Answer:
sin = opposite/hypotenuse
Step-by-step explanation:
16/20
Integrate by parts.
![\displaystyle \int 6x \sec(x) \tan(x) \, dx = uv - \int v\,du](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%206x%20%5Csec%28x%29%20%5Ctan%28x%29%20%5C%2C%20dx%20%3D%20uv%20-%20%5Cint%20v%5C%2Cdu)
with
![u = 6x \implies du = 6\,dx](https://tex.z-dn.net/?f=u%20%3D%206x%20%5Cimplies%20du%20%3D%206%5C%2Cdx)
![dv = \sec(x)\tan(x)\,dx \implies v = \sec(x)](https://tex.z-dn.net/?f=dv%20%3D%20%5Csec%28x%29%5Ctan%28x%29%5C%2Cdx%20%5Cimplies%20v%20%3D%20%5Csec%28x%29)
so that
![\displaystyle \int 6x \sec(x) \tan(x) \, dx = 6x\sec(x) - 6 \int \sec(x) \, dx \\\\ ~~~~~~~~ = \boxed{6x\sec(x) - 6 \ln|\sec(x) + \tan(x)| + C}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%206x%20%5Csec%28x%29%20%5Ctan%28x%29%20%5C%2C%20dx%20%3D%206x%5Csec%28x%29%20-%206%20%5Cint%20%5Csec%28x%29%20%5C%2C%20dx%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%20%5Cboxed%7B6x%5Csec%28x%29%20-%206%20%5Cln%7C%5Csec%28x%29%20%2B%20%5Ctan%28x%29%7C%20%2B%20C%7D)
where the last integral follows from
![\displaystyle \int \sec(x) \, dx = \int \frac{\sec(x)(\sec(x)+\tan(x))}{\sec(x)+\tan(x)} \, dx = \int \frac{d(\sec(x)+\tan(x))}{\sec(x)+\tan(x)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Csec%28x%29%20%5C%2C%20dx%20%3D%20%5Cint%20%5Cfrac%7B%5Csec%28x%29%28%5Csec%28x%29%2B%5Ctan%28x%29%29%7D%7B%5Csec%28x%29%2B%5Ctan%28x%29%7D%20%5C%2C%20dx%20%3D%20%5Cint%20%5Cfrac%7Bd%28%5Csec%28x%29%2B%5Ctan%28x%29%29%7D%7B%5Csec%28x%29%2B%5Ctan%28x%29%7D)
It would be 45 degrees.
Since a an octagon takes 8 turns to go around and there is a total of 360 degrees, you can divide 360 by 8 to get your answer of 45 degrees.
For answering questions like this you need to take 360 and divide it be the amount of sides to get the minimum amount of degrees of rotation.
48/4 because it is 3 times as long as the other piece thinking there are 4 pieces. So one piece has to be 12 because 48/4 is 12 and 12 x 3 is 36 so then we double check 12+36=48 so the pieces are 12inc and 36inc
The correct answer would be :
<span>complementary angles
adjacent angles.
(I took the test...)</span>