anonymous
  • anonymous
need some help with a difference equation
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
wats the question
anonymous
  • anonymous
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.
anonymous
  • anonymous
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

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amistre64
  • amistre64
recurrence equation tends to get better results... 'difference' equation is the same thing, but a less used term
anonymous
  • anonymous
alrightt
amistre64
  • amistre64
what was the original question? it seems like we are trying to jump into it halfway thru
anonymous
  • anonymous
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.
anonymous
  • anonymous
Maybe my whole difference equation is wrong....
amistre64
  • amistre64
it might be; 'btm' asks alot of these questions and may have some insight
anonymous
  • anonymous
what is btm?
anonymous
  • anonymous
I might be able to help
amistre64
  • amistre64
what is ncmtj?
amistre64
  • amistre64
imran is smart as well :)
anonymous
  • anonymous
alright thank you
anonymous
  • anonymous
thank you amistreo
anonymous
  • anonymous
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.
anonymous
  • anonymous
yn=8000-(.80n)800y_o ?
anonymous
  • anonymous
it should be 8000, no y_0
anonymous
  • anonymous
can you send me whole question
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
For my difference equation I have yn=〖.80〗_(yn-1) y0=8000
anonymous
  • anonymous
Okay, So I see you are solving for number of healthy trees. which is fine
anonymous
  • anonymous
yeah
anonymous
  • anonymous
Infected trees y0=8000 y1=8000*(.8) y2= 8000*(.8)^2
anonymous
  • anonymous
yn=8000*(.8)^2
anonymous
  • anonymous
*correction yn=8000*(.8)^n
anonymous
  • anonymous
Total - Healthy trees= infected trees 8000-8000*(.8)^n=7560
anonymous
  • anonymous
-8000*(.8)^n=7560-8000 -8000*(.8)^n=-440 (.8)^n=-440/-8000 .8^n=.055 log[.8,.555]=n n=12.99
anonymous
  • anonymous
ok. that does make much better sense. But how from 8000-8000(.80)^n, did u get 7560?
anonymous
  • anonymous
7560 was given in your problem "the number of infected elm trees will be 7560"
anonymous
  • anonymous
thats right ok.
anonymous
  • anonymous
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.
anonymous
  • anonymous
your difference equation is \[y_n=y_{n-1}*.8\]
anonymous
  • anonymous
alright that makes more sense now!!!
anonymous
  • anonymous
would y0 be 8000?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
This is a difference equation for healthy trees
anonymous
  • anonymous
any other question?
anonymous
  • anonymous
that should be it. Thank you so much for your help.
anonymous
  • anonymous
np
anonymous
  • anonymous
hey imranmean, one more question. on the -8000(.80)n= -440, how did it come to Log[.8,.055]?
anonymous
  • anonymous
\[-8000*(.8)^n=7560-8000\] \[-8000*(.8)^n=-440\] \[(.8)^n={-440\over-8000}\] .8^n=.055 log[.8,.555]=n n=12.99

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