now, there are 12 months in a year, so 18 months is really 18/12 of a year, thus
![~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$4000\\ P=\textit{original amount deposited}\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ t=years\to \frac{18}{12}\dotfill &\frac{3}{2} \end{cases} \\\\\\ 4000=P[1+(0.05)(\frac{3}{2})]\implies 4000=P(1.075) \\\\\\ \cfrac{4000}{1.075}=P\implies 3720.93\approx P](https://tex.z-dn.net/?f=~~~~~~%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%20%26%20%5C%244000%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5C%5C%20r%3Drate%5Cto%205%5C%25%5Cto%20%5Cfrac%7B5%7D%7B100%7D%5Cdotfill%20%260.05%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B18%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%204000%3DP%5B1%2B%280.05%29%28%5Cfrac%7B3%7D%7B2%7D%29%5D%5Cimplies%204000%3DP%281.075%29%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B4000%7D%7B1.075%7D%3DP%5Cimplies%203720.93%5Capprox%20P)
The mean of a frequency distribution is given by
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</span>
The slope of the tangent line to 
at 
is given by the derivative of at that point:
Factorize the numerator:
We have 
approaching -1; in particular, this means 
, so that
Then
and the tangent line's equation is
Area of a sector is x over 360 * pi r squares so in this case x would be 210 so it would be 210 over 360 and the pi r squared would be 24 pi and when we times that our answer is 14 pi
Answer:14 pi
Answer:
Option (D). G(x) = x³ - x
Step-by-step explanation:
Given function is the attachment,
G(x) = (x - 1.5)³ - (x - 1.5)
This function when translated by left by 1.5 units, rule for the translation is,
G(x) → G'(x + 1.5)
Therefore, translated function will be,
G'(x) = (x - 1.5 + 1.5)³ - (x - 1.5 + 1.5)
G'(x) = x³ - x
Therefore, Option (D) will be the answer.
