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Math HELP! Please reply quick

  1. Feb 6, 2006 #1
    I have the equation [(m^3)(n^-3)]^-1 over m^-5n
    So far i got 1 over m^6n^6 divide by m^-5n...
    Is it right or no?
     
  2. jcsd
  3. Feb 6, 2006 #2

    Tom Mattson

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    I've moved this from the Tutorials Forum into the Homework Help section.

    That's not an equation, it's an expression.

    No, it isn't. What steps did you take to manipulate the numerator of your original expression?
     
  4. Feb 6, 2006 #3
    First off i went to the basics I multiplied (m^3)(n^-2) and got m^6n^6
    Then i inversed it to get 1 over m^6n^6, so i can divide that with m^-5. Yet, i still don't know arithmetic.
     
  5. Feb 6, 2006 #4

    Tom Mattson

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    That's not right. You should be using the following rule:

    [tex](x^ay^b)^c=x^{ac}y^{bc},[/tex]

    which should be in your textbook.
     
  6. Feb 6, 2006 #5

    chroot

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    What are you trying to do? Simplify the expression?

    Note that [itex]m^3 \cdot n^{-3}[/itex] does not equal [itex]m^6 n^6[/itex].

    Where did the extra powers of m come from? Where did the negative powers of n go?

    - Warren
     
  7. Feb 6, 2006 #6
    So the exponent doesn't affect the base number
     
  8. Feb 6, 2006 #7

    chroot

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    No. There is no "base number" here anyway -- just the variables m and n.

    - Warren
     
  9. Feb 6, 2006 #8
    Im so confused
     
  10. Feb 6, 2006 #9
    Okay wait what happens to the exponents when its (m^3)(n^2)
     
  11. Feb 6, 2006 #10

    chroot

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    Did you see what Hurkyl posted? The general rule of "exponent distribution:"

    [tex](x^ay^b)^c=x^{ac}y^{bc}[/tex]

    You have an example of this here:

    [tex](m^3 n^{-3})^{-1} = m^{3 \cdot -1}n^{-3 \cdot -1} = m^{-3} n^{3}[/tex]

    Again, I'll ask you: what are you trying to do? Simplify the expression?

    - Warren
     
  12. Feb 6, 2006 #11

    chroot

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    Nothing at all "happens" to the exponents -- the bases are different, and thus they are completely unrelated to each other.

    - Warren
     
  13. Feb 6, 2006 #12
    Would it be mn^-6?
     
  14. Feb 6, 2006 #13
    Sorry i just read simplify. yes i need to simplify
     
  15. Feb 6, 2006 #14

    chroot

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    No. Now you're getting me a bit confused. First you told us the problem was this:

    (m^3)(n^-3)]^-1 over m^-5n

    and now you're talking about (m^3)(n^2). Which is it?

    Please repeat the problem, taking care to type it exactly as shown in your homework. Also, please tell us exactly what you're trying to do: are you just supposed to simplify the expression?

    - Warren
     
  16. Feb 6, 2006 #15

    chroot

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    No. Now you're getting me a bit confused. First you told us the problem was this:

    (m^3)(n^-3)]^-1 over m^-5n

    and now you're talking about (m^3)(n^2). Which is it?

    Please repeat the problem, taking care to type it exactly as shown in your homework. Also, please tell us exactly what you're trying to do: are you just supposed to simplify the expression?

    - Warren
     
  17. Feb 6, 2006 #16
    Nevermind i got it I'm simplifying it It's n^2m^5 over m^3n

    The question was simplify [(m^3)(n^-3)]^-1 over m^-5n
     
  18. Feb 6, 2006 #17
    I got another question I have to factor 3x^2 -13x-10
    So far i get this far
    3x^2 -13x-10
    3x^2+2x-15x-10
    x(3x+2)-5(3x+2)
    What do i do next...
     
  19. Feb 6, 2006 #18

    chroot

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    I'm afraid that

    [itex]\frac{(m^3 n^{-3})^{-1}}{m^{-5n}}[/itex]

    does not simplify to

    [itex]\frac{n^2 m^5}{m^{3n}}[/itex]

    - Warren
     
  20. Feb 6, 2006 #19
    Well i don't know any other way
     
  21. Feb 6, 2006 #20

    chroot

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    Let's do this in steps.

    First, simplify the numerator of

    [itex]\frac{(m^3 n^{-3})^{-1}}{m^{-5n}}[/itex]

    The -1 exponent "distributes", multiplying the exponents on m and n:

    [itex](m^3 n^{-3})^{-1} = m^{-3}n^3}[/itex]

    So the entire expression becomes:

    [itex]\frac{m^{-3}n^3}{m^{-5n}}[/itex]

    Now, factors with negative exponents can be "flipped" across the division line, as can be seen by multiplying both sides by the same factor with the exponent positive rather than negative:

    [itex]\frac{m^{-3}n^3}{m^{-5n}} = m^{-3}n^3 m^{5n}[/itex]

    One negative-exponent factor went up, while the other negative-exponent factor I left alone.

    Finally, multiplyiing two factors (with the same base) is the same as adding their exponents:

    [itex]n^3 m^{5n - 3}[/itex]

    If you do not understand any step of this, please let me know.

    - Warren
     
    Last edited: Feb 6, 2006
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