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anonymous

  • 5 years ago

The temperature of a body is measured as 104oF. It is observed that the amount the temperature changes for each period of two hours is -0.3 time the difference between the previous period's temperature and the room temperature, which is 70oF. a. Write a recurrence relation for tn, the temperature of the body at the end of n 2-hour time periods.

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  1. anonymous
    • 5 years ago
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    dT/dt = -0.3(T - 70)

  2. anonymous
    • 5 years ago
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    \[t_{0} = 104\]

  3. anonymous
    • 5 years ago
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    I had T(n) = -.03(104 - 70)

  4. anonymous
    • 5 years ago
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    so same thing

  5. anonymous
    • 5 years ago
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    u know differential eqns?

  6. anonymous
    • 5 years ago
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    oooo...somewhat...has been a while

  7. anonymous
    • 5 years ago
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    dT/(T-70) = -0.3dt

  8. anonymous
    • 5 years ago
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    \[\int\limits_{104}^{T}dT/(T-70) = -0.3\int\limits_{0}^{t}dt\]

  9. anonymous
    • 5 years ago
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    need to keep it basice with recurrence relations

  10. anonymous
    • 5 years ago
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    ln (t-70) = -0.3ln(34)t

  11. anonymous
    • 5 years ago
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    thanks though!

  12. amistre64
    • 5 years ago
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    so initial t{0} = 140 right?

  13. anonymous
    • 5 years ago
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    yes

  14. amistre64
    • 5 years ago
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    t{n} = -0.3*[ t{n-1} - 70]

  15. anonymous
    • 5 years ago
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    I havee that but can't get to work on calc

  16. amistre64
    • 5 years ago
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    i havent used the calulator :) plug it in google to create a table if need be

  17. anonymous
    • 5 years ago
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    gues i 'll do it by hand

  18. anonymous
    • 5 years ago
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    how on google...tried wolfram once...pretty cool

  19. anonymous
    • 5 years ago
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    Find a general and particular solution for the system and give the value of t(12), the temperature of the body after 24 hours.

  20. amistre64
    • 5 years ago
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    i just reiterete it over and over in the right place on google

  21. amistre64
    • 5 years ago
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    -.3*( 104 - 70) -.3(-.3*( 104 - 70)-70) -.3(-.3(-.3*( 104 - 70)-70)-70) like that

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