Explanation

(a) < XZY = 180° - (30° + 45°) = 105°

\(\frac{20}{\sin 105°} = \frac{YZ}{\sin 30°}\)

\(YZ = \frac{20 \sin 30°}{\sin 105°} = 10.35km\)

\(\approxeq 10.4km \) (to 3 sig. figs)

(ii) \(\frac{20}{\sin 75} = \frac{XZ}{\sin 45}\)

\(XZ = \frac{20 \sin 45}{\sin 75} = 14.64km\)

\(\approxeq 14.6km\) ( to 3 sig. figs)

(b) Let P and Q be two airfields.

Difference in latitude = 36° + 36° = 72°

(i) Distance between P and Q = \(\frac{72}{360} \times 2 \times \frac{22}{7} \times 6400\)

= \(\frac{22}{7} \times 2560\)

= \(8,045.71km \approxeq 8,050 km\) (3 sig. figs)

(ii) Time taken = \(\frac{8,045.71}{800} = 10.06 hours\)

\(\approxeq 10hours\) (to the nearest hour)

Correct Answer: B. y = -19

Explanation

\(\frac{y + 1}{2} - \frac{2y - 1}{3}\) = \(\frac{4}{1}\)

- \(\frac{3(y + 1) - 2(2y - 1)}{6} = \frac{4}{1}\)

3y + 3 - 4y + 2 = 24

- y + 5 = 24

- y = 24 - 5 = 19

y = - 19

Correct Answer: B. 109

Correct Answer: D. 0.88cm²S^-1

Correct Answer: A. \(\begin{pmatrix} -8 & -5 \\ 3 & 2 \end{pmatrix}\)

Explanation

Let \(\begin{pmatrix} a & b \\ c & d \end{pmatrix} = T^{-1}\)

\(T . T^{-1} = I\)

\(\begin{pmatrix} -2 & -5 \\ 3 & 8 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)

\(\implies -2a - 5c = 1\)

\(-2b - 5d = 0 \implies b = \frac{-5d}{2}\)

\(3a + 8c = 0 \implies a = \frac{-8c}{3}\)

\(3b + 8d = 1\)

\(-2(\frac{-8c}{3}) - 5c = \frac{16c}{3} - 5c = \frac{c}{3} = 1 \implies c = 3\)

\(3(\frac{-5d}{2}) + 8d = \frac{-15d}{2} + 8d = \frac{d}{2} = 1 \implies d = 2\)

\(b = \frac{-5 \times 2}{2} = -5\)

\(a = \frac{-8 \times 3}{3} = -8\)

\(\therefore T^{-1} = \begin{pmatrix} -8 & -5 \\ 3 & 2 \end{pmatrix}\)

Explanation

(a - d)

Correct Answer: C. 74

Explanation

common difference d = -1 - [-6] → -1 + 6

d = 5; a = -6

T\(_n\) = a + (n-1)d

: T\(_{17}\) = -6 + (17 - 1) 5

→ -6 + (16)5

→ -6 + 80 = 74

Correct Answer: D. Histogram

Correct Answer: C. (\(\frac{2}{3}, \frac{-4}{3}\))

Explanation

The equation for a circle with centre coordinates (a, b) and radius r is

\((x-a)^{2} + (y-b)^{2} = r^{2}\)

Expanding the above equation, we have

\(x^{2} - 2ax +a^{2} + y^{2} - 2by + b^{2} - r^{2} = 0\) so that

\(x^{2} - 2ax + y^{2} - 2by = r^{2} - a^{2} - b^{2}\)

Taking the original equation given, \(3x^{2} + 3y^{2} - 4x + 8y = 2\) and making the coefficients of \(x^{2}\) and \(y^{2}\) = 1,

\(x^{2} + y^{2} - \frac{4x}{3} + \frac{8y}{3} = \frac{2}{3}\), comparing, we have

\(2a = \frac{4}{3}; 2b = \frac{-8}{3}\)

\(\implies a = \frac{2}{3}; b = \frac{-4}{3}\)

Correct Answer: D. 2

Explanation

\(\frac{\log_{5} 8}{\log_{5} \sqrt{8}} = \frac{\log_{5} 8}{\log_{5} 8^{\frac{1}{2}}}\)

= \(\frac{\log_{5} 8}{\frac{1}{2}\log_{5} 8}\)

= \(\frac{1}{\frac{1}{2}} \)

= 2