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The function
b(t) gives the temperature (in degrees Celsius) of a pan of brownies by time
t (in minutes).
What does
int_(0)^(6)b^(')(t)dt=93 mean?
Choose 1 answer:
(A) At
t=6 minutes, the temperature of the brownies is 93 degrees Celsius.
B) At
t=6 minutes, the brownies have an instantaneous rate of temperature change of 93 degrees Celsius per minute.
(C) During the first six minutes, the temperature of the brownies increases an average of 93 degrees Celsius per minute.
(D) The temperature of the brownies increased by 93 degrees Celsius in the first six minutes.

The function $b(t)$ gives the temperature (in degrees Celsius) of a pan of brownies by time $t$ (in minutes). $\newline$ What does $\int_{0}^{6} b^{\prime}(t) d t=93$ mean? $\newline$ Choose $1$ answer: $\newline$ (A) At $t=6$ minutes, the temperature of the brownies is $93$ degrees Celsius. $\newline$ B) At $t=6$ minutes, the brownies have an instantaneous rate of temperature change of $93$ degrees Celsius per minute. $\newline$ (C) During the first six minutes, the temperature of the brownies increases an average of $93$ degrees Celsius per minute. $\newline$ (D) The temperature of the brownies increased by $93$ degrees Celsius in the first six minutes.

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Evaluate two-variable equations: word problems

Full solution

Q. The function

b(t)

gives the temperature (in degrees Celsius) of a pan of brownies by time

t

(in minutes).

\newline

What does

\int_{0}^{6} b^{\prime}(t) d t=93

mean?

\newline

Choose

1

answer:

\newline

(A) At

t=6

minutes, the temperature of the brownies is

93

degrees Celsius.

\newline

B) At

t=6

minutes, the brownies have an instantaneous rate of temperature change of

93

degrees Celsius per minute.

\newline

(C) During the first six minutes, the temperature of the brownies increases an average of

93

degrees Celsius per minute.

\newline

(D) The temperature of the brownies increased by

93

degrees Celsius in the first six minutes.

Given Integral Information: We are given the integral of the derivative of the temperature function $b(t)$ with respect to time $t$ , from $0$ to $6$ minutes, and it equals $93$ . This integral represents the total change in temperature over the time interval from $0$ to $6$ minutes.
Interpretation of Integral: The integral of a derivative function over an interval gives us the net change in the original function over that interval. In this case, it tells us how much the temperature of the brownies has changed from the start time $t=0$ to the end time $t=6$ minutes.
Calculation Explanation: The value of $93$ obtained from the integral does not represent the temperature at a specific time (which would be given by $b(t)$ itself, not its derivative). Nor does it represent an instantaneous rate of change at a specific time (which would be given by $b'(t)$ at that time). It also does not represent an average rate of change per minute over the interval, as that would require dividing the total change by the number of minutes.
Final Conclusion: Therefore, the correct interpretation of the integral is that the temperature of the brownies increased by a total of $93$ degrees Celsius over the first six minutes. This corresponds to option $(D)$ .

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