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\(\begin{vmatrix} 2 & 3 \\ 5 & 3x \end{vmatrix}\) = \(\begin{vmatrix} 4 & 1 \\ 3 & 2x \end{vmatrix}\) (2 x 3x) - (5 x 3) = (4 x 2x) - (3 x 1) 6x - 15 = 8x - 3 6x - 8x = 15 - 3 -2x = 12 x = \(\frac{12}{-2}\) = -6

\(\begin{vmatrix} 4 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix}\) 4 \(\begin{vmatrix} 3 & -1 \\ 1 & 3 \end{vmatrix}\) -2 \(\begin{vmatrix} 2 & -1 \\ -1 & 3\end{vmatrix}\) -1 \(\begin{vmatrix} 2 & 3 \\ -1 & 1 \end{vmatrix}\) 4[(3 x 3) - (-1 x 1)] -2 [(2x 3) - (-1 x -1)] -1 [(2 x 1) - (-1 x 3)] = 4[9 + 1] -2 [6 - 1] -1 [2 + 3] = 4(10) - 2(5) - 1(5) = 40 - 10 - 5 = 25

N = [2 3] N-1 = \(\frac{adj N}{|N|}\) adj N = \(\begin{vmatrix} 4 & -3 \\ -1 & 2 \end{vmatrix}\) |N| = (2 x4) - (1 x 3) = 8 - 3 =5 N-1 = \(\frac {1}{5}\) \(\begin{vmatrix} 4 & -3 \\ -1 & 2 \end{vmatrix}\)

Interior angle = (n - 2)180 but, n = 12 = (12 -2)180 = 10 x 180 = 1800 let each interior angle = x x = \(\frac{(n - 2)180}{n}\) x = \(\frac{1800}{12}\) = 150o

From Pythagoras theorem |OA|2 = |AN|2 + |ON|2 72 = |AN|2 + (5)2 49 = |AN|2 + 25 |AN|2 = 49 - 25 = 24 |AN| = \(\sqrt {24}\) = \(\sqrt {4 \times 6}\) = 2\sqrt6 cm |AN| = |NB| (A line drawn from the centre of a circle to a chord, divides the chord into two equal parts) |AN| + |NB| = |AB| 2\sqrt6 + 2\sqrt6 = |AB| |AB| = 4\sqrt6 cm

Volume of cube = L3 33 = 27cm3 volume of rectangular tank = L x B X h = 3 x 4 x 5 = 60cm3 volume of H2O the tank can now hold = volume of rectangular tank - volume of cube = 60 - 27 = 33cm3

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

P(x, y) Q(8, 6) midpoint = (5, 8) x + 8 = 5 \(\frac{y + 6}{2}\) = 8 x + 8 = 10 x = 10 - 8 = 2 y + 6 = 16 y + 16 - 6 = 10 therefore, P(2, 10)

2y = 5x + 4 (4, 2) y = \(\frac{5x}{2}\) + 4 comparing with y = mx + e m = \(\frac{5}{2}\) Since they are perpendicular m1m2 = -1 m2 = \(\frac{-1}{m_1}\) = -1 \(\frac{5}{2}\) = -1 x \(\frac{2}{5}\) The equator of the line is thus y = mn + c (4, 2) 2 = -\(\frac{2}{5}\)(4) + c \(\frac{2}{1}\) + \(\frac{8}{5}\) = c c = \(\frac{18}{5}\) y = -\(\frac{2}{5}\)x + \(\frac{18}{5}\) 5y = -2x + 18 or 5y + 2x - 18 = 0

tan\(\theta\) = \(\frac{3}{4}\) from Pythagoras tippet, the hypotenus is T i.e. 3, 4, 5. then sin \(\theta\) = \(\frac{3}{5}\) and cos\(\theta\) = \(\frac{4}{5}\) cos\(\theta\) - sin\(\theta\) \(\frac{4}{5}\) - \(\frac{3}{5}\) = \(\frac{1}{5}\)

tan\(\theta\) = \(\frac{100}{100}\) = 1 \(\theta\) = tan-1(1) = 45o The bearing of x from z is N45oE or 135o

2x + 1 \(\frac{d(3x + 1)}{\mathrm d x}\) + (3x + 1) \(\frac{d(2x + 1)}{\mathrm d x}\) 2x + 1 (3) + (3x + 1) (2) 6x + 3 + 6x + 2 = 12x + 5

Mode = L1 + (\(\frac{D_1}{D_1 + D_2}\))C D1 = frequency of modal class - frequency of the class before it D1 = 5 - 2 = 3 D2 = frequency of modal class - frequency of the class that offers it D2 = 5 - 3 = 2 L1 = lower class boundary of the modal class L1 = 5 - 5 C is the class width = 8 - 5.5 = 3 Mode = L1 + (\(\frac{D_1}{D_1 + D_2}\))C = 5.5 + \(\frac{3}{2 + 3}\)C = 5.5 + \(\frac{3}{5}\) x 3 = 5.5 + \(\frac{9}{5}\) = 5.5 + 1.8 = 7.3 \(\approx\) = 7

y = x3 + x2 - x + 1 \(\frac{dy}{dx}\) = \(\frac{d(x^3)}{dx}\) + \(\frac{d(x^2)}{dx}\) - \(\frac{d(x)}{dx}\) + \(\frac{d(1)}{dx}\) \(\frac{dy}{dx}\) = 3x2 + 2x - 1 = 0 \(\frac{dy}{dx}\) = 3x2 + 2x - 1 At the maximum point \(\frac{dy}{dx}\) = 0 3x2 + 2x - 1 = 0 (3x2 + 3x) - (x - 1) = 0 3x(x + 1) -1(x + 1) = 0 (3x - 1)(x + 1) = 0 therefore x = \(\frac{1}{3}\) or -1 For the maximum point \(\frac{d^2y}{dx^2}\) < 0 \(\frac{d^2y}{dx^2}\) 6x + 2 when x = \(\frac{1}{3}\) \(\frac{dx^2}{dx^2}\) = 6(\(\frac{1}{3}\)) + 2 = 2 + 2 = 4 \(\frac{d^2y}{dx^2}\) > o which is the minimum point when x = -1 \(\frac{d^2y}{dx^2}\) = 6(-1) + 2 = -6 + 2 = -4 -4 < 0 therefore, \(\frac{d^2y}{dx^2}\) < 0 the minimum point is 1/3

\(\int^{1}_{0}\)(3 - 2x)dx [3x - x\(^2\)]\(_{0} ^{1}\) [3(1) - (1)\(^2\)] - [3(0) - (0)\(^2\)] (3 - 1) - (0 - 0) = 2 - 0 = 2