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4. Use set notation to identify the shaded region.
(Note: The universe is not shaded)
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- anonymous

- jamiebookeater

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- anonymous

##### 1 Attachment

- AccessDenied

So, what sets are being shaded here? (For a moment, I'm only looking at which ones have ANY shading. We'll deal with parts in a second.)

- anonymous

Well A and part of C are shaded.

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- AccessDenied

Alright. So, we're going to have A combined with some part of C.
It appears here that the only part of C that is missing from the shading is the part where it intersects with B. So, if we remove B from C by subtraction,
C - B,
then we now are representing only the part of C not in B.
We then just add in all of A to this set: We originally removed a bit of the shaded part, but here we actually add it back in.
A u (C - B)
Would that make sense?

- AccessDenied

* I think some classes use a slanted mark "A \ B" for subtraction rather than a minus sign "A - B", so you may use whichever you are familiar with.

- anonymous

sorta my understanding of set notation is not all there. lol it really confusing
My teacher hasnt showed us anything with the / mark or -. we have just been using U and and upside down U and also '

- AccessDenied

Hmm... not sure how to define that region without a subtraction.
One thing to note about sets is that they don't accumulate unless you add unique elements to them. If you add in the same elements as that are in the set, they would just disappear since we already have that element.
U is like set addition. A U B takes the elements of A together with the elements of B. When they share elements, those elements don't repeat in the union.
n / the hill thing is like finding what two sets have in common. A n B gives you the elements in both A and B. I imagine it like putting A and B on top of each other where they have the same elements and then chopping off the parts that are falling off. :P
The set subtraction '-' or '\' is like taking all the parts of one set out of another. A - B means, 'take A and remove everything it shares with B.
I hope maybe that helps. Unless I'm just repeating stuff you knew. :P

- AccessDenied

* The complement thing A' is sort of like U \ A, taking your universal set and then excluding A. Literally, 'everything not A'

- anonymous

ok that helps some. lol i just feel really dumb when it comes to this kind of stuff.

- AccessDenied

Hmm, did you only begin stuff with sets recently? I think people tend to struggle with it at first since it's like a whole new system of rules with weird 'set' things rather than numbers. :P

- anonymous

Yes I'm in a freshman college pre-algebra course. This is the last problem I have on a math project due tomorrow.

- AccessDenied

Ah, okay. Cool. :)
Well, I am not able to think of another way of writing it other than A u (C - B) or something crazier involving subtractions. I think that is the best way of writing it.

- anonymous

that might be correct let me check over my note really quick and ill get back to you. thanks

- AccessDenied

You're welcome. :)

- anonymous

See if this helps. I'm just retarded and can't learn anything from this but this looks just like what were learning.
http://www.math.hawaii.edu/~williamdemeo/Math371-Summer2011/SetOperationsAndVenDiagrams.pdf

- anonymous

scroll to the bottom of that page.

- AccessDenied

Hmm... It looks like \(A \cup B^c\) contains the shaded region including the stuff outside our set, so we could intersect it with \( A \cup C \) to remove the excess outside the set:
\((A \cup B^c) \ \cap (A \cup C)\)

- AccessDenied

The \(A \cup B^c\) is shown on the second-to-last slide.

- anonymous

yeah i see that and that makes sense. so is that the answer?

- anonymous

this is not math! lol i hate this kind of stuff

- AccessDenied

You could probably pull the A out by reverse distribution of the union over intersections.
\( M \cup (N \cap P) = (M \cup N) \cap (M \cup P)\)
\( A \cup (B^c \cap C) = (A \cup B^c) \cap (A \cup C) \)
That'd look a little better I think.
In fact, it kind of looks like A u (C - B) like originally. I guess
A u (C n B') is the same thing as that subtraction. :P

- anonymous

yeah i guess it must be the same thing, who knows though. i guess ill find out when i turn it in tomorrow. lol

- AccessDenied

I'm thinking about it...
|dw:1352244687932:dw|
Yeah, it seems to work. :D
Set notation is so fun sometimes. lol

- anonymous

so is it A ∪ (C ∩ B')?

- AccessDenied

Yep. :)

- anonymous

thanks so much!!!!!!! you have been so much help!!!!!!!!!!!!!

- AccessDenied

You're welcome! I learned some things myself as well. I never knew about that complement trick instead of subtraction of sets. ;)

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