find the direct relationship between x and y and determine whether the parametric equations determine y as a function of x x=(1/3)t-1 y^2=t^2

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find the direct relationship between x and y and determine whether the parametric equations determine y as a function of x x=(1/3)t-1 y^2=t^2

Mathematics
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get t in terms of x from this one: x=(1/3)t-1
can you explain some more?
you have x=(1/3)t-1, ie an equation giving x in terms of t work it so that you have t in terms of x, ie t = .... use normal algebra

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is this correct so far? \[t=((1/3)-1)/x\]
y=3x+3, yes it's a function
isn't it \[x=\frac{1}{3}t-1\]??? all you have to do is move the t across to the RHS and the x across to the RHS, applying usual rules of maipulation
oh so just t= \[t=\frac{ 1 }{3 }x-1\]
Well, you have to find out if y is a function of x. first of all, if x=1/3t-1, then 3x=t-3, and t=3x+3. Then if y^2=t^2, y=t, so y=3x+3. Which is a function. I don't know where you got t=1/3x-1. That's not right
no, i said "applying usual rules of manipulation". you can't just lift them out(!!) and switch them around an example if \(x = \frac{2}{3} t + 4\) then \(x-4 = \frac{2}{3} t\) \(\frac{2}{3}t = x-4\) \(t = \frac{3}{2} x - 6\)
@whovianchick no, it is not a function but vinny has some work to do before we get there
Except that your math is wrong and you're forgetting y. Recheck your math. You can't just switch t and x, you have to solve for t. It's called 7th grade math. I'm in 11th grade math.
sorry his wording confused me its t=3x-3 correct?
not quite
sign on the 3??
positve
and when you have done that we are told that \(y^2 = t^2\) so stuff your new equation into that
ugh sorry im working on little sleep and silly mistakes are getting me'
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thanks whovian:-)
keep going vinny, we are nearly done!
y=3x=3
You're welcome. You had him working with the wrong equation, and I'm nitpicky like that
ah they are squared gh
yes so just square them, DO NOT multiply/expand out the x side,...!
"You had him working with the wrong equation," when?
if y^2 = t^2, then y=t Replace y for t in the equation: y=3x+3, which is a function
yeah thats what i did
thanks guys
no vinny it is NOT a function
No prob. Sorry @IrishBoy123 earlier he said something like 1/3x-1 = t and I thought it was you
you will see why if we finish this properly but if you are really tired, go to bed
@IrishBoy123 UMMMMMMMMMMMM y=3x+3 IS a function
it is a function, thatis ture, but that is not what we have here
"thatis ture" that is true
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That's the answer, I'm confused what you're saying
@vinny0515 are you done for the day?
this is what we have here |dw:1442362770822:dw|
No it's not, and both are functions
so the answer for the quesrion is y=3x+3
no it is not
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What is the answer, then? We've replaced y for t, and we're done. What the heck are you talking about?
y=3x+3 is the answer
how do i answer it then
The way I told you to. y=3x+3, and yes it's a function
\(y^2=(3x+3)^2\) is what you have if you want to simplify it you can say \(y= \color{red}\pm (3x+3)\) but whatever you do, x maps to 2 y values it is not a function
irsih boy is right
@IrishBoy123 I understand that. But it's saying as one line, is every solution a function. The answer is yes. Both of those equations are functions
it is not a function as you can get 2 y's for every x y = 3x+3 is a function y = -(3x+3) is a function this is both and is not a function |dw:1442363265605:dw|
and that's us done thank you both:p
so i write the answer as y is not a function
Y is a function. I think you're forgetting the concept of a function, @IrishBoy123 , it's ok if there are multiple y's for every x. What you don't want is multiply x's for every y. THAT is the definition of a function, and even with your skewed logic, it's STILL a function. Stop confusing @vinny0515. YOU ARE WRONG
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