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\(\begin{pmatrix}5 & -6 \\2 & -7\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix}7 \\-11\end{pmatrix}\) By matrices multiplication; 5x - 6y = 7 ........(1) 2x - 7y = -11 ......(2) 2 x (1): 10x - 12y = 14 .......(3) 5 x (2): 10x - 35y = -55 ......(4) (3) - (4): 23y = 69 y = 69/23 = 3

\(\begin{vmatrix}-x & 12 \\-1 & 4\end{vmatrix} = - 12\) -4x - (-1)12 = -12 -4x + 12 = -12 -4x = -12 - 12 -4x = - 24 x = 6

\(0 \begin{vmatrix}7 & 8 \\5 & 4\end{vmatrix} -3 \begin{vmatrix}1 & 8 \\0 & 4\end{vmatrix} +2 \begin{vmatrix}1 & 7 \\0 & 5\end{vmatrix}\) = 0(28 - 40) - 3(4 - 0) + 2(5 - 0) = 0(-12) - 3(4) + 2(5) = 0 - 12 + 10 = -2

If each interior angle of the polygon is 135^\circ, then each exterior angle is 180^\circ - 135^\circ = 45^\circ. Hence, number of sides = \(\frac{360^\circ}{\text{one exterior angle}}\) \(\frac{360^\circ}{45^\circ}\) = 8

Using \(V = \pi r^2 h\) 6160 = 22/7 x 28 x 28 x h \(h = \frac{6160}{22 \times 4 \times 28} \) \(h = 2.5m \)

Mid point of S(-5, 4) and T(-3, -2) is \([\frac{1}{2}(-5 + -3), \frac{1}{2}(4 + (-2))]\) \([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\) \([\frac{1}{2}(-8), \frac{1}{2}(2)]\) \([-4, 1]\)

\(\text{Gradient m} = \frac{y_2 - y_1}{x_2 - x_1}\) \(\frac{1}{2} = \frac{2 - 4}{1 - x}\) 1 - x = 2(2 - 4) 1 - x = 4 - 8 1 - x = -4 -x = -4 - 1 x = 5

y - 4x + 3 = 0 When y = 0, 0 - 4x + 3 = 0 Then -4x = -3 x = 3/4 So the line cuts the x-axis at point (3/4, 0). When x = 0, y - 4(0) + 3 = 0 Then y + 3 = 0 y = -3 So the line cuts the y-axis at the point (0, -3) Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is; \([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\) \([\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)]\) \([\frac{1}{2}(\frac{3}{4}), \frac{1}{2}(-3)]\) \([\frac{3}{8}, \frac{-3}{2}]\)

No expanded working has been attached for this question yet. The verified answer is Option B.

If y = 4x3 - 2x2 + x, then; \(\frac{\delta y}{\delta x}\) = 3(4x2) - 2(2x) + 1 = 12x2 - 4x + 1

y = cos 3x Let u = 3x so that y = cos u Now, \(\frac{\delta y}{\delta x} = 3\), \(\frac{\delta y}{\delta x} = -sin u\) By the chain rule, \(\frac{\delta y}{\delta x} = \frac{\delta y}{\delta u} \times \frac{\delta u}{\delta x}\) \(\frac{\delta y}{\delta x} = (-\sin u) (3)\) \(\frac{\delta y}{\delta x} = -3 \sin u\) \(\frac{\delta y}{\delta x} = -3 \sin 3x\)

y = x2 - 2x - 3, Then \(\frac{\delta y}{\delta x} = 2x - 2\) But at minimum point,\(\frac{\delta y}{\delta x} = 0\), Which means 2x - 2 = 0 2x = 2 x = 1. Hence the minimum value of y = x2 - 2x - 3 is; ymin = (1)2 - 2(1) - 3 ymin = 1 - 2 - 3 ymin = -4

\(\int \sin 2x dx = \frac{1}{2} (-\cos 2x) + k\) \(- \frac{1}{2} \cos 2x + k\)

\(\int (2x + 3)^{\frac{1}{2}} \delta x\) let u = 2x + 3, \(\frac{\delta y}{\delta x} = 2\) \(\delta x = \frac{\delta u}{2}\) Now \(\int (2x + 3)^{\frac{1}{2}} \delta x = \int u^{\frac{1}{2}}.{\frac{\delta x}{2}}\) \( = \frac{1}{2} \int u^{\frac{1}{2}} \delta u\) \( = \frac{1}{2} u^{\frac{3}{2}} \times \frac{2}{3} + k\) \( = \frac{1}{3} u^{\frac{3}{2}} + k\) \( = \frac{1}{3} (2x + 3)^{\frac{3}{2}} + k\)

Mean x = \(\frac{\sum x}{n}\) = [(2 - t) + (4 + t) + (3 - 2t) + (2 + t) + (t - 1)] \(\div\) 5 = [11 - 1 + 3t - 3t] \(\div\) 5 = 10 \(\div\) 5 = 2