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Start your solution by cross-multiplying, \(\frac{\sqrt{2}}{x+\sqrt{2}}=\frac{1}{x-\sqrt{2}}\) [x - \(\sqrt{2}\)]\(\sqrt{2}\) = x + \(\sqrt{2}\) where \(\sqrt{2}\) \times\(\sqrt{2}\) = 2 x\(\sqrt{2} - \sqrt{2} x \sqrt{2}\) = x + \(\sqrt{2}\) then collect like terms x\(\sqrt{2}\) - x = \(\sqrt{2}\) + 2 and factorize accordingly to get the unknown. x(\(\sqrt{2}\) - 1) = \(\sqrt{2}\) + 2 x = \(\frac{\sqrt{2}) + 2}{\sqrt{2} - 1}\) rationalize x = \(\frac{\sqrt{2} + 2}{\sqrt{2} - 1}\) * \(\frac{\sqrt{2} + 1}{\sqrt{2} + 1}\) x = \(\frac{\sqrt{4} + \sqrt{2} + 2\sqrt{2} + 2}{\sqrt{4} + \sqrt{2} - \sqrt{2} - 1}\) x = \(\frac{2 + 3\sqrt{2} + 2}{2 - 1}\) x = \(\frac{3\sqrt{2} + 4}{1}\) x = 3\(\sqrt{2}\) + 4

S = \(\begin{pmatrix} -1 & 1 \\ 1 & -2 \end{pmatrix}\) and ST = I (Identity matrix). This means T is the inverse of S Find the determinant of S \( \det(S)\) = (-1)(-2) - (1)(1) = 2 - 1 = 1 T = \(\frac{1}{\det(S)} \begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix}\) Since det = 1, T = \(\begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix}\) T = \(\begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix}\)

= \(\frac{(a^2)^{3/4}\times(b^{-3})^{3/4}\times c^{3/4}}{a^{-1} b^4 c^5}\) \(= a^p\times b^q\times c^r\) = \(\frac{a^{3/2}\times b^{-9/4} \times c^{3/4}}{a^{-1}b^4c^5}\) = \(a^p b^q c^r\) = \(a^{{3/4}+1} b^{{-9/4}-4} c^{{3/5}-5}\) = \(a^pb^q c^r\) = \(a^{5/2} b^{-25/4} c^{-17/4}\) = \(a^pb^qc^r\) therefore, p=5/2, q=-25/4 and r = -17/4 2a + q = \(\frac{5}{2} - 2\times\frac{-25}{4}\) = -20/2 = -10

Cost price (cp) = (100/5) x N1.20 =N24.00 Selling price (sp) = 100-20 = 80 (80/4) x N1.50 = N30.00 Gain = SP-CP = N30.00-N24.00 = N6.00 Gain% = (gain/CP) x 100 = 25%

Hint: Two methods can be used; 1. Direct division (if you know division in number bases) 2. Convert both sides to base 10, divide and convert your answer back to base 6. Your answer should be 356 2434\(_6\) to base 10 2 \(\times\) 6\(^3\) + 4 \(\times\) 6\(^2\) + 3 \(\times\) 6\(^1\) + 4 \(\times\) 6\(^0\) 2 \(\times\) 216 + 4 \(\times\) 36 + 3 \(\times\) 6 + 4 \(\times\) 1 432 + 144 + 18 + 4 = 598 And: 42\(_6\) to base 10 4 \(\times\) 6\(^1\) + 2 \(\times\) 6\(^0\) 4 \(\times\) 6 + 2 \(\times\) 1 24 + 2 = 26 Then divide 598 by 26 = 23 Then convert 23\(_{10}\) to base 6 6 23 6 3 r 5 6 0 r 3 23\(_{10}\) = 35\(_6\)

(2 x 9 \(^0\) ) (Y x 9 \(^1\)+ 3 x 9\(^0\)) = (3 x 5 \(^0\))(Y x 5\(^1\) + 3 x 5\(^0\)). (2 * 1)(Y *9 + 3) = (3 * 1)(Y * 5 + 3) 18y + 6 = 15y + 9 18y - 15y = 9 - 6 3y = 3 y = 1

Multiply appropriately to remove decimals on both numerator and denominators. Such that you have \(\sqrt{\frac{(23000 \times 750)}{(345 \times 125)}}\) Dividing, => \(\sqrt{400} = 20\).

\(m*n = \frac{m}{n}-\frac{n}{m} = \frac{-3}{4} - \frac{4}{-3} = \frac{-3}{4} + \frac{4}{3} = \frac{(-9+16)}{12} = \frac{7}{12}\)

Let the numbers be x and y x+y = 2p.....(i) x-y = p......(ii) 2x = 3p x = 3p/2

\(a*a^{-1} = aa^{-1} + a + a^{-1} = e\) if e = 0 \(2.2^{-1} + 2 + 2^{-1} = 0\) collecting like terms, we have: \(3.2^{-1} + 2 = 0\) = \(2^{-1}\) = -\(\frac{2}{3}\)

Expand the options and collect like terms to check which gives the presented expression. Expansion of Options A: (x+y+6)(x+y-3) → \(x^2+xy-3x + xy+y^2-3y +6x+6y-18\) Collect like terms: \(x^2 + xy + xy + y^2 +6x - 3x +6y - 3y - 18\) = \(x^2 + 2xy + y^2 + 3x + 3y - 18\)

l\(^2\) = 3 (k + m) Since they are consecutive positive numbers, we have l = k+1, m = k+2. \(\to\) (k+1)\(^2\) = 3(k + k + 2) k\(^2\) + 2k + 1 = 3(2k + 2) k\(^2\) + 2k + 1 = 6k + 6 k\(^2\) + 2k - 6k + 1 - 6 = 0 k\(^2\) - 4k - 5 = 0 k\(^2\) - 5k + k - 5 = 0 k(k - 5) + 1(k - 5) = 0 k = -1 or 5 Since k, l and m are positive, then k = 5. m = k + 2 = 5 + 2 = 7.

\(\frac{1}{x^{3} - 1}\) \(x^{3} - 1 = (x - 1)(x^{2} + x + 1)\) \(\frac{1}{x^{3} - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^{2} + x + 1}\) \(\frac{1}{x^{3} - 1} = \frac{A(x^{2} + x + 1) + (Bx + C)(x - 1)}{x^{3} - 1}\) Comparing the two sides of the equation, \(A + B = 0 ... (1)\) \(A - B + C = 0 ... (2)\) \(A - C = 1 ... (3)\) From (3), \(C = A - 1\), putting that in (2), \(A - B = -C \implies A - B = 1 - A\) \(2A - B = 1 ... (4)\) (1) + (4): \(3A = 1 \implies A = \frac{1}{3}\) \(A = -B \implies B = -\frac{1}{3}\) \(C = A - 1 \implies C = \frac{1}{3} - 1 = -\frac{2}{3}\) \(\therefore \frac{1}{x^{3} - 1} = \frac{1}{3(x - 1)} - \frac{x - 2}{3(x^{2} + x + 1)}\) = \(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x + 2)}{x^{2} + x + 1})\)

Let the common ratio be r so that the first term is 2r. Sum, s = \(\frac{a}{1-r}\) ie. 8 = \(\frac{2r}{1-r}\) 8(1-r) = 2r, 8 - 8r = 2r 8 = 2r + 8r 8 = 10r r = \(\frac{4}{5}\). where common ratio (r) = \(\frac{second term(n_2)}{first term(a)}\), r = \(\frac{n_2}{2r}\) r = \(\frac{4}{5}\) and a = 2r or \(\frac{8}{5}\) \(\frac{4}{5}\) * \(\frac{8}{5}\) = n\(_2\) \(\frac{32}{25}\) = n\(_2\) The sum of the first two terms = a + n\(_2\) = \(\frac{8}{5}\) + \(\frac{32}{25}\) = \(\frac{40 + 32}{25}\) = \(\frac{72}{25}\)

2x\(^2\) + x - 1 2x - 1 4x\(^3\) + 0x\(^2\) - 3x + 1 - 4x\(^3\) - 2x\(^2\) 2x - 1 + 2x\(^2\) - 3x + 1 - -x + 1 2x - 1 - 2x +1 - - 2x + 1 0 0 Thus, Divide 4x\(^3\) - 3x + 1 by 2x - 1 = 2x\(^2\) + x - 1