Correct Answer: B. 2[1-3m]

Explanation

\(\frac{2-18m^2}{1+3m}\) = \(\frac{2[1-9m^2]}{1+3m}\)

\(\frac{2[1-3m][1+3m]}{1+3m}\)

2[1-3m]

Correct Answer: B. 9/1000

Correct Answer: (b) 15sq.cm

Correct Answer: D. 45.5

Explanation

Cubic numbers are numbers taken to the third power. The first six cubic numbers are: 1

Correct Answer: A. -4 < X ≤ 5

Explanation

Explanation

(a) \(p(head) = p = \frac{1}{2}; p(tail) = q = \frac{1}{2}\)

\((p + q)^{8} = p^{8} + 8p^{7} q + 28p^{6} q^{2} + 56p^{5} q^{3} + 70p^{4} q^{4} + 56p^{3} q^{5} + 28p^{2} q^{6} + 8pq^{7} + q^{8}\)

(i) p(exactly 8 heads) = \(p^{8} = (\frac{1}{2})^{8} = \frac{1}{256}\)

(ii) p(5 heads) = \(56p^{5} q^{3} = 56(\frac{1}{2})^{5} (\frac{1}{2})^{3} = \frac{7}{32}\)

p(6 heads) = \(28p^{6} q^{2} = 28(\frac{1}{2})^{6} (\frac{1}{2})^{2} = \frac{7}{64}\)

p(7 heads) = \(8p^{7} q = 8(\frac{1}{2})^{7} (\frac{1}{2}) = \frac{1}{32}\)

p(at least 5 heads) = \(\frac{7}{32} + \frac{7}{64} + \frac{1}{32} + \frac{1}{256} = \frac{93}{256}\)

(iii) p(at most 1 head) = p(0 head) + p(1 head)

p(0 head) = \(q^{8} = (\frac{1}{2})^{8} = \frac{1}{256}\)

p(1 head) = \(8p q^{7} = 8(\frac{1}{2})(\frac{1}{2})^{7} = \frac{8}{256}\)

p(at most 1 head) = \(\frac{1}{256} + \frac{8}{256} = \frac{9}{256}\)

(b) The word SHEEP has 5 letters, two of which are identical.

(i) Hence 4 letters from the word SHEEP can be arranged in \(\frac{^{5} P_{4}}{2!}\) ways.

= \(\frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 60\).

(ii) The two Es are identical. If one of them are taken, there are 4 letters to be arranged in \(^{4} P_{4}\) ways.

= \(\frac{4!}{(4 - 4)!} = 4! = 24\).

Explanation

(a) \(m \begin{pmatrix} 2 \\ 1 \end{pmatrix} + n \begin{pmatrix} -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -4 \end{pmatrix}\)

\(2m - n = 5 ...(1)\)

\(m + 2n = -4 ... (2)\)

\((1) \times 2 : 4m - 2n = 10 ... (3)\)

\((2) + (3) : 5m = 6 \implies m = \frac{6}{5}\)

\(2(\frac{6}{5}) - n = 5 \implies n = \frac{12}{5} - 5 = \frac{-13}{5}\)

\(\therefore (m + n) = (\frac{6}{5} + \frac{-13}{5}) = -\frac{7}{5}\).

(b) A(-1, 3) = (-i + 3j)

B(2, -1) = (2i - j)

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\)

\(\overrightarrow{AB} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} - \begin{pmatrix} -1 \\ 3 \end{pmatrix}\)

= \(\overrightarrow{AB} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}\)

\(|\overrightarrow{AB}| = \sqrt{3^{2} + (-4)^{2}} = 5\)

Unit vector parallel to AB = \(\frac{1}{5} \begin{pmatrix} 3 \\ -4 \end{pmatrix}\)

\(\overrightarrow{OA} + \overrightarrow{AC} = \overrightarrow{OC}\)

\(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA}\)

= \(\begin{pmatrix} 5 \\3 \end{pmatrix} - \begin{pmatrix} -1 \\ 3 \end{pmatrix}\)

= \(\begin{pmatrix} 6 \\ 0 \end{pmatrix}\)

\(|\overrightarrow{AC}| = 6\)

Unit vector parallel to AC = \(\frac{1}{6} \begin{pmatrix} 6 \\ 0 \end{pmatrix}\)

(ii) Let the angle BAC = \(\theta\)

\(\overrightarrow{AB} \cdot \overrightarrow{AC} = |\overrightarrow{AB}||\overrightarrow{AC}| \cos \theta\)

\((3i - 4j) \cdot (6i + 0j) = |3i + 4j||6i + 0j| \cos \theta\)

\(18 = 5 \times 6 \times \cos \theta \implies \cos \theta = \frac{18}{30} = 0.6\)

\(\theta = \cos^{-1} (0.6) = 53.1°\)

Correct Answer: C. 101011₂

Correct Answer: A. 84.7