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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  • 8 months ago
    Favorite Answer

    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

    • Anonymous8 months agoReport

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

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  • ted s
    Lv 7
    8 months ago

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

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  • 8 months ago

     

    √(7x^7y) √(21xy)

    =  √(7*21x^8y^2)

    =  √(7^21) x^4y

    =  √147 x^4y

    = 12.1243557 x^4y

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  • ?
    Lv 7
    8 months ago

    √(7x⁷y) √(21xy) =

    √(7⁴ x⁸ y²) =

    7² x⁴ y =

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

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  • 8 months 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

  • Fuhr
    Lv 6
    8 months ago

    Seven                    .

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