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Consider the following problem:
The temperature of a cup of cocoa is decreasing at a rate of 
r(t)=-5.5e^(-0.09 t) degrees Celsius per minute (where 
t is the time in minutes). At time 
t=2, the temperature of the cocoa is 72 degrees Celsius. What is the temperature of the cocoa at 
t=10 minutes?
Which expression can we use to solve the problem?
Choose 1 answer:
(A) 
int_(2)^(10)r(t)dt+72
(B) 
int r(t)dt+72
(C) 
int r(t)dt
(D) 
int_(2)^(10)r(t)dt

Consider the following problem:\newlineThe temperature of a cup of cocoa is decreasing at a rate of r(t)=5.5e0.09t r(t)=-5.5 e^{-0.09 t} degrees Celsius per minute (where t t is the time in minutes). At time t=2 t=2 , the temperature of the cocoa is 7272 degrees Celsius. What is the temperature of the cocoa at t=10 t=10 minutes?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) 210r(t)dt+72 \int_{2}^{10} r(t) d t+72 \newline(B) r(t)dt+72 \int r(t) d t+72 \newline(C) r(t)dt \int r(t) d t \newline(D) 210r(t)dt \int_{2}^{10} r(t) d t

Full solution

Q. Consider the following problem:\newlineThe temperature of a cup of cocoa is decreasing at a rate of r(t)=5.5e0.09t r(t)=-5.5 e^{-0.09 t} degrees Celsius per minute (where t t is the time in minutes). At time t=2 t=2 , the temperature of the cocoa is 7272 degrees Celsius. What is the temperature of the cocoa at t=10 t=10 minutes?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) 210r(t)dt+72 \int_{2}^{10} r(t) d t+72 \newline(B) r(t)dt+72 \int r(t) d t+72 \newline(C) r(t)dt \int r(t) d t \newline(D) 210r(t)dt \int_{2}^{10} r(t) d t
  1. Integrate Rate of Change: To find the temperature of the cocoa at t=10t=10 minutes, we need to integrate the rate of temperature change from t=2t=2 to t=10t=10 and add this to the initial temperature at t=2t=2.
  2. Calculate Change in Temperature: The rate of temperature change is given by r(t)=5.5e(0.09t)r(t)=-5.5e^{(-0.09 t)}. To find the change in temperature from t=2t=2 to t=10t=10, we need to integrate r(t)r(t) with respect to tt from 22 to 1010.
  3. Find Total Temperature Change: The integral of r(t)r(t) from t=2t=2 to t=10t=10 will give us the total change in temperature over this time period. Since we are given the temperature at t=2t=2, we will add this change to 7272 degrees Celsius to find the temperature at t=10t=10.
  4. Use Correct Expression: The correct expression to use for this problem is the definite integral of r(t)r(t) from t=2t=2 to t=10t=10, plus the initial temperature at t=2t=2. This corresponds to choice (A) 210r(t)dt+72\int_{2}^{10}r(t)\,dt+72.