Is it necessary to use the formulas for special products of binomials to multiply these binomials?
1 answer:
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Answer:
n(A) = 3
n(B) = 6
-6 ∈ A => True
-14 ∈ A => False
k ∈ B=> True
Q∈B = >False
Step-by-step explanation:
Given
A is the set of integers greater than - 7 and less than - 3
B={c, h, j, k, v, y}
For set A:
The integers greater than -7 will be -6,-5 ....
As the set has integers less than -3, the set will be:
<u>Cardinalities:</u>
Cardinality is the number of elements in the set.
So,
Now for the statements:
-6 ∈ A True as -6 is a member of set A
-14 ∈ A False as -14 is not a member of A
k ∈ B True
Q∈B False
Hence,
n(A) = 3
n(B) = 6
-6 ∈ A => True
-14 ∈ A => False
k ∈ B=> True
Q∈B = >False