0 and 1
0 and 4
0 and 5
1 and 4
Correct answer is D
\(s = ut + \frac{at^{2}}{2}\)
This movement is against gravity, so it is negative.
\(s = ut - \frac{gt^{2}}{2}\)
\(s = 20m, u = 25ms^{-1}\)
\(20 = 25t - \frac{10t^{2}}{2} \implies 20 = 25t - 5t^{2}\)
\(5t^{2} - 25t + 20 = 0 \)
\(5t^{2} - 5t - 20t + 20 = 0 \implies 5t(t - 1) - 20(t - 1) = 0\)
\(5t - 20 = \text{0 or t - 1 = 0}\)
\(t = \text{1 or 4}\)
-4
-3
-1
2
Correct answer is B
\(P = begin{pmatrix} y - 2 & y - 1 \\ y - 4 & y + 2 \end{pmatrix}\)
\(|P| = (y - 2)(y + 2) - (y - 1)(y - 4) = (y^{2} - 4) - (y^{2} - 5y + 4) = -23\)
\(5y - 8 = -23 \implies 5y = -23 + 8 = -15\)
\(y = \frac{-15}{5} = -3\)
Given that \(\frac{\mathrm d y}{\mathrm d x} = \sqrt{x}\), find y.
\(2x^{\frac{3}{2}} + c\)
\(\frac{2}{3}x^{\frac{3}{2}} + c\)
\(\frac{3}{2}x^{\frac{3}{2}} + c\)
\(\frac{2}{3}x^{2} + c\)
Correct answer is B
\(\frac{\mathrm d y}{\mathrm d x} = \sqrt{x} = x^{\frac{1}{2}}\)
\(y = \int x^{\frac{1}{2}} \mathrm {d} x\)
= \(\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + c = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + c \)
= \(\frac{2}{3}x^{\frac{3}{2}} + c\)
Find the range of values of x for which \(x^{2} + 4x + 5\) is less than \(3x^{2} - x + 2\)
\(x > \frac{-1}{2}, x > 3\)
\(x < \frac{-1}{2}, x > 3\)
\(\frac{-1}{2} \leq x \leq 3\)
\(\frac{-1}{2} < x < 3\)
Correct answer is B
No explanation has been provided for this answer.
\(\frac{1}{3}\)
\(\frac{1}{2}\)
\(2\)
\(3\)
Correct answer is B
\(T_{n} = ar^{n - 1}\)
\(T_{4} = ar^{4 - 1} = ar^{3} = 192\)
\(T_{9} = ar^{9 - 1} = ar^{8} = 6\)
Dividing \(T_{9}\) by \(T_{4}\),
\(r^{8 - 3} = \frac{6}{192}\)
\(r^{5} = \frac{1}{32} = (\frac{1}{2})^{5}\)
\(r = \frac{1}{2}\)