19.5
17.0
14.5
12.5
Correct answer is C
\(\frac{dm}{dt}\) = 4 - 0.6t
\(\int\)dm = \(\int\)(4 - 0.6t)dt
m = \(4t - 0.3t^2 + c\), when t = 0, m = 2g
∴ c = 2
m = \(4t - 0.3t^2 + 2\), when t = 5 minutes
m = \(4(5) - 0.3(5)^2 + 2 = 20 - 7.5 + 2\)
= 14.5
If y = x sin x, Find \(\frac{d^2 y}{d^2 x}\)
2 cosx - x sinx
sinx + x cosx
sinx - x cosx
x sinx - 2 cosx
Correct answer is A
\(y = x \sin x\)
\(\frac{\mathrm d y}{\mathrm d x} = x \cos x + \sin x\)
\(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} = x (- \sin x) + \cos x + \cos x\)
= \(2 \cos x - x \sin x\)
Evaluate \(\lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4}\)
7
2
3
4
Correct answer is B
\(\lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4}\)
\(\frac{(x - 2)(x^{2} + 3x - 2)}{x^{2} - 4} = \frac{(x - 2)(x^{2} + 3x - 2)}{(x - 2)(x + 2)}\)
= \(\frac{(x^{2} + 3x - 2)}{x + 2}\)
\(\therefore \lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4} = \lim \limits_{x \to 2} \frac{x^{2} + 3x - 2}{x + 2}\)
= \(\frac{2^{2} + 3(2) - 2}{2 + 2}\)
= \(\frac{4 + 6 - 2}{4} = 2\)
Given that \(\theta\) is an acute angle and sin \(\theta\) = \(\frac{m}{n}\), find cos \(\theta\)
\(\frac{\sqrt{n^2 - m^2}}{m}\)
\(\frac{\sqrt{(n + m)(n - m)}}{n}\)
\(\frac{m}{\sqrt{n^2 - m^2}}\)
\(\sqrt{\frac{n}{n^2 - m^2}}\)
Correct answer is B
sin \(\theta\) = \(\frac{m}{n}\)
Opp = m; Hyp = n
Adj = \(\sqrt{n^{2} - m^{2}}\)
\(\cos \theta = \frac{\sqrt{n^{2} - m^{2}}}{n}\)
= \(\frac{\sqrt{(n + m)(n - m)}}{n}\)
Find then equation line through (5, 7) parallel to the line 7x + 5y = 12
5x + 7y = 120
7x + 5y = 70
x + y = 7
15x + 17y = 90
Correct answer is B
Equation (5, 7) parallel to the line 7x + 5y = 12
5Y = -7x + 12
y = \(\frac{-7x}{5}\) + \(\frac{12}{5}\)
Gradient = \(\frac{-7}{5}\)
∴ Required equation = \(\frac{y - 7}{x - 5}\) = \(\frac{-7}{5}\) i.e. 5y - 35 = -7x + 35
5y + 7x = 70