1. **Vector Magnitude Problem:** Given vectors \(\vec{OP} = 5\mathbf{i} - 3\mathbf{j}\) and \(\vec{OQ} = 3\mathbf{i} + 5\mathbf{j}\), find the magnitude of \(\vec{PQ}\). - \(\vec{PQ} = \vec{OQ} - \vec{OP} = (3 - 5)\mathbf{i} + (5 - (-3))\mathbf{j} = -2\mathbf{i} + 8\mathbf{j}\). - Magnitude \(|\vec{PQ}| = \sqrt{(-2)^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68} = 2\sqrt{17}\). 2. **Make \(n\) the subject:** Given \(\frac{r}{p} = \frac{m}{\sqrt{n - 1}}\), solve for \(n\). - Cross multiply: \(r \sqrt{n - 1} = pm\). - Isolate \(\sqrt{n - 1} = \frac{pm}{r}\). - Square both sides: \(n - 1 = \left(\frac{pm}{r}\right)^2\). - Solve for \(n = 1 + \frac{p^2 m^2}{r^2}\). 3. **Factorisation to solve quadratic:** Solve \(5x^2 + 45x + 90 = 0\). - Divide entire equation by 5: \(x^2 + 9x + 18 = 0\). - Factorise: \((x + 6)(x + 3) = 0\). - Solutions: \(x = -6\) or \(x = -3\). 4. **Variation problem:** Given \(P\) varies directly as \(Q^2\) and inversely as \(\sqrt{R}\). (i) Find constant \(k\) using \(P = k \frac{Q^2}{\sqrt{R}}\) with \(P=20, Q=5, R=9\). - \(20 = k \frac{5^2}{\sqrt{9}} = k \frac{25}{3}\). - \(k = \frac{20 \times 3}{25} = \frac{60}{25} = 2.4\). - Find \(P\) when \(Q=7, R=9\): - \(P = 2.4 \times \frac{7^2}{3} = 2.4 \times \frac{49}{3} = 2.4 \times 16.3333 = 39.2\). (ii) If \(Q\) increases by 20%, new \(Q = 1.2Q\); \(R\) decreases by 36%, new \(R = 0.64R\). - New \(P' = k \frac{(1.2Q)^2}{\sqrt{0.64R}} = k \frac{1.44 Q^2}{0.8 \sqrt{R}} = \frac{1.44}{0.8} k \frac{Q^2}{\sqrt{R}} = 1.8 P\). - Percentage increase in \(P = (1.8 - 1) \times 100\% = 80\%\). 5. **Compound interest:** Principal \(P=200000\), rate \(r=0.05\), time \(t=4\) years. - Amount \(A = P(1 + r)^t = 200000 (1.05)^4 = 200000 \times 1.21550625 = 243101.25\). - Interest accrued = \(A - P = 243101.25 - 200000 = 43101.25\). 6. **Convert recurring decimal:** Convert \(3.4\overline{7}\) to fraction. - Let \(x = 3.4\overline{7}\). - Multiply by 10: \(10x = 34.7\overline{7}\). - Multiply by 100: \(100x = 347.7\overline{7}\). - Subtract: \(100x - 10x = 347.7\overline{7} - 34.7\overline{7} = 313\). - \(90x = 313 \Rightarrow x = \frac{313}{90} = 3 \frac{43}{90}\). 7. **Arithmetic progression problem:** Sum of three numbers \(a, a+d, a+2d = 33\). - Sum: \(3a + 3d = 33 \Rightarrow a + d = 11\). - Sum of squares: \(a^2 + (a+d)^2 + (a+2d)^2 = 365\). - Expand: \(a^2 + (a+d)^2 + (a+2d)^2 = a^2 + (a^2 + 2ad + d^2) + (a^2 + 4ad + 4d^2) = 3a^2 + 6ad + 5d^2 = 365\). - Using \(a + d = 11 \Rightarrow a = 11 - d\). - Substitute: \(3(11 - d)^2 + 6(11 - d)d + 5d^2 = 365\). - Expand: \(3(121 - 22d + d^2) + 66d - 6d^2 + 5d^2 = 365\). - Simplify: \(363 - 66d + 3d^2 + 66d - d^2 = 365\). - \(363 + 2d^2 = 365 \Rightarrow 2d^2 = 2 \Rightarrow d^2 = 1 \Rightarrow d = \pm 1\). - If \(d=1\), then \(a = 11 - 1 = 10\), numbers: 10, 11, 12. - If \(d=-1\), then \(a = 11 + 1 = 12\), numbers: 12, 11, 10. 8. **Solve equation:** \(2(e^x)^2 - 3e^x + 3 = 0\). - Let \(y = e^x\), equation: \(2y^2 - 3y + 3 = 0\). - Discriminant \(\Delta = (-3)^2 - 4 \times 2 \times 3 = 9 - 24 = -15 < 0\), no real roots. - Hence, no real solution for \(x\). 9. **Simultaneous equations by substitution:** - Equations: \(3x + 5y = 21\), \(2x - 3y = -5\). - From second: \(2x = -5 + 3y \Rightarrow x = \frac{-5 + 3y}{2}\). - Substitute into first: \(3 \times \frac{-5 + 3y}{2} + 5y = 21\). - Multiply both sides by 2: \(3(-5 + 3y) + 10y = 42\). - \(-15 + 9y + 10y = 42 \Rightarrow 19y = 57 \Rightarrow y = 3\). - Substitute back: \(x = \frac{-5 + 3 \times 3}{2} = \frac{-5 + 9}{2} = 2\). 10. **Matrix transpose property:** Given \(A = \begin{bmatrix}5 & 4 \\ 6 & 3\end{bmatrix}\), \(B = \begin{bmatrix}7 & 8 \\ 2 & 4\end{bmatrix}\). - Compute \(AB = \begin{bmatrix}5 \times 7 + 4 \times 2 & 5 \times 8 + 4 \times 4 \\ 6 \times 7 + 3 \times 2 & 6 \times 8 + 3 \times 4\end{bmatrix} = \begin{bmatrix}43 & 52 \\ 48 & 60\end{bmatrix}\). - Then \((AB)^T = \begin{bmatrix}43 & 48 \\ 52 & 60\end{bmatrix}\). - Compute \(B^T = \begin{bmatrix}7 & 2 \\ 8 & 4\end{bmatrix}\), \(A^T = \begin{bmatrix}5 & 6 \\ 4 & 3\end{bmatrix}\). - Compute \(B^T A^T = \begin{bmatrix}7 & 2 \\ 8 & 4\end{bmatrix} \begin{bmatrix}5 & 6 \\ 4 & 3\end{bmatrix} = \begin{bmatrix}7 \times 5 + 2 \times 4 & 7 \times 6 + 2 \times 3 \\ 8 \times 5 + 4 \times 4 & 8 \times 6 + 4 \times 3\end{bmatrix} = \begin{bmatrix}43 & 48 \\ 52 & 60\end{bmatrix}\). - Since \((AB)^T = B^T A^T\), property is proven.