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magepker728

  • 3 years ago

how would you solve this

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  1. magepker728
    • 3 years ago
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    |dw:1349555414557:dw|

  2. Decart
    • 3 years ago
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    a=x

  3. magepker728
    • 3 years ago
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    how and why

  4. Decart
    • 3 years ago
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    this is an old eqaution from Issac Newton

  5. zzr0ck3r
    • 3 years ago
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    ((7-3(x+h)) - (7-3x))/h

  6. magepker728
    • 3 years ago
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    @zzr0ck3r you replace 7-3x with f(a) why is that?

  7. zzr0ck3r
    • 3 years ago
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    take f(x) and put in a+h for x f(a+h) = 7-3(a+h) then subtract f(a) from all that (7-3(a+h) - (7-3a)) then put that all over h (7-3(a+h) - (7-3a)) /h

  8. Decart
    • 3 years ago
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    This is called the differance quotient

  9. Decart
    • 3 years ago
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    \[\frac{ f(x+h)-f(x) }{ h }\]

  10. zzr0ck3r
    • 3 years ago
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    I dont think the name and its history is helping him much, as im sure they told him that in class.

  11. Decart
    • 3 years ago
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    you substitute 7-3x for all x values

  12. magepker728
    • 3 years ago
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    well thanks everyone...

  13. zzr0ck3r
    • 3 years ago
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    no, this is not what you do....

  14. zzr0ck3r
    • 3 years ago
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    Please stop @Decart you are saying many wrong things

  15. zzr0ck3r
    • 3 years ago
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    (7-3a-3h-(7-3a))/h

  16. Decart
    • 3 years ago
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    http://www.youtube.com/watch?v=ZsNiIX6gDf0

  17. Fgcbear16
    • 3 years ago
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    I'm with @zzr0ck3r. you didn't distribute your h @Decart

  18. Decart
    • 3 years ago
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    you are correct i did not srry

  19. Decart
    • 3 years ago
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    I had to refer to my precalculus text book it had been some time since I had this

  20. Decart
    • 3 years ago
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    \[\frac{ 7-3(a+h)-(7-3a) }{ h }\]

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