Answer:
a
Since the integral has an infinite discontinuity, it is a Type 2 improper integral
b
Since the integral has an infinite interval of integration, it is a Type 1 improper integral
c
Since the integral has an infinite interval of integration, it is a Type 1 improper integral
d
Since the integral has an infinite discontinuity, it is a Type 2 improper integral
Step-by-step explanation:
Considering a

Looking at this we that at x = 3 this integral will be infinitely discontinuous
Considering b

Looking at this integral we see that the interval is between
which means that the integral has an infinite interval of integration , hence it is a Type 1 improper integral
Considering c

Looking at this integral we see that the interval is between
which means that the integral has an infinite interval of integration , hence it is a Type 1 improper integral
Considering d

Looking at the integral we see that at x = 0 cot (0) will be infinity hence the integral has an infinite discontinuity , so it is a Type 2 improper integral
This is the number of combinations of 2 from 23
23C2 = 23! / 2! 21!
A quick way to do this is 23*22 / 2 = 253
Answer:
Step-by-step explanation:
The rule is that after the first term, every term after that is made out of
the smallest power of 2 that hasn't been used yet, with a zero after it.