Solve using the quadratic formula: (Standard form 1st!) 2x^2-2x=24
1 answer:
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Answer:
9). x = 36.9°
10). x = 18.3°
Step-by-step explanation:
9). By applying tangent rule in the given triangle,
tan(x) =
tan(x) =
x =
x = 36.87°
x = 36.9°
10). By applying sine rule in the given triangle,
sin(x) =
sin(x) =
x =
x = 18.32
x ≈ 18.3°
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)