1 1/40, because the 1 in front of the decimal is 1 whole, and the 0.025 is 1/40
![\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}](https://tex.z-dn.net/?f=%5Cmathfrak%7B%5Chuge%7B%5Corange%7B%5Cunderline%7B%5Cunderline%7BAnSwEr%3A-%7D%7D%7D%7D%7D)
Actually Welcome to the concept of Equilateral triangles.
Since, from the name itself, equilateral triangles have all sides and angles equal.
Hence the correct answer is
==> D. All sides are the same length.
let's check how much is it after 2 years firstly.
![\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &1900\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases} \\\\\\ A=1900\left(1+\frac{0.035}{1}\right)^{1\cdot 2}\implies A=1900(1.035)^2\implies A=2035.3275](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%20%5Ctextit%7BCompound%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%5Cleft%281%2B%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%261900%5C%5C%20r%3Drate%5Cto%203.5%5C%25%5Cto%20%5Cfrac%7B3.5%7D%7B100%7D%5Cdotfill%20%260.035%5C%5C%20n%3D%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%20%5Ctextit%7Byearly%2C%20thus%20once%7D%20%5Cend%7Barray%7D%5Cdotfill%20%261%5C%5C%20t%3Dyears%5Cdotfill%20%262%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D1900%5Cleft%281%2B%5Cfrac%7B0.035%7D%7B1%7D%5Cright%29%5E%7B1%5Ccdot%202%7D%5Cimplies%20A%3D1900%281.035%29%5E2%5Cimplies%20A%3D2035.3275)
Brian invested the money for 6 years, so now let's check how much is that for the remaining 4 years.
![\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &2035.3275\\ r=rate\to 4.9\%\to \frac{4.9}{100}\dotfill &0.049\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%20%5Ctextit%7BCompound%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%5Cleft%281%2B%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%262035.3275%5C%5C%20r%3Drate%5Cto%204.9%5C%25%5Cto%20%5Cfrac%7B4.9%7D%7B100%7D%5Cdotfill%20%260.049%5C%5C%20n%3D%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%20%5Ctextit%7Byearly%2C%20thus%20once%7D%20%5Cend%7Barray%7D%5Cdotfill%20%261%5C%5C%20t%3Dyears%5Cdotfill%20%264%20%5Cend%7Bcases%7D)
![\bf A=2035.3275\left(1+\frac{0.049}{1}\right)^{1\cdot 4}\implies A=2035.3275(1.049)^4 \\\\\\ A\approx 2464.54\implies \boxed{\stackrel{\textit{rounded up }}{A=2465}}](https://tex.z-dn.net/?f=%5Cbf%20A%3D2035.3275%5Cleft%281%2B%5Cfrac%7B0.049%7D%7B1%7D%5Cright%29%5E%7B1%5Ccdot%204%7D%5Cimplies%20A%3D2035.3275%281.049%29%5E4%20%5C%5C%5C%5C%5C%5C%20A%5Capprox%202464.54%5Cimplies%20%5Cboxed%7B%5Cstackrel%7B%5Ctextit%7Brounded%20up%20%7D%7D%7BA%3D2465%7D%7D)
Answer:
950 + 175
Step-by-step explanation:
to find the original price before the sale, you need to add.