Question

Help Help Help!!!!!!

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

******Attachment Below********

Answer 1

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.)

Answer 2

Well A and part of C are shaded.

Answer 3

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?

Answer 4

* 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.

Answer 5

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 '

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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.

Answer 12

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

Answer 13

You're welcome. :)

Answer 14

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

Answer 15

scroll to the bottom of that page.

Answer 16

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:
\((A \cup B^c) \ \cap (A \cup C)\)

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

I'm thinking about it...

Yeah, it seems to work. :D

Set notation is so fun sometimes. lol