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superhelp101

  • one year ago

X=(3t^2)/2 Y=4t-1 Using these two parametric equations convert it into rectangular form. Then u have to determine what type of equation the rectangular form describes I recall don't get this

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  1. superhelp101
    • one year ago
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    @satellite73 Do u think u can help?

  2. anonymous
    • one year ago
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    yes

  3. anonymous
    • one year ago
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    or better yes, solve \[y=4t-1\] for \(t\)

  4. anonymous
    • one year ago
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    you should get \(t=\frac{y-1}{4}\)

  5. superhelp101
    • one year ago
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    Yes

  6. anonymous
    • one year ago
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    then replace \(t\) in \[x=\frac{3t^2}{2}\] by \(\frac{y-1}{2}\)

  7. anonymous
    • one year ago
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    you will have \(x\) written in terms of \(y\)

  8. anonymous
    • one year ago
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    i get \[x=\frac{3(y-1)^2}{4}\] check my algebra

  9. superhelp101
    • one year ago
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    Yeah I got that too

  10. anonymous
    • one year ago
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    well it is wrong 1

  11. anonymous
    • one year ago
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    it should be \[x=\frac{3(y-1)^2}{8}\] i forgot to square the 2

  12. superhelp101
    • one year ago
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    Oops I think I did y+2 instead of y-1

  13. anonymous
    • one year ago
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    oh crap the whole thing is wrong must be getting late

  14. anonymous
    • one year ago
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    \[t=\frac{y+1}{4}\]right?

  15. anonymous
    • one year ago
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    so \[x=3\frac{\left(\frac{y+1}{4}\right)^2}{2}\]

  16. anonymous
    • one year ago
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    or \[x=\frac{3(y+1)^2}{32}\]

  17. superhelp101
    • one year ago
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    Yesss :))

  18. anonymous
    • one year ago
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    a parabola

  19. superhelp101
    • one year ago
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    Oh so that is rectangular form?

  20. anonymous
    • one year ago
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    yes

  21. anonymous
    • one year ago
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    with no parameter of \(t\) is all that means

  22. superhelp101
    • one year ago
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    Okay

  23. superhelp101
    • one year ago
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    Thxx :D

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