Six squared over six to the fifth
2 answers:
You might be interested in
Answer:
= {1,3,5,7,9,....}
= {1,2,4,5,7,8,10,11,13,14......}
= {1,2,3,4,5}
= {10,11,12,13,14,15,......}
Step-by-step explanation:
Solution:
It is given that, Universal set is the set of all Natural Numbers.
and
Set A, Set B, Set C, Set D are subsets of Universal set.
A = {x:x is even natural numbers} = {2,4,6,8,10,.....}
B = {x:x ∈ N and x is a multiple of 3} = {3,6,9,12,15,....}
C = {x:x∈ N and x < 5} = {6,7,8,9,10,.....}
D = {x:x ∈ N and x > 10} = {1,2,3,4,5,6,7,8,9}
U = {1,2,3,4,5,6,7,8,9,10,11,........}
Complements:
= U-A = {1,2,3,4,5,6,7,8,9,10,11,........} - {2,4,6,8,10,.....}
= {1,3,5,7,9,....}
= U-B = {1,2,3,4,5,6,7,8,9,10,11,........} - {3,6,9,12,15,....}
= {1,2,4,5,7,8,10,11,13,14......}
= U-C = {1,2,3,4,5,6,7,8,9,10,11,........} - {6,7,8,9,10,.....}
= {1,2,3,4,5}
= {1,2,3,4,5,6,7,8,9,10,11,........} - {1,2,3,4,5,6,7,8,9}
= {10,11,12,13,14,15,......}
Answer:
A. tan(2π/3) =
Answer: 91
Step-by-step explanation:
Given : The number of documentaries = 5
The number of comedies = 7
The number of mysteries = 4
The number of horror films =5
The total number of movies other than comedy = 14
Now, the number of possible combinations of 9 movies can he rent if he wants all 7 comedies is given by :-
Therefore, the number of possible combinations of 9 movies can he rent if he wants all 7 comedies is 91 .