If a cup of tea has an initial temperature of 203 degree F in a room whose temperature is 70 degreeF, then according to Newton's law of cooling the temperature of the tea after t minute have passed is

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If a cup of tea has an initial temperature of 203 degree F in a room whose temperature is 70 degreeF, then according to Newton's law of cooling the temperature of the tea after t minute have passed is

Mathematics
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T (t) =70+133e^(-t/5) what is the average temperature of the tea one hour after the initial time?? please help!
replace t by 1. \[T(1)=70+133e^{\frac{-1}{5}}\]
i get 178.89 to two decimal places

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should 133 be 203?
no
i see it now lol 203 - 70 right?
the room temp is 70, 203-70=133 and it is the difference that decays
how do you get that answer of 178.89 from that equation? b/c i got something different :(
yeah, so t = -60 minutes if that is accurate....
70 + (133 * (e^((-60) / 5))) = 70.0008172 is what google figures out
or is this a recursion equation.... and not a particular solution?
nah; particular it is; no place for a T{n-1} lol

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