There are only 1*26 and 2*13 so there is no answer unless you allow decimals.
The value of c to complete the square is one. In order to find the c value you divide the b value (-2) by 2 (-1) and then square that. -1 squared is 1. This can then be easily factored into (x-1)^2
The best in quality will probably be the 1470 but easiest to get i to is probably 1190 because if your sat score is way above their average they will think you are flaunting and deny your request
Plz mark as brainliest!
now, assuming is simple interest applied.
bearing in mind that 6 months is not even a year, is really 6/12 of a year.
![\bf ~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&1876.50\\ P=\textit{original amount deposited}\\ r=rate\to 8.5\%\to \frac{8.5}{100}\dotfill &0.085\\ t=years\to \frac{6}{12}\dotfill &\frac{1}{2} \end{cases} \\\\\\ 1876.50=P\left[ 1+(0.085)\left( \frac{1}{2} \right) \right]\implies 1876.5=P(1.0425) \\\\\\ \cfrac{1876.5}{1.0425}=P\implies 1800=P](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5Cdotfill%261876.50%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5C%5C%20r%3Drate%5Cto%208.5%5C%25%5Cto%20%5Cfrac%7B8.5%7D%7B100%7D%5Cdotfill%20%260.085%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B6%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B1%7D%7B2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%201876.50%3DP%5Cleft%5B%201%2B%280.085%29%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%20%5Cright%5D%5Cimplies%201876.5%3DP%281.0425%29%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B1876.5%7D%7B1.0425%7D%3DP%5Cimplies%201800%3DP)
Answer:
28x-84
Step-by-step explanation: