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Actual length of rope = 1.25m Measured length of rope = 1.26m error = (1.26 - 1.25)m - 0.01m Percentage error = \(\frac{error}{\text{actual length}}\) x 100% = \(\frac{0.01}{1.25}\) x 100% = 0.8%

\(\begin{vmatrix} 2 & 0 & 5 \\ 4 & 6 & 3 \\ 8 & 9 & 1 \end{vmatrix}\) = 2(6 - 27) - 0(4 - 24) + 5(36 - 48) = 2(-21) - 0 + 5(-12) = -42 + 5(-12) = -42 - 60 = -102

P = \(\begin{pmatrix} 2 & -3 \\ 1 & 1 \end{pmatrix}\) |P| = 2 - 1 x -3 = 5 P-1 = \(\frac{1}{5}\)\(\begin{pmatrix} 1 & 3 \\ -1 & 2 \end{pmatrix}\) = \(\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \\ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\)

If p : q = \(\frac{2}{3}\) : \(\frac{5}{6}\), then the sum S1 of ratio = \(\frac{2}{3}\) + \(\frac{5}{6}\) = \(\frac{9}{6}\) If q : r = \(\frac{3}{4}\) : \(\frac{1}{2}\), then the sum S2 of ratio = \(\frac{3}{4}\) + \(\frac{1}{2}\) = \(\frac{5}{4}\) Let p + q = T1, then q = (\(\frac{5}{6} \div \frac{9}{6}\))T1 = (\(\frac{5}{6} \times \frac{6}{9}\))T1 = \(\frac{5}{9}\)T1 Again, let q + r = T2, then q = (\(\frac{3}{4} \div \frac{5}{4}\))T2 = (\(\frac{3}{4} \times \frac{4}{5}\))T2 = \(\frac{3}{5}\)T2 Using q = q \(\frac{5}{9}\)T1 = \(\frac{3}{5}\)T2 5 x 5T1 = 9 x 3T2 \(\frac{T_1}{T_2}\) = \(\frac{9 \times 3}{5 x 5}\) = \(\frac{27}{25}\) Giving that, T1 = 27 and T2 = 25 P = (\(\frac{2}{3} \div S_1\))T1 = (\(\frac{2}{3} \div \frac{9}{6}\))T1 = (\(\frac{2}{3} \times \frac{6}{9}\))27 = 12 q = (\(\frac{5}{6} \div S_1\))T1 = (\(\frac{5}{6} \div \frac{9}{6}\))T1 = (\(\frac{5}{6} \times \frac{6}{9}\))27 = 15 and r = (\(\frac{1}{2} \div S_2\))T2 = (\(\frac{1}{2} \div \frac{5}{4}\))T2 = (\(\frac{1}{2} \times \frac{4}{5}\))25 = 10 Hence p : q : r = 12: 15 : 10

Using simple interest = \(\frac{P \times T \times R}{100}\), where: P denoted principal = N400 T denotes time = 3 years R denotes interest rate = ? 24 = \(\frac{400 \times 3 \times R}{100}\) 24 x 100 = 400 x 3 x R R = \(\frac{24 \times 100}{400 \times3}\) = 2%

x \(\alpha\) \(\frac{1}{y}\) .........(1) x = k x \(\frac{1}{y}\) .........(2) When x = 2\(\frac{1}{2}\) = \(\frac{5}{2}\), y = 2 (2) becomes \(\frac{5}{2}\) = k x \(\frac{1}{2}\) giving k = 5 from (2), x = \(\frac{5}{y}\) so when y =4, x = \(\frac{5}{y}\) = 1\(\frac{1}{4}\)

W \(\alpha\) U W = ku u = \(\frac{w}{k}\); \(\frac{2}{7}\) x \(\frac{3}{5}\) = \(\frac{6}{35}\)

p = \(\frac{M}{5}\)(X + Q) + 1 P - 1 = \(\frac{M}{5}\)(X + Q) \(\frac{5}{M}\)(p - 1) = X + Q \(\frac{5}{M}\)(p - 1)- x = Q Q = \(\frac{5(p -1) - Mx}{M}\) = \(\frac{5p - 5 - Mx}{M}\) = \(\frac{5p - Mx - 5}{M}\)

Given that \( 9x^2 + 6xy + 4y^2 \) is a factor of \( 27x^3 - 8y^3 \), we can express: \(27x^3 - 8y^3 = (3x - 2y)(9x^2 + 6xy + 4y^2)\) Thus, the other factor is: 3x - 2y

\(\frac{1}{2}\)x + \(\frac{1}{4}\) > \(\frac{1}{3}\)x + \(\frac{1}{2}\) Multiply through by through by the LCM of 2, 3 and 4 12 x \(\frac{1}{2}\)x + 12 x \(\frac{1}{4}\) > 12 x \(\frac{1}{3}\)x + 12 x \(\frac{1}{2}\) 6x + 3 > 4x + 6 6x - 4x > 6 - 3 2x > 3 \(\frac{2x}{2}\) > \(\frac{3}{2}\) x > \(\frac{3}{2}\)

-6 \(\leq\) 4 - 2x < 5 - x split inequalities into two and solve each part as follows: -6 \(\leq\) 4 - 2x = -6 - 4 \(\leq\) -2x -10 \(\leq\) -2x \(\frac{-10}{-2}\) \(\geq\) \(\frac{-2x}{-2}\) giving 5 \(\geq\) x or x \(\leq\) 5 4 - 2x < 5 - x -2x + x < 5 - 4 -x < 1 \(\frac{-x}{-1}\) > \(\frac{1}{-1}\) giving x > -1 or -1 < x Combining the two results, gives -1 < x \(\leq\) 5

(x\(^2\) - 1) > 0 x\(^2\) > 1 Taking the square root of both sides x >1 or x < -1

3rd term : a + 2d = -9 .......(1) 7th term : a + 6d = -29 ......(2) (2) - (1): 4d = -20 :. d = -20/4 = -5 From (1) : a + 2(-5) = -9 a - 10 = -9 :. a = -9 + 10 = 1 :. 10th term of A.P is a + 9d = 1 + 9 (-5) = 1 - 45 = -44

Using S\(\infty\) = \(\frac{a}{1 - r}\) r = \(\frac{0.05}{0.5}\) = \(\frac{1}{10}\) S\(\infty\) = \(\frac{0.5}{{\frac{1}{10}}}\) = \(\frac{0.5}{({\frac{9}{10}})}\) = \(\frac{0.5 \times 10}{9}\) = \(\frac{5}{9}\)

Total surface area of the cylindrical pipe = area of circular base + curved surface area = \(\pi\)r\(^2\) + 2\(\pi\)rh = \(\pi\) x 7\(^2\) + 2\(\pi\) x 7 x 50 = 49\(\pi\) + 700\(\pi\) = 749\(\pi\)m\(^2\)