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anonymous
 5 years ago
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
anonymous
 5 years ago
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0T (t) =70+133e^(t/5) what is the average temperature of the tea one hour after the initial time?? please help!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0replace t by 1. \[T(1)=70+133e^{\frac{1}{5}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i get 178.89 to two decimal places

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i see it now lol 203  70 right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the room temp is 70, 20370=133 and it is the difference that decays

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do you get that answer of 178.89 from that equation? b/c i got something different :(

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, so t = 60 minutes if that is accurate....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.070 + (133 * (e^((60) / 5))) = 70.0008172 is what google figures out

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0or is this a recursion equation.... and not a particular solution?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0nah; particular it is; no place for a T{n1} lol
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