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Rosy95

PreCal Help Please!

  • one year ago
  • one year ago

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

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

    • one year ago
  3. Rosy95
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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.

    • one year ago
  4. abb0t
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    Best of luck.

    • one year ago
  5. Rosy95
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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.

    • one year ago
  6. Outkast3r09
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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 \[3(\frac{-1}{4})-y=1\] \[\frac{-3}{4}-1=y\] \[\frac{-7}{4}=y \] \[\frac{-1}{4}+y=-2\] \[\frac{7}{4}=-y\] \[\frac{-7}{4}=y\]

    • one year ago
  7. Outkast3r09
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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 \[y-y_1=m(x-x_1)\] \[y-(-1.75)=1(x-(-.25))\] \[y+1.75=1x+.25\] \[y=1x-1.5=x-1.5\]

    • one year ago
  8. Outkast3r09
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    |dw:1358765159547:dw|

    • one year ago
  9. Rosy95
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    Thanks and my answer was right :)

    • one year ago
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