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3\sqrt42x = 16 this implies that (3\sqrt42x)3 = (16)3 42x = 42*3 42x = 46 ∴ 2x = 6 x = 3

Numerator: \(\frac{1}{10}\cdot\frac{2}{3} + \frac{1}{4} = \frac{1}{15} + \frac{1}{4} = \frac{4+15}{60} = \frac{19}{60}\) Denominator: \(\frac{\frac{1}{2}}{\frac{3}{5}} - \frac{1}{4} = \frac{1}{2}\cdot\frac{5}{3} - \frac{1}{4} = \frac{5}{6} - \frac{1}{4} = \frac{10-3}{12} = \frac{7}{12}\) Final: \(\frac{19}{60} \div \frac{7}{12}\) = \(\frac{19}{60} \times \frac{12}{7}\) = \(\frac{19}{35}\)

The shaded portion shows the elements in Q and R, but not in P, Q n R n P'

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

5000 : 5000 * 500 5000 : 2500000 5 : 2500 1 : 500

Convert each value to decimal (base 10) \(1011_{two}\) → 11 + \(1101_{two}\) → 13 + \(1001_{two}\) → 9 - \(111_{two}\) → 7 : 11 + 13 + 9 - 7 = 26 26 → \(11010_{two}\)

\((216^{\frac{1}{3}})^{-2}\) = \(216^{\frac{-2}{3}}\) = (3\sqrt216)\(^{-2}\) = (6)\(^{-2}\) = \(\frac{1}{6^2}\) =\(\frac{1}{36}\)

LCM of 15abc and 20a\(^2\)bc = 60a\(^2\)bc Factors of 15abc = 1, 3, 5,15, a, b, c and of 20a\(^2\)bc = 1, 2, 4, 5, 10, 20, a, b, c HCF of 15abc and 20a\(^2\)bc = 5abc → LCM \div HCF = 60a\(^2\)bc \div 5abc = 12a

|AB| = 4cm |BC| = 8cm In right-angled BAC; 8\(^2\) = 4\(^2\) + |AC|\(^2\) |AC|\(^2\) = 8\(^2\) - 4\(^2\) |AC\\(^2\) = 64 - 16 → 48 |AC| = \(\sqrt{48}\)cm → \(4\sqrt{3}cm\) The area of rectangle = L x B = |AB| x |AC| = \((4 \times 4\sqrt{3}cm^2\) =\(16\sqrt{3}cm^2\)

\(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div16\frac{1}{2}; \frac{15}{2} - \left(\frac{5}{2}+\frac{3}{1}\right)\div \frac{33}{2}\\ \frac{15}{2} - \left(\frac{5+6}{2}\right)\div \frac{33}{2}; \frac{15}{2} - \frac{11}{2} \div \frac{33}{2}\\ \(\frac{11}{2}\) \div \(\frac{33}{2}\) = \(\frac{11}{2}\) * \(\frac{2}{33}\) = \(\frac{11}{33}\) or \(\frac{1}{3}\) (when simplified) \(\frac{15}{2}\) - \(\frac{1}{3}\) = \(\frac{43}{6}\) 7.166 The nearest whole number is 7

\(202^2_{three}\)when converted to base ten \(=(202_3)^2\) 202_3 = 2 \(\times 3^2 + 0 \times 3^1 + 2\times 3^0 = 18 + 0 + 2\) =20\(_{ten}; (202_3)^2 = (20)^2_{ten} = 400\) 112\(^2_{three}\)when converted to base ten \(= (112_3)^2\) 112_3 = 1 \(\times 3^2 + 1 \times 3^1 + 2\times 3^0 = 9+3+2=14_{ten}\) (112_3)\(^2 = (14)^2_{ten} = 196_{ten}\) Evaluate \(\Longrightarrow 400-196 = 204\) Reconvert to base three 204 \div 3 = 68 remainder 0 68 \div 3 = 22 remainder 2 22 \div 3 = 7 remainder 1 7 \div 3 = 2 remainder 1 2 \div 3 = 0 remainder 2 = 21120\(_3\)

\(23_{five} = X_{ten}; X_{ten}\) = 2\(\times 5^1 + 3\times 5^0 = 10 + 3 = 13\) 101\(_{three}\) = P\(_{ten}\); P\(_{ten}\) = 1\(\times 3^2 + 0\times 3^1 + 1\times 3^0 = 9 + 0 + 1 = 10_{ten}\) Y = 13+10=23\(_{ten}\); Converting to base two \(\begin{matrix} 2 & 23\\ 2 & 11 &R1\\ 2 & 5 & R1\\ 2 & 2 & R1\\ 2 & 1 & R0\\ & 0& R1 \uparrow\\ \end{matrix}\) = y = 10111\(_2\)

Let the girls pocket money be rep. by x. The amount spent on books = \(\frac{1}{4}of\hspace{1mm}x = \frac{x}{4}\) The amount spent on dress \(=\frac{1}{3} of \hspace{1mm}x=\frac{x}{3}\) ∴The fraction that remains = \(\frac{x}{1}-\left(\frac{x}{4}+\frac{x}{3}\right)\\ \frac{3x+4x}{12}; = \frac{12x-7x}{12} = \frac{5x}{12} = \frac{5}{12}\)

\(\frac{PR}{QR} = \frac{QR}{QY} = \frac{QP}{YR}\) (SSS Congruence) \(\frac{QR}{4} = \frac{8}{5}\) \(QR = \frac{4 \times 8}{5}\) = 6.4 cm

\(x + y = 7 ------ I\\ 3x – y = 5 ------ II\) Add I and II; \(4x = 12 => x = \frac{12}{4} = 3\\ Y = 7 – 3 = 4, evaluate \hspace{1mm} \frac{y}{2} – 3\\ \frac{4}{2} – 3 => 2-3 = -1\)