Answer: the future value is $1748.4
Step-by-step explanation:
We would apply the formula for determining compound interest which is expressed as
A = P(1+r/n)^nt
Where
A = total amount in the account at the end of t years
r represents the interest rate.
n represents the periodic interval at which it was compounded.
P represents the principal or initial amount deposited
From the information given,
P = 1550
r = 4% = 4/100 = 0.04
n = 365 because it was compounded 365 times in a year.
t = 3 years
Therefore,.
A = 1550(1 + 0.04/365)^365 × 3
A = 1550(1+0.00011)^1095
A = 1550(1.00011)^1095
A = 1550 × 1.128
A = 1748.4
Answer:
0.3
33.3%
Step-by-step explanation:
One way to capture the domain of integration is with the set
![D = \left\{(x,y) \mid 0 \le x \le 1 \text{ and } -x \le y \le 0\right\}](https://tex.z-dn.net/?f=D%20%3D%20%5Cleft%5C%7B%28x%2Cy%29%20%5Cmid%200%20%5Cle%20x%20%5Cle%201%20%5Ctext%7B%20and%20%7D%20-x%20%5Cle%20y%20%5Cle%200%5Cright%5C%7D)
Then we can write the double integral as the iterated integral
![\displaystyle \iint_D \cos(y+x) \, dA = \int_0^1 \int_{-x}^0 \cos(y+x) \, dy \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciint_D%20%5Ccos%28y%2Bx%29%20%5C%2C%20dA%20%3D%20%5Cint_0%5E1%20%5Cint_%7B-x%7D%5E0%20%5Ccos%28y%2Bx%29%20%5C%2C%20dy%20%5C%2C%20dx)
Compute the integral with respect to
.
![\displaystyle \int_{-x}^0 \cos(y+x) \, dy = \sin(y+x)\bigg|_{y=-x}^{y=0} = \sin(0+x) - \sin(-x+x) = \sin(x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-x%7D%5E0%20%5Ccos%28y%2Bx%29%20%5C%2C%20dy%20%3D%20%5Csin%28y%2Bx%29%5Cbigg%7C_%7By%3D-x%7D%5E%7By%3D0%7D%20%3D%20%5Csin%280%2Bx%29%20-%20%5Csin%28-x%2Bx%29%20%3D%20%5Csin%28x%29)
Compute the remaining integral.
![\displaystyle \int_0^1 \sin(x) \, dx = -\cos(x) \bigg|_{x=0}^{x=1} = -\cos(1) + \cos(0) = \boxed{1 - \cos(1)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E1%20%5Csin%28x%29%20%5C%2C%20dx%20%3D%20-%5Ccos%28x%29%20%5Cbigg%7C_%7Bx%3D0%7D%5E%7Bx%3D1%7D%20%3D%20-%5Ccos%281%29%20%2B%20%5Ccos%280%29%20%3D%20%5Cboxed%7B1%20-%20%5Ccos%281%29%7D)
We could also swap the order of integration variables by writing
![D = \left\{(x,y) \mid -1 \le y \le 0 \text{ and } -y \le x \le 1\right\}](https://tex.z-dn.net/?f=D%20%3D%20%5Cleft%5C%7B%28x%2Cy%29%20%5Cmid%20-1%20%5Cle%20y%20%5Cle%200%20%5Ctext%7B%20and%20%7D%20-y%20%5Cle%20x%20%5Cle%201%5Cright%5C%7D)
and
![\displaystyle \iint_D \cos(y+x) \, dA = \int_{-1}^0 \int_{-y}^1 \cos(y+x) \, dx\, dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciint_D%20%5Ccos%28y%2Bx%29%20%5C%2C%20dA%20%3D%20%5Cint_%7B-1%7D%5E0%20%5Cint_%7B-y%7D%5E1%20%5Ccos%28y%2Bx%29%20%5C%2C%20dx%5C%2C%20dy)
and this would have led to the same result.
![\displaystyle \int_{-y}^1 \cos(y+x) \, dx = \sin(y+x)\bigg|_{x=-y}^{x=1} = \sin(y+1) - \sin(y-y) = \sin(y+1)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-y%7D%5E1%20%5Ccos%28y%2Bx%29%20%5C%2C%20dx%20%3D%20%5Csin%28y%2Bx%29%5Cbigg%7C_%7Bx%3D-y%7D%5E%7Bx%3D1%7D%20%3D%20%5Csin%28y%2B1%29%20-%20%5Csin%28y-y%29%20%3D%20%5Csin%28y%2B1%29)
![\displaystyle \int_{-1}^0 \sin(y+1) \, dy = -\cos(y+1)\bigg|_{y=-1}^{y=0} = -\cos(0+1) + \cos(-1+1) = 1 - \cos(1)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-1%7D%5E0%20%5Csin%28y%2B1%29%20%5C%2C%20dy%20%3D%20-%5Ccos%28y%2B1%29%5Cbigg%7C_%7By%3D-1%7D%5E%7By%3D0%7D%20%3D%20-%5Ccos%280%2B1%29%20%2B%20%5Ccos%28-1%2B1%29%20%3D%201%20-%20%5Ccos%281%29)
Answer:
5 is the answer
Step-by-step explanation:
hope this helps