\begin{questions} \Question[3] A bobbin for an industrial knitting machine is in the shape of a truncated cone. The diameter of the top is \(4 \mathrm{~cm}\), t diameter of the base is \(6 \mathrm{~cm}\) and the length of the slant is \(10 \mathrm{~cm}\). Find the height of the bobbin. \begin{center} \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-09} \end{center} \begin{solutionordottedlines}[2cm] Let the height of the bobbin be \(h\), the radius of the top be \(r_1 = 2 \mathrm{~cm}\), and the radius of the base be \(r_2 = 3 \mathrm{~cm}\). By the Pythagorean theorem, we have: \[l^2 = h^2 + (r_2 - r_1)^2\] Substituting the given values: \[10^2 = h^2 + (3 - 2)^2\] \[100 = h^2 + 1\] \[h^2 = 99\] \[h = \sqrt{99}\] \[h \approx 9.95 \mathrm{~cm}\] The height of the bobbin is approximately \(9.95 \mathrm{~cm}\). \end{solutionordottedlines} \Question[4] Rationalise the following: \begin{parts}\begin{multicols}{2} \part \[\frac{1}{\sqrt{3}+2}\] \begin{solutionordottedlines}[2cm] Multiply the numerator and denominator by the conjugate of the denominator: \[\frac{1}{\sqrt{3}+2} \cdot \frac{\sqrt{3}-2}{\sqrt{3}-2} = \frac{\sqrt{3}-2}{(\sqrt{3}+2)(\sqrt{3}-2)} = \frac{\sqrt{3}-2}{3-4} 2-\sqrt{3}\] \end{solutionordottedlines} \part \[\frac{1}{\sqrt{3}+\sqrt{2}}\] \begin{solutionordottedlines}[2cm] Multiply the numerator and denominator by the conjugate of the denominator: \[\frac{1}{\sqrt{3}+\sqrt{2}} \cdot \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{\sqrt{3}-\sqrt{2}}{3-2} = \sqrt{3}-\sqrt{2}\] \end{solutionordottedlines} \end{multicols}\end{parts} \Question[3] Find the integers \(p\) and \(q\) such that \[\frac{\sqrt{5}}{\sqrt{5}-2}=p+q \sqrt{5}\]. \begin{solutionordottedlines}[2cm] Multiply the numerator and denominator by the conjugate of the denominator: \[\frac{\sqrt{5}}{\sqrt{5}-2} \cdot \frac{\sqrt{5}+2}{\sqrt{5}+2} = \frac{5+2\sqrt{5}}{5-4} = \frac{5+2\sqrt{5}}{1} = 5+2\sqrt{5}\ Thus, \(p = 5\) and \(q = 2\). \end{solutionordottedlines} \Question[9] Simplify the following: \begin{parts}\begin{multicols}{2} \part \(\frac{3}{\sqrt{5}-2}+\frac{2}{\sqrt{5}+2}\) \begin{solutionordottedlines}[2cm] Multiply each term by the conjugate of its denominator: \[\frac{3}{\sqrt{5}-2} \cdot \frac{\sqrt{5}+2}{\sqrt{5}+2} + \frac{2}{\sqrt{5}+2} \cdot \frac{\sqrt{5}-2}{\sqrt{5}-2}\] \[= \frac{3(\sqrt{5}+2)}{5-4} + \frac{2(\sqrt{5}-2)}{5-4}\] \[= 3\sqrt{5} + 6 + 2\sqrt{5} - 4\] \[= 5\sqrt{5} + 2\] \end{solutionordottedlines} \part \(\frac{5}{(\sqrt{7}-\sqrt{2})^{2}}\) \begin{solutionordottedlines}[2cm] Expand the denominator and simplify: \[\frac{5}{7 - 2\sqrt{14} + 2}\] \[= \frac{5}{9 - 2\sqrt{14}}\] Multiply the numerator and denominator by the conjugate of the denominator: \[\frac{5}{9 - 2\sqrt{14}} \cdot \frac{9 + 2\sqrt{14}}{9 + 2\sqrt{14}}\] \[= \frac{45 + 10\sqrt{14}}{81 - 56}\] \[= \frac{45 + 10\sqrt{14}}{25}\] \[= \frac{9}{5} + 2\sqrt{14}\] \end{solutionordottedlines} \part \(0.0 \mathrm{i} \dot{6}\) \begin{solutionordottedlines}[2cm] This is a repeating decimal. Let \(x = 0.0\dot{6}\), then: \[10x = 0.\dot{6}\] \[10x - x = 0.\dot{6} - 0.0\dot{6}\] \[9x = 0.6\] \[x = \frac{0.6}{9}\] \[x = \frac{2}{30}\] \[x = \frac{1}{15}\] \end{solutionordottedlines} \part \(0.3 \dot{2} \dot{4}\) \begin{solutionordottedlines}[2cm] This is a repeating decimal. Let \(x = 0.3\dot{2}\dot{4}\), then: \[1000x = 324.\dot{2}\dot{4}\] \[1000x - x = 324.\dot{2}\dot{4} - 0.3\dot{2}\dot{4}\] \[999x = 324\] \[x = \frac{324}{999}\] \[x = \frac{108}{333}\] \[x = \frac{36}{111}\] \[x = \frac{12}{37}\] \end{solutionordottedlines} \part \(0.51 \dot{2} \dot{6}\) \begin{solutionordottedlines}[2cm] This is a repeating decimal. Let \(x = 0.51\dot{2}\dot{6}\), then: \[1000x = 512.\dot{2}\dot{6}\] \[1000x - 10x = 512.\dot{2}\dot{6} - 5.1\dot{2}\dot{6}\] \[990x = 507.1\] \[x = \frac{507.1}{990}\] \[x = \frac{5071}{9900}\] \[x = \frac{5071}{9900}\] Simplifying the fraction, we get: \[x = \frac{5071}{9900}\] \[x = \frac{563}{1100}\] \[x = \frac{563}{1100}\] \end{solutionordottedlines} \part \(0.001 \dot{1}\) \begin{solutionordottedlines}[2cm] This is a repeating decimal. Let \(x = 0.001\dot{1}\), then: \[1000x = 1.\dot{1}\] \[1000x - x = 1.\dot{1} - 0.001\dot{1}\] \[999x = 1.1\] \[x = \frac{1.1}{999}\] \[x = \frac{11}{9990}\] \[x = \frac{1}{909}\] \end{solutionordottedlines} \end{multicols}\end{parts} \Question[6] \textbf{Challenge:} \begin{parts} \part Show that \[\frac{138}{19}=7+\frac{1}{3+\frac{1}{1+\frac{1}{4}}}\] \begin{solutionordottedlines}[2cm] Start by evaluating the continued fraction from the innermost fraction outward: \[1 + \frac{1}{4} = 1 + \frac{1}{4} = \frac{5}{4}\] \[3 + \frac{1}{\frac{5}{4}} = 3 + \frac{4}{5} = \frac{15}{5} + \frac{4}{5} = \frac{19}{5}\] \[7 + \frac{1}{\frac{19}{5}} = 7 + \frac{5}{19} = \frac{133}{19} + \frac{5}{19} = \frac{138}{19}\] Therefore, the continued fraction equals \(\frac{138}{19}\). \end{solutionordottedlines} \part Express \[\frac{153}{11}\] as a continued fraction with all numerators 1 . \begin{solutionordottedlines}[2cm] Perform the Euclidean algorithm to find the continued fraction: \[\frac{153}{11} = 13 + \frac{10}{11}\] \[\frac{11}{10} = 1 + \frac{1}{10}\] \[\frac{10}{1} = 10\] So, \[\frac{153}{11} = 13 + \frac{1}{1 + \frac{1}{10}}\] Therefore, the continued fraction is \(13 + \frac{1}{1 + \frac{1}{10}}\). \end{solutionordottedlines} \end{parts} \end{questions}