The complete question in the attached figure
we know that
<span>The measure of the external angle is the semidifference of the arcs that it covers.
so
</span><span>m∠BED =(1/2)*[mAC-mBD]-------> solve for mBD
mBD=mAC-2</span>m∠BED
mBD=80-2*25--------> mBD=30°
the answer ismBD=30°
Recall that to get the x-intercepts, we set the f(x) = y = 0, thus
![\bf \stackrel{f(x)}{0}=-4cos\left(x-\frac{\pi }{2} \right)\implies 0=cos\left(x-\frac{\pi }{2} \right) \\\\\\ cos^{-1}(0)=cos^{-1}\left[ cos\left(x-\frac{\pi }{2} \right) \right]\implies cos^{-1}(0)=x-\cfrac{\pi }{2} \\\\\\ x-\cfrac{\pi }{2}= \begin{cases} \frac{\pi }{2}\\\\ \frac{3\pi }{2} \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bf%28x%29%7D%7B0%7D%3D-4cos%5Cleft%28x-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%20%5Cright%29%5Cimplies%200%3Dcos%5Cleft%28x-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%20%5Cright%29%0A%5C%5C%5C%5C%5C%5C%0Acos%5E%7B-1%7D%280%29%3Dcos%5E%7B-1%7D%5Cleft%5B%20cos%5Cleft%28x-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%20%5Cright%29%20%5Cright%5D%5Cimplies%20cos%5E%7B-1%7D%280%29%3Dx-%5Ccfrac%7B%5Cpi%20%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0Ax-%5Ccfrac%7B%5Cpi%20%7D%7B2%7D%3D%0A%5Cbegin%7Bcases%7D%0A%5Cfrac%7B%5Cpi%20%7D%7B2%7D%5C%5C%5C%5C%0A%5Cfrac%7B3%5Cpi%20%7D%7B2%7D%0A%5Cend%7Bcases%7D)
15x^3+4x^2+14x+12
Multiply 3x by every value in the second bracket. Then multiply 2 by every value in the second bracket. Add like terms. (Only answers with the same square are like terms.)
Answer:
65 degrees
Step-by-step explanation:
The angles will sum up to 360, just do 360 - all known angles
360 - 90 - 128 - 77 = 65