Further Mathematics Questions and Answers

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 $t = \frac{4}{3} seconds$.

No explanation has been provided for this answer.

$\frac{x^{3} + 5x + 1}{x^{3}} \equiv  1 + \frac{5}{x^{2}} + \frac{1}{x^{3}}$

$\equiv \int (1 + \frac{5}{x^{2}} + \frac{1}{x^{3}}) \mathrm {d} x = \int (1 + 5x^{-2} + x^{-3}) \mathrm {d} x$

= $(x - 5x^{-1} - \frac{1}{2}x^{-2} + c)$

= $x - \frac{5}{x} - \frac{1}{2x^{2}} + c$.

$\begin{vmatrix} 1+2x & -1 \\ 6 & 3-x  \end{vmatrix} =  -3 \implies (1+2x)(3-x) - (-6) = -3$

$3 - x + 6x - 2x^{2} + 6 = -3$

$-2x^{2} + 5x + 3 + 6 + 3 = 0$

Multiplying through with -1, 

$2x^{2} - 5x -12 = 0$

$(2x + 3)(x - 4) = 0 \implies x = \frac{-3}{2} , 4$

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) = $\frac{56 - 16}{90} = \frac{4}{9}$