Question

Can someone help me with part b? I've been stuck on it for an hour :(

Answer 1

Answer:

Check explanation

Step-by-step explanation:

You're given the equation

$A=-0.6t^2 + 37.2t + 243$

where t is the number of year after 1980.

I guess you'd have learned the average rate of change equation.

$r=\frac{y_2-y_1}{x_2-x_1}=\frac{f(x_2)-f(x_1)}{x_2-x_1}$

You're asked to find average rate of change from

$t_1=2000-1980=20$

$t_2=2015-1980=35$

So lets plug that in

$r=\frac{A(t_2)-A(t_1)}{t_2-t_1}$

$r=\frac{A(35)-A(20)}{35-20}$

$r=\frac{[-0.6(35)^2 + 37.2(35) + 243]-[-0.6(20)^2 + 37.2(20) + 243]}{35-20}$

$r=\frac{[-0.6(35)^2 + 37.2(35) + 243]-[-0.6(20)^2 + 37.2(20) + 243]}{15}$

plug that in your calculator to get the answer. I dont have a calculator with me so I can't find the answer.

Answer 2

Answer:

The average rate of change is 4.2 billion kilowatt hours per year.  

In other words, the amount of nuclear energy increased, on average, by 4.2 billion kilowatt hours per year from 2000 to 2015.

Step-by-step explanation:

Recall that to find the average rate of change between two points, we simply need to find the slope between the two points.

Hence, to find the average rate of change from 2000 to 2015, find the slope between the points A(20) and A(35):

$\displaystyle \begin{aligned} m_\text{avg} & = \frac{A(35)-A(20)}{(35)-(20)} \\ \\ & = \frac{(810)-(747)}{15} \\ \\ & = \frac{63}{15} = 4.2\end{aligned}$

In conclusion, the amount of nuclear energy increased, on average, by 4.2 billion kilowatt hours per year from 2000 to 2015.