1-70) 3 500 6) A pizza is baked at 450 and is removed from the oven at 5:00. The
ID: 2912955 • Letter: 1
Question
1-70) 3 500 6) A pizza is baked at 450 and is removed from the oven at 5:00. The temperature in the room is To 5 minutes, the temperature of the pizza is 300. Use the model (t T(Uo-T)ekt, where U(e) represents the temperature of the pizza after t minutes, T is the room temperature, Uo is the initial temperature of the pizza, and k is some cooling constant. a. Find the cooling constant k b. What will be the temperature of the pizza at S:157 c. What will be the temperature at 6:00? d. What time will the pizza be 200? e. What time will the pizza be 135*? 7) The Richter scale measures the intensity of an earthquake. It is modeled by the function logio! where I is the intensity of the shock wave. In 1906, there was an earthquake in San Francisco that measured 8.6 on the Richter scale. There was another earthquake in the same area in 1989 that measured 7.7. Compare the intensities of these earthquakes. 8) Formerly living (a.k.a dead) organisms can be dated by the amount of carbon-14 present at time t compared to the amount present when the organism was alive. The half-life of carbon-1 is 5730 years. Suppose an organism is known to have contained 85 grams of carbon-14 when it was alive. How much of the carbon would remain after 100 years? How much would remain after 1000 years? How long before the sample decays to 50 grams? How long before only 10 grams remain? a. b. c. d. Suppose a culture of bacteria in a lab grows continuously, doubling every 4 hours a. Write a function modeling this situation. b. If the initial amount is 100 bacteria, how many will there be in 12 hours? c. If the initial amount is 100 bacteria, how many will there be in 18 hours? d. When will there be 3,000 bacteria? e. When will there be 100,000 bacteria?Explanation / Answer
T = temp of surroundings
U0 = initial temp
So, we have
Temp = 70 + (450 - 70)e^(kt)
Temp = 70 + 380e^(kt)
When t = 5, T = 300 :
300 = 70 + 380e^(5k)
k = -0.1004183887594472296 ---> ANS
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b)
Eqn becomes
T = 70 + 380e^(-0.10042t)
At 5:15, plug in t = 15 :
T = 70 + 380e^(-0.10042*15)
154.26 --> ANS
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c)
6 :00 :
Plug in t = 60
70.918 degrees
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d)
200 = 70 + 380e^(-0.10042t)
130/380 = e^(-0.10042t)
ln(13/38) = -0.10042t
t = 10.6815 minutes
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e)
135 = 70 + 380e^(-0.10042t)
e^(-0.10042t) = 13/76
-0.10042t = -1.7657839828247943
t = 17.583 min --> ANS
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