Hey there!!
Let's take the first number as ' x '
Then, the second number would be ' x + 1 '
Sum of these = 183
Hence,





Hence, the small integer is 91
Hope my answer helps!
Answer:
I accidently clicked answer and it won't let me quit
Step-by-step explanation:
I'm not even in high school yet D: sorry
Answer:
52 if im wrong sorry :(
Step-by-step explanation:
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