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MetalPen

  • 2 years ago

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

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  1. slaaibak
    • 2 years ago
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    We will be here to answer your questions :)

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

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

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

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

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

  7. Akshay_Budhkar
    • 2 years ago
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    yea slaaibak got it

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

  9. Akshay_Budhkar
    • 2 years ago
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    it is simple

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

  11. MetalPen
    • 2 years ago
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    No.

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

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

  14. MetalPen
    • 2 years ago
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    Okay.

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

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

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

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

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

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

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

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

  23. slaaibak
    • 2 years ago
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    type over

  24. MetalPen
    • 2 years ago
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    ...

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

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

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

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

  29. slaaibak
    • 2 years ago
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    yeah sure

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

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

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

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

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

  35. MetalPen
    • 2 years ago
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    Lol yeah.

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

  37. Akshay_Budhkar
    • 2 years ago
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    i am not lol

  38. MetalPen
    • 2 years ago
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    =.=

  39. MetalPen
    • 2 years ago
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    ?

  40. slaaibak
    • 2 years ago
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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: \[{1 \over y^5} * {8 \over x^2 * y^4}\]

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

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

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

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

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

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

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

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

  49. MetalPen
    • 2 years ago
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    OH okay.

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

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

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

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

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

  55. MetalPen
    • 2 years ago
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    Yes.

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

  57. Akshay_Budhkar
    • 2 years ago
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    1/900

  58. MetalPen
    • 2 years ago
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    Sure...

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

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

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

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

  63. Akshay_Budhkar
    • 2 years ago
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    it is my pleasure

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