Further Mathematics Questions and Answers

\(\log \frac{1}{8} + \log \frac{1}{2} = \log 8^{-1} + \log 2^{-1}\)

= \(\log 2^{-3} + \log 2^{-1}\)

= \(-3 \log 2 - 1 \log 2 = -4 \log 2\)

\(a^{\frac{5}{6}} \times a^{\frac{-1}{n}} = 1\)

\(\implies a^{\frac{5}{6} + \frac{-1}{n}} = a^{0}\)

Equating bases, we have

\(\frac{5}{6} - \frac{1}{n} = 0\)

\(\frac{5n - 6}{6n} = 0\)

\(5n - 6 = 0 \implies 5n = 6\)

\(n = \frac{6}{5} = 1.20\)

\(\sin \theta = \tan \theta \implies \frac{\sin \theta}{1} = \frac{\sin \theta}{\cos \theta}\)

Equating, we have

\(\cos \theta = 1 \implies \theta = \cos^{-1} 1\)

= \(0°\)

\(x * y = x + y - xy\)

Let \(x^{-1}\) be the inverse of x, so that

\(x * x^{-1} = x + x^{-1} - x(x^{-1}) = 0\)

\(x + x^{-1} - x(x^{-1}) = 0 \implies x(x^{-1}) - x^{-1} = x\)

\(x^{-1}(x - 1) = x \implies x^{-1} = \frac{x}{x - 1}\)

= \(\frac{x}{-(1 - x)} = \frac{-x}{1 - x}, x \neq 1\) 

\(F = \begin{pmatrix} F_{x} \\ F_{y} \end{pmatrix} = \begin{pmatrix} F\cos \theta \\ F\sin \theta \end{pmatrix}\)

\((14N, 240°) = \begin{pmatrix} 14\cos 240 \\ 14\sin 240 \end{pmatrix}\)

= \(\begin{pmatrix} 14 \times -0.5 \\ 14 \times \frac{-\sqrt{3}}{2} \end{pmatrix}\)

= \(\begin{pmatrix} -7 \\ -7\sqrt{3} \end{pmatrix}\)