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- mathmath333

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- mathmath333

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

- vera_ewing

A' is the set of all triangles with all angles 60°.

- mathmath333

how

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## More answers

- vera_ewing

A is the set of all triangles with at least one different angle from 60°.

- vera_ewing

So that means that A' is the set of all equilateral triangles.

- mathmath333

lol u wrote the same thing

- vera_ewing

http://www.meritnation.com/ask-answer/question/let-u-be-the-set-of-all-triangles-in-a-plane-if-a-is-the/sets/2453760

- mathmath333

i neeed A compliment

- mathmath333

\(\Huge A'\)

- ParthKohli

Jesus. I just typed everything out and it removed it.

- ParthKohli

OK, so \(A'\) *is* the set of equilateral triangles.

- mathmath333

type it on notepad first then copy

- Empty

Or you could use a better site such as www.peeranswer.com

- ParthKohli

\(A \cup A' = T\) 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.

- mathmath333

no i have faith in OP.com

- ParthKohli

Read that as "I have faith in OP's mom"

- imqwerty

OP.com??????

- mathmath333

OPenstudy.com

- imqwerty

ohhhhh :O

- mathmath333

lol

- ganeshie8

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"

- mathmath333

hmm this is confujing

- mathmath333

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 = . . .

- ganeshie8

whats the negation of below statement :
at least of the gifts is a car

- mathmath333

car is not in the gifts

- ganeshie8

right, another
whats the negation of below statement :
at least of the gifts is not a car

- mathmath333

Their might be car as gift more than once.

- ganeshie8

try again

- mathmath333

at least of the gifts is a car

- ganeshie8

```
at least of the gifts is not a car
```
means, there exist one or more gifts that are not cars

- ganeshie8

whats the negation of that ?

- mathmath333

this_>? at least of the gifts is a car

- ganeshie8

nope

- ganeshie8

negation of "at least one" is "none"

- mathmath333

i mean negation of which sentence

- ganeshie8

```
at least of the gifts is not a car
```
negation of above statement would be :
```
none of the gifts is not a car
```

- ganeshie8

that means, there are no gifts that are not cars,
which is same as saying that all the gifts are cats

- mathmath333

ok i see my english is poor

- ganeshie8

it has nothing to do with english, it is logic, you just need to see it once

- mathmath333

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 = . . .

- ganeshie8

\(A'\) is everything that is not \(A\)
therefore, combining\(A\) with \(A'\) gives you everything :
\[A\cup A' = U\]

- ganeshie8

may be lets work this with venn diagrams

- mathmath333

|dw:1438798363617:dw|

- ganeshie8

Yes, that looks good. can you show me A and A' in above venn diagram

- mathmath333

|dw:1438798432076:dw|

- ganeshie8

as expected, you can see that combining A and A' covers the entire U

- mathmath333

ok

- ganeshie8

what about \(A\cap A'\) ?
is there anything common between A and A' ?

- mathmath333

nothing common

- ganeshie8

since there is nothing common, we say it equals `emptyset`, \(\varphi\)

- ganeshie8

\[A\cap A' = \varphi\]

- mathmath333

ok

- mathmath333

this one \(\emptyset′ ∩ A\) =

- ganeshie8

whats the compliment of emptyset ?

- ganeshie8

make a guess

- mathmath333

\( \emptyset\)

- ganeshie8

try again

- ganeshie8

by definition, complement is different from the original

- ganeshie8

emptyset and universal set are complements of each other

- ganeshie8

\(\varphi' = U\)
\(U' = \varphi\)

- mathmath333

ok i see

- ganeshie8

so \(\emptyset′ ∩ A=U\cap A = ?\)

- mathmath333

A

- ganeshie8

Yep, try the last one

- mathmath333

\[\emptyset\]

- ganeshie8

Yes \(U′ ∩ A = \varphi \cap A = \varphi\)

- mathmath333

thnx

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