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0.00001 = 10\(^{-5}\), so, log\(_{10}\)(10\(^{-5}\)) = - 5. Then -(-5) = 5.

log\(_{10}\)x = 2.3675 ⇒ x = 10\(^{2.3675}\) ⇒ x ≈ 0.02331 log\(_{10}\)y = 2.9738 ⇒ y = 10\(^{2.9738}\) ⇒ y ≈ 0.09414 ∴ x + y = 0.02331 + 0.09414 = 0.11745 = 0.118 (3 s.f.)

\(\sqrt{75} - \sqrt{12} + \sqrt{27} = y\sqrt{3}\) Simplify each term: - \(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\) - \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\) - \(\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}\) Substitute: \(5\sqrt{3} - 2\sqrt{3} + 3\sqrt{3} = y\sqrt{3}\) Combine like terms: \((5 - 2 + 3)\sqrt{3} = y\sqrt{3}\) \(6\sqrt{3} = y\sqrt{3}\) Thus, y = 6

\(\sqrt[4]{0.16} = (0.16)^{\frac{1}{4}} = (16 \times 10^{-2})^{\frac{1}{4}} = 2 \times 10^{-\frac{1}{2}}\)

Let the salary be \( S \). Amount spent on shirts: \(\text{Shirts} = \frac{1}{5} S\) Remainder after shirts: \(\text{Remainder} = S - \frac{1}{5} S = \frac{4}{5} S\) Amount spent on transport: \(\text{Transport} = \frac{1}{3} \left( \frac{4}{5} S \right) = \frac{4}{15} S\) Remainder after transport: \(\text{Remainder} = \frac{4}{5} S - \frac{4}{15} S = \frac{12}{15} S - \frac{4}{15} S = \frac{8}{15} S\) Thus, the fraction left is: \(\frac{8}{15}\)

Two or more sets are said to be equivalent when they contain equal elements, denoted by ≡ Original: {1,2,3,4,5}. ≡ {4,3,1,5,2}

The Venn diagram has three overlapping circles labeled P, Q, and R. The shaded region has cross-hatching and covers only the part that belongs to Q but not to P or R (the exclusive part of Q outside the overlaps with P and R). In set notation, the shaded region is \( Q \cap (P \cup R)^c \)

Let the matrix be \(A = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}\) The determinant of \(A\) is calculated as follows: \(\text{Determinant} = 1 \cdot 5 - 2 \cdot 3 = 5 - 6 = -1\) To find the inverse of \(A\), we first determine the adjoint matrix: \(\text{Adjoint} = \begin{bmatrix} 5 & -2 \\ -3 & 1 \end{bmatrix}\) The inverse is given by \(A^{-1} = \frac{1}{\text{Determinant}} \cdot \text{Adjoint} = \frac{1}{-1} \begin{bmatrix} 5 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -5 & 2 \\ 3 & -1 \end{bmatrix}\) Next, we calculate the sum of the elements in the inverse matrix: \(\text{Sum} = -5 + 2 + 3 + (-1) = -1\)

The 2\times2 identity matrix is \(I_2 = \begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}\) Determinant = (1)(1) - (0)(0) = 1.

Given the matrix \(A = \begin{pmatrix} -2 & 3 & 1 \\ p & 2 & 1 \\ 1 & 4 & 2 \end{pmatrix}\) the determinant is set to \(-5\): \(\text{det}(A) = -2 \begin{vmatrix} 2 & 1 \\ 4 & 2 \end{vmatrix} - 3 \begin{vmatrix} p & 1 \\ 1 & 2 \end{vmatrix} + 1 \begin{vmatrix} p & 2 \\ 1 & 4 \end{vmatrix}\) Calculating the \(2 \times 2\) determinants: 1. \( \begin{vmatrix} 2 & 1 \\ 4 & 2 \end{vmatrix} = 0 \) 2. \( \begin{vmatrix} p & 1 \\ 1 & 2 \end{vmatrix} = 2p - 1 \) 3. \( \begin{vmatrix} p & 2 \\ 1 & 4 \end{vmatrix} = 4p - 2 \) So, \(\text{det}(A) = -3(2p - 1) + (4p - 2) = -6p + 3 + 4p - 2 = -2p + 1\) Setting this equal to \(-5\): \(-2p + 1 = -5\) Solving for \( p \): \(-2p = -6 \implies p = 3\) Thus, the value of \( p \) is 3

Given \(P = \begin{pmatrix} 1 & 3 \\ 2 & -5 \end{pmatrix}, \quad Q = \begin{pmatrix} 3 & -7 \\ 1 & 2 \end{pmatrix}\) we need to find \(P + 2Q\). First, calculate \(2Q\): \(2Q = 2 \begin{pmatrix} 3 & -7 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 2 \cdot 3 & 2 \cdot -7 \\ 2 \cdot 1 & 2 \cdot 2 \end{pmatrix} = \begin{pmatrix} 6 & -14 \\ 2 & 4 \end{pmatrix}\) Now add \(P\) and \(2Q\): \(P + 2Q = \begin{pmatrix} 1 & 3 \\ 2 & -5 \end{pmatrix} + \begin{pmatrix} 6 & -14 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 1 + 6 & 3 + (-14) \\ 2 + 2 & -5 + 4 \end{pmatrix}\) Calculating each entry, we find: \(= \begin{pmatrix} 7 & -11 \\ 4 & -1 \end{pmatrix}\) Thus, the result of \(P + 2Q\) is \(\begin{pmatrix} 7 & -11 \\ 4 & -1 \end{pmatrix}\)

Total cost = 270,000 + 70,000 = 340,000. Profit = 490,000 - 340,000 = 150,000. Percentage profit = (\(\frac{150,000}{340,000}\)) \times 100 = (\(\frac{15}{34}\)) \times 100 ≈ 44.12%. The closest option is 45%

Using \(A = P(1 + rt)\) Substituting known values: \(600,000 = y \left(1 + 0.05 \times 4\right)\) \(600,000 = y \times 1.20\) Solving for \( y \): \(y = \frac{600,000}{1.20} = 500,000\) Thus, the principal amount is # 500,000

Form: y = a + b x. 7 = a + 3b. - - - - - - - 1 11 = a + 5b. - - - - - - 2 Solve both equations using the elimination method Subtract: 4 = 2b ⇒ b = 2. put b = 2 into eqn 1 a + 6 = 7 ⇒ a = 1. Constants a = 1 (constant part), b = 2 (variation constant).

A ∝ b\(^2\) so A = k b\(^2\). Substitute: 4 = k · 1\(^2\), k = 4. When b = 2, A = 4 · 4 = 16.