This appears to be a composite function situation where you find the value of t(x) and sub that in for the x in c(x). If x = 4, then t(4) = 4(4) + 3 and t(4) = 19. Sub that 19 in for x in c(x) to get 2(19) + 5 which is 43. So 43
Since V = (4/3) * pi * R^3
If R is halved, V' will reduce by a ratio of (1/2)^3 = 1/8
So V' = (1/8)V
Minimal grade to barely pass would be a C, but you should review and go for at least a B or B+ if you want to pass with a C in the class as a whole.
Look at the attached picture for the correct process.
Answer:
Third item in the list
Step-by-step explanation:
Third function: f(x) = x + 9. If we substitute 14 for x, we get f(14) = 14 + 9 = 23.