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3 years ago

6

Which equation shoes how the properties of numbers extend to negative whole numbers by commutative property of addition A) -2+4=

4+(-2) B)-2(4)=4(-2). C) (-2+4)+3=-2+(4+3. D)-2(4+3)=(2×4)+(-2×3)

1 answer:

Vaselesa [24]3 years ago

7 0

In option A the order of the two numbers being added changes. That is what the commutative property is, so the answer is A.

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By inclusion/exclusion,

$n(P' \cup Q) = n(P') + n(Q) - n(P' \cap Q)$

We have

$n(\xi) = n(P) + n(P') \implies n(P') = 28 - n(P)$

so that

$n(P' \cup Q) = (28 - n(P)) + n(Q) - 2n(P) = 28 - 3n(P) + n(Q)$

Now,

$n(\xi) = 28 \implies n(P \cup Q) = 28 - 7 = 21$

and by inclusion/exclusion,

$n(P \cup Q) = n(P) + n(Q) - n(P \cap Q)$

Decompose $Q$ into the union of two disjoint sets:

$Q = (P \cap Q) \cup (P' \cap Q)$

Since they're disjoint,

$n(Q) = n(P\cap Q) + n(P'\cap Q) \implies n(Q) = n(P\cap Q) + 2n(P)$

$\implies n(P \cup Q) = n(P) + (n(P\cap Q) + 2n(P)) - n(P \cap Q)$

$\implies 21 = 3n(P)$

$\implies n(P) = 7$

From the Venn diagram, we see there are 3 elements unique to $P$ - by the way, this is the set $P \cap Q'$ - so $n(P\cap Q) = 7-3 = 4$ , and it follows that

$n(Q) = n(P\cap Q) + 2n(P) = 4 + 2\times7 = 18$

Finally, we get for (a)

$n(P' \cup Q) = 28 - 3n(P) + n(Q) = 28 - 3\times7 + 18 = \boxed{25}$

For (b), we have by inclusion/exclusion that

$n(P \cup Q') = n(P) + n(Q) - n(P \cap Q') = 7 + 18 - 3 = \boxed{22}$

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