Answer:
A(0,3) B(3,4) C(2,1)
Step-by-step explanation:
when reflecting you just the change the sign if either the X or the Y. in this case it's the Y
Daaaaaaaaaammmmm that’s hard
7 1/5th is the anwser but thats easy your just being lazy i think
Answer:
Converges at -1
Step-by-step explanation:
The integral converges if the limit exists, if the limit does not exist or if the limit is infinity it diverges.
We will make use of integral by parts to determine:
let:
![dv=e^(2x)\cdot{dx}](https://tex.z-dn.net/?f=dv%3De%5E%282x%29%5Ccdot%7Bdx%7D)
![v=2\cdot{e^(2x)}](https://tex.z-dn.net/?f=v%3D2%5Ccdot%7Be%5E%282x%29%7D)
![\int\limits^a_b {u} \, dv = uv -\int\limits^a_b {v} \, du](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7Bu%7D%20%5C%2C%20dv%20%3D%20uv%20-%5Cint%5Climits%5Ea_b%20%7Bv%7D%20%5C%2C%20du)
![\int\limits^a_b {x\cdot{e^2^x} \, dx =2xe^2^x- \int\limits^a_b {2e^2^x} \, dx](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7Bx%5Ccdot%7Be%5E2%5Ex%7D%20%5C%2C%20dx%20%3D2xe%5E2%5Ex-%20%5Cint%5Climits%5Ea_b%20%7B2e%5E2%5Ex%7D%20%5C%2C%20dx)
![\int\limits^a_b {xe^2^x} \, dx = 2xe^2^x-2e^2^x-C](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7Bxe%5E2%5Ex%7D%20%5C%2C%20dx%20%3D%202xe%5E2%5Ex-2e%5E2%5Ex-C)
We can therefore determine that if x tends to 0 the limit is -1
![\lim_{x \to \0} 2xe^2^x-2e^2^x=0-1=-1](https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%20%5C0%7D%202xe%5E2%5Ex-2e%5E2%5Ex%3D0-1%3D-1)