Well Tristan better get it together
Answer:
3
Step-by-step explanation:
We can see that the graph is constant.=> the path of the dot/ the pattern is constant or not changing.
The balance was 18 when there were no visits.
It was 0 after 6 visits.
=>the amount spent on each visit is
Balance/number of visits= 18/6= $3
This can clearly be seen on the graph.
It may not work with a graph where the patter is not constant/ keeps on changing.
Answer:

Step-by-step explanation:
I will work with radians.
![$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$](https://tex.z-dn.net/?f=%24%5Cfrac%20%7B%5Ccos%5E2%20%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B2%7D-x%20%5Cright%29%2B%5Csin%28-x%29-%5Csin%5E2%20%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B2%7D-x%20%5Cright%29%2B%5Ccos%20%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B2%7D-x%20%5Cright%29%7D%20%7B%5B%5Csin%28%5Cpi%20-x%29%2B%5Ccos%28-x%29%5D%20%5Ccdot%20%5B%5Csin%282%5Cpi%20%2Bx%29%5Ccos%282%5Cpi-x%29%5D%7D%24)
First, I will deal with the numerator

Consider the following trigonometric identities:




Therefore, the numerator will be

Once



Now let's deal with the numerator
![[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]](https://tex.z-dn.net/?f=%5B%5Csin%28%5Cpi%20-x%29%2B%5Ccos%28-x%29%5D%20%5Ccdot%20%5B%5Csin%282%5Cpi%20%2Bx%29%5Ccos%282%5Cpi-x%29%5D)
Using the sum and difference identities:





Therefore,
![[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]](https://tex.z-dn.net/?f=%5B%5Csin%28%5Cpi%20-x%29%2B%5Ccos%28-x%29%5D%20%5Ccdot%20%5B%5Csin%282%5Cpi%20%2Bx%29%5Ccos%282%5Cpi-x%29%5D%20%5Cimplies%20%5B%5Csin%28x%29%2B%5Ccos%28x%29%5D%20%5Ccdot%20%5B%5Csin%28x%29%5Ccos%28x%29%5D)
![\implies [p+4] \cdot [p \cdot 4]=4p^2+16p](https://tex.z-dn.net/?f=%5Cimplies%20%5Bp%2B4%5D%20%5Ccdot%20%5Bp%20%5Ccdot%204%5D%3D4p%5E2%2B16p)
The final expression will be

Answer:
attached
Step-by-step explanation:
You should be able to do the 2nd problem yourself.