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
\(F = \begin{pmatrix} F_{x} \\ F_{y} \end{pmatrix} = \begin{pmatrix} F\cos \theta \\ F\sin \theta \end{pmatrix}\)
\((14N, 240°) = \begin{pmatrix} 14\cos 240 \\ 14\sin 240 \end{pmatrix}\)
= \(\begin{pmatrix} 14 \times -0.5 \\ 14 \times \frac{-\sqrt{3}}{2} \end{pmatrix}\)
= \(\begin{pmatrix} -7 \\ -7\sqrt{3} \end{pmatrix}\)
Evaluate \(\frac{\tan 120° + \tan 30°}{\tan 120° - \tan 60°}\)
\(\sqrt{3} + \sqrt{2}\)
\(\frac{2}{3}\)
\(\frac{1}{3}\)
\(-2\sqrt{3}\)
Correct answer is C
No explanation has been provided for this answer.
Find \(\lim\limits_{x \to 3} (\frac{x^{3} + x^{2} - 12x}{x^{2} - 9})\)
\(\frac{7}{2}\)
\(0\)
\(\frac{-7}{2}\)
\(-7\)
Correct answer is A
\(\lim\limits_{x \to 3} (\frac{x^{3} + x^{2} - 12x}{x^{2} - 9}) = \lim\limits_{x \to 3} (\frac{x^{3} - 3x^{2} + 4x^{2} - 12x}{(x - 3)(x + 3)}\)
\(\lim\limits_{x \to 3} (\frac{(x^{2} + 4x)(x - 3)}{(x - 3)(x + 3)} = \lim\limits_{x \to 3} (\frac{x^{2} + 4x}{x + 3})\)
=\(\frac{3^{2} + 4(3)}{3 + 3} = \frac{21}{6} = \frac{7}{2}\)
Find the distance between the points (2, 5) and (5, 9).
4 units
5 units
12 units
14 units
Correct answer is B
Distance between two points (a, b) and (c, d) = \(\sqrt{(d - b)^{2} + (c - a)^{2}}
Distance between (2, 5) and (5, 9) = \(\sqrt{(9-5)^{2} + (5-2)^{2}}\)
= \(\sqrt{16 + 9} = \sqrt{25} = 5 units\)
15.25m
13.25m
11.25m
10.25m
Correct answer is C
Maximum height (H) = \(\frac{u^{2}}{2g}\)
= \(\frac{15^{2}}{2 \times 10} = \frac{225}{20}\)
= \(11.25m\)