Correct Answer: D. 15°

Explanation

Sin\(\theta = \frac{20}{75}\)

\(\theta = sin^{-1} (\frac{20}{75})\)

= 15\(^o\)

Correct Answer: B. 3/5

Explanation

Total number of marbles = 5; 1st pick = 2/5 2nd pick = 3/4; 3rd pick = 3/5; 4th pick = 2/4

? Probability of picking two marbles of different colors = (2/5 x 3/4) + (3/5 + 2/4) = 12/20 = 3/5

Correct Answer: C. 25

Correct Answer: D. 1 and 4

Explanation

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

This movement is against gravity, so it is negative.

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

\(s = 20m, u = 25ms^{-1}\)

\(20 = 25t - \frac{10t^{2}}{2} \implies 20 = 25t - 5t^{2}\)

\(5t^{2} - 25t + 20 = 0 \)

\(5t^{2} - 5t - 20t + 20 = 0 \implies 5t(t - 1) - 20(t - 1) = 0\)

\(5t - 20 = \text{0 or t - 1 = 0}\)

\(t = \text{1 or 4}\)

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. - \(\frac{4}{5}\)

Explanation

Where sinθ = \(\frac{opp}{hyp}\) → \(\frac{-3}{5}\)

opp = -3, hyp = 5

using pythagoras formula

hyp\(^2\) = adj\(^2\) + opp\(^2\)

adj\(^2\) = hyp\(^2\) - opp\(^2\)

adj\(^2\) = 5\(^2\) - 3\(^2\) → 25 - 9

adj\(^2\) = 16

adj = 4

cosθ = \(\frac{adj}{hyp}\) → \(\frac{4}{5}\)

In third quadrant: cosθ is negative → - \(\frac{4}{5}\)

Explanation

(a) \(a = 3; d = 4 ; n = 28\)

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

\(S_{28} = \frac{28}{2}(2(3) + (28 - 1) 4)\)

= \(14(6 + 108)\)

= \(14(114)\)

= \(1596\)

(b) \(x = \frac{2m}{1 - m^{2}} ; y = \frac{2m}{1 + m}\)

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

= \(\frac{4m}{1 - m^{2}} - \frac{2m}{1 + m}\)

= \(\frac{4m - 2m(1 - m)}{1 - m^{2}}\)

= \(\frac{4m - 2m + 2m^{2}}{1 - m^{2}}\)

= \(\frac{2m^{2} + 2m}{1 - m^{2}}\)

= \(\frac{2m(m + 1)}{(1 + m)(1 - m)}\)

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

(c) Sum of interior angles = (2n - 4)90°

When n = 5 (pentagon),

= \((2(5) - 4) \times 90°\)

= \(540°\)

\(x + 2x + 3x + 2x + (3x - 10) = 11x - 10\)

= \(11x - 10 = 540°\)

= \(11x = 550°\)

= \(x = 50°\)

Correct Answer: B. 12.27

Correct Answer: A. 4

Correct Answer: D. \(\frac{3}{4}\)

Explanation

\(\frac{\log \sqrt{27}}{\log \sqrt{81}}\) = \(\frac{\log 27\frac{1}{2}}{81\frac{1}{2}}\)

= \(\frac{\log 3\frac{1}{2}}{\log 3^2}\)

\(\frac{\frac{3}{2} \log 3}{2 \log 3} = \frac{3}{2} \div \frac{2}{1}\)

= \(\frac{3}{2} \times \frac{1}{2}\)

= \(\frac{3}{4}\)