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Math Help - Assessment Help

  1. #1
    xgunwingx
    Guest

    Assessment Help

    so far, g'(t) = 95e^-0.25t * [1+19e^-0.25t]^-2

    where g'(t) is the growth rate formula

    the thing i have to do is find the mean height of the tree's and how many years that i expect the process would occur.

    mean = 2.821

    Integrate g'(t) = 95e^-0.25t * [1+19e^-0.25t]^-1

    95e^-0.25t = -380e^-0.25t+C

    [1+19e^-0.25t]^-1 = -4.75e^-0.25t(1+19e^-0.25t)^-1 +c
    -4.75e(sigma ) [1+19e^-0.25t]^-1.dt = 20(1+19e^-0.25t)^-1+c
    at this point you are to find c
    let g=0 & t=0

    0(0) = 20(1+19e^-0.25t)^-1+c
    0(0) = 20(20)^-1+c
    0(0) = 1+c
    therefore c=-1

    at this time you are to find the value of t when g=2.821

    2.821(t [dont know if this has to be here]) = 20(1+19e^-0.25t)^(-2or-1) -1 (dunno if its supposed to be -2or-1)

    3.821(t) = 20(1+19e^-0.25t)^(-2or-1)

    0.19105(t)=(1+19e^-0.25t)^(-2or-1)

    0.19105(t)= 1 / (1+19e^-0.25t)^(2or1)

    if ^2 = 1/(1+19e^-0.25t)x(1+19e^-0.25t)
    if ^1 = 1/1+19e^-0.25t

    Either way im lost, please help!
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  2. #2
    Mysteroius
    Guest

    evil trees -1

    0.191=1/(1+19e^-0.25t)
    1+19e^-0.25x=1/0.191
    1+19e^-.25x=5.24
    19e^-0.25x=4.24
    e^-0.25x=0.223
    ln0.223=-0.25x
    -1.5=0.25x
    x=6
    6years to rech 2.82
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