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) = \(\frac{u^{2}}{2g}\)
= \(\frac{15^{2}}{2 \times 10} = \frac{225}{20}\)
= \(11.25m\)
t = -2 and 3
t = 2 and -3
t = 2 and 3
t = -2 and -3
Correct answer is C
For collinear points (points on the same line), the slopes are equal for any 2 points on the line.
Given (-1, t - 1), (t, t - 3), (t - 6, 3),
\(slope = \frac{(t-3) - (t-1)}{t - (-1)} = \frac{3 - (t-3)}{(t-6) - t} = \frac{3 - (t-1)}{(t-6) - (-1)}\)
Taking any two of the equations above, solve for t.
\(\frac{t - 3 - t + 1}{t + 1} = \frac{6 -t}{-6}\)
\(12 = (6 - t)(t + 1)\)
\(-t^{2} + 5t + 6 - 12 = 0 \implies t^{2} - 5t + 6 = 0\)
Solving, we have t = 2 and 3.