457 irk sorry I don’t do this
H = 2s + 75 is the answer
Answer:

Option D is correct option.
Step-by-step explanation:
We are given: 
We need to find f(x+h)
For finding f(x+h) we need to replace x by x+h

Finding f(x+h) we get:

Option D is correct option.
Given a pre-image, you could use any single reflection, rotation, or translation, or combination of these transformations to re-position the pre-image on top of the image to confirm congruence. You would be able to visually see corresponding side lengths and corresponding angles measures are the same.