Math question... ?

 √7x^7y√21xy

I'm having trouble with this one. By '^7' I mean to the 7th power. 

6 Answers

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  • ted s
    Lv 7
    3 weeks ago

    7 x^4 | y | √3.....the absolute value symbol needed since both x & y could be negative

  • 3 weeks ago

     

    √(7x^7y) √(21xy)

    =  √(7*21x^8y^2)

    =  √(7^21) x^4y

    =  √147 x^4y

    = 12.1243557 x^4y

  • ?
    Lv 7
    3 weeks ago

    √(7x⁷y) √(21xy) =

    √(7⁴ x⁸ y²) =

    7² x⁴ y =

    49 x⁴ y .......................ANS

  • 3 weeks ago

    too vague, I can interpret this dozens of ways. 

    but following the laws of precedence

    (√7)(x^7)y(√21)xy

    (√(7•7•3)x⁷yxy

    x⁸y²(7√3)

    that is far as it can be simplified. 

    edit: your comment: √7x^7y is all under its radical and √21xy is all under its radical.

    so we have

    √(7x⁷y)√(21xy)

    √((7x⁷y)(21xy))

    √(7*7*3x⁸y²)

    7x⁴y√3

    • billrussell42
      Lv 7
      3 weeks agoReport

      why is it difficult to type parans ? they are on every keyboard.

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  • 3 weeks ago

    How much of that is under the radical?

    As written, you  have:

    √7 x⁷y √21 xy

    Multiply the radicals and multiply the variables together to get:

    x⁸y²√147

    147 has 49 as a factor, so:

    7x⁸y²√3

    --------------

    If that is supposed to be:

    √(7x⁷y) √(21xy)

    Then multiply both radicals together and simplify what's left:

    √(147x⁸y²)

    Pulling out perfect square factors, the 49 from before is still in play but now we have all of the variables as being perfect square factors that we can pull out, leaving us with:

    7x⁴y √3

    • Anonymous3 weeks agoReport

      Thank you!!! The second was what I meant. It's was hard trying to type in on here, sorry. But thanks again!

  • Fuhr
    Lv 6
    3 weeks ago

    Seven                    .

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