1. You gathered data and realized that the temperature drop followed the function T=85 (0.3)^t/2 where T is the temperature in Celsius and t is the time in hours the water was left out for.
A. What was the initial temperature of the water when it was placed outside?
B. What was the temperature of the water after 3 hours?
C. How long does it take for the water to reach 47 Celsius?
D. What is an appropriate domain and range for this scenario?
To determine the answers to the questions, we can use the given function: T = 85(0.3)^(t/2).
A. To find the initial temperature of the water when it was placed outside, we can substitute t = 0 into the function:
T = 85(0.3)^(0/2)
T = 85(0.3)^0
T = 85(1)
T = 85
Therefore, the initial temperature of the water when it was placed outside is 85 degrees Celsius.
B. To find the temperature of the water after 3 hours, we can substitute t = 3 into the function:
T = 85(0.3)^(3/2)
T = 85(0.3)^(1.5)
Calculating this value will give you the answer.
C. To find how long it takes for the water to reach 47 degrees Celsius, we can rearrange the equation:
47 = 85(0.3)^(t/2)
Taking the logarithm of both sides will allow us to isolate t and find the answer.
D. In terms of the domain, t represents time in hours, so it cannot be negative or zero (since it is a measure of time passed). Therefore, the domain for this scenario would be t > 0.
In terms of the range, the function T = 85(0.3)^(t/2) shows that the temperature T will always be positive since it is multiplied by a positive factor (0.3) raised to a positive power (t/2). Hence, the range for this scenario would be T > 0, meaning the temperature cannot be negative.