Answer:
Step-by-step explanation:
a)

this can be done by substitute lnx = u
dx/x = du
When x =1, u =0 and when x =2, u = ln 2
So integral = 
We find that this integral value is not definid for p =1
Hence for values of p other than 1, this converges.
When we substitute limits

and converges for p ≠1
b) 
So not converging for p =1
But ln x is defined only for x >0
So integral 0 to 1 makes this integral not valid and hence not convergent.
Answer:
A.) 1^3
Step-by-step explanation:
2^5/2^2
= 1^5 - 2
= 1^3
No, because 32 is a composite number, and prime numbers cannot fit into composite numbers. Hope this helps.
Answer:
- 2x + y
- 3 x - 2y
- -1
2(3 x - 2y) = -6x + 4y
2x + y + -6x + 4y + -1 = -4x + 5y -1
I think the answer is 76 because the tree is halfway between building and stake
or it could be 57 either one but 70% it is 76