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svet-max [94.6K]
4 years ago
15

f(x) = 500(1.05)x What was the average rate of change of the value of Sophia's investment from the second year to the fourth yea

r?
Mathematics
1 answer:
hjlf4 years ago
6 0
\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\
f(x)=500(1.05)^x   \qquad 
\begin{cases}
x_1=2\\
x_2=4
\end{cases}\implies \cfrac{f(4)-f(2)}{4-2}
\\\\\\
\cfrac{[500(1.05)^4]~-~[500(1.05)^2]}{4-2}\implies \cfrac{56.503125}{2}\implies 28.2515625
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