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

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

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

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

\(\sqrt{30}\) \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

25% of N80 = 25/100 × 80 = N20.00

2.04 x 3.7 = 204 x 37 ----- 7548 = 7.548 = 7.55(2dp)

5÷ (2 − √3) Using conjugate surds [5 ÷ (2 − √3)]×(2 + √3) ÷ (2 + √3)] [5(2+√3)÷((2− √3))2] [5(2 +√3) ÷ (2)2− (√3)2] [5(2 +√3) ÷(4− 3)] = 5(2 + √3)

The trinomial = \( a^2 − 4a + 4 \\ a^2 − b^2− 4a + 4 = (a^2 − 4a + 4) − b^2 \\ (a^2 − 2a − 2a + 4) − b^2 \\ a(a − 2)− 2(a − 2)] − b^2 \\ (a − 2)2 − b^2 \\ (a − 2 + b)(a − 2 − b )\)

use trial and error method, By substituting any of x = 2, 5, 15, 18 into 7 + x = 4 (mod 9) any value of x that gives 7 + x = 4 (mod 9) is the correct value of x In this case, x = 15.

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\(\frac{27^x \times 3^{1 - x}}{9^{2x}} = 1\) \(\frac{3^{3x} \times 3^{1 - x}}{3^{2(2 - x)}} = 3^0\) \(3^{3x} \times 3^{1 - x} \div 3^{4x} = 3^0\) \(3^{(3x + 1 - x - 4x)} = 3^0\) \(3^{(1 - 2x)} = 3^0\) since the bases are equal, 1 - 2x = 0 - 2x = -1 x = \(\frac{1}{2}\)

If 2n = y, then, 2\(^{(2 + \frac{n}{3})}\) = 22 x 2\(^\frac{n}{3}\) = 4 x (2n)\(^{\frac{1}{3}}\) But y = 2n, hence 2\(^{(2 + \frac{n}{3})}\) = 4 x y\(^{\frac{1}{3}}\) = 4y\(^\frac{1}{3}\)