Answer:
(-3,1) dhdjdjdjdjejdhdjehehdhdjddjdjddh
(-3,1)
120/1500=x/100, x=8
So, you make 8% commision, meaning you now have to do this:
x/1300=8/100, so x=$104
![n = - 1](https://tex.z-dn.net/?f=n%20%3D%20%20-%201)
<h2><em>hope</em><em> it</em><em> helps</em></h2>
Answer: option C is the correct answer
Step-by-step explanation:
The system of linear equations is
10x + 7y = 12 - - - - - - - 1
8x + 7y = 18 - - - - - - - 2
Since the coefficient of y is the same in equation 1 and equation 2, we will eliminate y by subtracting equation 2 from equation 1, it becomes
10x - 8x + 7y - 7y = 12 - 18
2x = -6
x = - 6/2 = - 3
Substituting x = - 3 into equation 1, it becomes
10×-3 + 7y = 12
-30 + 7y = 12
Let the constants be on the right hand side and the term containing y be on the left hand side. It becomes
7y = 12 + 30
7y = 42
y = 42/7
y = 6
C) (−3, 6)
Answer:
The value of this investment at the end of the 5 years is of $662.5.
Step-by-step explanation:
Compound interest:
The compound interest formula is given by:
![A(t) = P(1 + \frac{r}{n})^{nt}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20P%281%20%2B%20%5Cfrac%7Br%7D%7Bn%7D%29%5E%7Bnt%7D)
Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per year and t is the time in years for which the money is invested or borrowed.
Dina invests $600 for 5 years at a rate of 2% per year compound interest.
This means that
. Thus
![A(t) = P(1 + \frac{r}{n})^{nt}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20P%281%20%2B%20%5Cfrac%7Br%7D%7Bn%7D%29%5E%7Bnt%7D)
![A(t) = 600(1 + \frac{0.02}{1})^{t}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20600%281%20%2B%20%5Cfrac%7B0.02%7D%7B1%7D%29%5E%7Bt%7D)
![A(t) = 600(1.02)^t](https://tex.z-dn.net/?f=A%28t%29%20%3D%20600%281.02%29%5Et)
Calculate the value of this investment at the end of the 5 years.
This is A(5). So
![A(5) = 600(1.02)^5 = 662.5](https://tex.z-dn.net/?f=A%285%29%20%3D%20600%281.02%29%5E5%20%3D%20662.5)
The value of this investment at the end of the 5 years is of $662.5.