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If P = \(\begin{pmatrix}3 & -2 & 4\\5 & 0 & 6\\7 & 5 & -1\end{pmatrix}\) Then -2P = -2 x \(\begin{pmatrix}3 & -2 & 4\\5 & 0 & 6\\7 & 5 & -1\end{pmatrix}\) = \(\begin{pmatrix}-6 & 4 & -8\\-10 & 0 & -12\\-14 & -10 & 2\end{pmatrix}\)

a = N 100.00, d = N 20.00, Un = N 5800.00, n = ? Un = \(\frac{n}{2}\)(2a + (n - 1)d) 5800 = \(\frac{n}{2}\)( 2 x 100 + ( 20n - 20) 11600 = 200n + 20n\(^2\) - 20n 20n\(^2\) + 180n - 11600 = 0 divide through by 20 n\(^2\) + 9n - 580 = 0 solve using the quadratic formula n = 20 0r - 29 n = 20 years Years cannot be calculated in - Ve

n = 3, U\(_3\) = 4x - 2y, U\(_n\) = a + (n-1)d, 4x - 2y = a + 2d... (1) U\(_9\) = 10x - 8y, 10x - 8y = a + 8d... (2) Solving (1) and (2), subtract eqn i from ii a-a + 8d - 2d = 10x -4x - 8y- (-2y) 6d = 6x - 8y+ 2y. 6d= 6x - 6y 6d = 6( x-y) d = x - y

This is a polynomial of the 3rd order, thus x should have three answers. Use the factors given to get values of x as 1, -1 and -2. Form three equations, and carry out elimination and subsequent substitution to get a = -1/2, b = 1, and c = 1/2 Equate each factor to zero to get the values of x x-1 = 0, x = 1 x +1 = 0, x = -1 x - 2 = 0, x = 2 substitute these value of x in the given equation \(ax^3 + bx^2\) + cx - 1 when x = 1 \(ax^3 + bx^2\) + cx - 1 = a + b + c - 1= 0 ------------------------(1) when x = -1 \(ax^3 + bx^2\) + cx - 1 = -a + b - c - 1 = 0-------------------------(2) when x = 2 \(ax^3 + bx^2\) + cx - 1 = 8a + 4b + 2c - 1 = 0 ----------------------(3) add equations 1 and 2 a + b + c - 1= 0 + -a + b - c - 1 = 0 b = 1 put b = 1 into equation (1) a + b + c - 1= 0 = a + 1 + c -1 = 0 a = - c put a = - c into equation(3) 8a + 4b + 2c - 1 = 0, -8c + 4 + 2c - 1 = 0 ( b = 1 remember) -6c = -3 c = \(\frac{-3}{-6}\) = \(\frac{1}{2}\) put c = \(\frac{1}{2}\) and b = 1 into equation (1) a + b + c - 1= 0 a + 1 + \(\frac{1}{2}\) - 1 = 0 a + \(\frac{1}{2}\) = 0 then a = \(\frac{-1}{2}\)

2 - x > x\(^2\) 2 - x - x\(^2\) > 0 (x+2)(x-1) < 0 Either (x + 2) > 0 and (x - 1) < 0. x + 2 > 0 \(\implies\) x > -2. -2 < x < 1

1/\alpha + 1/\beta = \beta+\alpha/\alpha\beta 3x\(^2\) + 5x - 2 = 0 x\(^2\) + 5x/3 - 2/3 = 0 \alpha\beta = -2/3 \beta+\alpha = -5/3 Thus; \beta+\alpha/\alpha\beta = -5/3 \div -2/3 = 5/2

Grad of 3y = 4x - 1 y = \(\frac{4x}{3}\) - \(\frac{1}{3}\) Grad = \(\frac{4}{3}\) Grad of Ky = x + 3 y = \(\frac{\text{x}}{\text{k}}\) + \(\frac{3}{4}\) Grad = \(\frac{1}{\text{k}}\) Since the two lines are perpendicular, \(\frac{1}{\text{k}}\) = -\(\frac{3}{4}\) -3k = 4 k = -\(\frac{4}{3}\)

At x = 1, substituting x = 1 in the equation: ax2 + bx + c = 5; f(1) => a + b + c = 5 .....(1) Taking the first derivative of f(x) in the original equation gives dy/dx = 2ax + b = 2x + 1 (given)....(2) From (2),=> b = 1, and 2ax = 2x, => a = 1. ( comparing coefficients) substituting into (1) 1 + 1 + c = 5, => c = 5 - 2 = 3 Thus a = 1, b = 1 and c = 3

3y\(^2\) = 27y 3y\(^2\) - 27y = 0 3y(y - 9) = 0 3y = 0 or y - 9 = 0 y = 0 or y = 9 y = 0 or 9

P varies directly as q means p = kq some constant k q varies inversely as r means q = \(\frac{\text{m}}{\text{r}}\) for some constant m Substituting the expression for q into the equation for p: p =\(\frac{\text{km}}{\text{r}}\) This shows that p varies inversely as r.

Equation with roots -8 and 5: (x + 8)(x - 5) = 0 \(x^2 - 5x + 8x - 40 = 0\) \(x^2 + 3x - 40 = 0\)

3 + 5x - 2x\(^2\) = 0 3 + 6x - x - 2x\(^2\) = 0 3(1 + 2x) - x(1 + 2x) = 0 (3 - x)(1+2x) = 0 3 - x = 0 or 1 + 2x = 0 x = 3 or x = \(\frac{-1}{2}\)

more than means greater than WHILE not more than means less than or equal to therefore the answer is 18 < d \le 23 meaning d is more than 8 but can be equal to or less than 23

S = \(\begin{pmatrix} -1 & 1 \\ 1 & -2 \end{pmatrix}\) and ST = I (Identity matrix). This means T is the inverse of S Find the determinant of S \( \det(S)\) = (-1)(-2) - (1)(1) = 2 - 1 = 1 T = \(\frac{1}{\det(S)} \begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix}\) Since det = 1, T = \(\begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix}\) T = \(\begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix}\)

The locus of a point P(x,y) such that PV = PW where V = (1,1) and W = (3,5). This means that the point P moves so that its distance from V and W are equidistance. PV = PW \(\sqrt{(x-1)^{2} + (y-1)^{2}} = \sqrt{(x-3)^{2} + (y-5)^{2}}\). Squaring both sides of the equation, (x-1)2 + (y-1)2 = (x-3)2 + (y-5)2. x2-2x+1+y2-2y+1 = x2-6x+9+y2-10y+25 Collecting like terms and solving, x + 2y = 8.