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Perimeter. of a rectangle = 2(L + B) = 24 Area of a rectangle = L x B = 35 From above, L + B = 12 L = 12 - B L x B = 35(given) 12 - B)B = 35 12B - B\(^2\) = 35 B\(^2\) - 12B + 35 = 0 B\(^2\) - 7B - 5B + 35 = 0 B(B - 7) - 5(B - 7) = 0 (B-5)(B -7) = 0 B = 5m OR 7m

Let his assistant work for x days ∴ his master worked (x + 10) day. Amount received by master = 40(x + 10), amount got by his assistance = 10x Total amount collected = N2000.00 ∴ 40(x + 10) + 10x = 2000 = 40x + 400 + 10x = 2000 50x + 400 = 2000 50x = 2000 - 400 50x = 1600 x = \(\frac{1600}{50}\) x = 32 days

\(\frac{x}{x + y}\) + \(\frac{y}{x - y}\) - \(\frac{x^2}{x^2 - y^2}\) \(\frac{x}{x + y}\) + \(\frac{y}{x - y}\) - \(\frac{x^2}{(x + y)(x - y}\) = \(\frac{x(x - y) + y(x + y) - x^2}{(x + y)(x - y}\) = \(\frac{x^2 + xy + xy + y^2 - x^2}{(x + y)(x - y}\) = \(\frac{y^2}{(x + y)(x - y)}\) = \(\frac{y^2}{(x^2 - y^2)}\)

Given that x2 + y2 + z2 = 194, calculate z if x = 7 and \(\sqrt{y}\) = 3 x = 7 ∴ x2 = 49 \(\sqrt{y}\) = 3 ∴ y2 = 81 = x2 + y2 + z2 = 194 49 + 81 + z2 = 194 130 + z2 = 194 z2 = 194 - 130 = 64 z = \(\sqrt{64}\) = 8

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

3 + 6 + 12 + .....18thy term 1st term = 3, common ratio \(\frac{6}{3}\) = 2 n = 18, sum of GP is given by Sn = a\(\frac{(r^n - 1)}{r - 1}\) s18 = 3\(\frac{(2^{18} - 1)}{2 - 1}\) = 3(2^18 - 1)

x\(^2\) + x + 1 \(\frac{dy}{dx}\) = 2x + 1 At the turning point, \(\frac{dy}{dx}\) = 0 2x + 1 = 0 x = -\(\frac{1}{2}\)

Cos \(\theta\) = \(\frac{12}{13}\) x\(^2\) + 12\(^2\) = 13\(^2\) x\(^2\) = 169- 144 = 25 x = \(\sqrt{25}\) = 5 Hence, tan\(\theta\) = \(\frac{5}{12}\) and cos\(\theta\) = \(\frac{12}{13}\) If cot\(^2\)\(\theta\) = 1 + \(\frac{1}{tan^2\theta}\) = 1 + \(\frac{1}{\frac{(5)^2}{12^2}}\) = 1 + \(\frac{1}{\frac{25}{144}}\) = 1 + \(\frac{144}{25}\) = \(\frac{25 + 144}{25}\) = \(\frac{169}{25}\)

\(\frac{h_1}{h_2}\) = \(\frac{2}{3}\) h2 = \(\frac{2h_1}{3}\) \(\frac{r_1}{r_2}\) = \(\frac{9}{8}\) r2 = \(\frac{9r_1}{8}\) v1 = \(\pi\)(\(\frac{9r_1}{8}\))2(\(\frac{2h_1}{3}\)) = \(\pi\)r1 2h1 x \(\frac{27}{32}\) v = \(\frac{\pi r_1 2h_1 \times \frac{27}{32}}{\pi r_1 2h_1}\) = \(\frac{27}{32}\) v2 : v1 = 27 : 32

\(\begin{array}{c|c} Scores & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{No. of students} & 1 & 4 & 5 & 6 & x & 2\end{array}\) Average = 3.5 3.5 = \(\frac{(1 \times 1) + (2 \times 4) + (3 \times 5) + (4 \times 6) + 5x + (6 \times 2)}{1 + 4 + 5 + 6 + x + 2}\) \(\frac{3.5}{1}\) = \(\frac{1 + 8 + 15 + 24 + 5x + 12}{18 + x}\) \(\frac{3.5}{1}\) = \(\frac{60 + 5x}{18 + x}\) 60 + 5x = 3.5(18 \(\div\) x) 60 + 5x = = 63 + 1.5x 5x - 1.5x = 63 - 60 1.5x = 3 x = \(\frac{3}{1.5}\) \(\frac{30}{15}\) = 2

The required equation is y = x2 - x - 2 i.e. B where the graph touches the graph touches the x-axis y = 0 x2 - x - 2 = 0 = (x + 1)(x - 2) = 0 Hence roots of the equation are -1 and 2 as shown in the graph

Since PS = QS = RS S is the centre of circle passing through P, Q, R /PS/ = /RS/ = radius QPR \(\pm\) \(\frac{100^o}{2}\) = 50o

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

yzx + 105o + 30o = 180o yzx = 180 - 155 = 45o \(\frac{yz}{sin 30^o} = \frac{8}{sin 45^o}\) yz = \(\frac{8 \sin 30}{sin 45}\) = 8(\(\frac{1}{2}) = \frac{8}{1} \times \frac{1}{2} \times \frac{\sqrt{2}}{1}\) = 4 \(\div\) \(\frac{1}{\sqrt{2}}\) = 4\(\sqrt{2}\)cm

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