Find the equation of the line
1 answer:
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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)
Answer:
There are no simple factors to this expression. You can however solve for x, giving you:
x = -1 ± √(y + 3)
Step-by-step explanation:
There is no simple pair of factors to this equation, so we'll have to go through the full process. You can see that by observing that there is no pair of rational numbers that add to make 2 and multiply to make -2. We can however solve for x: