Simplify this maths problem?

I tried it and got 81/y^3 (0.04y^8)

What have I done wrong?

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9 Answers

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  • Como
    Lv 7
    4 months ago

    --

    ( 18/y^4 ) x (y^8/25)

    ( 18/25 ) y^4

  • 4 months ago

    y8/y⁴ (81/25 ) = y ^(8-4) * 81/25

    =81/25 y⁴

  • JOHN
    Lv 7
    4 months ago

    Answer

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  • sepia
    Lv 7
    4 months ago

    (-3/y)^4 (1/5 y^4)^2

    = 81y^8 / (25 y^4)

    = 81y^4 / 25

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  • 4 months ago

    81/y^4 (0.04y^8) is ok as far as it goes, but it could go further

    81/y^4 * y^8 / 25

    81/25 y⁴

    • Lôn
      Lv 7
      4 months agoReport

      You've misread it...it's y^4 /5 NOT 1/(5y^4)

  • 4 months ago

    [(-3/y)^4][(y^4)/5]^2

    =

    [81/y^4][(y^8)/25]

    =

    (81y^4)/25

  • 4 months ago

    (-3/y)^4(1/5 y^4)^2

    = (81/y^4)(y^8/25)

    = 81y^4/25

  • 4 months ago

    First, keep fractions as fractions.

    Then your y's can be combined from the first factor and the second factor.

    You started correctly, but needed to finish things up.

  • Lôn
    Lv 7
    4 months ago

    Do each separately first....

    (81/y^4) x (y^8/25) ......now multiply

    81y^8 / 25y^4

    (81y^4)/25

    • Lôn
      Lv 7
      4 months agoReport

      Minus 3 to the power of 4 is 81
      y to the power of 4 is y^4

      y^4 squared is y^8

      5 squared is 25.....,

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