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(25)\(^{\frac{-1}{2}}\) x (27)\(^{\frac{1}{3}}\) + (121)\(^{\frac{-1}{2}}\) x (625)\(^{\frac{-1}{4}}\) 5\(^{2 \times \frac{-1}{2}}\) x 3\(^{3 \times \frac{1}{3}}\) + 11\(^{2 \times \frac{-1}{2}}\) x 5\(^{4 \times \frac{-1}{4}}\) 5\(^{-1}\) x 3\(^1\) x 11\(^{-1}\) x 5\(^{-1}\) \(\frac{1}{5} \times \frac{3}{1} + \frac{1}{11} \times \frac{1}{5}\) \(\frac{3}{5} + \frac{1}{55} = \frac{33+1}{55}\) = \(\frac{34}{55}\)

3, - 6, +12, -24 a = 3, r = -2 \(Sn = \frac{a(1-r^n)}{1-r}\\ ∴1-2^8 = \frac{3(1-(-2^{n}))}{1-(-2)}\\ 1-2^8 = \frac{3(1-(-2^{n}))}{3}\) 1 - 2\(^8\) = 1 - (-2)\(^n\) Since 1 - 2\(^8\) = 1 - (-2)\(^n\)

\(\frac{25 \frac{2}{3} \div 25 \frac{1}{6}}{( \frac{1}{5})^{-\frac{7}{6}} \times ( \frac{1}{5})^{\frac{1}{6}}}\) = \(\frac{25^{4 - \frac{1}{6}}}{(\frac{1}{5})^{-7 + \frac{1}{6}}}\) = \(\frac{25^{\frac{1}{2}}}{(\frac{1}{5})^{-1}}\) = \(\frac{(5^2)^{\frac{1}{2}}}{(5^{-1})^{-1}}\) = \(\frac{5}{5}\) = 1

y = k\(\sqrt{x}\) when y = 1, x = 4 k = \(\frac{y}{\sqrt{x}}\) = \(\frac{1}{\sqrt{4}}\) = \(\frac{1}{2}\) = 0.5

From y = \(\frac{1}{2} \sqrt{x}\) when y = 2.5 or \(\frac{5}{2}\), x = P \(\frac{5}{2} \times \frac{1}{2} \sqrt{P}\) \(\sqrt{P} = \frac{10}{2} = 5\) P = \(5^2\) = 25

loga270 - loga10 + loga\(\frac{1}{3}\) = 2 loga270 + loga\(\frac{1}{3}\) - loga10 = 2 loga(270\(\frac{\frac{1}{3}}{10}\)) = 2 loga9 = 2 aa = 9 a = \(\sqrt{9}\) = 3

\(\frac{20}{5 \sqrt{4 \times 7} - 2\sqrt{9 \times 7}}\) \(\frac{20}{10 \sqrt{7} - 6\sqrt{7}}\) = \(\frac{20 \sqrt{7}}{4 \sqrt{7}}\) = \(\frac{5\sqrt{7}}{7}\)

tan 6o = \(\frac{|PR|}{|QR|}\) \(\sqrt{3} = \frac{|PR|}{5}\) = |PR| = \(5 \sqrt{3}\)cm sin 45 = \(\frac{|PR|}{|PS|}\) \(\frac{1}{\sqrt{2}}\) = \(\frac{5 \sqrt{3}}{|PS|}\) |PS| = \(5 \sqrt{3}\) x \(\sqrt{2}\) = 5\(\sqrt{6}\)cm

(a) \(\frac{4\frac{2}{9} - 1\frac{13}{15}}{2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3}}\) \(4\frac{2}{9} - 1\frac{13}{15} = \frac{38}{9} - \frac{28}{15}\) = \(\frac{190 - 84}{45}\) = \(\frac{106}{45}\) \(2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3} = \frac{11}{5} + (\frac{4}{7} \times \frac{7}{3})\) = \(\frac{11}{5} + \frac{4}{3}\) = \(\frac{33 + 20}{15}\) = \(\frac{53}{15}\) \(\therefore \frac{4\frac{2}{9} - 1\frac{13}{15}}{2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3}} = \frac{106}{45} \div \frac{53}{15}\) = \(\frac{106}{45} \times \frac{15}{53}\) = \(\frac{2}{3}\) (b) \(\frac{7\sqrt{5}}{\sqrt{7}}\) = \(\frac{7\sqrt{5}}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}\) = \(\frac{7\sqrt{35}}{7}\) = \(\sqrt{35}\)

\(\frac{321_4}{23_4}\) = \(\frac{(3\times4^{2})+(2\times4^{1})+(1\times4^{0})}{(2\times4^{0})+(3\times4^{0})}\) = \(\frac{3\times16+2\times4+1\times1}{2\times4+3\times1}\) = \(\frac{48+8+1}{8+3}\) =\(\frac{57}{11}\) = 5\(\hspace{1mm}remainder\hspace{1mm}2\) ∴ r = 2\(_{10}\) Now\(\hspace{1mm}convert\hspace{1mm}2_{10} \hspace{1mm}to\hspace{1mm}base\hspace{1mm}4\frac{4}{2}\) = 2\(\frac{4}{0}\) = 0\(\hspace{1mm}or\hspace{1mm}2\) ∴ r = 2

\(\frac{81^{\frac{3}{4}}-27^{\frac{1}{3}}}{3 \times 2^3}\) = \(\frac{(3^{3-\frac{3}{4}}-3^{3-\frac{3}{4}})}{3\times 2^3}\) = \(\frac{3^3 - 3}{3 \times 8}\) = \(\frac{27-3}{24}\\=\frac{24}{24}\) =1

\(\frac{(\sqrt{12}-\sqrt{3})}{(\sqrt{12}+\sqrt{3})}\) = \(\frac{\sqrt{4\times 3} - \sqrt{3}}{\sqrt{4\times 3}+\sqrt{3}}\) = \(\frac{2\sqrt{3}-\sqrt{3}}{2\sqrt{3}+\sqrt{3}}\) = \(\frac{\sqrt{3}}{3\sqrt{3}}\) = \(\frac{1}{3}\)

\(\frac{5}{\sqrt{3}}-\frac{3}{\sqrt{2}}=\frac{5\sqrt{2}-3\sqrt{3}}{\sqrt{6}}\\ =\frac{5\sqrt{2}-3\sqrt{3}}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}Rationalize\\ \frac{\sqrt{6}(5\sqrt{2}-3\sqrt{3})}{6}\\ \frac{5\sqrt{12}-3\sqrt{18}}{6}=\frac{10\sqrt{3}-9\sqrt{2}}{6}\\ \frac{1}{6}(10\sqrt{3}-9\sqrt{2})\)

\(log_{10}18 - log_{10}2.88+log_{10}16\\ =log_{10}18 - log_{10}\left(\frac{288}{100}\right)+log_{10}16 = log_{10}\left(\frac{18\times 16}{1}\times \frac{100}{288}\right)\\ =log_{10}\left(\frac{288\times 100}{288}\right)=log_{10}100=log_{10}10^2=2log_{10}10=2\)

Using Pythagoras theorem \(|PR|^2=|PO|^2+|OR|^2\\ |PR|^2=5^2+12^2\\ |PR|=\sqrt{25+144}=\sqrt{169}=13km\)