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P = {1,2,u,v,w,x} Q = {2,3,u,v,w,5,6,y} R = {2,3,4,v,x,y} P - Q = {1,x} (P - Q) ∩ R = {1,x} ∩ {2,3,4,v,x,y} = {x} NB: The set P−Q consists of elements that are in P but not in Q.

1st year, Population = 240,000 x (\(\frac{2}{100}\)) = 4800. Being the 2nd year population = 240,000 + 4800 = 244800. Increase in Pop. in 2nd year = 244800 x (\(\frac{2}{100}\)) = 4896 Jan 2000, Pop. = 244800 + 4896 = 249,696

Rationalize \(\frac{(2\sqrt{3}-\sqrt{2})}{(\sqrt{3}+2\sqrt{2})}\) and equate to \(m +n\sqrt{6}\). Such that m = -2, and n = 1. Use \(\sqrt{3}-2\sqrt{2}\) as the conjugate for Rationalization \(\frac{(2\sqrt{3}-\sqrt{2})}{(\sqrt{3}+2\sqrt{2})}\) X \(\frac{(\sqrt{3}-2\sqrt{2})}{(\sqrt{3}-2\sqrt{2})}\) \(\frac{6 - 4\sqrt{6} - \sqrt{6} + 4}{3 - 2\sqrt{6} + 2\sqrt{6} - 8}\) =\(\frac{10 - 5\sqrt{6}}{-5}\) = -2 + \(\sqrt{6}\) = \(m +n\sqrt{6}\) ∴ m = -2 and n = 1

\(\frac{2.813 \times 10^{-3} \times 1.063}{5.637 \times 10^{-2}}\) = \(\frac{0.002813 \times 1.063}{0.05637}\) \(\approxeq \frac{0.0028 \times 1.1}{0.056}\) = \(0.055\)

Amount A = P(1+r)n; A = N150,000, r = 25%, n = 3. 150,000 = P(1+0.25)3 = P(1.25)3 P = 150,000/1.253 =N76,800.00

\(314_{10} - 256_7 = 340_x\) Convert 2567 and 340x to base 10, such that: 314 - 139 = 3x\(^2\) + 4x => \(3x^2\) + 4x - 175 = 0 (quadratic) Factorising, (x - 7) (3x + 25) = 0, either x = 7 or x = -25/3 ( but x cannot be negative) Therefore, x = 7.

Start by expanding \(\frac{3(2^{n+1}) - 4(2^{n-1})}{2^{n+1} - 2^n}\): \(\frac{3 \times 2^n \times 2^1 - 2^2 \times 2^n \times 2^{-1}}{2^n \times 2 - 2^n}\) NUMERATOR : 2\(^n\) ( 3\(^1\) X 2\(^1\) - 2\(^2\) X 2\(^{-1}\) ) --> 2\(^n\) ( 3 X 2 — 4 X \(\frac{1}{2}\) ) --> 2\(^n\) ( 6 - 2 ) --> 2\(^n\) (4) DENOMINATOR : 2\(^n\) ( 2\(^1\) - 1 ) --> 2\(^n\) ( 2 - 1) --> 2\(^n\) => \(\frac{2^n ( 4)}{2^n}\) = 4

Convert everything to base 10 and collect like terms, such that \(210P - 42P = 434 + 406\) \(168P = 840\) \(P = 840/168 = 5\)

a * b = a\(^b\) a * 2 = a\(^2\) = 2 - a a\(^2\) + a - 2 = 0 a\(^2\) + 2a - a - 2 = 0 a(a+2) - 1(a+2) = 0 (a-1)(a+2) = 0 a = 1, -2

If P\(^{-1}\) is the inverse of P and O is the identity, Then P\(^{-1}\) * P = P * P\(^{-1}\) = 0 i.e. P\(^{-1}\) + P - P\(^{-1}\).P = 0 P\(^{-1}\) - P\(^{-1}\).P = -P P\(^{-1}\)(1 - P) = -P P\(^{-1}\) =\(\frac{-P}{(P-1)}\) = \(\frac{P}{(P-1)}\)

S\(_{\infty}\) = \(\frac{\text{a}}{(1-r)}\), where a = \(\frac{1}{2}\), r = -\(\frac{1}{2}\). S\(_{\infty}\) = \(\frac{\frac{1}{2}}{1 - \frac{-1}{2}}\) = \(\frac{\frac{1}{2}}{\frac{3}{2}}\) = \(\frac{1}{2}\) x \(\frac{2}{3}\) \(\frac{1}{3}\) - 1 = -\(\frac{2}{3}\)

The lateral view of the frustrum \(\tan \theta = \frac{6}{1.5} = 4\) \(\theta = \tan^{-1} 4 = 75.96^\circ\) Extending to the full height of the pyramid, we have The height of the pyramid which formed the frustrum = (6 + x)m, \(\tan 75.96 = \frac{6 + x}{2.5}\) \(6 + x = 2.5 \times 4 = 10.0\) \(\therefore \text{The height of the pyramid} = 10.0m\)

∫30 \pi(y+1) dy = \pi[y\(^2 + y]^3_0\) = \pi(9/2 + 3) = 15\pi/2

Number of letter in MATHEMATICS = 11 Number of letter M = 2 Number of letter T = 2 Number of letter A = 2 Arrangement = 11!/(2! 2! 2!) ways

\Delta PQR is isosceles ∴∠PRQ = 30o Also in \Delta PQR ^Q = 180 -(30 + 30) = 120o PQRS is a cyclic quad ∴^S + ^Q = 180(opp ∠s of a cyclic quad) S + 120 = 180 S = 60o In \DeltaPSR x + 50 + 60 = 180(sum of ∠s of a \Delta) x + 110 = 180 x = 180 - 110 x = 70o