Mathematics Questions and Answers

y = $\sqrt[3]{x^2(2x - x^2)} = x^{2/3} (2x - x^2)$

= $2x^{5/3} - x^{8/3}$

Now, we can differentiate the function

$\therefore \frac {dy}{dx} = \frac {10x^{2/3}}{3} - \frac {8x^{5/3}}{3}$

$3x + 2y + 4 = 0$

Rearrange:

$2y = -3x - 4$

Divide both sides by 2

y = $\frac {-3 \times - 4}{2}$

y = $\frac {-3}{2} \times - 2$

∴ the gradient of the line 3x + 2y + 4 = 0 is $\frac {-3}{2}$

If two lines are perpendicular to each other ∴ $m_1 x m_2$ = -1

Let $m_1 = \frac {-3}{2} \therefore m_2 = \frac {-1}{m_1} = \frac {-1}{-3/2} = \frac {2}{3}$

From the equation of a line which is given as m = $\frac {y - y_1}{x - x_1} where (x_1, y_1) = (2,3)$

$\therefore \frac {2}{3} = \frac {y - 3}{x - 2}$

=3(y - 3) = 2(x - 2)

=3y - 9 = 2 x -4

=3y = 2 x -4 + 9

∴ 3y = 2x + 5

Sample space = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}

Total number of possible outcomes = 2 x 2 x 2 = 8

favourable number of outcomes = 7

∴Pr(at least one head) = $\frac {7}{8}$

tan θ = $\frac {opp}{adj} = \frac {RQ}{QP} = \frac {6}{8}$

tan θ = 0.75

θ = tan$^{-1} (0.75) = 36.87^o$

∴ The bearing of R from P = 360$^o$ - 36.87$^o$ = 323$^o$ (to the nearest degree)

$\begin{bmatrix}2& -1&3\\4&1&2\\1&-3&1\\\end{bmatrix}$

= 2[(1 x 1) - (2 x -3)] - (-1)[(4 x 1) - (2 x 1)] + 3[(4 x -3) - (1 x 1)]

= 2(1 - (-6)) + 1(4 - 2) + 3(-12 - 1)

= 2(1 + 6) + 1(2) + 3(-13)

= 2(7) + 1(2) + 3(-13)

= 14 + 2 - 39

∴ |D| = -23