$\bf \qquad \textit{Amount for Exponential Decay}\\\\
A=I(1 - r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\to &730\\
r=rate\to 27.6\%\to \frac{27.6}{100}\to &0.276\\
t=\textit{elapsed time}\to &12\\
\end{cases}
\\\\\\
A=730(1-0.276)^{12}\implies A=730(0.724)^{12}$

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3 years ago

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Find the x-coordinate of the point on the graph of y=x^2 where the tangent line is parallel to the secant line that cuts the cur

jek_recluse [69]

First find the secant line. The slope of the secant line through $(-1,1)$ (when $x=-1$ ) and $(2,4)$ (when $x=2$ ) is the average rate of change of $y=x^2$ over the interval $[-1,2]$ :

$\text{slope}_{\text{secant}}=\dfrac{2^2-(-1)^2}{2-(-1)}=\dfrac33=1$

The tangent line to $y=x^2$ will have a slope determined by the derivative:

$y=x^2\implies y'=2x$

Both the secant and tangent will have the same slope when $2x=1$ , or when $x=\dfrac12$ .

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3 years ago