h(t)=(t+3) 2 +5 h, left parenthesis, t, right parenthesis, equals, left parenthesis, t, plus, 3, right parenthesis, squared, plu
lesya692 [45]
Answer:
1
Step-by-step explanation:
If I understand the question right, G(t) = -((t-1)^2) + 5 and we want to solve for the average rate of change over the interval −4 ≤ t ≤ 5.
A function for the rate of change of G(t) is given by G'(t).
G'(t) = d/dt(-((t-1)^2) + 5). We solve this by using the chain rule.
d/dt(-((t-1)^2) + 5) = d/dt(-((t-1)^2)) + d/dt(5) = -2(t-1)*d/dt(t-`1) + 0 = (-2t + 2)*1 = -2t + 2
G'(t) = -2t + 2
This is a linear equation, and the average value of a linear equation f(x) over a range can be found by (f(min) + f(max))/2.
So the average value of G'(t) over −4 ≤ t ≤ 5 is given by ((-2(-4) + 2) + (-2(5) + 2))/2 = ((8 + 2) + (-10 + 2))/2 = (10 - 8)/2 = 2/2 = 1
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Start by distributing the exponent to each of the terms in 
. This will become 
, and 
. Now, the expression is:
.
We can now simplify the bottom to read: 
because when multiplying variables raised to an exponent, we add the exponents. The expression now looks like:
The 2/8 simplifies to 1/4:
Now, we have two u terms on the top and bottom. When dividing variables raised to an exponent, we subtract the exponents. However, since 3-5=-2, the term will be on the bottom to avoid the negative exponent. The final answer is:
Hope this makes sense!!
