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(a) first is to sketch the diagram Using Pythagoras theorem, x\(^2\) = 17\(^2\) - 15\(^2\) = 64 and x = 8 m. Then, multiplying by 2 we have 16m as the length of the fence within the reach of the donkey. (b)(i) PR = \(\sqrt{15^2 + 15^2}\) = 21.2132 cm and PO = OR = \(\frac{21.2132}{2}\) = 10.61 cm. Then height of the pyramid is 32\(^2\) = h\(^2\) + (10.6066)\(^2\) and h yields h = 30.2 cm. (b)(ii) Volume = \(\frac{1}{3}\) x 15 x 15 x 30.19106 = 2264 cm\(^2\) (correct to the nearest cm\(^3\)).

(a) Applying the laws of logarithm, log(y + 2) = log ax which amplies that y + 2 = ax and finding x to get x = \(\frac{y + 2}{a}\) (b)(i) The pie is shown (b)(i) The percentage of timber produced that year that was from Amenfi = \(\frac{900}{7200}\) x 100% = 12.5% (ii) Difference in revenue = $560.00(1800 - 600) = $672,000.00

(a) The quantity of shares bought \(\frac{8400}{2.5}\) = 3360 shares. Therefore, the dividend the man received is 3360 X 0.25 = N840.00. (b) Clearing fractions and simplifying they got the quadratic equations 9x\(^2\) - 9x + 2 = 0. Next, dividing by 9 and adding the square of the coefficient of x to both sides yields (x - \(\frac{1}{2}\))\(^2\) = \(\frac{1}{4} - \frac{2}{9}\). Then, taking square root and solving for x To obtain x = \(\frac{2}{3}\) or x = \(\frac{1}{3}\)

(a) By the use of Pythagoras theorem, the adjacent side of the right – angled triangle is 4. Then, cos⁡(X = \(\frac{4}{5}\)) and substituting to get 5(cos X)\(^2\) – 3 = 5(\(\frac{4}{5}\))\(^2\) - 3 = \(\frac{16}{5}\) - 3 = \(\frac{(16 - 15)}{5}\) = \(\frac{1}{5}\) (b) find Volume of pyramid = \(\frac{1}{3}\) \times 42 \times 11 \times 12 = 1,848 (cm)\(^2\) Volume of conical container = \(\frac{1}{3}\) \times \(\frac{22}{7}\) x r\(^2\) \times 12 = 1,848 and simplifying to get r\(^2\) = 147 so that r =7\sqrt3 cm as required.

(a). Completing the table as follows \(\oplus\) 2 3 4 2 4 0 1 3 0 1 2 4 1 2 3 \(\otimes\) 2 3 4 2 4 1 3 3 1 4 2 4 3 2 1 (a) (ii) (a), they obtained the value of e as 1. (a) (ii) (b), the value of n was 2. (b), they couldn't answer this question correctly. If p = q, then: (i) q \(\Rightarrow\) p, this is invalid (ii) \(sin\) q \(\Rightarrow\) \(sin\) p, this is valid (iii) \(sin\) p \(\Rightarrow\) \(sin\) q, this is invalid

(a) (i) The diagram can be represented as: (ii) first they were expected to convert from litres to cm\(^3 \) by using the fact that 1000 cm\(^3\) = 1 litre, then 4.158 litres = 4158 cm\(^3\). If the volume and height of the container were V\(_1\) and h\(_1\) respectively, then, V\(_1\) = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 21 x 21 x h\(_1\) = 4,158 which simplified to 462h\(_1\) = 4,158. Therefore, h\(_1\) = 9cm. (b) let h\(_2\) be the height of the water remaining in the container. Then by similar triangles, \(\frac{h_2}{9}\) = \(\frac{14}{21}\), h\(_2\) = 6 cm. If we let V\(_2\) be the volume of water remaining in the container, V\(_2\) = \(\frac{1}{3} \times \frac{22}{7}\) x 14 x 14 x 6 = 1,232 cm\(^2\). Volume of water drawn out = (4,158 — 1,232) cm\(^3\) = 2,926 cm\(^3\)

2x + m = 180 x + m = 112 x = 122 - m 2(112 - m) + m = 180 224 - 2m + m = 180 224 - m = 180 224 - 180 = m m = 44\(^o\)

x = 10 ; 10 = \(\frac{x}{25}\) x = 250 x = \(\frac{250 + 12.4}{26}\) x = 10.09 x = 10.1 years

d = \(\sqrt{\frac{6}{x} - \frac{y}{2}}\) \(d^2 = \frac{6}{x} - \frac{y}{2}\) \(2xd^2 = 12 - xy\) \(2xd^2 + xy = 12\) x = \(\frac{12}{2d^2 + y}\)

T = {evenn numbers from 1 to 12} N = {common factors of 6,8 and 12} Find T ∩ N T = {2, 4, 6, 8, 10, 12} N = {2} T ∩ N = {2} i.e value common to T & N

Curved surface area of cylinder = 2πrh 110 = 2 × \(\frac{22}{7}\) × r × 5 r = \(\frac{110 \times 7}{44 \times 5}\) = 3.5cm

Y = Px2 + q Y = 2x2 - 1 Px2 + q = 2x2 - 1 Px2 = 2x2 - 1 - q p = \(\frac{2x^2 - 1 - q}{x^2}\) at x = 2 P = \(\frac{2(2)^2 - 1 - q}{2^2}\) = \(\frac{2(4) - 1 -q}{4}\) = \(\frac{8 - 1 - q}{4}\) P = \(\frac{7 - q}{4}\)

hypotenuse sin = \(\frac{1}{2}\) \(\sin45 = \frac{1}{\sqrt{2}}\) = \(\frac{2}{2}\) ∴ (sin45 + sin30) = \(\frac{1}{\sqrt{2}} + \frac{1}{2}\) = \(\frac{\sqrt{2}}{2}\) + \(\frac{1}{2}\) = \(\frac{\sqrt{2} + 1}{2}\) = \(\frac{1 + \sqrt{2}}{2}\)

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

M = N \(\sqrt{\frac{SL}{T}}\), make T subject of formula square both sides M\(^{2}\) = \(\frac{N^2SL}{T}\) TM\(^{2}\) = N\(^{2}\)SL T = \(\frac{N^2SL}{M^2}\)