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From the constraint y < 3 and y > 0, the possible values for y are 1 and 2 When y = 1, 3y + 5x \(\leq\) 15 3 + 5x \(\leq\) 15 ⇒ 5x \(\leq\) 12 ⇒ 2.4 since x must be an integer greater than 0 but less than or equal to 2.4, the possible values of x are 1 and 2 When y = 2, 3y + 5x \(\leq\) 15 ⇒ 6 + 5x \(\leq\) 15 ⇒ 6 \(\leq\) 15 - 6 = 11 ⇒ x = 2.2 Again, Since x must be an integer greater than 0 and less than or equal to 2.2, the only possible integer value of x is 1 So, the pairs (x, y) that satisfy all the conditions are ( 1, 1), (1, 2) and (2, 1).

\(\begin{vmatrix}-1 & -1 & -1\\3 & 1 & -1\\1 & 2 & 1\end{vmatrix}\) = -\(\begin{vmatrix}1 & -1 \\2 & 1\end{vmatrix}\) + \(\begin{vmatrix}3 & -1\\1 & 1\end{vmatrix}\) - \(\begin{vmatrix}3 & 1\\1 & 2\end{vmatrix}\) = -( 1 + 2) + (3 + 1) - ( 6 - 1) = -3 + 4 - 5 = 4 - 8 = -4.

K = \(\begin{pmatrix}2 & 1\\3 & 4\end{pmatrix}\) K\(^2\) = K = \(\begin{pmatrix}2 & 1\\3 & 4\end{pmatrix}\) x K = \(\begin{pmatrix}2 & 1\\3 & 4\end{pmatrix}\) = \(\begin{pmatrix}7 & 6\\18 & 19\end{pmatrix}\) I = \(\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\) K\(^2\) + K + I = \(\begin{pmatrix}7 & 6\\18 & 19\end{pmatrix}\) + \(\begin{pmatrix}2 & 1\\3 & 4\end{pmatrix}\) + \(\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\) = \(\begin{pmatrix}10 & 7\\21 & 24\end{pmatrix}\)

If P = \(\begin{pmatrix}3 & -2 & 4\\5 & 0 & 6\\7 & 5 & -1\end{pmatrix}\) Then -2P = -2 x \(\begin{pmatrix}3 & -2 & 4\\5 & 0 & 6\\7 & 5 & -1\end{pmatrix}\) = \(\begin{pmatrix}-6 & 4 & -8\\-10 & 0 & -12\\-14 & -10 & 2\end{pmatrix}\)

a = N 100.00, d = N 20.00, Un = N 5800.00, n = ? Un = \(\frac{n}{2}\)(2a + (n - 1)d) 5800 = \(\frac{n}{2}\)( 2 x 100 + ( 20n - 20) 11600 = 200n + 20n\(^2\) - 20n 20n\(^2\) + 180n - 11600 = 0 divide through by 20 n\(^2\) + 9n - 580 = 0 solve using the quadratic formula n = 20 0r - 29 n = 20 years Years cannot be calculated in - Ve

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

x = no. of ppl that like none. no. of ppl that like both Traditional and Modern music, which is equal to 3x Modern Music = 60 - 3x Traditional Music = 50 - 3x 60-3x + 50 - 3x + 3x + x = 94 110 - 3x + x = 94 -2x = 94 - 110 =>-2x = -16, this x = 8. Members that like only one music: = 60 - 3x + 50 - 3x = 60 - 3[8] + 50 - 3[8] = 60 - 24 + 50 - 24 = 36 + 26 = 62 Members that like only one type of music = 62

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

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 \(210P - 42P = 434 + 406\) \(168P = 840\) \(P = 840/168 = 5\)

Fresh water is less dense than the salt water, hence when the ship sails in fresh water, it will sink more, thus the fraction of the ship above the water surface will decrease.