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Rosy95

  • 2 years ago

PreCal Help Please!

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  1. Rosy95
    • 2 years ago
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    Would it be.... \[(3\sqrt{2}-\sqrt{10})x-(\sqrt{10}+\sqrt{2})y-2\sqrt{10}-\sqrt{2}=0\]

  2. Rosy95
    • 2 years ago
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    @Hero do you know if I am right? or check to make sure? if not thanks anyways :)

  3. Rosy95
    • 2 years ago
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    abb0t...thanks for trying to help but that was taken off of yahoo answers and if you continue reading what the person who wrote this said, you will see that none of the answer choices provided were correct. So again thanks, but it doesn't help too much. I just want to know If I go the right answer.

  4. abb0t
    • 2 years ago
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    Best of luck.

  5. Rosy95
    • 2 years ago
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    Thanks and again thanks for trying but I had already seen that post and it didn't help. That is why I posted here.

  6. Outkast3r09
    • 2 years ago
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    Alright this is new to me but this is what It says to do \[3x-y=1\] \[x+y=-2\] \[(3+1)x+(1-1)y=-2+1\] \[4x=-1\] \[x=\frac{-1}{4}\] to make sure this is correct test for each for y \[3(\frac{-1}{4})-y=1\] \[\frac{-3}{4}-1=y\] \[\frac{-7}{4}=y \] \[\frac{-1}{4}+y=-2\] \[\frac{7}{4}=-y\] \[\frac{-7}{4}=y\]

  7. Outkast3r09
    • 2 years ago
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    intersection point is \[[-.25,-1.75]\] which is a point on your bisector line also next find the slopes and take the arimetic mean line 1 \[3x-1=y\] \[m=3\] \[x+y=-2\] \[-x-2=y\] \[-1=m\] \[\frac{-1+3}{2}=1\] using that withhin the point slope equation \[y-y_1=m(x-x_1)\] \[y-(-1.75)=1(x-(-.25))\] \[y+1.75=1x+.25\] \[y=1x-1.5=x-1.5\]

  8. Outkast3r09
    • 2 years ago
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    |dw:1358765159547:dw|

  9. Rosy95
    • 2 years ago
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    Thanks and my answer was right :)

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