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anonymous

  • 5 years ago

need some help with a difference equation

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  1. anonymous
    • 5 years ago
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    wats the question

  2. anonymous
    • 5 years ago
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    yn=8000-(.80n)800o. Trying to solve for N. I know the answer I just need help with finding out how to get N alone and such. The answer should be like 13.

  3. anonymous
    • 5 years ago
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    Im not sure if I need to divide by another number. I have another number and its 7560. Do I need to put the 7560 where the yn is

  4. amistre64
    • 5 years ago
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    recurrence equation tends to get better results... 'difference' equation is the same thing, but a less used term

  5. anonymous
    • 5 years ago
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    alrightt

  6. amistre64
    • 5 years ago
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    what was the original question? it seems like we are trying to jump into it halfway thru

  7. anonymous
    • 5 years ago
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    Ok, it says to use the solved difference equation to find in what calender year, the number of infected elm trees will be 7560. You must solve the solved difference equation: you may not guess and check. Get an equation for n and solve it.

  8. anonymous
    • 5 years ago
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    Maybe my whole difference equation is wrong....

  9. amistre64
    • 5 years ago
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    it might be; 'btm' asks alot of these questions and may have some insight

  10. anonymous
    • 5 years ago
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    what is btm?

  11. anonymous
    • 5 years ago
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    I might be able to help

  12. amistre64
    • 5 years ago
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    what is ncmtj?

  13. amistre64
    • 5 years ago
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    imran is smart as well :)

  14. anonymous
    • 5 years ago
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    alright thank you

  15. anonymous
    • 5 years ago
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    thank you amistreo

  16. anonymous
    • 5 years ago
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    I can send you more information about the problem to maybe be able to help you more. There is like five questions to be answered.

  17. anonymous
    • 5 years ago
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    yn=8000-(.80n)800y_o ?

  18. anonymous
    • 5 years ago
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    it should be 8000, no y_0

  19. anonymous
    • 5 years ago
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    can you send me whole question

  20. anonymous
    • 5 years ago
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    yes.Suppose a region has 8000 healthy elm trees. In 1950, bark beetles harboring the fungus arrive in the region. In each subsequent yera, the number of elm trees that contract the disease is 20% of the number of healthy trees at the end of the previous year. Once a tree is infected, it stays infected. Find the difference equation.

  21. anonymous
    • 5 years ago
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    Also, to solve the difference equation, and use the solved difference equatoin to find in what calender year, the number of infected elm trees will be 7560. You must solve the solved difference equation.

  22. anonymous
    • 5 years ago
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    For my difference equation I have yn=〖.80〗_(yn-1) y0=8000

  23. anonymous
    • 5 years ago
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    Okay, So I see you are solving for number of healthy trees. which is fine

  24. anonymous
    • 5 years ago
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    yeah

  25. anonymous
    • 5 years ago
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    Infected trees y0=8000 y1=8000*(.8) y2= 8000*(.8)^2

  26. anonymous
    • 5 years ago
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    yn=8000*(.8)^2

  27. anonymous
    • 5 years ago
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    *correction yn=8000*(.8)^n

  28. anonymous
    • 5 years ago
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    Total - Healthy trees= infected trees 8000-8000*(.8)^n=7560

  29. anonymous
    • 5 years ago
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    -8000*(.8)^n=7560-8000 -8000*(.8)^n=-440 (.8)^n=-440/-8000 .8^n=.055 log[.8,.555]=n n=12.99

  30. anonymous
    • 5 years ago
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    ok. that does make much better sense. But how from 8000-8000(.80)^n, did u get 7560?

  31. anonymous
    • 5 years ago
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    7560 was given in your problem "the number of infected elm trees will be 7560"

  32. anonymous
    • 5 years ago
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    thats right ok.

  33. anonymous
    • 5 years ago
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    In my book the normal difference equations are written as for example, yn=-.8yn-1 +9, yo=10. Do I need to put any of those into this new difference equation. The equation seems bare.

  34. anonymous
    • 5 years ago
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    your difference equation is \[y_n=y_{n-1}*.8\]

  35. anonymous
    • 5 years ago
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    alright that makes more sense now!!!

  36. anonymous
    • 5 years ago
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    would y0 be 8000?

  37. anonymous
    • 5 years ago
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    Yes

  38. anonymous
    • 5 years ago
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    This is a difference equation for healthy trees

  39. anonymous
    • 5 years ago
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    any other question?

  40. anonymous
    • 5 years ago
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    that should be it. Thank you so much for your help.

  41. anonymous
    • 5 years ago
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    np

  42. anonymous
    • 5 years ago
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    hey imranmean, one more question. on the -8000(.80)n= -440, how did it come to Log[.8,.055]?

  43. anonymous
    • 5 years ago
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    \[-8000*(.8)^n=7560-8000\] \[-8000*(.8)^n=-440\] \[(.8)^n={-440\over-8000}\] .8^n=.055 log[.8,.555]=n n=12.99

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