Express log18+log12 in terms of log2
3 log 2
4 log 2
-3 log 2
-4 log 2
Correct answer is D
log18+log12=log8−1+log2−1
= log2−3+log2−1
= −3log2−1log2=−4log2
Given that a56×a−1n=1, solve for n
-6.00
-1.20
0.83
1.20
Correct answer is D
a56×a−1n=1
⟹a56+−1n=a0
Equating bases, we have
56−1n=0
5n−66n=0
5n−6=0⟹5n=6
n=65=1.20
200°
90°
60°
0°
Correct answer is D
sinθ=tanθ⟹sinθ1=sinθcosθ
Equating, we have
cosθ=1⟹θ=cos−11
= 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}