anonymous
  • anonymous
A roast turkey is taken from an oven when its temperature has reached 190 Fahrenheit and is placed on a table in a room where the temperature is 75 Fahrenheit. (a) If the temperature of the turkey is 155 Fahrenheit after half an hour, what is its temperature after 45 minutes? (b) When will the trukey have cooled to 100 Fahrenheit?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1332461105993:dw|
apoorvk
  • apoorvk
actually i think this is about linear maths and not newton's law of cooling...
anonymous
  • anonymous
the detail is in satellite73's answer, we trust him

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More answers

anonymous
  • anonymous
initial difference in temperature is 190 - 75 = 115 difference after half an hour is 155 - 75 = 80 so you can model this as the temperature difference is \[115\times (\frac{80}{115})^{\frac{t}{30}}\] where t is time in minutes
anonymous
  • anonymous
see? nice job, dude
anonymous
  • anonymous
maybe even \[115\times (\frac{16}{23})^{\frac{t}{30}}\]
anonymous
  • anonymous
replace t by 45 get \[115\times (\frac{16}{23})^{\frac{3}{2}}=66.72 \] rounded so after 45 minutes it will be \[75+67=142\] aobut
anonymous
  • anonymous
when it is 100 the difference in temperature will be 100- 75 = 25 degress so set \[115(\frac{16}{23})^{\frac{t}{30}}=25\] and solve for t you got that?
anonymous
  • anonymous
142 didnt work
anonymous
  • anonymous
does "45 minutes" mean 45 minutes from the start, or after the 30 minutes that had already passed?
anonymous
  • anonymous
or maybe it needs more decimals?
anonymous
  • anonymous
If the temperature of the turkey is 155 Fahrenheit after half an hour, what is its temperature after 45 minutes? looks like it means after 45 mins
anonymous
  • anonymous
so answer seems reasonable. you could try 141.72
anonymous
  • anonymous
or try for 45 minues after the half an hour that has already passed
anonymous
  • anonymous
that would be about 121.4
anonymous
  • anonymous
yea that works what about second part
anonymous
  • anonymous
thanks btw
anonymous
  • anonymous
yw hold on one minute
anonymous
  • anonymous
\[25=115\times (\frac{16}{23})^{\frac{t}{30}}\] \[\frac{5}{23}=(\frac{16}{23})^{\frac{t}{30}}\]
anonymous
  • anonymous
\[\frac{t}{30}=\frac{\ln(\frac{5}{23})}{\ln(\frac{16}{23})}\] \[t=\frac{30\ln(\frac{5}{23})}{\ln(\frac{16}{23})}\]
anonymous
  • anonymous
t= 78.20. my software says no
anonymous
  • anonymous
satellite...you there?
anonymous
  • anonymous
hold on i will check
anonymous
  • anonymous
ok
anonymous
  • anonymous
i get 126 or so http://www.wolframalpha.com/input/?i=30*ln%2825%2F115%29%2Fln%2816%2F23%29
anonymous
  • anonymous
that doesnt work as well
anonymous
  • anonymous
i did it a different way and still got 126.153 so i do not have another suggestion, sorry
anonymous
  • anonymous
ok thats fine. do you know much about beams and moment of inertia

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