five more than three times the number is one-third more than the sum of the number and itself. 5(3n)=1/3n n 5(3n)=1/3 n 3n 5=(n
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Answer:
x = 4 or x = -3
Step-by-step explanation:
Take square both the sides,
x² - x - 12 = 0
x² - 4x + 3x - 3*4 =0
x*(x - 4) + 3*(x - 4)= 0
(x- 4 )(x+3) =0
x -4 =0 or x + 3 = 0
x = 4 or x = -3
Annuities whose payments are made at the end of the period are "ordinary annuity" type
![\bf \qquad \qquad \textit{Future Value of an ordinary annuity} \\\\ A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right]](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Cqquad%20%5Ctextit%7BFuture%20Value%20of%20an%20ordinary%20annuity%7D%0A%5C%5C%5C%5C%0AA%3Dpymnt%5Cleft%5B%20%5Ccfrac%7B%5Cleft%28%201%2B%5Cfrac%7Br%7D%7Bn%7D%20%5Cright%29%5E%7Bnt%7D-1%7D%7B%5Cfrac%7Br%7D%7Bn%7D%7D%20%5Cright%5D)
![\bf \qquad \begin{cases} A= \begin{array}{llll} \textit{accumulated amount} \end{array} \\ pymnt=\textit{periodic payments}\to &100\\ r=rate\to 14\%\to \frac{14}{100}\to &0.14\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{weekly, thus} \end{array}\to &52\\ t=years\to &5 \end{cases} \\\\\\ A=100\left[ \cfrac{\left( 1+\frac{0.14}{52} \right)^{52\cdot 5}-1}{\frac{0.14}{52}} \right]](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Baccumulated%20amount%7D%0A%5Cend%7Barray%7D%0A%5C%5C%0Apymnt%3D%5Ctextit%7Bperiodic%20payments%7D%5Cto%20%26100%5C%5C%0Ar%3Drate%5Cto%2014%5C%25%5Cto%20%5Cfrac%7B14%7D%7B100%7D%5Cto%20%260.14%5C%5C%0An%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%0A%5Ctextit%7Bweekly%2C%20thus%7D%0A%5Cend%7Barray%7D%5Cto%20%2652%5C%5C%0At%3Dyears%5Cto%20%265%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D100%5Cleft%5B%20%5Ccfrac%7B%5Cleft%28%201%2B%5Cfrac%7B0.14%7D%7B52%7D%20%5Cright%29%5E%7B52%5Ccdot%205%7D-1%7D%7B%5Cfrac%7B0.14%7D%7B52%7D%7D%20%5Cright%5D)