Express \(\log \frac{1}{8} + \log \frac{1}{2}\) in terms of \(\log 2\)
3 log 2
4 log 2
-3 log 2
-4 log 2
Correct answer is D
\(\log \frac{1}{8} + \log \frac{1}{2} = \log 8^{-1} + \log 2^{-1}\)
= \(\log 2^{-3} + \log 2^{-1}\)
= \(-3 \log 2 - 1 \log 2 = -4 \log 2\)
Given that \(a^{\frac{5}{6}} \times a^{\frac{-1}{n}} = 1\), solve for n
-6.00
-1.20
0.83
1.20
Correct answer is D
\(a^{\frac{5}{6}} \times a^{\frac{-1}{n}} = 1\)
\(\implies a^{\frac{5}{6} + \frac{-1}{n}} = a^{0}\)
Equating bases, we have
\(\frac{5}{6} - \frac{1}{n} = 0\)
\(\frac{5n - 6}{6n} = 0\)
\(5n - 6 = 0 \implies 5n = 6\)
\(n = \frac{6}{5} = 1.20\)
Solve: \(\sin \theta = \tan \theta\)
200°
90°
60°
0°
Correct answer is D
\(\sin \theta = \tan \theta \implies \frac{\sin \theta}{1} = \frac{\sin \theta}{\cos \theta}\)
Equating, we have
\(\cos \theta = 1 \implies \theta = \cos^{-1} 1\)
= \(0°\)
\(\frac{-x}{1 - x}, x \neq 1\)
\(\frac{1}{1 - x}, x \neq 1\)
\(\frac{-1}{1 - x}, x \neq 1\)
\(\frac{x}{1 - x}, x \neq 1\)
Correct answer is A
\(x * y = x + y - xy\)
Let \(x^{-1}\) be the inverse of x, so that
\(x * x^{-1} = x + x^{-1} - x(x^{-1}) = 0\)
\(x + x^{-1} - x(x^{-1}) = 0 \implies x(x^{-1}) - x^{-1} = x\)
\(x^{-1}(x - 1) = x \implies x^{-1} = \frac{x}{x - 1}\)
= \(\frac{x}{-(1 - x)} = \frac{-x}{1 - x}, x \neq 1\)
Express (14N, 240°) as a column vector.
\(\begin{pmatrix} -7 \\ -7\sqrt{3} \end{pmatrix}\)
\(\begin{pmatrix} 7\sqrt{3} \\ 7\sqrt{3} \end{pmatrix}\)
\(\begin{pmatrix} -7\sqrt{3} \\ -7 \end{pmatrix}\)
\(\begin{pmatrix} 7 \\ -7\sqrt{3} \end{pmatrix}\)
Correct answer is A
\(F = \begin{pmatrix} F_{x} \\ F_{y} \end{pmatrix} = \begin{pmatrix} F\cos \theta \\ F\sin \theta \end{pmatrix}\)
\((14N, 240°) = \begin{pmatrix} 14\cos 240 \\ 14\sin 240 \end{pmatrix}\)
= \(\begin{pmatrix} 14 \times -0.5 \\ 14 \times \frac{-\sqrt{3}}{2} \end{pmatrix}\)
= \(\begin{pmatrix} -7 \\ -7\sqrt{3} \end{pmatrix}\)