algebra

According to Newton's law of cooling, the rate at which an object cools is directly proportional to the difference in temperature between the object and the surrounding medium. The face of a household iron cools from 135° to 107.5° in 30 minutes in a room that remains at a constant temperature of 80°. From calculus, the temperature f(t) of the face after t hours of cooling is given by f(x) = 55(2)^(−2t) + 80.

Assuming t = 0 corresponds to 1:00 P.M., approximate to the nearest tenth of a degree the temperature of the face at 2:00 P.M., 3:30 P.M., and 4:00 P.M.

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1. f(t) = 55 (2)^(-2t) + 80
now you tell me that if it starts at 135 it will be at 107.5 in 0.5 hours
let me check that
yes, it is 135 at t = 0
now at t = 0.5
it is
55 (2)^(-1) + 80
= 27.5 + 80
= 107.5 sure enough I believe you
now the problem
say at 2 pm..... then t = 1 hour after 1 pm
so
f(t) = 55 (2)^(-2t) + 80
= 55 * 2^-2 + 80
= 55/4 + 80
= 93.75 at 2 pm

then use t = 2.5 and t = 3 for 3:30 and 4

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Damon

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