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anonymous

  • 2 years ago

I bet no one can solve this equation.

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  1. anonymous
    • 2 years ago
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    5x^-2y^10 over 2x^-1(-3x^-3y^-1)^-2

  2. myininaya
    • 2 years ago
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    Because there is no equation to solve. :p

  3. myininaya
    • 2 years ago
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    Do you mean to simplify the expression.

  4. anonymous
    • 2 years ago
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    Yes, straight from an SAT study book

  5. DawnR
    • 2 years ago
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    this is http://media.tumblr.com/tumblr_memy8mw86I1qc9tqw.gif

  6. myininaya
    • 2 years ago
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    I'm going to phrase this using latex just to make sure I understand what you have written: \[\frac{5x^{-2}y^{10}}{2x^{-1}(-3x^{-3}y^{-1})^{-2}}\]

  7. anonymous
    • 2 years ago
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    @DawnR XD hahaha

  8. myininaya
    • 2 years ago
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    It is very important to recall the law of exponents here. I will write some useful ones here regarding this problem: \[x^{-n}=\frac{1}{x^n}\] \[\frac{1}{x^{-n}}=x^{n}\] \[x^{1}=x\] \[x^0=1 \text{ note: } x \neq 0\] \[(a^rb^tc^s)n=a^{rn}b^{tn}c^{sn}\]

  9. myininaya
    • 2 years ago
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    also \[x^nx^m=x^{m+n} \text{ and } \frac{x^n}{x^m}=x^{n-m}\]

  10. myininaya
    • 2 years ago
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    These are good lows to recall for this problem.

  11. myininaya
    • 2 years ago
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    also that one law is a little broken let me rewrite it

  12. myininaya
    • 2 years ago
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    \[(a^rb^tc^s)^n=a^{rn}b^{tn}c^{sn}\]

  13. myininaya
    • 2 years ago
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    can you use this last law here on the problem anywhere that you can see?

  14. anonymous
    • 2 years ago
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    I'm not asking this question because I need help, I am asking to see if someone for actually get it

  15. anonymous
    • 2 years ago
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    Can*

  16. myininaya
    • 2 years ago
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    |dw:1403479310371:dw| I will give you a hint: look in the circle below:

  17. myininaya
    • 2 years ago
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    |dw:1403479439985:dw|

  18. myininaya
    • 2 years ago
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    that looks exactly like that law on the left hand side

  19. myininaya
    • 2 years ago
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    now use the right hand side to rewrite this thing in the circle

  20. myininaya
    • 2 years ago
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    I do get it And I can help you

  21. anonymous
    • 2 years ago
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    I don't need help, I wanted to see if anyone can actually get it. Its not HW or anything, just a fun equation

  22. myininaya
    • 2 years ago
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    Ok, well I do get it...

  23. anonymous
    • 2 years ago
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    I never said you didn't...

  24. myininaya
    • 2 years ago
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    I guess I don't understand what you mean then.

  25. anonymous
    • 2 years ago
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    Ok solve the equation then

  26. myininaya
    • 2 years ago
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    I will help you to do just that.

  27. myininaya
    • 2 years ago
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    If you are posting this as a challenge, I think I will let someone else do it.

  28. myininaya
    • 2 years ago
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    This problem is not challenging to me. I'm sorry.

  29. mathmale
    • 2 years ago
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    Drummer: If you want feedback on the solution of this problem, please share with the rest of us the work that you yourself have done. if you're not willing to provide such feedback, then, frankly, it sounds as that you may be trying to get someone else to do your work for you...and that'd be true regardless of whether you're doing this problem for the challenge or for an assignment.

  30. anonymous
    • 2 years ago
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    Do you want the website I got this from?? I just wanted to see if people could actual do it

  31. mathmale
    • 2 years ago
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    Again, please try doing it yourself first. If you want feedback on your work after you've done that, fine. To paraphrase you, I just want to see whether YOU can actually solve this problem at hand, before I invest time and effort in this problem solving myself. Isn't that fair?

  32. anonymous
    • 2 years ago
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    Again, this is just a friendly question. It does not come from a quiz, worksheet, or homework of any kind. I simply goggled "Very hard math equations" and copy and pasted one from a website to give a fun challenge to people.

  33. mathmale
    • 2 years ago
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    Fine, Drummin. Take the challenge and do the problem yourself, as far as you can. Then I'd be happy to give you feedback on how I think we might finish solving the problem. Remember that everyone here on OpenStudy is here voluntarily, and thus is free to help or not help, get involved or not get involved, as he or she wishes. I've made my wants clear: show what you have done and list any questions you may have, before I get involved. Glad you're curious and motivated and eager to solve challenging questions. Still, I feel more obligated to help those here on OpenStudy who try their best to get started out in their problem solving and share what they have done with me up front.

  34. anonymous
    • 2 years ago
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    I have no clue how to do this, I don't even know where to start

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