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

Explanation

There are 3 triangles: \(\bigtriangleup\)MRS, \(\bigtriangleup\) and \(\bigtriangleup\) PRQ

Area PRMS has 2 triangles: \(\bigtriangleup\) MRS and \(\bigtriangleup\) RSP

the fraction = \(\frac{2}{3}\)

Correct Answer: A. N48.75K

Explanation

(a) The second term is a d= x - 1 while the fourth term is a + 3d = x + 1. Subtracting the second tam from the fourth term yielded 2d = 2 so that d = 1.

(b). Substituting for d in the expression for the sixth term to get a + 5 = 7 and solving for a to obtain a = 2.

(c). Substituted for a and d in the expression for the second term and found x = 2 + 1 + 1 = 4.

Explanation

F\(_1\) = (10N, 090\(^o\)), F\(_2\) = (20N, 210\(^o\))

F\(_3\) = (4N, 33\(^o\))

F\(_3\) component x - axis

= - 4 cos 60\(^o\) = - 2N

F\(_2\) along y - axis

= 20 x cos 30\(^o\) = -1.7320

\(\frac{8N}{-1.732N}\)

Resultant force

= \(\sqrt{8^2 + (-1.732)^2}\)

= \(\sqrt{65.999}\)

= 8.185N

= 8.2N

Explanation

(a)(i) \(\log_{10} 35 = \log_{10} (7 \times 5) = \log_{10} 7 + \log_{10} 5\)

= \(0.8451 + 0.6990 = 1.5441\)

(ii) \(\log_{10} 2.8 = \log_{10} (\frac{4 \times 7}{10} = \log_{10} 4 + \log_{10} 7 - \log_{10} 10\)

= \(\log_{10} 2^{2} + \log_{10} 7 - 1\)

= \(2\log_{10} 2 + \log_{10} 7 - 1 = 2(0.3010) + 0.8451 - 1\)

= \(0.4471\)

(b) \(N^{0.8942} = 2.8\)

\(0.8942 \log N = \log 2.8\)

\(\log N = \frac{\log 2.8}{0.8942} = \frac{0.4471}{0.8942}\)

\(\log N = 0.5\)

Explanation

(a) \(3y^{2} - 5y + 2 = 0\)

\(\frac{3y^{2}}{3} - \frac{5y}{3} = \frac{-2}{3}\)

\(y^{2} - \frac{5}{3}y = \frac{-2}{3}\)

\(\frac{1}{2} of \frac{-5}{3} = \frac{-5}{6}\)

\(y^{2} - \frac{5}{3}y + (-\frac{5}{6})^{2} = \frac{-2}{3} + (\frac{-5}{6})^{2}\)

\((y - (\frac{5}{6}))^{2} = \frac{1}{36}\)

Taking square root of both sides,

\(\sqrt{(y - (\frac{5}{6}))^{2}} = \sqrt{\frac{1}{36}}\)

\(y - \frac{5}{6} =\pm {\frac{1}{6}}\)

\(y = \frac{5}{6} \pm \frac{1}{6}\)

\(y = \frac{5}{6} + \frac{1}{6} = 1.00\) or \(y = \frac{5}{6} - \frac{1}{6} = \frac{2}{3} = 0.67\).

(b) \(M = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}, N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}\)

\(MN = \begin{pmatrix} m + 2n & x + 2y \\ 4m + 2n & 4x + 2y \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}\)

\(m + 2n = 2 ... (1)\)

\(4m + 2n = 3 ... (2)\)

Multiplying (1) by 4, we have

\(4m + 8n = 8 ... (3)\)

(3) - (1) : \(8n - 2n = 8 - 2 = 6\)

\(6n = 6 \implies n = 1\)

\(m + 2n = m + 2(1) = m + 2 = 2\)

\(m = 2 - 2 = 0\)

\(x + 2y = 1 ... (1)\)

\(4x + y = 4 ... (2)\)

Multiply (1) by 4,

\(4x + 8y = 4 ... (3)\)

(3) - (2) : \(8y - y = 4 - 4 = 0\)

\(\implies y = 0\)

\(x + 2y = x + 2(0) = x + 0 = 1\)

\(x = 1 - 0 = 1\)

\(\therefore N = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).

Correct Answer: D. 210

Explanation

For choosing, its different 'combinations'

Options - 7 men, 5 women

To pick - 2 men, 3 women

∴ The number of ways to choose a committee of 3 women and 2 men from a group of 7 men and 5 women is:

=\(^5C_3 \times ^7C_2\)

=\(\frac{5\times4\times3}{3\times2}\times\frac{7\times6}{2\times1}\)

=\(10\times21\)

=210

Correct Answer: B. 3

Correct Answer: A. least value of x is -3.2

Correct Answer: C. {6,7}