Answer:
see below
Step-by-step explanation:
l -4 +2x l -3 > 9
Add 3 to each side
l -4 +2x l -3+3 > 9+3
l -4 +2x l > 12
We Separate into two inequalities, remembering to flip the negative inequality
-4+2x>12 or -4+2x<-12
Add 4 to each side
-4+4+2x>12+4 or -4+4+2x<-12 +4
2x> 16 or 2x< -8
Divide each side by 2
2x/2 > 16/2 or 2x/2 < -8/2
x > 8 or x < -4
To graph we have open circles at 8 going to the right and at -4 going to the left
Answer:
Step-by-step explanation:
<u>Given expression</u>
4x³ + 24x² - 288x
<u>Factorize 4x out from the expression</u>
4x · x² + 4x · 6x - 4x · 72
4x (x² + 6x - 72)
<u>Cross multiply to factorize the remaining polynomial expression</u>
<em>The meaning is to allow the factored product of the constant to add up to the first-degree term</em>
x 12
x -6
<u>Combine the result</u>
Hope this helps!! :)
Please let me know if you have any questions
Answer:
a. {1, 2, 3, 4, 5, 6, 7}
b. {1, 2, 3, 4, 5, 6, 8}
c. {4, 5, 7}
d. {1, 4, 5}
Step-by-step explanation:
The union of two sets is the list of elements in either set. The intersection of two sets is the list of elements in both sets. The complement of a set is the list of elements in the universal set that are not in the set being complemented. The complement of a set can be indicated with an apostrophe: A' is the complement of set A, for example.
<h3>a. AU(BUC)</h3>B∪C = {2, 5, 6, 7} ∪ {3, 4, 6, 7} = {2, 3, 4, 5, 6, 7}
A∪(B∪C) = {1, 4, 5, 7} ∪ {2, 3, 4, 5, 6, 7} = {1, 2, 3, 4, 5, 6, 7}
<h3>b. (AN(BNC))'</h3>B∩C = {2, 5, 6, 7} ∩ {3, 4, 6, 7} = {6, 7}
A ∩ (B∩C) = {1, 4, 5, 7} ∩ {6, 7} = {7}
(A∩(B∩C))' = {7}' = {1, 2, 3, 4, 5, 6, 8}
<h3>c. (ANB)U(ANC)</h3>A∩Β = {1, 4, 5, 7} ∩ {2, 5, 6, 7} = {5, 7}
Α∩C = {1, 4, 5, 7} ∩ {3, 4, 6, 7} = {4, 7}
(A∩B)∪(A∩C) = {5, 7} ∪ {4, 7} = {4, 5, 7}
<h3>d. (ANB')U(ANC')</h3>(A∩B')∪(A∩C') = A∩(B'∪C') = A∩(B∩C)'
(B∩C)' = {6, 7}' = {1, 2, 3, 4, 5, 8}
A∩(B∩C)' = {1, 4, 5, 7} ∩ {1, 2, 3, 4, 5, 8} = {1, 4, 5}
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<em>Additional comment</em>
In part (d) we made use of De Morgan's law for sets:
B'∪C' = (B∩C)'
We also made use of the distributive property for sets:
A∩(B∪C) = (A∩B)∪(A∩C)
