A because we all know the intersection of sin is always in (0,0) but in this case it is different
Answer:
Step-by-step explanation:
2^0 is less than or equal to 1!, because 1<= 1
if 2^n <= (n+1)!, we wish to show that 2^(n+1) <= (n+2)!, since
(n+2)! = (n+1)! * (n+2), and (n+1)!>= 2^n, then we want to prove that n+2<=2, which is always true for n>=0
As an integer and a raitonal number as well as a real number.
Use an online math calculator for more accurate answers just plug in the variables
Answer: The correct answer is Tyrell, Yen, Hana....its C
Step-by-step explanation: Hope this helped :}
