Correct Answer: D. 9.46

Explanation

log x = \(\bar{2}.3675\) ; log y = 0.9750

\(x = 10^{\bar{2}.3675} = 0.02331 \)

\(y = 10^{0.9750} = 9.441 \)

\(x + y = 9.4641 \approxeq 9.46\)

Explanation

(a) \(p(Kunle) = \frac{1}{3} ; p(\text{not Kunle}) = \frac{2}{3}\)

\(p(Tayo) = \frac{1}{5} ; p(\text{not Tayo}) = \frac{4}{5}\)

\(p(\text{only one solve the question}) = p(\text{Kunle and not Tayo}) + p(\text{Tayo and not Kunle})\)

= \((\frac{1}{3} \times \frac{4}{5}) + (\frac{1}{5} \times \frac{2}{3})\)

= \(\frac{4}{15} + \frac{2}{15}\)

= \(\frac{6}{15} = \frac{2}{5}\)

(b) 10 persons to choose 8

Number of ways = \(^{10}C_{8}\)

= \(\frac{10!}{(10 - 8)! 8!}\)

= \(\frac{10 \times 9}{2}\)

= 45 ways.

Correct Answer: A. \(\frac{7}{2}\)

Explanation

\(\lim\limits_{x \to 3} (\frac{x^{3} + x^{2} - 12x}{x^{2} - 9}) = \lim\limits_{x \to 3} (\frac{x^{3} - 3x^{2} + 4x^{2} - 12x}{(x - 3)(x + 3)}\)

\(\lim\limits_{x \to 3} (\frac{(x^{2} + 4x)(x - 3)}{(x - 3)(x + 3)} = \lim\limits_{x \to 3} (\frac{x^{2} + 4x}{x + 3})\)

=\(\frac{3^{2} + 4(3)}{3 + 3} = \frac{21}{6} = \frac{7}{2}\)

Correct Answer: a. 800,000 votes

Correct Answer: B. (18 + \(\frac{9\pi}{5}\))cm

Correct Answer: C. 128°

Explanation

x + 52^o = 90

x = 90 - 52

x = 38^o

k = opposite angle Z

k = 38^o

y + k = 90^o

y + 38^o = 90^o

y = 90^o - 38^o

y = 52^o

y = x = 180^o(sum of angles on straight line)

52 + x = 180^o

x = 180 - 52

x = 128

Explanation

(a)

< ABC = 100°

\(\therefore b^{2} = 8^{2} + 13^{2} - 2(13)(8) \cos 100°\)

= \(64 + 169 - (208 \times - 0.1736)\)

= \(233 + 36.12\)

\(b^{2} = 269.12 \implies b = \sqrt{269.12} = 16.405 km\)

\(\approxeq 16.4 km\) (3 significant figures)

(b) \(\frac{\sin B}{b} = \frac{\sin A}{a}\)

\(\frac{\sin 100}{16.4} = \frac{\sin A}{13}\)

\(\sin A = \frac{13 \times \sin 100}{16.4}\)

\(\sin A = \frac{12.803}{16.4}\)

\(\sin A = 0.7806\)

\(A = \sin^{-1} (0.7806) = 51.32°\)

\(\therefore\) Bearing of C from A = 180° - (50° + 51.32°)

= 180° - 101.32°

= 78.68° \(\approxeq\) 78.7° (3 significant figure).

(c) \(\cos 40 = \frac{BD}{13}\)

\(BD = 13 \cos 40\)

= \(13 \times 0.7660\)

= 9.959 km

\(\approxeq\) 9.96 km east of B.

Explanation

The sum of exterior angles = 360°

\(60° + 2(90 - m)° + 2(30 + 2m)° = 360°\)

\(60° + 180° - 2m° + 60° + 4m° = 360°\)

\(300° + 2m = 360°\)

\(\implies 2m = 360° - 300° = 60°\)

\(m = 30°\)

(b)

In \(\Delta QRS, \cos Q = \frac{\sqrt{3}}{2}\)

\(\cos Q = 0.8660 \implies Q = 30°\)

\(\therefore < SQR = 30°^\)

\(< PQS = 180° - 30° \) (angles on a straight line = 180°)

\(< PQS = 150°\)

Correct Answer: D. -2 + 2\(\sqrt2\) m

Explanation

Perimeter of a triangle = sum of all sides

⇒ \(P = y + x + x = 2\)

⇒ \(y + 2x = 2\)

⇒ \(y= 2 - 2x\)-----(i)

Using Pythagoras theorem

\(y^2 = x^2 + x^2\)

⇒ \(y^2 = 2x^2\)

⇒ \(y = \sqrt2x^2\)

⇒ \(y = x\sqrt2\)-----(ii)

Equate \(y\)

⇒ \(2 - 2x = x\sqrt2\)

Square both sides

⇒ \((2 -2x) ^2 = (x\sqrt2)^2\)

⇒ \(4 - 8x + 4x^2 = 2x^2\)

⇒ \(4 - 8x + 4x^2 - 2x^2 = 0\)

⇒ \(2x^2 - 8x + 4 = 0\)

⇒ \(x = \frac{-(-8)\pm\sqrt(-8)^2 - 4\times2\times4}{2\times2}\)

⇒ \(x = \frac{8\pm\sqrt32}{4}\)

⇒ \(x = \frac{8\pm4\sqrt2}{4}\)

⇒ \(x = 2\pm\sqrt2\)

⇒ \(x = 2 + \sqrt2\) or \(2 - \sqrt2\)

∴ \(x = 2 - \sqrt2\) (for \(x\) has to be less than its perimeter)

∴ \(y = 2 - 2x = 2 - 2(2 - \sqrt2) = -2 + 2 \sqrt2\)

∴ The length of the longer side = -2 + 2\(\sqrt2\)m

Correct Answer: B. -3x² + 9x - 1