Explanation

\(s = ut + \frac{1}{2} gt^{2}\)

\(45 = 20t + \frac{1}{2} (10)(t^{2})\)

\(45 = 20t + 5t^{2}\)

\(5t^{2} + 20t - 45 = 0\)

\(\equiv t^{2} + 4t - 9 = 0\)

= \(\frac{-4 \pm \sqrt{(4^{2}) - 4(1)(-9)}}{2(1)}\)

= \(\frac{-4 \pm \sqrt{16 + 36}}{2}\)

= \(\frac{-4 \pm 2\sqrt{13}}{2}\)

= \(\frac{-2 \pm \sqrt{13}}\)

\(t = -2 + \sqrt{13} \text{ or } t = -2 - \sqrt{13}\)

The time cannot be negative therefore, \(t = - 2 + \sqrt{13} = 1.61s\).

(b) \(v^{2} = u^{2} + 2gs\)

\(v^{2} = 20^{2} + 2(10)(45)\)

\(v^{2} = 400 + 900\)

\(v^{2} = 1300\)

\(v = \sqrt{1300}\)

\(v = \sqrt{100 \times 13} = 10\sqrt{13} ms^{-1}\)

Correct Answer: C. 16.5

Explanation

The third quartile can be found in the \((\frac{3N}{4})^{th}\) position

= \((\frac{3 \times 100}{4})^{th}\) position

This is found in the class 14 - 16.

The upper class boundary = 16.5

Correct Answer: D. 8.26N

Correct Answer: A. (3 N, 020°)

Correct Answer: B. 0.08 mm/s

Explanation

Given: \(\frac{\mathrm d A}{\mathrm d t} = 4 mm^{2}/s\)

\(\frac{\mathrm d A}{\mathrm d t} = (\frac{\mathrm d A}{\mathrm d r})(\frac{\mathrm d r}{\mathrm d t})\)

\(A = \pi r^{2} \implies \frac{\mathrm d A}{\mathrm d r} = 2\pi r\)

\(\implies 4 = 2\pi r \times \frac{\mathrm d r}{\mathrm d t}\)

\(\frac{\mathrm d r}{\mathrm d t} = \frac{4}{2\pi r} = \frac{4 \times 7}{2 \times 22 \times 8}\)

= \(0.07954 mm/s \approxeq 0.08 mm/s\)

Explanation

\(r = (4t^{3} - t^{2})i - (2t^{2} - t)j\)

\(v = \frac{\mathrm d r}{\mathrm d t} = (12t^{2} - 2t)i - (4t - 1)j\)

\(a = \frac{\mathrm d v}{\mathrm d t} = (24t - 2)i - 4j\)

\(t = 2 : a = (24(2) - 2)i - 4j\)

\(a = 46i - 4j\)

\(F = ma = 3(46i - 4j)\)

= \(138i - 12j)N\)

Correct Answer: A. \(\frac{\sqrt33}{4}\)

Explanation

Standard Form equation of a circle (Center-Radius Form): \((x − a)^2 + (y − b)^2 = r^2\)

Where "a" and "b" are the coordinates of the center and "r" is the radius of the circle

\(2x^2+2y^2-4x+5y+1=0\)

Divide through by 2

= \(x^2+y^2-2x+\frac{5}{2}y+\frac{1}{2}= 0\)

=\(x^2-2x+y^2+\frac{5}{ 2}y=-\frac{1}{ 2}\)

=\(x^2-2x+1^2+y^2+\frac{5}{ 2}y+(\frac{5}{ 4})^2-1-\frac{25}{16}=-\frac{1}{2}\)

=\((x-1)^2+(y+\frac{5}{4})^2=-\frac{1}{2}+1+\frac{25}{16}\)

=\((x-1)^2+(y-(-\frac{5}{4}))^2=\frac{33}{16}\)

=\((x-1)^2+(y-(-\frac{5}{4}))^2=(\frac{\sqrt33}{4})^2\)

\(\therefore a = 1, b = - \frac{5}{4} and\) \(r \frac{\sqrt33}{4} (answer)\)

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. (-2,-3)

Explanation

y= 3x\(^2\) + 11x +7;

dy/dx = 6x + 11= -1;

6x = -12;

x=-2

y= 3x\(^2\) + 11x + 7

At x-2,

y= 3(-2)\(^2\) +11(-2) + 7

= 12 - 22 + 7= -3

p(x, y) = (-2,-3)

Explanation

(a) From Paschal's triangle, coefficients of the power of 5 are

1, 5, 10, 10, 5, 1

\((2 - \frac{1}{2}x)^{5} = (1)(2^{5}) + 5(2^{4})(-\frac{1}{2}x) + 10(2^{3})(-\frac{1}{2}x)^{2} + 10(2^{2})(-\frac{1}{2}x)^{3} + ...\)

= \(32 - 40x + 20x^{2} - 5x^{3} + ...\)

(b) \((1.99)^{5} = (2 - 0.01)^{5} = (2 - \frac{1}{2}(0.02))^{5}\)

Using the expansion above, \(32 - 40x + 20x^{2} - 5x^{3} + ...\)

\((1.99)^{5} \approxeq 32 - 40(0.02) + 20(0.02)^{2} - 5(0.02)^{3} \)

\(\approxeq 32 - 0.08 + 0.008\)

= \(31.928\)

= \(31.93\) (to 2 decimal place)