Take a factor out of the square root: 48x^2 , where x≤0

Question

Take a factor out of the square root: 48x^2 , where x≤0

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Nevaeh 3 months 2022-02-15T05:57:28+00:00 2 Answers 0 views 0

Answers ( )

    0
    2022-02-15T05:58:43+00:00

    Answer: [tex]4/x/\sqrt{3}[/tex]

    Step-by-step explanation:

    [tex]\sqrt{48x^{2} }[/tex]

    this could also be written as :

    [tex]\sqrt{48}[/tex] x [tex]\sqrt{x^{2}}[/tex]

    [tex]\sqrt{x^{2}}[/tex] = ±[tex]x[/tex] , so we have

    [tex]\sqrt{48}/x/[/tex]

    [tex]\sqrt{48}[/tex] could be written as: [tex]\sqrt{16}[/tex] x [tex]\sqrt{3}[/tex]

    which will result into [tex]4\sqrt{3}[/tex]

    combining all together , we have

    [tex]4/x/\sqrt{3}[/tex]

    Therefore , one of the factor of [tex]\sqrt{48x^{2} }[/tex] is [tex]4/x/\sqrt{3}[/tex]

    0
    2022-02-15T05:59:25+00:00

    Question:

    Which expression is equivalent to [tex]\sqrt{48x^{5} }[/tex] , if x > 0?

    Answer:

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

    Step-by-step explanation:

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45:7+7-4:2-5:5*4+35:2 =? ( )