JAMB Mathematics Past Questions & Answers - Page 13

\(3x + 2y + 4 = 0\)

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

If the majority are women implies that:

3 Women and 2 Men Or 4 Women and 1 Man

= \(^4C_3 \times^6C_2+^4C_4\times^6C_1\)

= \(4 \times 15 + 1 \times 6\)

= 60 + 6 = 66