A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Did i do this right? Simplify each expression. Use Positive exponents m3 n^-6 p^0. 1n^-6/1 = 1/n^-6 now we put the positive n^6 back in it's original place. m^3 n^6 p

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    uhhhh i think so i belive so

  2. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The rule for negative exponents is: \(\Large\color{black}{ \displaystyle \color{blue}{\rm a}^{-\color{red}{\rm b}}=\frac{1 }{\color{blue}{\rm a}^{\color{red}{\rm b}}} }\)

  3. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so, your expression is: \(m^3n^{-6}p^0\) is that correct?

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i may be wrong my b

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay 1n^-6/1 = 1/n^-6 = 1/n^6 now we put the positive n^6 back in it's original place. m^3 n^6 p How about now

  6. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you are still doing 2 parts incorrectly, not just the n^(-6), but p^0 is incorrect as well-:(

  7. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    lets start from \(p^0\), ok?

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay.

  9. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The rule for an exponent of 0 is: \(\Large\color{black}{ \displaystyle \color{blue}{\rm x}^{\color{red}{\rm 0}}=1 }\) This is true for any value of x, but provided that x is NOT zero. ------------------------------------------------------ ADDITIONALLY: The prove for this rule is: \(\Large\color{black}{ \displaystyle \color{blue}{\rm x}^{\color{red}{\rm 0}}=\color{blue}{\rm x}^{w-w} }\) (i am using w, but you can replace w with 1, 3, or any other number and the statement will still hold) \(\Large\color{black}{ \displaystyle \color{blue}{\rm x}^{\color{red}{\rm 0}}=\color{blue}{\rm x}^{w-w} =\dfrac{x^w}{x^w}=1}\) (anything divided by itself gives a result of 1, but if x was 0 then we have 0/0 which is not defined. This is why I said every x but NOT x=0)

  10. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So if any number (except 0) raised to the power of 0, gives a result of 1, then \(p^0={\rm what?}\)

  11. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ((Another NOTE: they didn't state that \(p\ne0\), it is their fault, but you have to make this assumption, to get the answer, or else it would be indeterminate))

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Alright.

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So p^1?

  14. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    p^1 ? you don't have that in your expression, do you?

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No, so do i leave the p^0 alone?

  16. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you have p^0, and you know that: (anything)\(^0\) = ?

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    x?

  18. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    (anything)\(^0\)=1 (as long as this anything is NOT zero)

  19. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    therefore, you can say that p\(^0\)=?

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    But p has a zer exponent?

  21. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes, so p\(^0\)= what?

  22. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Zero* And is it 1

  23. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes, p\(^0\)=1

  24. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So lets re-write our expression \({\rm m}^3\cdot {\rm n}^{-6} \cdot {\rm p}^0\) this kis what it is used to be, but we know p\(^0\)=1, so we can say: \({\rm m}^3\cdot {\rm n}^{-6} \cdot 1\) and multiplying times 1, is not changing the value, so we can leave the "•1" part out. (right?) We have: \({\rm m}^3\cdot {\rm n}^{-6} \)

  25. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    is everything making sense till this point?

  26. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Alright. Yeah so far.

  27. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Because the p is = to 1 right?

  28. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes, because p\(^0\)=1

  29. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So yeah i get it not sure why we need to leave the 1 out after solving the problem though. Is it because it's exponent was 0?

  30. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    why we have to leave the "•1" out? Not that we must, but it does matter. For any number, if you multiply it times 1 you still have that same number. Multiplying times 1 is same, and just as good as, not multiplying by anything at all.

  31. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Makes sense. Thanks for your help.

  32. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    we aren't done yet.

  33. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Alright

  34. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So our new expression is: \({\rm m}^3\cdot {\rm n}^{-6} \) And, this can be simplified more. ------------------------------ Now, if we can recall, I mentioned another important property, and it refers to negative exponents. This property is: \(\Large\color{black}{ \displaystyle \color{blue}{\rm a}^{-\color{red}{\rm b}}=\frac{1 }{\color{blue}{\rm a}^{\color{red}{\rm b}}} }\)

  35. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    according to this property, \(\Large\color{black}{ \displaystyle \color{blue}{\rm n}^{-\color{red}{\rm 6}}=\frac{1 }{\color{blue}{\rm n}^{\color{red}{\rm 6}}} }\) right?

  36. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah

  37. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    same, but n instead of a, and 6 instead of b. good...

  38. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    n^-6/1/n^6

  39. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    n\(^{-6}\) = 1 / n\(^6\) like this, you mean?

  40. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah forgot to put in my equal.

  41. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    lol.

  42. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :P

  43. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so, we had: \({\rm m}^3\cdot {\rm n}^{-6} \) and now we know that: \( {\rm n}^{-6}=\dfrac{1}{{\rm n}^6} \) So, we can re-write our expression the following way: \({\rm m}^3\cdot\dfrac{1}{{\rm n}^{6}} \)

  44. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    then you can make it as 1 fraction, just the following way: \({\rm m}^3\cdot\dfrac{1}{{\rm n}^{6}} \) \(\dfrac{{\rm m}^3}{1}\cdot\dfrac{1}{{\rm n}^{6}} \) \(\dfrac{{\rm m}^3\cdot 1}{1\cdot {\rm n}^{6}} \) \(\dfrac{{\rm m}^3}{{\rm n}^{6}} \)

  45. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    if you have any questions regarding any of the rules or steps that were aplied or mentioned, please ask...

  46. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No questions, because you timing it by one which would make it the same number.

  47. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    alright. G☼☼d luck

  48. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks :)

  49. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.