Hmmm, as I read it, in year 1, but it doesn't say at the beginning of the year or end, so we'll be assuming is at the beginning of year 1, the amount is 850, how much will it be at the beginning of year 6? namely, 5 years later.
![\bf ~~~~~~ \textit{Simple Interest Earned Amount}\\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to& \$850\\ r=rate\to 8\%\to \frac{8}{100}\to &0.08\\ t=years\to &5 \end{cases} \\\\\\ A=850[1+0.08(5)]\implies A=850(1.4)](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%5C%5C%5C%5C%0AA%3DP%281%2Brt%29%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%0AP%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cto%26%20%5C%24850%5C%5C%0Ar%3Drate%5Cto%208%5C%25%5Cto%20%5Cfrac%7B8%7D%7B100%7D%5Cto%20%260.08%5C%5C%0At%3Dyears%5Cto%20%265%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D850%5B1%2B0.08%285%29%5D%5Cimplies%20A%3D850%281.4%29)
What fraction of an hour is 18 minutes.
1 answer:
You might be interested in
Answer:
or ×
Step-by-step explanation:
In this type of context "of" means to multiply
First divide both sides by 2:
Move the constant term to the right side:
Since
we want to have
Then
So in our equation, we need to add 49/16 to both sides in order to complete the square: