Let A be some subset of a universal set U. The "complement of A" is the set of elements in U that do not belong to A.
For example, if U is the set of all integers {..., -2, -1, 0, 1, 2, ...} and A is the set of all positive integers {1, 2, 3, ...}, then the complement of A is the set {..., -2, -1, 0}.
Notice that the union of A and its complement make up the universal set U.
In this case,
U = {1, 2, 3, 6, 10, 13, 14, 16, 17}
The set {3, 10, 16} is a subset of U, since all three of its elements belong to U.
Then the complement of this set is all the elements of U that aren't in this set:
{1, 2, 6, 13, 14, 17}
The sets are (1,-1) or (-1,1)
Lets see
Take until 4terms
#1
(1,-1)
(-1,1)
#2
(1,-1)
(-1,1)
Hence verified
Another set can be (0,0)
Proof:-


Hence verified.
Answer:
7212960
Step-by-step explanation:
You have to round up since the number in the ones place is a 5.
Answer: a(n) = 5 - 3n
the sequence has:
a1 = 2
a2 = -1
a3 = -7
.........
we can see that: a2 - a1 = a3 - a2 = -3
=> the sequence is a arithmetic sequence
=> the distance between the numbers is d = -3
because this sequence is a arithmetic sequence
=> a(n) = a1 + (n - 1)d = 2 + (n - 1).(-3) = 5 - 3n
Step-by-step explanation:
The table would be y = 6x
Divide the Y values by the X values and they all equal 6, so you multiply the X value by 6 to get y.
Look at the dots on the graph ( 1,5) (2,10) (3,15) (4,20)
Divide the Y value by the X and they all equal 5, soy = 5x