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Ber [7]
2 years ago
13

In their yard, Larry’s parents decided to plant tomatoes in a certain area. They chose an area that was of a yard long and of a

yard wide. Choose the model that shows the area they’ll use for tomatoes.
18/18
15/18
8/15
10/18
Mathematics
1 answer:
Marianna [84]2 years ago
3 0

the answer is 18/18 which is the first one

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Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam
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  • X-(A\cup B)=(X-A)\cap(X-B)

I'll assume the usual definition of set difference, X-A=\{x\in X,x\not\in A\}.

Let x\in X-(A\cup B). Then x\in X and x\not\in(A\cup B). If x\not\in(A\cup B), then x\not\in A and x\not\in B. This means x\in X,x\not\in A and x\in X,x\not\in B, so it follows that x\in(X-A)\cap(X-B). Hence X-(A\cup B)\subset(X-A)\cap(X-B).

Now let x\in(X-A)\cap(X-B). Then x\in X-A and x\in X-B. By definition of set difference, x\in X,x\not\in A and x\in X,x\not\in B. Since x\not A,x\not\in B, we have x\not\in(A\cup B), and so x\in X-(A\cup B). Hence (X-A)\cap(X-B)\subset X-(A\cup B).

The two sets are subsets of one another, so they must be equal.

  • X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices i\in I.

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