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)
18.00 + 8.50 + 8.50 + 8.50 + 8.50 = 52.00
So that's 4 Friends
1. 7+ (-9) = 7-9=-2
2.(-8) + (+5)= -8+5= -(8-5)=-3
3.(+9) - (-3) = 9+3=12
4.(+12) - (-1) = 12+1=13
5. (-7) - (-5) = -7+5==(7-5)=-2
<span>(-14) - (+2) = -14-2=-(14+2)=-16</span>
Answer:
Perimeter of A'B'C'D' = 9 units
Step-by-step explanation:
Given:
Perimeter of ABCD = 27 units
Scale factor = 1 / 3
Find:
Perimeter of A'B'C'D'
Computation:
Perimeter of A'B'C'D' = 1/3[Perimeter of ABCD]
Perimeter of A'B'C'D' = 1/3[27]
Perimeter of A'B'C'D' = 9 units
