That'd be true only if the value of "s" is the exact same one for both
namely if sec(s) = cos(s)
then solving for "s"
thus
![\bf sec(s)=cos(s)\qquad but\implies sec(\theta)=\cfrac{1}{cos(\theta)} \\\\\\ thus\cfrac{1}{cos(s)}=cos(s)\implies 1=cos^2(s)\implies \pm \sqrt{1}=cos(s) \\\\\\ \pm 1=cos(s)\impliedby \textit{now taking }cos^{-1}\textit{ to both sides} \\\\\\ cos^{-1}(\pm 1)=cos^{-1}[cos(s)]\implies cos^{-1}(\pm 1)=\measuredangle s](https://tex.z-dn.net/?f=%5Cbf%20sec%28s%29%3Dcos%28s%29%5Cqquad%20but%5Cimplies%20sec%28%5Ctheta%29%3D%5Ccfrac%7B1%7D%7Bcos%28%5Ctheta%29%7D%0A%5C%5C%5C%5C%5C%5C%0Athus%5Ccfrac%7B1%7D%7Bcos%28s%29%7D%3Dcos%28s%29%5Cimplies%201%3Dcos%5E2%28s%29%5Cimplies%20%5Cpm%20%5Csqrt%7B1%7D%3Dcos%28s%29%0A%5C%5C%5C%5C%5C%5C%0A%5Cpm%201%3Dcos%28s%29%5Cimpliedby%20%5Ctextit%7Bnow%20taking%20%7Dcos%5E%7B-1%7D%5Ctextit%7B%20to%20both%20sides%7D%0A%5C%5C%5C%5C%5C%5C%0Acos%5E%7B-1%7D%28%5Cpm%201%29%3Dcos%5E%7B-1%7D%5Bcos%28s%29%5D%5Cimplies%20cos%5E%7B-1%7D%28%5Cpm%201%29%3D%5Cmeasuredangle%20s)
That’s weird for the answer i got 10/33
Answer: C
Step-by-step explanation: For a function, each x-coordinate corresponds to exactly one y-coordinate.
To determine whether the graph shown here
is a function, we can use the vertical line test.
The vertical line test tells us that if each x-coordinate on the graph corresponds to exactly one y-coordinate, then any vertical line that we draw on the graph should hit the graph at only one point.
For the graph show here, any vertical line that you draw with hit the graph at only one point which means it does pass the vertical line test.
So this graph is a <em>function</em>.
Answer:
3a^2+8/6a^2
Step-by-step explanation:
there ya go hope this helps
I believe the answer would be C. 4. hope that helped
