\(26 ms^{-2}\)
\(18 ms^{-2}\)
\(17 ms^{-2}\)
\(16 ms^{-2}\)
Correct answer is D
\(v(t) = 3t^{2} - 2t + 1\)
\(\frac{\mathrm d v}{\mathrm d t} = a(t) = 6t - 2\)
\(a(3) = 6(3) - 2 = 18 - 2 = 16 ms^{-2}\)
13 metres
15 metres
18 metres
23 metres
Correct answer is A
\(v^{2} = u^{2} + 2as\)
\(v^{2} = u^{2} - 2gs\)
\(0 = 10^{2} - 2(10)s \implies -100 = -20s\)
\(s = 5 m + 8 m = 13m\) (8m is the height from where it was thrown)
Which of the following is the semi- interquartile range of a distribution?
\(Mode - Median\)
\(\text{Highest score - Lowest score}\)
\(\frac{1}{2}(\text{Upper quartile - Median})\)
\(\frac{1}{2}(\text{Upper quartile - Lower quartile})\)
Correct answer is D
No explanation has been provided for this answer.
If \(P = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\), find \((P^{2} + P)\).
\(\begin{vmatrix} 4 & 3 \\ 6 & 1 \end{vmatrix}\)
\(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\)
\(\begin{vmatrix} 2 & 2 \\ 6 & 2 \end{vmatrix}\)
\(\begin{vmatrix} 3 & 2 \\ 6 & 4 \end{vmatrix}\)
Correct answer is B
\( P^{2} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)
\(\begin{vmatrix} 1 \times 1 + 1 \times 2 & 1 \times 1 + 1 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 1 + 1 \times 1 \end{vmatrix}\)
= \(\begin{vmatrix} 3 & 2 \\ 4 & 3 \end{vmatrix} + \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)
= \(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\)
Evaluate \(\int_{1}^{2} [\frac{x^{3} - 1}{x^{2}}] \mathrm {d} x\)
0.5
1.0
1.5
2.0
Correct answer is B
\(\frac{x^{3} - 1}{x^{2}} \equiv x - \frac{1}{x^{2}} = x - x^{-2}\)
\(\int_{1}^{2} (x - x^{-2}) \mathrm {d} x = (\frac{x^{2}}{2} + \frac{1}{x})|_{1}^{2}\)
= \((\frac{2^{2}}{2} + \frac{1}{2}) - (\frac{1^{2}}{2} + \frac{1}{1})\)
= \((2 + \frac{1}{2}) - (\frac{1}{2} + 1)\)
= \(2\frac{1}{2} - 1\frac{1}{2}\)
= \(1.0\)