Maddy1251
  • Maddy1251
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'
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
yes, the second is a cirgcle
anonymous
  • anonymous
lol circle
anonymous
  • anonymous
\[x=|y|\] looks like |dw:1449716799576:dw|this

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anonymous
  • anonymous
so no for the circle, no for \(x=|y|\) if that is what it is
Maddy1251
  • Maddy1251
@satellite73 its y=|x| :P
anonymous
  • anonymous
\[x=y^2\] is a parabola that opens right so not for that one too
Maddy1251
  • Maddy1251
Which makes it |dw:1449716885691:dw|
anonymous
  • anonymous
\[y=|x|\] is certainly fa function, because each number has only one absolute value
anonymous
  • anonymous
is the first one \[x=|y|\]
Maddy1251
  • Maddy1251
No, they are brackets.
anonymous
  • anonymous
not sure what that means greatest integer function?
anonymous
  • anonymous
anyways it is probably not a function since one x can correspond to two different y values
Maddy1251
  • Maddy1251
maybe, it doesn't specify. it just asks what a function. I was stuck between |dw:1449717075369:dw|
anonymous
  • anonymous
no clue
anonymous
  • anonymous
\[y=|x|\] yes the other i do not know what it means it may mean the greatest integer , in which case the answer is no
Maddy1251
  • Maddy1251
I don't know which one is correct, honestly. I might go with x=[y]
Maddy1251
  • Maddy1251
I mean, it is constant and linear. So, maybe that's it.
anonymous
  • anonymous
hold the phone
Maddy1251
  • Maddy1251
Lol okay :P
anonymous
  • anonymous
\[y=|x|\] is absolutely a function (pun intended)
anonymous
  • anonymous
so if you have only one choice, certainly pick that one
anonymous
  • anonymous
oh, you are holding a phone!!
Maddy1251
  • Maddy1251
@satellite73 I will try it, I think. :P It's just weird they both produce a function
anonymous
  • anonymous
i don't know what the other one is, but i doubt it is a function because it is written as x = something
anonymous
  • anonymous
function should be y = some expression in x
Maddy1251
  • Maddy1251
well, that is a good point there
anonymous
  • anonymous
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
  • Maddy1251
@satellite73 it produces a circle, anyways.
anonymous
  • anonymous
true enough
Maddy1251
  • Maddy1251
@satellite73 thanks again, you have saved me from several grey hairs.

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