Annually The amount after 10 years = $ 7247.295
quarterly compound after 10 years = $7393.5
Continuously interest =$7,419
Given:
P = the principal amount
r = rate of interest
t = time in years
n = number of times the amount is compounding.
Principal = $4500
time= 10 year
Rate = 5%
To find: The amount after 10 years.
The principal amount is, P = $4500
The rate of interest is, r = 5% =5/100 = 0.05.
The time in years is, t = 10.
Using the quarterly compound interest formula:
A = P (1 + r / 4)4 t
A= 4500(1+.05/4)40
A= 4500(4.05/4)40
A= 4500(1.643)
Answer: The amount after 10 years = $7393.5
Using the Annually compound interest formula:
A = P (1 + r / 100) t
A= 4500(1+5/100)10
A= 4500(105/100)10
Answer: The amount after 10 years = $ 7247.295
Using the Continuously compound interest formula:
e stands for Napier’s number, which is approximately 2.7183

A= $2,919
Answer: The amount after 10 years = $4500+$2,919=$7,419
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Answer:
3sqrt(2) ................
Answer:
(1.5, -6)
Step-by-step explanation:
To find the x coordinate of the midpoint
(x1+x2)/2 = (8+-5)/2 = 3/2 = 1.5
To find the y coordinate of the midpoint
(y1+y2)/2 = (-3+-9)/2 = -12/2 = -6
Answer:convergent
Step-by-step explanation:
Given
Improper Integral I is given as


integration of
is 
![I=1000\times \left [ e^x\right ]^{0}_{-\infty}](https://tex.z-dn.net/?f=I%3D1000%5Ctimes%20%5Cleft%20%5B%20e%5Ex%5Cright%20%5D%5E%7B0%7D_%7B-%5Cinfty%7D)
![I=1000\times I=\left [ e^0-e^{-\infty}\right ]](https://tex.z-dn.net/?f=I%3D1000%5Ctimes%20I%3D%5Cleft%20%5B%20e%5E0-e%5E%7B-%5Cinfty%7D%5Cright%20%5D)
![I=1000\times \left [ e^0-\frac{1}{e^{\infty}}\right ]](https://tex.z-dn.net/?f=I%3D1000%5Ctimes%20%5Cleft%20%5B%20e%5E0-%5Cfrac%7B1%7D%7Be%5E%7B%5Cinfty%7D%7D%5Cright%20%5D)

so the integration converges to 1000 units
Answer:
-2
Step-by-step explanation:
1. Distribute: -3x + 6x + 3 = x-1
2. Combine like terms: 3x + 3 = x-1
3. Subtract 3 on both sides: 3x = x-4
4. Subtract x on both sides: 2x = -4
5. Divide each side by 2: x = -2
6. Check by plugging in -2 to the equation: