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Q + 2P = Q + 2(Q + P - Q) = 2(Q + P) - Q Q + P = \(\begin{pmatrix} 7 & -2 \\ 6 & 8 \end{pmatrix}\) 2(Q + P) = \(\begin{pmatrix} 14 & -4 \\ 12 & 16 \end{pmatrix}\) Q + 2P = 2(Q + P) - Q = \(\begin{pmatrix} 14-6 & -4-0 \\ 12-4 & 16-5 \end{pmatrix} = \begin{pmatrix} 8 & -4 \\ 8 & 11 \end{pmatrix}\) Determinant: |Q + 2P| = 8 \(\times\) 11 - (-4)\(\times\) 8 = 88 + 32 = 120.

Total number of mangoes = m Amina ate 3 mangoes \(\implies\) Remainder = m - 3 Shared equally with Uche \(\implies \frac{m - 3}{2}\) \(\frac{m - 3}{2} \geq 10\)

the line graph shows that \(x\le -2\) and \(x\ge1\)

x = -m \(\implies\) x + m = 0; x = 2n \(\implies\) x - 2n = 0. \(\implies (x + m)(x - 2n) = 0\) \(x^2 + mx - 2nx - 2mn = 0\) \(x^2 + x(m - 2n) - 2mn = 0\)

2x + 3 < 5x 3 < 3x \(\frac{3}{3}\) < x 1 < x x > 1

In \(\bigtriangleup\)WOP, b2 = r2 - (4.502...........(i) In \(\bigtriangleup\)WOT, b\(\bigtriangleup\) + (4.5 + x)2 = a2...........(ii) In \(\bigtriangleup\)ZOT, a2 = 62 + r2 ..........(iii) subst. equation (iii) and (i) in eqn (ii) r2 - (4.5)2 + (4.5 + x)2 = 36 = 20.25 = (4.5 + x)2 = \(\sqrt{56.25}\) = (4.5 + x) = 56.25 = 4.5 + x = 7.5 x = 3.0m x =

(a) \(3y - 2x = 21 ... (1)\) \(4y + 5x = 5 .... (2)\) Multiply (1) by 4 and (2) by 3 so we have, \(12y - 8x = 84 ... (3)\) \(12y + 15x = 15 ... (4)\) (3) - (4) : \(-8x - 15x = 69 \implies -23x = 69\) \(x = \frac{69}{-23} = -3\) Put x = -3 in (1), \(3y - 2(-3) = 21\) \(3y + 6 = 21 \implies 3y = 21 - 6 = 15\) \(y = \frac{15}{3} = 5\) \(x, y = -3, 5\). (b)(i) Probability of losing Probability of winning = \(\frac{1}{6}\) \(\therefore\) Probability of losing = \(1 - \frac{1}{6} = \frac{5}{6}\). (ii) Losing $20.00 after two picks = picking the card numbered 1 twice. = \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)

x 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ 210^\circ y 9 10.3 8.8 5 -0.2 -5.3 -9 -10.3 (b) (c)(i) From the graph, \(9\cos x + 5\sin x = 0 \implies x = 119^\circ\) (d) Maximum value of y = 10.3.

Given: \(\frac{^6C_r}{^6P_r} = \frac{1}{6}\) Recall the formulas: \(^nC_r = \frac{n!}{r!(n-r)!}, \quad ^nP_r = \frac{n!}{r!}\) Substitute \( n = 6 \): \(\frac{^6C_r}{^6P_r} = \frac{\dfrac{6!}{r!(6-r)!}}{\dfrac{6!}{r!}} = \frac{6!}{r!(6-r)!} \times \frac{r!}{6!} = \frac{1}{(6-r)!}\) So the equation simplifies to: \(\frac{1}{(6-r)!} = \frac{1}{6}\) Therefore: (6-r)! = 6 We know that 3! = 6, so: 6 - r = 3 \(\implies\) r = 3

2x\(^2\) - 3x - 2 = 0 2x\(^2\) - 4x + x - 2 = 0 2x(x - 2) + 1(x - 2) = 0 (2x + 1)(x - 2) = 0 x = \(\frac{-1}{2}\) or x = 2

x\(^2\) - (sum of given roots)x + (product of given roots ) = 0 x\(^2\) - (\(\frac{2}{3} + \frac{-1}{4}\))x + (\(\frac{2}{3} * \frac{-1}{4}\)) = 0 x\(^2\) - (\(\frac{8 - 3}{12}\))x + (\(\frac{- 2}{12}\)) x\(^2\) - \(\frac{5}{12}\)x + \(\frac{- 2}{12}\) multiply through by the LCM 12 12 * x\(^2\) - 12 * \(\frac{5}{12}\)x + 12 * \(\frac{- 2}{12}\) 12x\(^2\) - 5x - 2 = 0

When n = 1, n = 2, n = 3 3 x 22 - 1 = 6 3 x 22 - 2 = 3 3 x 22 - 3 = 3/2

5x2 - 4x - 1 = 0 5x2 5x + x -1 = 0 5x(x - 1) + (x - 1) = 0 (5x + 1)(x -1) = 0 x = -1/5 or 1 = 0

\(V = \frac{K}{\sqrt{T-S}}\) square both sides of the equation \(V^2 = \frac{K^2}{T-S}\) cross multiply \(V^2 (T - S) = K^2\) \(T - S = \frac{K^2}{V^2}\) \(T = \frac{K^2}{V^2} + S\) \(T = \frac{K^2}{V^2} + S\)

We are given the following simultaneous equations: \(4x - 3y = 7 \quad \text{(Equation 1)}\) \(3x - 2y = 5 \quad \text{(Equation 2)}\) To eliminate \(y\), we multiply Equation 1 by 2 and Equation 2 by 3: \(2(4x - 3y) = 2(7) \implies 8x - 6y = 14 \quad \text{(Equation 3)}\) \(3(3x - 2y) = 3(5) \implies 9x - 6y = 15 \quad \text{(Equation 4)}\) Now we subtract Equation 3 from Equation 4: \((9x - 6y) - (8x - 6y) = 15 - 14\) \(x = 1\) Substituting \(x = 1\) into Equation 2: \(3(1) - 2y = 5\) \(3 - 2y = 5\) \(-2y = 2 \implies y = -1\) Now we have \(x = 1\) and \(y = -1\). Thus, \(X - Y = x - y = 1 - (-1) = 1 + 1 = 2\) Therefore, the value of \(X - Y\) is: 2.