So we have an equation:
2w=12-4w
Firstly, let's simplify, and divide all terms by 2:
w=6-2w
Now, we add 2w over to the other side:
3w=6
And finally, we divide both sides by 3:
w=2
Hope this helps!
Answer:
Step-by-step explanation:
If you need a fraction just think of what equals 24 like this
Ex.
6/25 = .24
-45 + 150 = 105
Carol has $105 in her account after receiving her paycheck.
For the 1st fraction, since 14 × 1 = 14,
9
14
=
9 × 1
14 × 1
=
9
14
Likewise, for the 2nd fraction, since 7 × 2 = 14,
6
7
=
6 × 2
7 × 2
=
12
14
Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction
9
14
<
12
14
or
9
14
<
6
7
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)
