Correct Answer: B. 50( \(\sqrt{3}\) - \(\sqrt{3}\))m

Correct Answer: D. pair of bisectors of the angles between the two lines

Explanation

The locus of a point equidistant from two intersecting lines is pair of bisectors of the angles between the two lines.

Explanation

A.P ; \(S_{12} = 168 ; T_{3} = 7\)

\(S_{n} = \frac{n}{2}(2a + (n - 1) d)\)

\(T_{n} = a + (n - 1)d\)

\(168 = \frac{12}{2}(2a + (12 - 1)d \implies 168 = 6(2a + 11d)\)

\(168 = 12a + 66d \implies 84 = 6a + 33d ..... (1)\)

\(7 = a + 2d .... (2)\)

From (2), a = 7 - 2d. Put into (1), we have

\(84 = 6(7 - 2d) + 33d \)

\(84 = 42 - 12d + 33d \implies 84 - 42 = 21d\)

\(21d = 42 \implies d = 2\)

\(a = 7 - 2(2) = 7 - 4 = 3\)

\(a(\text{first term}) = 3 ; d(\text{common difference}) = 2\)

Correct Answer: D. {-9, -5, -1, 3, 7}

Explanation

The elements of x are {-2, -1, 0, 1, 2}

\(y = 4x - 1\) = 4(-2) - 1 = -9; 4(-1) - 1 = -5; 4(0) - 1 = -1; 4(1) - 1 = 3; 4(2) - 1 = 7.

The range of x is {-9, -5, -1, 3 7}.

Correct Answer: B. 210^o

Correct Answer: D. 70, 500

Correct Answer: B. 24.50°

Explanation

\(p . q = |p||q|\cos \theta\)

\(156 + 70 = (\sqrt{13^{2} + 14^{2}})(\sqrt{12^{2} + 5^{2}}) \cos \theta\)

\(226 = (\sqrt{365})(13) \cos \theta\)

\(\frac{226}{13\sqrt{365}} = \cos \theta\)

\(\cos \theta = 0.9099\)

\(\theta = 24.50°\)

Explanation

(a) \(5 \cos (x + 8.5)° - 1 = 0\)

\(5 \cos (x + 8.5)° = 1\)

\(\cos (x + 8.5)° = \frac{1}{5} = 0.2\)

\((x + 8.5)° = \cos^{-1}(0.2) = 78.463°\)

\(x = 78.463° - 8.5 = 69.963°\)

\(\approxeq 70°\) (to the nearest degree).

(b)(i)

(ii) By the cosine rule,

\(|QR|^{2} = 32^{2} + 24^{2} - 2 \times 32 \times 24 \times \cos 45\)

\(|QR|^{2} = 1024 + 576 - 1536 \cos 45\)

= \(1600 - 1086.1056\)

\(|QR|^{2} = 513.8944\)

\(|QR| = \sqrt{513.8944} = 22.669 km\)

\(\approxeq 22.67 km\) ( 2 decimal place)

(iii) By the sine rule,

\(\frac{32}{\sin \alpha} = \frac{22.67}{\sin 45}\)

\(\sin \alpha = \frac{32 \times \sin 45}{22.67}\)

= \(0.9981\)

\(\alpha = \sin^{-1} (0.9981) = 86.4787°\)

The diagram below shows all the angles at Q;

The bearing of R from Q is given by the reflex angle NQR. Thus

reflex < NQR = 360° - (86.47° + 30°) = 360° - 116.47°

= 243.53°

Hence, the bearing of R from Q = 244° (to the nearest degree).

(a) \(5 \cos (x + 8.5)° - 1 = 0\)

\(5 \cos (x + 8.5)° = 1\)

\(\cos (x + 8.5)° = \frac{1}{5} = 0.2\)

\((x + 8.5)° = \cos^{-1}(0.2) = 78.463°\)

\(x = 78.463° - 8.5 = 69.963°\)

\(\approxeq 70°\) (to the nearest degree).

Correct Answer: C. \(\frac{1}{8} (2x + 1)^4 + k\)

Explanation

\(\int(2x + 1)^3 dx\)

Using substitution method, Let \(u = 2x + 1\)

\(\frac{du}{dx}=2==>du=2dx==>dx=\frac{du}{2}\)

=\(\int\frac{u^3}{2} du = \frac{1}{2}\int u^3 du\)

=\(\frac{1}{2}(\frac{u^4}{4})=\frac{u^4}{8}\)

\(\therefore\frac{1}{8} (2x + 1)^4 + k\)

Explanation

Spearmann's rank correlation coefficient = \(1 - \frac{6\sum D^{2}}{N(N^{2} - 1)}\)

= \(1 - \frac{6(36)}{8(64 - 1)}\)

= \(1 - \frac{216}{504}\)

= \(1 - 0.4286\)

= \(0.5714\)

\(\approxeq 0.57\).