acceleration is a(t) = 3t and velocity is v(t) = 1 at t = 1, what is the displacement from t = 2 to t = 4
1 answer:
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Answer:
is 200 times larger than
Step-by-step explanation:
To compare both values we divide the exponents
6 divide by 3 is 2
for simplify exponents we use exponential property
a^m / a^n = a^(m-n)
is 200 times larger than
Answer:
The average rate of change of the function over the interval is of 6.
Step-by-step explanation:
Average rate of change:
The average rate of change of a function h(x) over an interval [a,b] is given by:
In this question:
Over the interval [-9,-2], so
The function is:
Then
The average rate of change of the function over the interval is of 6.