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MetalPen

Could somebody please help me study for a pre-algebra test? I have some questions.

  • 2 years ago
  • 2 years ago

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  1. slaaibak
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    We will be here to answer your questions :)

    • 2 years ago
  2. MetalPen
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    Okay thanks! I really appreciate it.

    • 2 years ago
  3. MetalPen
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    Okay. sooo, this is really broad, but I don't understand negative exponents.... At all.

    • 2 years ago
  4. Akshay_Budhkar
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    So you got some specific questions or do you want me or us to explain you negative exponents from scratch?

    • 2 years ago
  5. MetalPen
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    Ummmmmmm. Why does a negative exponent make a fraction? I don't understand the basic concept.

    • 2 years ago
  6. slaaibak
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    Negative exponents simply means this: \[a^{-x} = {1 \over a^x}\]

    • 2 years ago
  7. Akshay_Budhkar
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    yea slaaibak got it

    • 2 years ago
  8. MetalPen
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    Yes but why are they defined that way?

    • 2 years ago
  9. Akshay_Budhkar
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    it is simple

    • 2 years ago
  10. Akshay_Budhkar
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    do you know \[a^x \times a^y= a^{x+y}\]

    • 2 years ago
  11. MetalPen
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    No.

    • 2 years ago
  12. Akshay_Budhkar
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    now say we want \[a^x \times a^y = 1\]

    • 2 years ago
  13. MetalPen
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    Well yeah I do sorry.

    • 2 years ago
  14. MetalPen
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    Okay.

    • 2 years ago
  15. MetalPen
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    You'd made one negative!!! AHHH I GET IT!

    • 2 years ago
  16. Akshay_Budhkar
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    so when we solve that we get \[a^{x+y}= 1 = a^0\]

    • 2 years ago
  17. Akshay_Budhkar
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    thats good! You get it :D

    • 2 years ago
  18. MetalPen
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    Teehee! Wow that just clicked. My math teacher didn't explain it that way.......

    • 2 years ago
  19. Akshay_Budhkar
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    its my pleasure helping you click it :D

    • 2 years ago
  20. MetalPen
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    So another thing, how do you deal with #s with exponents in a fraction?

    • 2 years ago
  21. slaaibak
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    Give an example please?

    • 2 years ago
  22. MetalPen
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    Ummmm. How do you make fractions with this equation thing?

    • 2 years ago
  23. slaaibak
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    type over

    • 2 years ago
  24. MetalPen
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    ...

    • 2 years ago
  25. Akshay_Budhkar
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    3 over 4 \[3 \over 4\]

    • 2 years ago
  26. slaaibak
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    Use this code: {1 over 2} would be \[{1 \over 2}\]

    • 2 years ago
  27. MetalPen
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    ... Um can I just say it?

    • 2 years ago
  28. Akshay_Budhkar
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    yea what is your doubt? i do not comprehend it

    • 2 years ago
  29. slaaibak
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    yeah sure

    • 2 years ago
  30. MetalPen
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    Okay so if you had y^-5 times 2^3 over y^4 times x^2.

    • 2 years ago
  31. MetalPen
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    Or something to that effect.

    • 2 years ago
  32. Akshay_Budhkar
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    \[ y^-5 \times 2^3 \over y^4 \times x^2.\]

    • 2 years ago
  33. slaaibak
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    \[{ y^{-5} * 2^3 \over y^4 * x^2}\]

    • 2 years ago
  34. Akshay_Budhkar
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    is that your query?

    • 2 years ago
  35. MetalPen
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    Lol yeah.

    • 2 years ago
  36. slaaibak
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    You're faster than me xD

    • 2 years ago
  37. Akshay_Budhkar
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    i am not lol

    • 2 years ago
  38. MetalPen
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    =.=

    • 2 years ago
  39. MetalPen
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    ?

    • 2 years ago
  40. slaaibak
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    If there's multiplication, you can take y^-5 and but it below the line, \[y^{-5} = {1 \over y^5}\] You can immediately write 2^3 as 8. So now you have: \[{1 \over y^5} * {8 \over x^2 * y^4}\]

    • 2 years ago
  41. slaaibak
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    put*, not but

    • 2 years ago
  42. slaaibak
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    Then, you can multiply similar variables with eachother and add the powers. it becomes: \[8 \over y^9 * x^2\]

    • 2 years ago
  43. MetalPen
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    What are you talking about?

    • 2 years ago
  44. Akshay_Budhkar
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    about the question you asked

    • 2 years ago
  45. slaaibak
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    I made a spelling mistake in my post.

    • 2 years ago
  46. MetalPen
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    Yeah but how did you get y^9?

    • 2 years ago
  47. MetalPen
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    I know you have to add... Or subtract the exponents or something.

    • 2 years ago
  48. slaaibak
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    \[y^4 * y^5 = y^{5 + 4}\]

    • 2 years ago
  49. MetalPen
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    OH okay.

    • 2 years ago
  50. Akshay_Budhkar
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    the main aim of such questions is to convert your negative exponents into positive by taking its reciprocal

    • 2 years ago
  51. MetalPen
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    Um one more thing. How would you get {1 over 9} % of 90. Or something like that.

    • 2 years ago
  52. MetalPen
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    Aw shoot. I messed up the thing.

    • 2 years ago
  53. MetalPen
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    \[{1 \over 9}\]

    • 2 years ago
  54. Akshay_Budhkar
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    percent is always divided by 100

    • 2 years ago
  55. MetalPen
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    Yes.

    • 2 years ago
  56. Akshay_Budhkar
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    so 1/9 percent is \[(1/9 ) \over 100 = 1\over900\]

    • 2 years ago
  57. Akshay_Budhkar
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    1/900

    • 2 years ago
  58. MetalPen
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    Sure...

    • 2 years ago
  59. Akshay_Budhkar
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    of 90 is 1/900 of 90.. of implies multiplication

    • 2 years ago
  60. Akshay_Budhkar
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    so it is 1/900 times 90 = 90/900 = 1/10

    • 2 years ago
  61. MetalPen
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    Yes yes I get it.

    • 2 years ago
  62. MetalPen
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    That makes sense! Okay that's all! And I have to go. Thank you so much. ^.^

    • 2 years ago
  63. Akshay_Budhkar
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    it is my pleasure

    • 2 years ago
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