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Mathematics 13 Online
OpenStudy (anonymous):

Use a graphing utility to confirm your result. Then eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve. x=3-2t y=2+3t

OpenStudy (anonymous):

how is it a curve? its just a linear line isn't it. heres the graph

OpenStudy (anonymous):

i did that originally and my teacher said it was wrong. he said to eliminate the parameters. how do i do that?!

OpenStudy (anonymous):

you make the x equation equal t then sub that into the y equation then simply it

OpenStudy (anonymous):

so y = 3x - 5 i think

OpenStudy (anonymous):

sorry 2y = 3x - 5

OpenStudy (anonymous):

so would it be t = (3-x)/2 y = (13-3x)/2

OpenStudy (anonymous):

hmm

OpenStudy (anonymous):

i am getting \(t=\frac{3-x}{2}\) and so \(y=2+3(\frac{3-x}{2})=\frac{13}{2}-\frac{3}{2}x\)

OpenStudy (anonymous):

y = (3/2)x ?

OpenStudy (anonymous):

SO what about these equations? x=sqrt(t) y=1-t

OpenStudy (anonymous):

x^2 = t y = 1 - x^2

OpenStudy (anonymous):

and yeah @satellite73 is correct with the first one. I forgot the minus in the x equation

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