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)
Well you put it into parts...
so that u end up with 6= 16+x
Answer:assuming it’s not a 3D object then, 2x10=20
If it is, then do the base 2x10 then times the height
Step-by-step explanation:
Answer:
Sec (x) - 2 tan (x)
Step-by-step explanation:
3/4, 0.075, 0.75%. Best of luck!