Answer:
x= 1 7/12
Step-by-step explanation:
x+1/2+x+2/3-x+3/4=2
1. We know that x+x-x=x
2. 1/2+2/3-3/4=6/12+8/12-9/12=5/12
3. x+5/12=2
4. x=2-5/12
x=1 7/12

Taking

gives

, so that the integral becomes





When

, we have


and from here we can substitute

to proceed from here.
Quick note: When we set

, we are implicitly enforcing

just so that the substitution can be undone later via

. But note that over this domain, we automatically guarantee that

, so the absolute value bars can be dropped immediately.
Answer:
B.
Step-by-step explanation:
h -234 = 1,192
-234 is a negative and if you add a negative number with a positive you have subtract and the remaining is your answer.
1,192+234=1,426
1,426-234=1,192
h=1,426
Answer:
15
Step-by-step explanation:
