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03 Oct 2006, 17:43
Kudos
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The temperature of a cup of coffee 10 minutes after it was poured was 120 degrees F. If the temp (F) of the cup of coffee t minutes after it was poured can be determined by F=120(2^(at)) + 60 , where a is a constant, then what was the temperature 30 minutes after it was poured?
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03 Oct 2006, 19:12
Kudos
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F=120(2^(at)) + 60
120=120(2^(a10)) + 60
1/2=2^(a10)
1/2=1/2^(-a10) ---> so we need that (-a*10)=1 so a=-1/10
For 30 min
F=120(2^((-1/10)*30)) + 60
F=120*(1/2^3)+60
F=15+60
F=75
120=120(2^(a10)) + 60
1/2=2^(a10)
1/2=1/2^(-a10) ---> so we need that (-a*10)=1 so a=-1/10
For 30 min
F=120(2^((-1/10)*30)) + 60
F=120*(1/2^3)+60
F=15+60
F=75
