Explanation

\(\frac{3x^2 + 3x - 2}{(x - 1)(x + 1)}\) = P + \(\frac{Q}{x - 1} + \frac{R}{x - 1}\)

Long division

\(\frac{6x^{-2}}{(x - 1)(x + 1)}\)

= \(\frac{Q}{x - 1} + \frac{R}{x + 1}\)

6x - 2 = Q (x + 1) + R(x - 1)

Let x = -1

6(-1) - 2 = Q(-1 + 1) + R (-1 - 1)

-6 - 2 = -2R

\(\frac{-8}{-2} = \frac{-2R}{-2}\)

R = 4

From 6x - 2 = Q(x - 1) + R(x - 1)

Let x = 1

6(1) - 2 = Q(1 + 1) + R(1 - 1)

6 - 2 = 2Q

\(\frac{4}{2} = \frac{2Q}{2}\)

= Q = 2

Correct Answer: B. 87°

Explanation

\(a . b = |a||b| \cos \theta\)

\(a = 3i - 4j; b = 6i + 4j\)

\(18 - 16 = (\sqrt{3^{2} + (-4)^{2}})(\sqrt{6^{2} + 4^{2}}) \cos \theta\)

\(2 = 5\sqrt{52} \cos \theta\)

\(\cos \theta = \frac{2}{5\sqrt{52}} = 0.0555\)

\(\theta = 86.8° \approxeq 87°\)

Correct Answer: C. 7

Explanation

7km has the highest frequency(14)

Explanation

Using Lami's Theorem;

(a) \(\frac{T}{\sin 90^o} = \frac{120}{\sin 135^o}\) and T = 169.71N

(b) \(\frac{R}{\sin 135^o} = \frac{120}{\sin 135^o}\) and R = 120N

Explanation

60% boys, 40% girls.

p(boy) = p = 60% = 0.6

p(girl) = q = 40% = 0.4

The binomial distribution function is

\((p + q)^{5} = p^{5} + 5p^{4} q + 10p^{3} q^{2} + 10p^{2} q^{3} + 5pq^{4} + q^{5}\)

(a) p(exactly 3 boys) = \(10p^{3} q^{2}\)

\(10(\frac{3}{5})^{3} (\frac{2}{5})^{2}\)

= \(10(\frac{27}{125})(\frac{4}{25})\)

= \(\frac{1080}{3125}\)

= \(\frac{216}{625}\)

(b) p(at least 3 are girls) = 1 - [p(0 girls) + p(1 girl) + p(2 girls)].

= \(1 - [p^{5} + 5p^{4}q + 10p^{3} q^{2}]\)

= \(1 - [(\frac{3}{5})^{5} + 5(\frac{3}{5})^{4} (\frac{2}{5}) + 10(\frac{3}{5})^{3} (\frac{2}{5})^{2}]\)

= \(1 - [\frac{243}{3125} + \frac{810}{3125} + \frac{1080}{3125}]\)

= \(1 - [\frac{2133}{3125}]\)

= \(\frac{992}{3125}\).

Correct Answer: A. \(x^{3} - 4x - 9\)

Explanation

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 4\)

\(y = \int (3x^{2} - 4) \mathrm {d} x = x^{3} - 4x + c\)

y = 6 when x = 3

\(6 = 3^{3} - 4(3) + c \implies 6 = 27 - 12 + c\)

\(c = 6 - 15 = -9\)

\(y = x^{3} - 4x - 9\)

Explanation

P : 5r, 3b, 2w

Q : 3r, 2b, 4w

For Mr P: \(P(r) = \frac{1}{2}; P(b) = \frac{3}{10}; P(w) = \frac{1}{5}\)

For Mr Q: \(P(r) = \frac{1}{3}; P(b) = \frac{2}{9}; P(w) = \frac{4}{9}\)

\(P(\text{ bus of the same color}) = \frac{1}{2} \times \frac{1}{3} + \frac{3}{10} \times \frac{2}{9} + \frac{1}{5} \times \frac{4}{9}\)

= \(\frac{29}{90}\)

\(P(\text{not of the same color}) = 1 - \frac{29}{90} = \frac{61}{90}\)

(b) 6 airlines, 4 shipping lines

If 1 airline is used from Nigeria to Ghana, there will be 5 out of 6 for the return journey. This can be done in \(^{6}C_{5}\) ways. Similarly, if 1 shipping line is used from Ghana to Liberia, there will be 3 shipping lines left for his return journey. This can be done in \(^{4}C_{3}\) ways.

\(\therefore \text{The total number of ways =} ^{6}C_{5} + ^{4}C_{3}\)

= \(6 + 4 = 10\) ways.

Correct Answer: B. \(\frac{7x^{6}}{16}\)

Explanation

\((\frac{x}{2} - 1)^{8} = ^{8}C_{8}(\frac{x}{2})^{8}(-1)^{0} + ^{8}C_{7}(\frac{x}{2})^{7}(-1)^{1} + ^{8}C_{6}(\frac{x}{2})^{6}(-1)^{2} + ...\)

\(\text{The third term in the expansion =} ^{8}C_{6}(\frac{x}{2})^{6}(-1)^{2}\)

= \(\frac{8!}{6!2!}(\frac{x^{6}}{64})(1) \)

= \(28 \times \frac{x^{6}}{64} = \frac{7x^{6}}{16}\)

Correct Answer: B. \(\frac{1}{10}\)

Explanation

Let the number of mixed schools be S

The number of boys only = 42 - S

The number of girls only = 35 - S

(42 - S) + S + (35 - S) = 77 - S = 70

\(\implies\) S = 7

\(Probability = \frac{7}{70} = \frac{1}{10}\)

Correct Answer: B. 22\(cm^{2}s^{-1}\)

Explanation

With radius = 7cm, \(Area = \pi r^{2} = \frac{22}{7} \times 7^{2}\)

= \(154cm^{2}\)

The next second, radius = 7.5cm, \(Area = \pi r^{2} = \frac{22}{7} \times 7.5^{2}\)

= \(176cm^{2}\)

Change in area = \((176 - 154)cm^{2} = 22cm^{2}\)

\(\therefore\) The rate of increase = \(22cm^{2}s^{-1}\)

OR

\(Area (A) = \pi r^{2} \implies \frac{\mathrm d A}{\mathrm d r} = 2\pi r\)

Given \(\frac{\mathrm d r}{\mathrm d t} = 0.5\)

\(\frac{\mathrm d A}{\mathrm d r} \times \frac{\mathrm d r}{\mathrm d t} = \frac{\mathrm d A}{\mathrm d t}\)

\(\frac{\mathrm d A}{\mathrm d t} = 2\pi r \times 0.5 = 2 \times \frac{22}{7} \times 7 \times 0.5\)

= \(22cm^{2}s^{-1}\)