3 years ago

HELP ME PLEASE 40 POINTS!!!!!!!

Mathematics

1 answer:

Solnce55 [7]3 years ago

6 0

Answer:

The answer is The first one -10/27

You might be interested in

Solve the equation sin theta = 0.3 to the nearest tenth. Use the restrictions 90degrees < theta < 180degrees.

Alexxandr [17]

We first do:
∅ = sin⁻¹(0.3)
∅ = 17.5°

To go into the second quadrant, we add 90°.
∅ = 17.5 + 90
= 107.5°
The answer is A.

5 0

3 years ago

Find an exact value cos(19pi/12)

dmitriy555 [2]

$cos\left ( \frac{9\pi}{12}\right )=\frac{\sqrt{2-\sqrt{3}}}{2}$

Step-by-step explanation:

We know that

cos 2x = 2cos²x - 1

So we have

$cosx=\sqrt{\frac{1+cos2x}{2}}$

Substituting $x=\frac{19\pi}{12}$

$cos\left ( \frac{19\pi}{12}\right )=\sqrt{\frac{1+cos2\left ( \frac{19\pi}{12}\right )}{2}}\\\\cos\left ( \frac{19\pi}{12}\right )=\sqrt{\frac{1+cos\left ( \frac{19\pi}{6}\right )}{2}}\\\\cos\left ( \frac{9\pi}{12}\right )=\sqrt{\frac{1+cos570}{2}}\\\\cos\left ( \frac{9\pi}{12}\right )=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\\\\cos\left ( \frac{9\pi}{12}\right )=\sqrt{\frac{2-\sqrt{3}}{4}}\\\\cos\left ( \frac{9\pi}{12}\right )=\frac{\sqrt{2-\sqrt{3}}}{2}$