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course Mth 173
The Celsius temperature of a hot potato placed in a room is given by the function T = 40* 2- .007 t + 24 , where t is clock time in seconds and T is temperature in Celsius. At what average rate is the temperature of the potato changing between clock times t = 6.8 and t = 6.9 seconds seconds?
40*2^.007(6.8)+24 40*2^.007(6.9)+24
40*2^-.0476+24 40*2^-.0483+24
38.70178+24 38.68300+24
62.70178deg C 62.68300deg C
62.68300-62.70178=-.01878degC
6.9-6.8=.1sec
-.01878/.1=-.1878degC/sec
At what average rate is the temperature of the potato changing between clock times t = 6.8 and t = 6.81 seconds?
40.2^-.007(6.81)+24
40.2^-.04767+24
38.69990+24
62.69990deg C
62.69990-62.70178=.001875667deg C
6.81-6.8=.01sec
-.001875667/.01=.1875667deg C/sec
At what average rate is the temperature of the potato changing between clock times t = 6.8 and t = 6.801 seconds?
T=40*2^-.007(6.801)+24
40*2^-.047607+24
62.70159
6270159-62.70178=-.000185672deg C
6.801-6.8=.001sec
-.000185672/.001=-.1856722degC/sec
What do you estimate is the rate at which temperature is changing at clock time t = 6.8 seconds?
-.18degC/sec
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Good.
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The rate at which the Celsius temperature of a hot potato placed in a room is given by Rate = .041 * 2- .007 t, where R is rate of change in Celsius degrees per second and t is clock time in seconds. How much temperature change do you estimate would occur between t = 6.8 and t = 13.6 seconds?
13.6-6.8=6.8
.041*2^-.007(6.8)
-.039669
.041*2^-.007(13.6)
-.0078064
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You have calculated most of the information needed to find the answer to the question, but you have not yet found the estimated change in temperature.
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