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I known the answer, i just need help with showing the work, i need to show how i found the scale favor and what it was before

Posted on October 23, 2021 By Jay1041 5 Comments on I known the answer, i just need help with showing the work, i need to show how i found the scale favor and what it was before

i known the answer, i just need help with showing the work, i need to show how i found the scale favor and what it was before it was simplified. plz help


[tex]i known the answer, i just need help with showing the work, i need to show how i found the scale fa[/tex]

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Comments (5) on “I known the answer, i just need help with showing the work, i need to show how i found the scale favor and what it was before”

  1. melkumathurin says:
    October 23, 2021 at 3:26 pm

    Step-by-step explanation:

    x + 3y + 2z = 7

    2x − y + z = 9

    -x + 4y + 5z = 14

    First add the first and third equations to eliminate x.

    7y + 7z = 21

    Simplify:

    y + z = 3

    Solve for y:

    y = 3 − z

    Substitute into the first two equations:

    x + 3(3 − z) + 2z = 7

    2x − (3 − z) + z = 9

    Simplify:

    x + 9 − 3z + 2z = 7 → x − z = -2

    2x − 3 + z + z = 9 → 2x + 2z = 12 → x + z = 6

    Add the equations to eliminate z.

    2x = 4

    x = 2

    Therefore, z = 4 and y = -1.

    Reply
  2. reagybear says:
    October 23, 2021 at 5:04 pm

    Yes. By SAS (side angle side)

    Step-by-step explanation:

    Reply
  3. poptropic7623 says:
    October 23, 2021 at 7:02 pm

    Just use photomath that’s easier

    Reply
  4. kendall984 says:
    October 23, 2021 at 7:18 pm

    Dvcxxxdccvfggeveff dc gf

    Reply
  5. jessicachichelnitsky says:
    October 23, 2021 at 11:06 pm

    see explanation

    Step-by-step explanation:

    Given

    [tex]\sqrt{\frac{4x^2}{3y} }[/tex]

    = [tex]\frac{\sqrt{4x^2} }{\sqrt{3y} }[/tex]

    = [tex]\frac{2x}{\sqrt{3y} }[/tex]

    Rationalise the denominator by multiplying the numerator/ denominator by [tex]\sqrt{3y}[/tex]

    Note that [tex]\sqrt{a}[/tex] × [tex]\sqrt{a}[/tex] = a

    = [tex]\frac{2x}{\sqrt{3y} }[/tex] × [tex]\frac{\sqrt{3y} }{\sqrt{3y} }[/tex]

    = [tex]\frac{2x\sqrt{3y} }{3y}[/tex]

    Reply

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