\(\frac{1}{3} seconds\)
\(\frac{3}{4} seconds\)
\(\frac{4}{3} seconds\)
\(2 seconds\)
Correct answer is C
Given \(a(t) = 3t - 2\), \(v(t) = \int (a(t)) \mathrm {d} t\)
= \(\int (3t - 2) \mathrm {d} t = \frac{3}{2}t^{2} - 2t\)
\(\frac{3}{2}t^{2} - 2t = 0 \implies t(\frac{3}{2}t - 2) = 0\)
\(t = \text{0 or t} = \frac{3}{2}t - 2 = 0 \implies t = \frac{4}{3}\)
The time t = 0 was the starting point, The next time v = 0 m/s is at \(t = \frac{4}{3} seconds\).
\((5\frac{2}{5}, \frac{2}{5})\)
\((\frac{2}{5}, 5\frac{2}{5})\)
\((\frac{-2}{5}, 5\frac{2}{5})\)
\((\frac{2}{5}, 4\frac{1}{5})\)
Correct answer is D
No explanation has been provided for this answer.
Find \(\int \frac{x^{3} + 5x + 1}{x^{3}} \mathrm {d} x\)
\(x^{2} + 10x + c\)
\(x + \frac{5}{3}x^{3} + x^{4} + c\)
\(x - 5x^{2} - 2x^{3} + c\)
\(x - \frac{5}{x} - \frac{1}{2x^{2}} + c\)
Correct answer is D
\(\frac{x^{3} + 5x + 1}{x^{3}} \equiv 1 + \frac{5}{x^{2}} + \frac{1}{x^{3}}\)
\(\equiv \int (1 + \frac{5}{x^{2}} + \frac{1}{x^{3}}) \mathrm {d} x = \int (1 + 5x^{-2} + x^{-3}) \mathrm {d} x\)
= \((x - 5x^{-1} - \frac{1}{2}x^{-2} + c)\)
= \(x - \frac{5}{x} - \frac{1}{2x^{2}} + c\).
If \(\begin{vmatrix} 1+2x & -1 \\ 6 & 3-x \end{vmatrix} = -3 \), find the values of x.
\(x = 3, -2\)
\(x = 4, \frac{-2}{3}\)
\(x = -4, \frac{3}{2}\)
\(x = 4, \frac{-3}{2}\)
Correct answer is D
\(\begin{vmatrix} 1+2x & -1 \\ 6 & 3-x \end{vmatrix} = -3 \implies (1+2x)(3-x) - (-6) = -3\)
\(3 - x + 6x - 2x^{2} + 6 = -3\)
\(-2x^{2} + 5x + 3 + 6 + 3 = 0\)
Multiplying through with -1,
\(2x^{2} - 5x -12 = 0\)
\((2x + 3)(x - 4) = 0 \implies x = \frac{-3}{2} , 4\)
\(\frac{28}{45}\)
\(\frac{4}{9}\)
\(\frac{8}{45}\)
\(\frac{1}{9}\)
Correct answer is B
Let the number of candidates that speak both Igbo and Hausa = x
The number that speaks Igbo only = 56 - x
The number that speaks Hausa only = 50 - x
56 - x + x + 50 - x = 90
106 - x = 90
x = 16
P(Igbo only) = \(\frac{56 - 16}{90} = \frac{4}{9}\)