Answer:
F(-7,3) -> F'(-7,-3)
G(2,6) -> G'(2,-6)
H(3,5) ->H'(3,-5)
Step-by-step explanation:
If you are taking point (a,b) and reflecting it across the x-axis (the horizontal axis), your x value is going to stay the same because you want the point on the same vertical line as (a,b). The y-coordinate is going to be opposite because you want a reflection and the opposite of b will this give you the same distance from the x-axis as b.
So the transformation is this: (a,b) -> (a,-b).
All this means is leave x the same and take the opposite of y.
F(-7,3) -> F'(-7,-3)
G(2,6) -> G'(2,-6)
H(3,5) ->H'(3,-5)
Answer:
Great job!
Step-by-step explanation:
As we know that remote exterior angle can be given by
m<z=m<x+m<y
151-5n=4n-18+n+9
151+18-9=5n+4n+n
160=10n
n=16
m<z=151-5*16
m<z=151-80=71
A. It's a composite function, so basically, wherever you see a p, replace it with 5t, because we are given that information. So, your answer is:
![A[p(t)] = 5t \pi 2=10t \pi](https://tex.z-dn.net/?f=A%5Bp%28t%29%5D%20%3D%205t%20%5Cpi%202%3D10t%20%5Cpi%20)
B. Let's use the function we created, and just plug in 2 for t:
![A[p(2)] = 10(2) \pi](https://tex.z-dn.net/?f=A%5Bp%282%29%5D%20%20%3D%2010%282%29%20%5Cpi%20)
![A[p(2)] = 62.83](https://tex.z-dn.net/?f=A%5Bp%282%29%5D%20%3D%2062.83)
So, your answer is
(approximately)
62.83 units².
