Define A = {x : 0 ≤ x ≤ 1}, B = {x : 0 ≤ x ≤ 3}, and C = {x : −1 ≤ x ≤ 2}. Draw diagrams showing each of the following sets of p
oints: (a) AC ∩ B ∩C (b) AC ∪(B ∩C) (c) A ∩ B ∩CC (d) [(A ∪ B)∩CC ] C
1 answer:
Answer: (a) {0,1}; (b) {0,1, 2}; (c) {0,1}; (d) {0, 1, 2}
Step-by-step explanation: Defining the sets A,B and C:
A = {0,1}; B = {0, 1, 2, 3}; C = {-1, 0, 1, 2}
(a) A ∩ B ∩ C = {0, 1}, because it represents the intersection of the sets, the items that appears in all the three sets.
(b) A ∪ (B ∩ C) = {0, 1} ∪ {0, 1, 2} = {0, 1, 2}. In this case, it represents the union of the sets A with the intersection of sets B and C.
The (c) and (d) are a combination between union and intersection of the sets.
The Venn Diagrams are the representations of the sets and their interactions and are represented in the attachments
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9. ±1, ±2, ±3, ±6
11. ±1, ±2, ±3, ±4, ±6, ±12
Step-by-step explanation:
The possible rational roots are (plus or minus) the divisors of the constant term, divided by the divisors of the leading coefficient.
Here, the leading coefficient is 1 in each case, so the possible rational roots are plus or minus a divisor of the constant term.
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9. The constant is -6. Divisors of 6 are 1, 2, 3, 6. The possible rational roots are ...
±{1, 2, 3, 6}
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11. The constant is 12. Divisors of 12 are 1, 2, 3, 4, 6, 12. The possible rational roots are ...
±{1, 2, 3, 4, 6, 12}
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A graphing calculator is useful for seeing if any of these values actually are roots of the equation. (The 4th-degree equation will have 2 complex roots.)
