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}
I believe that the correct answer would be A
Hope this helped love <3
We will use the distributive property. -4 times 9z=-36z -4 times -2=8 so we have -36z+8 as our answer
Answer:
yes it's (A)
Step-by-step explanation:
Its A because you aways have to add the one under the (N) and add (1) and multiple them and you got your answee
Answer:
x = ±2, 3 are the critical points of the given inequality.
Step-by-step explanation:
The given inequality is 
To find the critical points we will equate the numerator and denominator of the inequality to zero.
For numerator,
(x - 2)(x + 2) = 0
x = ±2
For denominator,
x² - 5x + 6 = 0
x² - 3x -2x + 6 = 0
x(x - 3) -2(x - 3) = 0
(x - 3)(x - 2) = 0
x = 2, 3
Therefore, critical points of the inequality are x = ±2, 3 where the sign of the inequality will change.
