Hmm if I'm not mistaken, is just an "ordinary" annuity, thus
![\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%0A%5C%5C%5C%5C%5C%5C)
![\bf \begin{cases} A= \begin{array}{llll} \textit{original amount}\\ \textit{already compounded} \end{array} & \begin{array}{llll} \end{array}\\ pymnt=\textit{periodic payments}\to &400\\ r=rate\to 10\%\to \frac{10}{100}\to &0.1\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, so once} \end{array}\to &1\\ t=years\to &14 \end{cases} \\\\\\ A=400\left[ \cfrac{\left( 1+\frac{0.1}{1} \right)^{1\cdot 14}-1}{\frac{0.1}{1}} \right] ](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0AA%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Boriginal%20amount%7D%5C%5C%0A%5Ctextit%7Balready%20compounded%7D%0A%5Cend%7Barray%7D%20%26%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%0A%5Cend%7Barray%7D%5C%5C%0Apymnt%3D%5Ctextit%7Bperiodic%20payments%7D%5Cto%20%26400%5C%5C%0Ar%3Drate%5Cto%2010%5C%25%5Cto%20%5Cfrac%7B10%7D%7B100%7D%5Cto%20%260.1%5C%5C%0An%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%0A%5Ctextit%7Bannually%2C%20so%20once%7D%0A%5Cend%7Barray%7D%5Cto%20%261%5C%5C%0A%0At%3Dyears%5Cto%20%2614%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D400%5Cleft%5B%20%5Ccfrac%7B%5Cleft%28%201%2B%5Cfrac%7B0.1%7D%7B1%7D%20%5Cright%29%5E%7B1%5Ccdot%2014%7D-1%7D%7B%5Cfrac%7B0.1%7D%7B1%7D%7D%20%5Cright%5D%0A)
Assuming the numbers are exponents expect the coefficient of 2, top expression, -12x^3y^6, multiply bottom
and= 72x^4y^8
Answer:
14°
Step-by-step explanation:
(7x-1)°+(6x-1)°=180°(sum of angles in a straight line/being linear pair)
or,7x-1°+6x-1°=180°
or,13x-2°=180°
or,13x=182°
or,x=182°/13
or,x=14°
Answer:
0.5
Step-by-step explanation:
Ok, so it's asking for what (1/(x-1) - 2/(x^2-1)) approaches as x approaches 1. Before we deal with the limit, let's simplify the inside.
We want to combine the two fractions into one fraction. Therefore, we need a common denominator.
1/(x-1) is equal to (x+1)/((x+1)(x-1) is equal to (x+1)/(x^2-1).
the inside expression is therefore (x+1)/(x^2-1) - 2/(x^2-1)
which simplifies to (x-1)/(x^2-1).
and that simplifies further to 1/(x+1).
Now this is a continuous function when x = 1, so to find the limit as x approaches 1 of this function, we can by definition just plug 1 in.
limx->1 (1/(x+1)) = 1/2.
The reason why we didn't just plug 1 in at the beginning is because the function wasn't continuous when x was 1.
Answer:
Jaime = 11
Sammy = 44
Step-by-step explanation:
It is given that Sammy is eleven less than five times Jaime's age. Use parameter (S) to represent Sammy's age, and parameter (J) to represent Jaime's age. Form an equation based on this given information:
S = 5J - 11
It is given that the sum of the two girl's ages is (55), use an equation to represent this.
S + J = 55
Substitute in the expression for the value of (S) in terms of (J),
(5J - 11) + J = 55
Simplify,
6J - 11 = 55
Inverse operations,
6J - 11 = 55
6J = 66
J = 11
Backsolve for the value of (S), substitute in the value of (J) into the equation for (S) and solve,
S = 5(J) - 11
S = 5(11) -11
S = 55 - 11
S = 44
