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anonymous

  • 5 years ago

how would you solve this problem in simplest form: x with power of 3 divided by 2y power of 2 times 6y power of 4 divided by xy?

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  1. anonymous
    • 5 years ago
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    \[x ^{3}\div(2y)^{2} * (6y)^{4}\div xy\]

  2. anonymous
    • 5 years ago
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    But written as two fractions... is that what you mean?

  3. anonymous
    • 5 years ago
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    or \[2y ^{2}\]

  4. anonymous
    • 5 years ago
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    and \[6y ^{4}\]

  5. anonymous
    • 5 years ago
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    in the 2y problem the y is to the power of 2 and in 6y the y is to the power of 4

  6. anonymous
    • 5 years ago
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    OK.. So here is a long way of thinking about this first.

  7. anonymous
    • 5 years ago
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    x^3 = x * x * x 2y^2 = 2 * y * y 6y^4 = 2 * 3 * y * y * y * y xy = x * y Do you agree with this?

  8. anonymous
    • 5 years ago
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    yes

  9. anonymous
    • 5 years ago
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    So we have x * x * x 2 * 3 * y * y * y * y -------- * ---------------- 2 * y * y x * y OK?

  10. anonymous
    • 5 years ago
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    ok

  11. anonymous
    • 5 years ago
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    When everything is being multiplied... You can cancel out anything on the top with anything on the bottom...

  12. anonymous
    • 5 years ago
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    So I would put a slash thru the 2 on the top and then put a slash thru the 2 on the bottom. What else do you see?

  13. amistre64
    • 5 years ago
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    cancel out any "like" terms top to bottom ;)

  14. anonymous
    • 5 years ago
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    oops you are so right amistre64.. I was thinking it but forgot to type it.

  15. anonymous
    • 5 years ago
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    are you talking about just the 2's I'm confused.

  16. anonymous
    • 5 years ago
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    No... what else is alike... I see an x on the bottom and an x on top... put slashes thru those. I see three y's on the bottom that can cancel with three y's on top. Do you see that?

  17. anonymous
    • 5 years ago
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    that should leave you with 3x^2y x^2 is x to the 2nd power.

  18. anonymous
    • 5 years ago
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    yes I just did that now do I cross multiply then divide to get the answer?

  19. anonymous
    • 5 years ago
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    Well. you don't cross multiply... You basically cancel like terms top and bottom and then whatever is left over on top stays on top and whatever is left over on the bottom stays on the bottom (what I mean by left over is that it doesn't get cancelled out) There is a simpler way... but I don't know if you are ready for it yet. Do you have another problem like this?

  20. anonymous
    • 5 years ago
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    I have nine more problems to do on a worksheet.

  21. anonymous
    • 5 years ago
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    OK

  22. anonymous
    • 5 years ago
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    I am going to try to do them by myself since I got you to help me I may have to ask you again later if that is okay?

  23. anonymous
    • 5 years ago
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    Ooops.. Any algebraic Equation, which requires a solution should (must) have two sides, I mean exactly one "equals to (=)". Sorry. But this question is not complete.

  24. anonymous
    • 5 years ago
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    Yes... I will stay on this post.

  25. amistre64
    • 5 years ago
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    simplifying does not require an (=) sign. the results of simplifying do equal, but to simplify or reduce does not necessitate the need for an (=) sign :)

  26. anonymous
    • 5 years ago
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    Okay. then it is 324*x^2*y. As I simplified (starting from the very first responce by blexting: [{x^3}/{4y^2}] * [{1296y^4}/xy] 324{x^2}y

  27. amistre64
    • 5 years ago
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    that would be a good result :) now all we need to know is if we have the right "problem" ;)

  28. anonymous
    • 5 years ago
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    wpgaurav.. if you notice, Angelia said that the (2y)^4 was actually 2y^4 and (6y)^4 was actually 6y^4.... I wrote what I thought she was saying, but it wasn't.

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