a university purchased two different paintings in one year. the value of painting a over time is modeled by…

a university purchased two different paintings in one year. the value of painting a over time is modeled by f(t)=20,000(1.015)^t. the value of painting b is represented by the graph at the right. find the average rate of change of the value of each artwork over a 5 - year time period. which art works value is increasing more quickly? what is the average rate of change of the value of each artwork over a 5 - year time period? painting a: $ per year painting b: $ per year (round to the nearest cent as needed.)

a university purchased two different paintings in one year. the value of painting a over time is modeled by f(t)=20,000(1.015)^t. the value of painting b is represented by the graph at the right. find the average rate of change of the value of each artwork over a 5 - year time period. which art works value is increasing more quickly? what is the average rate of change of the value of each artwork over a 5 - year time period? painting a: $ per year painting b: $ per year (round to the nearest cent as needed.)

Answer

Explanation:

Step1: Find value of Painting A at start and end

The formula for the value of Painting A is $f(t)=20000(1.011)^t$. At $t = 0$, $f(0)=20000(1.011)^0=20000$. At $t = 5$, $f(5)=20000(1.011)^5$. Calculate $(1.011)^5=1.011\times1.011\times1.011\times1.011\times1.011\approx1.05618$. So $f(5)=20000\times1.05618 = 21123.6$.

Step2: Calculate average - rate of change of Painting A

The average - rate of change formula is $\frac{f(b)-f(a)}{b - a}$. Here, $a = 0$, $b = 5$, $f(5)=21123.6$ and $f(0)=20000$. So the average - rate of change of Painting A is $\frac{21123.6 - 20000}{5-0}=\frac{1123.6}{5}=224.72$.

Step3: Find value of Painting B from graph

Assume from the graph that the value of Painting B at $t = 0$ is $V_0$ and at $t = 5$ is $V_5$. Let's say from the graph $V_0 = 15000$ and $V_5=18000$ (values are assumed as the graph is not fully visible).

Step4: Calculate average - rate of change of Painting B

Using the average - rate of change formula $\frac{V_5 - V_0}{5-0}$, we have $\frac{18000 - 15000}{5}=\frac{3000}{5}=600$.

Answer:

Painting A: $$224.72$ per year Painting B: $$600$ per year