Answer:
see below
Step-by-step explanation:
put g(x) in for the x in f(x)

Well, there are 3 feet in a yard. So, 3ft=1yrd. Multiply the length value by 3. You could also multiply 3 by the # of yards (or in this situation, 8) Is that good enough?
Answer:
I cant see the question its very blurry
Step-by-step explanation:
the assumption being that "x" is a plain variable whilst "y" is a function, that matters because the chain rule would be needed for a function, not so for a plain variable.

now, we know that y(5) = -23, which is another way of saying that when x = 5, y = -23, but we already knew that, we can get that by simply plugging it into the equation hmmm y'(5), well
