
Look at the square root like "x", then
Answer:converge at 
Step-by-step explanation:
Given
Improper Integral I is given as

integration of
is -
![I=\left [ -\frac{1}{x}\right ]^{\infty}_3](https://tex.z-dn.net/?f=I%3D%5Cleft%20%5B%20-%5Cfrac%7B1%7D%7Bx%7D%5Cright%20%5D%5E%7B%5Cinfty%7D_3)
substituting value
![I=-\left [ \frac{1}{\infty }-\frac{1}{3}\right ]](https://tex.z-dn.net/?f=I%3D-%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B%5Cinfty%20%7D-%5Cfrac%7B1%7D%7B3%7D%5Cright%20%5D)
![I=-\left [ 0-\frac{1}{3}\right ]](https://tex.z-dn.net/?f=I%3D-%5Cleft%20%5B%200-%5Cfrac%7B1%7D%7B3%7D%5Cright%20%5D)

so the value of integral converges at 
Answer:0.25 tons
Step-by-step explanation:
I looked it up on the internet
Answer:
10 pages
Step-by-step explanation:
1 and 1/3 pages multiplies by 7.5 brings us to the answer of ten pages
20 minutes goes into 2 and 1/2 hours 7.5 times
The domain is all the x-values you can use with this function. When dealing with a function like this, you need to ask yourself, what would make the bottom equal to 0? Whatever those numbers are, you do NOT want them in your domain.
A fundamental rule is that you cannot divide by 0 or have 0 as a denominator.
So, you solve x^2+6x=0 to find that x=0 or x=-6 would make the denominator 0.
The domain is everything BUT those two values, because they essentially break your function.