8 5\) as follows: \(\log_8 5 = \log_8 \left( \frac{10}{2} \right) = \log_8 10 - \log_8 2\) Using the change of base formula, we find: \(\log_8 2 = \frac{\log 2}{\log 8} = \frac{1}{3}\) Thus, substituting back, we get: \(\log_8 5 = X - \log_8 2 = X - \frac{1}{3}\)

Start your solution by cross-multiplying, \(\frac{\sqrt{2}}{x+\sqrt{2}}=\frac{1}{x-\sqrt{2}}\) [x - \(\sqrt{2}\)]\(\sqrt{2}\) = x + \(\sqrt{2}\) where \(\sqrt{2}\) \times\(\sqrt{2}\) = 2 x\(\sqrt{2} - \sqrt{2} x \sqrt{2}\) = x + \(\sqrt{2}\) then collect like terms x\(\sqrt{2}\) - x = \(\sqrt{2}\) + 2 and factorize accordingly to get the unknown. x(\(\sqrt{2}\) - 1) = \(\sqrt{2}\) + 2 x = \(\frac{\sqrt{2}) + 2}{\sqrt{2} - 1}\) rationalize x = \(\frac{\sqrt{2} + 2}{\sqrt{2} - 1}\) * \(\frac{\sqrt{2} + 1}{\sqrt{2} + 1}\) x = \(\frac{\sqrt{4} + \sqrt{2} + 2\sqrt{2} + 2}{\sqrt{4} + \sqrt{2} - \sqrt{2} - 1}\) x = \(\frac{2 + 3\sqrt{2} + 2}{2 - 1}\) x = \(\frac{3\sqrt{2} + 4}{1}\) x = 3\(\sqrt{2}\) + 4

S = \(\begin{pmatrix} -1 & 1 \\ 1 & -2 \end{pmatrix}\) and ST = I (Identity matrix). This means T is the inverse of S Find the determinant of S \( \det(S)\) = (-1)(-2) - (1)(1) = 2 - 1 = 1 T = \(\frac{1}{\det(S)} \begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix}\) Since det = 1, T = \(\begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix}\) T = \(\begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix}\)

= \(\frac{(a^2)^{3/4}\times(b^{-3})^{3/4}\times c^{3/4}}{a^{-1} b^4 c^5}\) \(= a^p\times b^q\times c^r\) = \(\frac{a^{3/2}\times b^{-9/4} \times c^{3/4}}{a^{-1}b^4c^5}\) = \(a^p b^q c^r\) = \(a^{{3/4}+1} b^{{-9/4}-4} c^{{3/5}-5}\) = \(a^pb^q c^r\) = \(a^{5/2} b^{-25/4} c^{-17/4}\) = \(a^pb^qc^r\) therefore, p=5/2, q=-25/4 and r = -17/4 2a + q = \(\frac{5}{2} - 2\times\frac{-25}{4}\) = -20/2 = -10

Cost price (cp) = (100/5) x N1.20 =N24.00 Selling price (sp) = 100-20 = 80 (80/4) x N1.50 = N30.00 Gain = SP-CP = N30.00-N24.00 = N6.00 Gain% = (gain/CP) x 100 = 25%

Hint: Two methods can be used; 1. Direct division (if you know division in number bases) 2. Convert both sides to base 10, divide and convert your answer back to base 6. Your answer should be 356 2434\(_6\) to base 10 2 \(\times\) 6\(^3\) + 4 \(\times\) 6\(^2\) + 3 \(\times\) 6\(^1\) + 4 \(\times\) 6\(^0\) 2 \(\times\) 216 + 4 \(\times\) 36 + 3 \(\times\) 6 + 4 \(\times\) 1 432 + 144 + 18 + 4 = 598 And: 42\(_6\) to base 10 4 \(\times\) 6\(^1\) + 2 \(\times\) 6\(^0\) 4 \(\times\) 6 + 2 \(\times\) 1 24 + 2 = 26 Then divide 598 by 26 = 23 Then convert 23\(_{10}\) to base 6 6 23 6 3 r 5 6 0 r 3 23\(_{10}\) = 35\(_6\)

(2 x 9 \(^0\) ) (Y x 9 \(^1\)+ 3 x 9\(^0\)) = (3 x 5 \(^0\))(Y x 5\(^1\) + 3 x 5\(^0\)). (2 * 1)(Y *9 + 3) = (3 * 1)(Y * 5 + 3) 18y + 6 = 15y + 9 18y - 15y = 9 - 6 3y = 3 y = 1

Multiply appropriately to remove decimals on both numerator and denominators. Such that you have \(\sqrt{\frac{(23000 \times 750)}{(345 \times 125)}}\) Dividing, => \(\sqrt{400} = 20\).

\(m*n = \frac{m}{n}-\frac{n}{m} = \frac{-3}{4} - \frac{4}{-3} = \frac{-3}{4} + \frac{4}{3} = \frac{(-9+16)}{12} = \frac{7}{12}\)