to get the inverse of any expression, we start off by doing a quick switcheroo on the variables and then solve for "y".

If you're using the app, try seeing this answer through your browser: brainly.com/question/2762144_______________
Let


because that is the range of the inverse cosine funcition.
Also,
![\mathsf{cos\,\theta=cos\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]}\\\\\\ \mathsf{cos\,\theta=\dfrac{4}{5}}\\\\\\ \mathsf{5\,cos\,\theta=4}](https://tex.z-dn.net/?f=%5Cmathsf%7Bcos%5C%2C%5Ctheta%3Dcos%5C%21%5Cleft%5Bcos%5E%7B-1%7D%5C%21%5Cleft%28%5Cdfrac%7B4%7D%7B5%7D%5Cright%29%5Cright%5D%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7Bcos%5C%2C%5Ctheta%3D%5Cdfrac%7B4%7D%7B5%7D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B5%5C%2Ccos%5C%2C%5Ctheta%3D4%7D)
Square both sides and apply the fundamental trigonometric identity:



But

which means

lies either in the 1st or the 2nd quadrant. So

is a positive number:
![\mathsf{sin\,\theta=\dfrac{3}{5}}\\\\\\ \therefore~~\mathsf{sin\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]=\dfrac{3}{5}\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%5Cmathsf%7Bsin%5C%2C%5Ctheta%3D%5Cdfrac%7B3%7D%7B5%7D%7D%5C%5C%5C%5C%5C%5C%0A%5Ctherefore~~%5Cmathsf%7Bsin%5C%21%5Cleft%5Bcos%5E%7B-1%7D%5C%21%5Cleft%28%5Cdfrac%7B4%7D%7B5%7D%5Cright%29%5Cright%5D%3D%5Cdfrac%7B3%7D%7B5%7D%5Cqquad%5Cquad%5Ccheckmark%7D)
I hope this helps. =)
Tags: <em>inverse trigonometric function cosine sine cos sin trig trigonometry</em>
Which is the most appropriate to describe a quantity decreasing at a steady rate?
With a linear equation.
Answer:
7
Step-by-step explanation:
The text of this expression is:
"If 3 is added to twice a number, the result is 17. Find the number"
Let's call
the number we want to find.
Twice the number is equal to

So, if we add 3 to 2x, we get

And we know that this is equal to 17:

Now we can solve the equation to find x: first, we subtract 3 from both sides

Now we divide both sides by 2:
