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labwork [276]
3 years ago
6

According to the distributive property, 6 (a + b) =

Mathematics
2 answers:
HACTEHA [7]3 years ago
7 0
Its b

because this is how the distributive property works....
i'm going to use 2 and 4 for example
6(2+4)= 36 because you add 2 and 4(or a and b) then multiply by 6 because when using the <span>distributive you use pemdas which is an abbreviation for parenthesis, exponents, multiplication, division, addition and subtraction and that is the order you do it in 

o answer b you multiply 6 by a (or 2) then multiply 6 by b (or 4) after doing that you add them together and it equals 36 just like 6(2+4) (or 6(a+b)..)
and that is why it is b</span><span />
Helen [10]3 years ago
6 0
The answer is option B. In the distributive property you need to multiply the constant that is outside of the parenthesis, with the terms that are inside :)
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Given:

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Now,

Taking, b= 8.6 and h = 10.9 we get,

A=\frac{1}{2}(8.6)(10.9) sq yd

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Hence,

The area of the given triangle is 46.87 sq yd.

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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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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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4 0
3 years ago
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