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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\(\large \color{black}{\begin{align}& \normalsize \text{Let U be the set of all triangles in a plane.}\hspace{.33em}\\~\\ & \normalsize \text{If A is the set of all triangles with at}\hspace{.33em}\\~\\ & \normalsize \text{least one angle different from 60°, what is A′?}\hspace{.33em}\\~\\ \end{align}}\)
A' is the set of all triangles with all angles 60°.
how

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A is the set of all triangles with at least one different angle from 60°.
So that means that A' is the set of all equilateral triangles.
lol u wrote the same thing
http://www.meritnation.com/ask-answer/question/let-u-be-the-set-of-all-triangles-in-a-plane-if-a-is-the/sets/2453760
i neeed A compliment
\(\Huge A'\)
Jesus. I just typed everything out and it removed it.
OK, so \(A'\) *is* the set of equilateral triangles.
type it on notepad first then copy
Or you could use a better site such as www.peeranswer.com
\(A \cup A' = T\) where \(T\) is the set of all triangles. Thus, \(A'\) is the set of all triangles that do not satisfy the criterion for \(A\) (that at least one angle should be different from 60). The only way to violate that? Have none of them different from 60, which pretty much means that all of them are 60.
no i have faith in OP.com
Read that as "I have faith in OP's mom"
OP.com??????
OPenstudy.com
ohhhhh :O
lol
negation of "at least one" is "none" so the negation of "at least one angle different from 60" is "none of the angles different from 60", which is same as "all angles are 60"
hmm this is confujing
Fill in the blanks to make each of the following a true statement : (i) A ∪ A′ = . . . (ii) φ′ ∩ A = . . . (iii) A ∩ A′ = . . . (iv) U′ ∩ A = . . .
whats the negation of below statement : at least of the gifts is a car
car is not in the gifts
right, another whats the negation of below statement : at least of the gifts is not a car
Their might be car as gift more than once.
try again
at least of the gifts is a car
``` at least of the gifts is not a car ``` means, there exist one or more gifts that are not cars
whats the negation of that ?
this_>? at least of the gifts is a car
nope
negation of "at least one" is "none"
i mean negation of which sentence
``` at least of the gifts is not a car ``` negation of above statement would be : ``` none of the gifts is not a car ```
that means, there are no gifts that are not cars, which is same as saying that all the gifts are cats
ok i see my english is poor
it has nothing to do with english, it is logic, you just need to see it once
oh, hmm ok i have one morre question Fill in the blanks to make each of the following a true statement : (i) A ∪ A′ = . . . (ii) φ′ ∩ A = . . . (iii) A ∩ A′ = . . . (iv) U′ ∩ A = . . .
\(A'\) is everything that is not \(A\) therefore, combining\(A\) with \(A'\) gives you everything : \[A\cup A' = U\]
may be lets work this with venn diagrams
|dw:1438798363617:dw|
Yes, that looks good. can you show me A and A' in above venn diagram
|dw:1438798432076:dw|
as expected, you can see that combining A and A' covers the entire U
ok
what about \(A\cap A'\) ? is there anything common between A and A' ?
nothing common
since there is nothing common, we say it equals `emptyset`, \(\varphi\)
\[A\cap A' = \varphi\]
ok
this one \(\emptyset′ ∩ A\) =
whats the compliment of emptyset ?
make a guess
\( \emptyset\)
try again
by definition, complement is different from the original
emptyset and universal set are complements of each other
\(\varphi' = U\) \(U' = \varphi\)
ok i see
so \(\emptyset′ ∩ A=U\cap A = ?\)
A
Yep, try the last one
\[\emptyset\]
Yes \(U′ ∩ A = \varphi \cap A = \varphi\)
thnx

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