Answer:
Here we have the function:
S(t) = 500 - 400*t^(-1)
Then the rate of change at the value t, will be:
S'(t) = dS(t)/dt
This differentiation will be:
S'(t) = -400/t^2
Then:
a) the rate of change at t = 1 is:
S'(1) = -400/1^2 = -400
The rate of change after one year is -400
b) t = 10
S'(10) = -400/10^2 = -400/100 = -4
The rate of change after 10 years is -4, it reduced as the years passed, as expected.
I would say approx 110ft. I’m not certain.
I hope this helped!
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Answer:
The answer to your question is D 64
Step-by-step explanation:
hope this helps you have a wonderful day.
Answer:
7.2 + 0.8w
Step-by-step explanation:
Combine like terms.
Here Like terms: i) 2.2w and (-1.4w)
II) 4.8 and 2.4
4.8 + 2.2w - 1.4w + 2.4 = 4.8 + 2.4 + 2.2w - 1.4w
= 7.2 + 0.8w