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natka813 [3]
3 years ago
7

ah i only need to know if its AND or OR! i already have the graphing done, if you could, please explain how you got AND/OR! than

ks!

Mathematics
1 answer:
GuDViN [60]3 years ago
3 0

Answer:

  AND

Step-by-step explanation:

<em>As a rule, a compound inequality such as this is an AND inequality</em>. That is confirmed by the fact that ...

  (a) the comparison symbols both point the same way, and

  (b) the way they point is correct for the values on either end:

     -2 < 3

__

If we were to reverse the comparisons so they were ...

  -2 ≥ x > 3

we know right away that -2 > 3 is FALSE, so there will be no single value of x that will satisfy both inequalities. That is, if we interpret this as an AND inequality, its solution is the empty set.

On the other hand, if we interpret this as an OR inequality, then the solution set will be the union ...

  (-2 ≥ x) ∪ (x > 3)

Because -2 > 3 is false, this form of the inequality (-2 ≥ x > 3) is <em>not one you will usually see</em>.*

__

As a rule, inequalities that have disjoint solution sets will not be written in this way. They will be written as separate inequalities, clearly identified as OR.

_____

* I like to use this form because it is compact. When I use it, I do so with the understanding <em>it is technically incorrect</em>, and that the solution will be the union of the two solution sets. Not everyone who sees this "work" will understand this interpretation.

For example, |x| > 1 can be resolved to -1 > x > 1. This is technically incorrect, but can be seen as equivalent to two inequalities whose logic is OR. A technically correct resolution would identify this as equivalent to -x > 1 OR x > 1.

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Answer:

Step-by-step explanation:

Given:

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(a) Verify A(BnB)=(AUB)(AUB)

A U (B n B)

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verifies

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=  {2,4,6,8,10,12,14} n {3,5,6,9,10,12,15}

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= ( {2,4,6,8,10,12,14} n {3,6,9,12,15} ) U ( {2,4,6,8,10,12,14} n {5,10,15} )

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(c) Verify AU(BUC)=(AUB)U(AUC)​

AU(BUC)

=  {2,4,6,8,10,12,14}U{3,6,9,12,15}U{5,10,15}

= {2,3,4,5,6,8,9,10,12,14,15}

(AUB)U(AUC)​

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= {2,3,4,6,8,9,10,12,14,15}U{2,4,5,6,8,10,12,14,15}

= {2,3,4,5,6,8,9,10,12,14,15}

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AU(BUC)=(AUB)U(AUC)​

verifies

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