Confused on this problem because I am getting two functions?
Which of the following represents a function?
x=[y] (straight, linear line)
x^2+y^2=49 (this is a circle, so no.)
y=|x| (makes a 'v')
x=y^2(parabola, so no.)

Now, if it means one-to-one, which it doesn't specify, it would be x=[y], which is what I picked. But I don't know since y=|x| is in the shape of a 'v'

Question

Confused on this problem because I am getting two functions?
Which of the following represents a function?
x=[y] (straight, linear line)
x^2+y^2=49 (this is a circle, so no.)
y=|x| (makes a 'v')
x=y^2(parabola, so no.)

Now, if it means one-to-one, which it doesn't specify, it would be x=[y], which is what I picked. But I don't know since y=|x| is in the shape of a 'v'

anonymous · Answer

yes, the second is a cirgcle

anonymous · Answer

lol circle

anonymous · Answer

$$x=|y|$$ looks like |dw:1449716799576:dw|this

anonymous · Answer

so no for the circle, no for $x=|y|$ if that is what it is

maddy1251 · Answer

@satellite73 its y=|x| :P

anonymous · Answer

$$x=y^2$$ is a parabola that opens right so not for that one too

maddy1251 · Answer

Which makes it |dw:1449716885691:dw|

anonymous · Answer

$$y=|x|$$ is certainly fa function, because each number has only one absolute value

anonymous · Answer

is the first one $$x=|y|$$

maddy1251 · Answer

No, they are brackets.

anonymous · Answer

not sure what that means 
greatest integer function?

anonymous · Answer

anyways it is probably not a function since one x can correspond to two different y values

maddy1251 · Answer

maybe, it doesn't specify. it just asks what a function. I was stuck between |dw:1449717075369:dw|

anonymous · Answer

no clue

anonymous · Answer

$$y=|x|$$ yes 
the other i do not know what it means
it may mean the greatest integer , in which case the answer is no

maddy1251 · Answer

I don't know which one is correct, honestly. I might go with x=[y]

maddy1251 · Answer

I mean, it is constant and linear. So, maybe that's it.

anonymous · Answer

hold the phone

maddy1251 · Answer

Lol okay :P

anonymous · Answer

$$y=|x|$$ is absolutely a function (pun intended)

anonymous · Answer

so if you have only one choice, certainly pick that one

anonymous · Answer

oh, you are holding a phone!!

maddy1251 · Answer

@satellite73 I will try it, I think. :P It's just weird they both produce a function

anonymous · Answer

i don't know what the other one is, but i doubt it is a function because it is written as x = something

anonymous · Answer

function should be y =  some expression in x

maddy1251 · Answer

well, that is a good point there

anonymous · Answer

you could solve $$x^2+y^2=49$$ for $y$ but you would get (if you did it correctly)
$$y=\pm\sqrt{49-x^2}$$ and the $\pm$ tells you it is not a function

maddy1251 · Answer

@satellite73 it produces a circle, anyways.

anonymous · Answer

true enough

maddy1251 · Answer

@satellite73 thanks again, you have saved me from several grey hairs.