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anonymous

  • 4 years ago

Find the average rate of change of f from 0 to (11pi)/12. f(x)=tan(2x)

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  1. myininaya
    • 4 years ago
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    \[\frac{f(0)-f(\frac{11 \pi}{12})}{0-\frac{11 \pi}{12}}\] it would have been nice if you didn't repost this twice so i wouldn't have to type that again

  2. anonymous
    • 4 years ago
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    last time i did not include f(x)=tan(2x)

  3. myininaya
    • 4 years ago
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    \[\frac{\tan(2 \cdot 0)-\tan(2 \cdot \frac{11 \pi}{12})}{-\frac{11 \pi}{12}}\] \[\frac{0-\tan(\frac{11 \pi}{6})}{\frac{-11 \pi}{12}}\] it is the same thing though i mean it is the same formula to use

  4. myininaya
    • 4 years ago
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    \[\frac{ \tan(\frac{11 \pi}{6})}{\frac{11 \pi}{12}} \cdot \frac{12}{12}=\frac{ 12 \tan(\frac{11 \pi}{6})}{11 \pi}\]

  5. anonymous
    • 4 years ago
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    ((-11pi)sqrt3)/36

  6. myininaya
    • 4 years ago
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    \[\frac{12 \tan(\frac{6 \pi}{6}+\frac{5 \pi}{6})}{11 \pi}\] \[\frac{12 \tan( \frac{5\pi}{6}+\pi)}{11 \pi}\]

  7. myininaya
    • 4 years ago
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    |dw:1327879891939:dw|

  8. myininaya
    • 4 years ago
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    |dw:1327879937416:dw|

  9. myininaya
    • 4 years ago
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    \[\frac{12 (\frac{-1}{\sqrt{3}})}{11 \pi}\]

  10. myininaya
    • 4 years ago
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    does that help? do you understand?

  11. myininaya
    • 4 years ago
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    \[\frac{12 (\frac{-1}{\sqrt{3}})}{11 \pi} \cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{12 (-1)}{11 \pi \sqrt{3}}\]

  12. myininaya
    • 4 years ago
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    \[\frac{-12}{11 \pi \sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{-12 \sqrt{3}}{11 \pi (3)}\]

  13. myininaya
    • 4 years ago
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    \[\frac{-12 \sqrt{3}}{33 \pi}\]

  14. myininaya
    • 4 years ago
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    \[\frac{-4 \sqrt{3}}{11 \pi}\]

  15. anonymous
    • 4 years ago
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    ok i need to look over everything you have been laying out for me very slowly

  16. anonymous
    • 4 years ago
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    thank you for taking the time to write this out

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