Answer:
22: 45 and 126
Step-by-step explanation:
The value of the derivative at the maximum or minimum for a continuous function must be zero.
<h3>What happens with the derivative at the maximum of minimum?</h3>
So, remember that the derivative at a given value gives the slope of a tangent line to the curve at that point.
Now, also remember that maximums or minimums are points where the behavior of the curve changes (it stops going up and starts going down or things like that).
If you draw the tangent line to these points, you will see that you end with horizontal lines. And the slope of a horizontal line is zero.
So we conclude that the value of the derivative at the maximum or minimum for a continuous function must be zero.
If you want to learn more about maximums and minimums, you can read:
brainly.com/question/24701109
The picture won't load, but I would surely help!
![\bf \qquad \textit{Simple Interest Earned}\\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\to& \$690.65\\ r=rate\to 7.3\%\to \frac{7.3}{100}\to &0.073\\ t=years\to &\frac{15+31+30+6}{365} \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7BSimple%20Interest%20Earned%7D%5C%5C%5C%5C%0AI%20%3D%20Prt%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0AI%3D%5Ctextit%7Binterest%20earned%7D%5C%5C%0AP%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cto%26%20%5C%24690.65%5C%5C%0Ar%3Drate%5Cto%207.3%5C%25%5Cto%20%5Cfrac%7B7.3%7D%7B100%7D%5Cto%20%260.073%5C%5C%0At%3Dyears%5Cto%20%26%5Cfrac%7B15%2B31%2B30%2B6%7D%7B365%7D%0A%5Cend%7Bcases%7D)
the penalty he'll incurred into, since July 6 is after the deadline of April 15, is I = Prt
now "t" is in years, how many days after April 15 to July 6? well, 15 + 31 + 30 +6, to convert to years, divide by 365