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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.

The statement "P is partly constant and varies partly as Q" means P consists of a constant term plus a term directly proportional to Q. Thus, the relationship is: P = a + bQ, where a is the constant part and b is the constant of proportionality. Given: P = 32 when Q = 16 → a + 16b = 32 P = 20 when Q = 12 → a + 12b = 20 Subtract the second equation from the first: (a + 16b) - (a + 12b) = 32 - 20 4b = 12 b = 3 Substitute b = 3 into the second equation: a + 12(3) = 20 a + 36 = 20 a = -16 Thus, P = -16 + 3Q When Q = 28: P = -16 + 3(28) = -16 + 84 = 68

Given: 2x + 3 > 5x + 8 2x + 3 - 5x - 8 > 0 ⇒ -3x - 5 > 0 ⇒ -3x > 5 ⇒ x < \(\frac{-5}{3}\) = -1 \(\frac{2}{3}\) (inequality reverses on dividing by) x < -1\(\frac{2}{3}\)

The sequence 3, 5, 7, 9, ... is an arithmetic progression with: - First term \(a_1 = 3\) - Common difference \(d = 2\) General \(k^{th}\) term: \(a_k = 3 + (k-1) \cdot 2 = 2k + 1\) For the \((n-2)^{th}\) term, substitute \(k = n-2\): \(a_{n-2} = 2(n-2) + 1 = 2n - 4 + 1 = 2n - 3\) The expression is \(2n - 3\)

To compute 1 * 3 using the given binary operation * where a * b = ab + a + b: Substitute a = 1 and b = 3 into the formula: 1 * 3 = (1)(3) + 1 + 3 = 3 + 1 + 3 = 7. Thus, 1 * 3 = 7.

To calculate the interior angle of a regular polygon, we use the formula: \(\text{Interior Angle} = \frac{(n - 2) \times 180^\circ}{n}\) where \( n \) is the number of sides. For a 5-sided polygon (pentagon): \(\text{Interior Angle} = \frac{(5 - 2) \times 180^\circ}{5} = \frac{3 \times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ\) Thus, the interior angle of a regular pentagon is \( 108^\circ \).

The triangle XYZ is a right-angled triangle at Y, with opposite side to \theta (YZ) = 11 cm and hypotenuse (XZ) = 13 cm. The length of XY = \(\sqrt{(13^2 - 11^2)}\) = \(\sqrt{(169 - 121)}\) = \(\sqrt{48}\) cm (or 4\(\sqrt{3}\) cm in simplified form). Cot \theta = \(\frac{\text{adjacent}}{\text{opposite}}\) = \(\frac{\text{XY}}{\text{YZ}}\) = \(\frac{\sqrt{48}}{11}\)

In classical constructions with compass and straightedge, 60^\circ (equilateral), 90^\circ (square), and 135^\circ (90^\circ + 45^\circ) are constructible. 145^\circ is not constructible in general (not a multiple resulting from bisections of constructible angles).