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tester [92]
3 years ago
5

Would y^10/y^10 become 1?

Mathematics
1 answer:
mote1985 [20]3 years ago
5 0

Answer:

yes

Step-by-step explanation:

If you are dividing it will become 1 because 10 minus 10 equals 0. Anything to the power of 0 is 1 then you keep the base which is y so the answer is y

If you are multipling the answer will be y^10  because 10 plus 10 is 20 and you keep the base which is y

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3. C(5,5) : n=5

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5. C(18,17) : r=3

Step-by-step explanation:

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3 years ago
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.

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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.

Proof of one direction for example:

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Answer: 249

Step-by-step explanation: 210+698-812+1453-1300

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