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[Return ÷ Investment] as a ratio ; i.e The Ratio is Return : Investmen [(Return1÷ Investment1 ) = (Return2÷ Investment2)] R1 = N 25000 R2 =? I1 = N450,000, I1 = N 700000 (25000 ÷ 450000) = (R2 ÷ 700000) R2 = [(25000 × 700000 ) ÷ 450000] = N38,888.90K ∴ The return on a investment of Y = N38888.90K

log717 = [log 17 ÷ log7] = [1.2304 ÷ 0.8451] [100.0899 ÷ 101.9270] = 1.455(antilog)

Convert the binary to base 10 and they convert back to base two 100112 + xxxxx2 + 111002 + 1012 = 10011112 (1 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 1 × 20) + xxxxx2 +(1 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 0 × 20) + (1 × 22 + 0 × 21 + 1 × 20) = (16 + 0 + 0 + 2 + 1) + xxxxx2 + (16 + 8 + 4 + 0 + 0 ) + (4 + 0 + 1) =(64 + 0 + 0 + 8 + 4 + 2 + 1) 19 + xxxxx2 + 33 = 79 xxxxx2 + 52 = 79 xxxxx2 = 79 − 52 xxxxx2 = 2710 \( \begin{array}{c|c} 2 & 27 \\ \hline 2 & 13 \text{ rem 1}\\ 2 & 6 \text{ rem 1}\\ 2 & 3 \text{ rem 0}\\ 2 & 1 \text{ rem 1}\\ & 0 \text{ rem 1}\\ \end{array}\uparrow \) 2710 = 110112 Therefore xxxxx2 = 2710 = 110112

\(\log_{5}(y^{2} x^{5} \div 125b^{\frac{1}{2}})\) = \(\log_{5} y^{2} + \log_{5} x^{5} - [\log_{5} 125 + \log_{5} b^{\frac{1}{2}}\) = \(2\log_{5} y + 5\log_{5} x - \log_{5} 5^{3} - \frac{1}{2} \log_{5} b\) = \(2\log_{5} y + 5\log_{5} x - 3 - \frac{1}{2}\log_{5} b\)

\(\int (5x^{3} + 7x^{2} - 2x + 5) \mathrm d x\) = \(\frac{5x^{4}}{4} + \frac{7x^{3}}{3} - \frac{2x^{2}}{2} + 5x + c\) = \(\frac{5x^{4}}{4} + \frac{7x^{3}}{3} - x^{2} + 5x + c\)

Find the slant height \( l^2 = h^2 + r^2(h = 4cm,r = 3cm)\) \( l^2 = 4^2 + 3^2 = 16 + 9 = 25 \) \( l^2 = √ 25 \) Squaring both sides l = 5cm The area of curved surface (s) =π(3)(5) 15π = 15 × 3.14 = 47.1cm2

[1 \div (x+1)] + [1 \div (x − 1)] = ((x − 1) + [(x + 1)) \div (x+1)(x − 1)] Using the L.C.M. = (x − 1 + x + 1) \div (x + 1)(x − 1) = (x + 2 − 1 + 1) \div (x + 1)(x − 1) = 2x \div (x + 1)(x − 1) =2x \div (x + 1)(x − 1)

Area of a circle (A) = πr2 Radius = 4cm π = 3.142 A = 3.142 × (4)2 = 3.142 × 16 = 50.272 A = 50.3cm2 (1dp)

To find the minor axis of an ellipse: Given: - Area \(A = 132 \, \text{cm}^2\) - Major axis = \(14 \, \text{cm}\) Semi-major axis: \(a = \frac{14}{2} = 7 \, \text{cm}\) Area formula: \(A = \pi \times a \times b \implies 132 = \pi \times 7 \times b\) Solving for \(b\): \(b = \frac{132}{\pi \times 7} \approx \frac{132}{21.98} \approx 6 \, \text{cm}\) Length of minor axis: \(\text{Minor axis} = 2b \approx 2 \times 6 = 12 \, \text{cm}\)

V = ⅓ πr2 h Volume of a cone (Vc) = ⅓ × 3.14 × (5)2 × 6 = 156.93 157cm3

Prob.(A outcome) = \(\frac{1}{6}\), Prob.(B outcome) = ¼ P(A ∪ B) = 1/12, Prob.(A ∩ B) Prob.(A ∩ B) = Prob.(A) + Prob.(B) − Prob.(A ∪ B) 1/6 + 1/4 − 1/12 = (2 + 3 − 1) ÷ 12 = 4/12 = 1/3 Prob.(A ∩ B) = 1/3

Z has 4 elements, power = number of subset = 2p Z = 2n = 24 = 16

\(27^x ÷ 81^{(x + 2y)} = 9 \\ (27)x = 9 × 81^{(x+2y)} \\ (3^3 )^x =32 \times 3^{4(x + 2y)} \\ =3^{(2 + 4x + 8y)}\\ 3^{3x} = 3^{ (2 + 4x + 8y)}\\ 3x = 2 + 4x + 8y\\ 3x − 4x − 8y = 2 … … … (1)\\ x + 4y = 0 … … … (2)\\ − 4y = 2\\ y = (− 2) ÷ 4 = − ½\\ y = − ½\\ \) Substitute the value of y into equation (2) i.e x + 4y = 0 x + 4( − 1/2) = 0 x − 2 = 0 x = 2 ∴ x = 2,y = − ½) Method II \( 27^x ÷ 31^{(x + 2y) }= 9\\ 3^{3x} × 3^{( − 4x − 8y)} = 32\\ 3^{(3x − 8y)} = 32\\ − x − 8y=2 ……… (1)\\ x + 4y = 0 ……… (2)\\ − 4 = 2\\ y= 2/4 = ½\\ y = ½ \) Substitute the value of y into equation 2 x + 4y=0 x + 4 (− 1) ÷ 2) = 0 x − 2 = 0 x = 2 x = 2, y = ½

\(\sqrt{30}\) \times \(\sqrt{40}\) = \(\sqrt{(30 \times 40)}\) \(\sqrt{(3 \times 10 \times 4 \times 10)}\) \(\sqrt{(400 \times 3)}\) \(\sqrt{400}\) \times \(\sqrt{3}\) 20 \times \(\sqrt{3}\) 20\(\sqrt{3}\)

√(2x + 2 ) − √x = 1 √(2x + 2) = 1 + √x Square both sides √(2x + 2)2 = (1 + √x)2 ((2x + 2)2)½ = 1 + √x + √x+ x 2x + 2 = 1 + 2√x +x Collect the like term together 2x − x + 2 − 1 = 2 √x x + 1 = 2√x Square both sides (x + 1)2 = (2√x)2 x2 + 2x + 1 = 4x x2 + 2x − 4x + 1 = 0 (x2− x)(x + 1)= 0 x(x − 1) −1(x − 1) (x − 1)(x −1) Either (x − 1) = 0 or (x − 1) = 0 x = 1 or 1 x = 1 twice