G(t)=−(t−1) 2 +5g, left parenthesis, t, right parenthesis, equals, minus, left parenthesis, t, minus, 1, right parenthesis, squared, plus, 5 what is the average rate of change of ggg over the interval -4\leq t\leq 5−4≤t≤5minus, 4, is less than or equal to, t, is less than or equal to, 5?

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

the 8 in 8,420 is in the thousands place so it's value is 8 thousand.

the 8 in 6,870 is in the hundreds place so it's value is 8 hundred.

divide 8000 by 800 to find the answer:

8000 / 800 = 10

the 8 in 8,420 is 10 times greater than the 8 in 6,870