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OpenStudy (mathmath333):

question

OpenStudy (mathmath333):

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

OpenStudy (vera_ewing):

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

OpenStudy (mathmath333):

how

OpenStudy (vera_ewing):

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

OpenStudy (vera_ewing):

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

OpenStudy (mathmath333):

lol u wrote the same thing

OpenStudy (mathmath333):

i neeed A compliment

OpenStudy (mathmath333):

\(\Huge A'\)

Parth (parthkohli):

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

Parth (parthkohli):

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

OpenStudy (mathmath333):

type it on notepad first then copy

OpenStudy (empty):

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

Parth (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.

OpenStudy (mathmath333):

no i have faith in OP.com

Parth (parthkohli):

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

imqwerty (imqwerty):

OP.com??????

OpenStudy (mathmath333):

OPenstudy.com

imqwerty (imqwerty):

ohhhhh :O

OpenStudy (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"

OpenStudy (mathmath333):

hmm this is confujing

OpenStudy (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

OpenStudy (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

OpenStudy (mathmath333):

Their might be car as gift more than once.

ganeshie8 (ganeshie8):

try again

OpenStudy (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 ?

OpenStudy (mathmath333):

this_>? at least of the gifts is a car

ganeshie8 (ganeshie8):

nope

ganeshie8 (ganeshie8):

negation of "at least one" is "none"

OpenStudy (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

OpenStudy (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

OpenStudy (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

OpenStudy (mathmath333):

|dw:1438798363617:dw|

ganeshie8 (ganeshie8):

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

OpenStudy (mathmath333):

|dw:1438798432076:dw|

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