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16\(^x\) = 0.25 16\(^x\) = \(\frac{25}{100} = \frac{1}{4}\) 4(\(^2)^x\) = 4\(^{-1}\) 2x = -1 x = \(\frac{-1}{2}\)

\(\frac{ 5 + \sqrt{7}}{3 + \sqrt{7}}\) multiply the denominator and the numerator by the conjugate of 3 + \(\sqrt{7}\) → 3 - \(\sqrt{7}\) \(\frac{ 5 + \sqrt{7}}{3 + \sqrt{7}}\) x \(\frac{3 - \sqrt{7}}{3 - \sqrt{7}}\) \(\frac{15 - 5\sqrt{7} + 3\sqrt{7} - 7}{3^2 - (\sqrt{7})^2}\) \(\frac{8 - 2\sqrt{7}}{9 - 7}\) = \(\frac{ 8 - 2\sqrt{7}}{2}\) = 4 - \(\sqrt{7}\)

65 = 30 -x + x + 45 - x 65 - 75 = 10 = x ( number of students offering both subjects) Pr( \(\text{students offering both subjects}\)) = \(\frac{10}{65}\) = \(\frac{2}{13}\)

P = {Primes numbers b/w 0 and 10} = { 2, 3, 5, 7} Q = { Odd numbers b/w 0 and 10} = {1, 3, 5, 7, 9} Let W = { Union of elements in P not in Q and elements in Q not in P} = {1, 2, 9}

For a set with n elements, the total number of subsets is given by: 2\(^n\) where n = the number of members of the set = 5 so, 2\(^5\) = 32 (every set is an improper subset of itself)

144 = 42 - x + 30 - x + 34 + 4x + 3x + 6 + x + 2x 144 = 112 + 8x 8x = 144 - 112 = 32 x = 4. The number of students that registered for physics and mathematics = 4x + 2x = 4 x 4 + 2 x 4 = 24.

M = \(\begin{bmatrix}3 & 4 & 6\\2 & 1 & -1\\-1 & 3 & 5\end{bmatrix}\) |M| = 3\(\begin{bmatrix}1 & -1\\3 & 5\end{bmatrix}\) - 4\(\begin{bmatrix}2 & -1\\-1 & 5\end{bmatrix}\) + 6\(\begin{bmatrix}2 & 1\\-1 & 3\end{bmatrix}\) 3(5 + 3) - 4(10 - 1) + 6(6 + 1) = 24 - 36 + 42 = 30

A = \(\begin{pmatrix}2 & -4 & 3\\5 & 1 & 0\end{pmatrix}\), B = \(\begin{pmatrix}1 & 4 & -2\\-3 & 3 & -1\end{pmatrix}\) 2B = 2 x \(\begin{pmatrix}1 & 4 & -2\\-3 & 3 & -1\end{pmatrix}\) = \(\begin{pmatrix}2 & 8 & -4\\-6 & 6 & -2\end{pmatrix}\) A + 2B = \(\begin{pmatrix}2 & -4 & 3\\5 & 1 & 0\end{pmatrix}\) + \(\begin{pmatrix}2 & 8 & -4\\-6 & 6 & -2\end{pmatrix}\) = \(\begin{pmatrix}4 & 4 & -1\\-1 & 7 & -2\end{pmatrix}\)

Given that A = \(\begin{pmatrix}3 & 2 & 1\\4 & 2 & -1\end{pmatrix}\) and B = \(\begin{pmatrix}1 & 4\\0 & 1\\3 & 2\end{pmatrix}\) A\(^T\) = \(\begin{pmatrix}3 & 4\\2 & 2\\1 & -1\end{pmatrix}\) Find A\(^T\) + B = \(\begin{pmatrix}3 & 4\\2 & 2\\1 & -1\end{pmatrix}\) + \(\begin{pmatrix}1 & 4\\0 & 1\\3 & 2\end{pmatrix}\) = \(\begin{pmatrix}4 & 8\\2 & 3\\4 & 1\end{pmatrix}\)

The ratio of men to women = 3:1 Number of committee members = 20 number of men in the committee = \(\frac{3}{4} \times 20 = 15\) Number of women in the committee = \(\frac{1}{4} \times 20 = 5\) New ratio = 3:2, and let the number of women to be added to make the ratio 3:2 be x 3:2 = 15:5+x \(\frac{3}{2} = \frac{15}{5 + x }\) 15 + 3x = 30 3x = 30 - 15 = 15 x = \(\frac{15}{3}\) = 5 Therefore, the number of women to be added to make the ratio of men to women 3:2 = 5.

Amount = principal + Interest I = \(\frac{ P\times R \times T}{100}\) P = #200,000, R = 5%, T = 3 yrs I = = \(\frac{ 200,000\times 5 \times 3}{100}\) = 2000 x 15 = #30,000 A = 200,000 + 30,000 = #230,000.

The boy bought 5 oranges for #24.00. Therefore, the cost price per orange is: c.p = \(\frac{24}{5}\) = #4.800 The boy sold 4 oranges for #30.00. Therefore, the selling price per orange is: s.p = \(\frac{30}{4}\) = # 7.50 Profit per orange is calculated as: Profit per orange = SP per orange − CP per orange = =7.50 − 4.80 = 2.70 If the boy sold 4 oranges, the total profit from selling 4 oranges is: Total Profit = 4 \times 2.70 = #10.80

y ∝ \(\frac{1}{x}\) y = \(\frac{k}{x}\) y = 4, x = \(\frac{1}{2}\), find x when y = 10 4 = \(\frac{k}{\frac{1}{2}}\) k = 2 Therefore, x = \(\frac{2}{10} = \frac{1}{5}\)

U = K\(\sqrt{V}\) 24 = K\(\sqrt{9}\) K = 8 U = 8\(\sqrt{V}\) 16 = 8\(\sqrt{V}\) 2 = \(\sqrt{V}\) V = 4

B = \(\frac{K}{C^{\frac{1}{3}}}\) 2 = \(\frac{K}{27^{\frac{1}{3}}}\) K = 2 x 3 = 6