mathmath333
  • mathmath333
question
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmath333
  • 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
  • vera_ewing
A' is the set of all triangles with all angles 60°.
mathmath333
  • mathmath333
how

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vera_ewing
  • vera_ewing
A is the set of all triangles with at least one different angle from 60°.
vera_ewing
  • vera_ewing
So that means that A' is the set of all equilateral triangles.
mathmath333
  • mathmath333
lol u wrote the same thing
vera_ewing
  • 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
  • mathmath333
i neeed A compliment
mathmath333
  • mathmath333
\(\Huge A'\)
ParthKohli
  • ParthKohli
Jesus. I just typed everything out and it removed it.
ParthKohli
  • ParthKohli
OK, so \(A'\) *is* the set of equilateral triangles.
mathmath333
  • mathmath333
type it on notepad first then copy
Empty
  • Empty
Or you could use a better site such as www.peeranswer.com
ParthKohli
  • 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
  • mathmath333
no i have faith in OP.com
ParthKohli
  • ParthKohli
Read that as "I have faith in OP's mom"
imqwerty
  • imqwerty
OP.com??????
mathmath333
  • mathmath333
OPenstudy.com
imqwerty
  • imqwerty
ohhhhh :O
mathmath333
  • mathmath333
lol
ganeshie8
  • 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
  • mathmath333
hmm this is confujing
mathmath333
  • 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
  • ganeshie8
whats the negation of below statement : at least of the gifts is a car
mathmath333
  • mathmath333
car is not in the gifts
ganeshie8
  • ganeshie8
right, another whats the negation of below statement : at least of the gifts is not a car
mathmath333
  • mathmath333
Their might be car as gift more than once.
ganeshie8
  • ganeshie8
try again
mathmath333
  • mathmath333
at least of the gifts is a car
ganeshie8
  • ganeshie8
``` at least of the gifts is not a car ``` means, there exist one or more gifts that are not cars
ganeshie8
  • ganeshie8
whats the negation of that ?
mathmath333
  • mathmath333
this_>? at least of the gifts is a car
ganeshie8
  • ganeshie8
nope
ganeshie8
  • ganeshie8
negation of "at least one" is "none"
mathmath333
  • mathmath333
i mean negation of which sentence
ganeshie8
  • 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
  • ganeshie8
that means, there are no gifts that are not cars, which is same as saying that all the gifts are cats
mathmath333
  • mathmath333
ok i see my english is poor
ganeshie8
  • ganeshie8
it has nothing to do with english, it is logic, you just need to see it once
mathmath333
  • 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
  • ganeshie8
\(A'\) is everything that is not \(A\) therefore, combining\(A\) with \(A'\) gives you everything : \[A\cup A' = U\]
ganeshie8
  • ganeshie8
may be lets work this with venn diagrams
mathmath333
  • mathmath333
|dw:1438798363617:dw|
ganeshie8
  • ganeshie8
Yes, that looks good. can you show me A and A' in above venn diagram
mathmath333
  • mathmath333
|dw:1438798432076:dw|
ganeshie8
  • ganeshie8
as expected, you can see that combining A and A' covers the entire U
mathmath333
  • mathmath333
ok
ganeshie8
  • ganeshie8
what about \(A\cap A'\) ? is there anything common between A and A' ?
mathmath333
  • mathmath333
nothing common
ganeshie8
  • ganeshie8
since there is nothing common, we say it equals `emptyset`, \(\varphi\)
ganeshie8
  • ganeshie8
\[A\cap A' = \varphi\]
mathmath333
  • mathmath333
ok
mathmath333
  • mathmath333
this one \(\emptyset′ ∩ A\) =
ganeshie8
  • ganeshie8
whats the compliment of emptyset ?
ganeshie8
  • ganeshie8
make a guess
mathmath333
  • mathmath333
\( \emptyset\)
ganeshie8
  • ganeshie8
try again
ganeshie8
  • ganeshie8
by definition, complement is different from the original
ganeshie8
  • ganeshie8
emptyset and universal set are complements of each other
ganeshie8
  • ganeshie8
\(\varphi' = U\) \(U' = \varphi\)
mathmath333
  • mathmath333
ok i see
ganeshie8
  • ganeshie8
so \(\emptyset′ ∩ A=U\cap A = ?\)
mathmath333
  • mathmath333
A
ganeshie8
  • ganeshie8
Yep, try the last one
mathmath333
  • mathmath333
\[\emptyset\]
ganeshie8
  • ganeshie8
Yes \(U′ ∩ A = \varphi \cap A = \varphi\)
mathmath333
  • mathmath333
thnx

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