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Fletchnuts
 2 years ago
Help Help Help!!!!!!
4. Use set notation to identify the shaded region.
(Note: The universe is not shaded)
******Attachment Below********
Fletchnuts
 2 years ago
Help Help Help!!!!!! 4. Use set notation to identify the shaded region. (Note: The universe is not shaded) ******Attachment Below********

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AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1So, 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.)

Fletchnuts
 2 years ago
Best ResponseYou've already chosen the best response.0Well A and part of C are shaded.

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1Alright. 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
 2 years ago
Best ResponseYou've already chosen the best response.1* 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.

Fletchnuts
 2 years ago
Best ResponseYou've already chosen the best response.0sorta 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
 2 years ago
Best ResponseYou've already chosen the best response.1Hmm... 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
 2 years ago
Best ResponseYou've already chosen the best response.1* The complement thing A' is sort of like U \ A, taking your universal set and then excluding A. Literally, 'everything not A'

Fletchnuts
 2 years ago
Best ResponseYou've already chosen the best response.0ok that helps some. lol i just feel really dumb when it comes to this kind of stuff.

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1Hmm, 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

Fletchnuts
 2 years ago
Best ResponseYou've already chosen the best response.0Yes I'm in a freshman college prealgebra course. This is the last problem I have on a math project due tomorrow.

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1Ah, 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.

Fletchnuts
 2 years ago
Best ResponseYou've already chosen the best response.0that might be correct let me check over my note really quick and ill get back to you. thanks

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1You're welcome. :)

Fletchnuts
 2 years ago
Best ResponseYou've already chosen the best response.0See 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/Math371Summer2011/SetOperationsAndVenDiagrams.pdf

Fletchnuts
 2 years ago
Best ResponseYou've already chosen the best response.0scroll to the bottom of that page.

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1Hmm... 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
 2 years ago
Best ResponseYou've already chosen the best response.1The \(A \cup B^c\) is shown on the secondtolast slide.

Fletchnuts
 2 years ago
Best ResponseYou've already chosen the best response.0yeah i see that and that makes sense. so is that the answer?

Fletchnuts
 2 years ago
Best ResponseYou've already chosen the best response.0this is not math! lol i hate this kind of stuff

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1You 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

Fletchnuts
 2 years ago
Best ResponseYou've already chosen the best response.0yeah i guess it must be the same thing, who knows though. i guess ill find out when i turn it in tomorrow. lol

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1I'm thinking about it... dw:1352244687932:dw Yeah, it seems to work. :D Set notation is so fun sometimes. lol

Fletchnuts
 2 years ago
Best ResponseYou've already chosen the best response.0so is it A ∪ (C ∩ B')?

Fletchnuts
 2 years ago
Best ResponseYou've already chosen the best response.0thanks so much!!!!!!! you have been so much help!!!!!!!!!!!!!

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1You'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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