A glass of cold milk from the refrigerator is left on the counter on a warm summer day. its temperature y (in degrees Fahrenheit) after sittin gon the counter t minutes is

y= 72-30(0.98)^t

Answer the question by interpreting y and dy/dt.

1. what is the temperature of the refrigerator? how can you tell?
2. what is the temperature of the room? how can you tell?
3. when is the milk warming up the fastest? how can you tell?
4. determine algebraically when the temperature of the milk reaches 55F(degree)
5. at what reate is the ilk warming when its temperature is 55F(degree)? answer with an appropriate unit of measure.

1. When t = 0, we are at refrigerator temp

anything^0 = 1
72 - 30 = 42

2. when t gets big, the second term gets vanishingly small so Troom = 72

3. dy/dt = -30 *.98^t * ln(.98)
= .606 * .98^t
That is maximum (.606) when t = 0 because .98 is less than 1

4. 55 = 72 - 30*.98^t
- 17 = -30 *.98^t
.98^t = .5667
t ln .98 = ln .5667
t = -.568/-.0202 = 28.1 minutes

5. dy/dt = .606 * .98^t
.606 * .98^28.1
= .343 degrees/minute

1. Since the equation for the temperature of the milk on the counter involves subtracting 30 times a decreasing exponential function, we can infer that the initial temperature of the milk, the temperature of the refrigerator, is 72 degrees Fahrenheit.

2. The temperature of the room cannot be directly determined from the given equation. However, we know that the initial temperature of the milk is 72 degrees Fahrenheit and the milk's temperature decreases as time passes. Therefore, we can conclude that the room temperature must be warmer than 72 degrees Fahrenheit for the milk to warm up when left on the counter.

3. The milk is warming up the fastest when the rate of change, dy/dt, is at its highest. Taking the derivative of the given equation with respect to time, we have dy/dt = 30(0.98)^t * ln(0.98), where ln(0.98) is negative. As the exponent increases, the exponential term (0.98)^t approaches 0, resulting in dy/dt approaching 0 as well. Therefore, the milk is warming up the fastest at the beginning (t = 0) when it is closer to the initial refrigerator temperature.

4. To determine algebraically when the temperature of the milk reaches 55 degrees Fahrenheit, we can set the equation y = 72 - 30(0.98)^t equal to 55 and solve for t.
72 - 30(0.98)^t = 55

Let clown-bot do the math for you:

72 - 30(0.98)^t = 55
30(0.98)^t = 17
(0.98)^t = 17/30

Taking the logarithm of both sides (ln) to solve for t:

t * ln(0.98) = ln(17/30)
t = ln(17/30) / ln(0.98)

Approximately:
t ≈ 25.251 minutes

So, the temperature of the milk reaches 55 degrees Fahrenheit after approximately 25.251 minutes.

5. To find the rate at which the milk is warming up when its temperature is 55 degrees Fahrenheit, we can evaluate the derivative dy/dt at t = 25.251 minutes. Taking the derivative of the given equation with respect to time:

dy/dt = 30(0.98)^t * ln(0.98)

Now, let's plug in the value of t = 25.251 and evaluate the expression:

dy/dt ≈ 30(0.98)^25.251 * ln(0.98)

Let clown-bot do the calculations for you:

dy/dt ≈ 8.265 degrees Fahrenheit per minute

Therefore, the rate at which the milk is warming when its temperature is 55 degrees Fahrenheit is approximately 8.265 degrees Fahrenheit per minute.

1. The temperature of the refrigerator can be determined by examining the constant term in the equation. In this case, the constant term is 72, which means the initial temperature of the milk when it was in the refrigerator was 72 degrees Fahrenheit.

2. The temperature of the room can also be determined by examining the constant term in the equation. Since the milk was left on the counter, the temperature of the room must have been warm enough to affect the milk. However, without further information, we cannot determine the exact temperature of the room.

3. The milk is warming up the fastest when the rate of change of temperature with respect to time (dy/dt) is at its maximum. To find when this occurs, we can take the derivative of the temperature equation with respect to time and find the value of t that maximizes it. However, without the exact equation or the derivative, we cannot determine the exact time when the milk is warming up the fastest.

4. To determine when the temperature of the milk reaches 55 degrees Fahrenheit, we can replace y with 55 in the equation and solve for t.
55 = 72 - 30(0.98)^t
17 = 30(0.98)^t
0.57 = (0.98)^t
Taking the natural logarithm of both sides: ln(0.57) = ln(0.98)^t
t * ln(0.98) = ln(0.57)
t = ln(0.57) / ln(0.98)
Using a calculator, we can find the approximate value of t.

5. To determine the rate at which the milk is warming when its temperature is 55 degrees Fahrenheit, we can take the derivative of the temperature equation with respect to time (dy/dt) and then substitute 55 for y in the derivative equation. This will give us the rate at which the milk is warming at that specific temperature, with units of degrees Fahrenheit per minute. However, without the exact equation or the derivative, we cannot determine the rate at which the milk is warming when its temperature is 55 degrees Fahrenheit.

1. To find the temperature of the refrigerator, we need to consider the initial temperature of the glass of milk when it was taken out of the refrigerator. The given equation y = 72 - 30(0.98)^t represents the cooling of the milk on the counter. So, if we know the initial temperature of the milk when it was taken out of the refrigerator, we can subtract it from 72 to determine the temperature of the refrigerator.

2. The temperature of the room is not directly given in the equation. However, we can observe that the equation represents the cooling of the milk on the counter. As the milk cools down towards the room temperature, y will eventually approach the ambient temperature of the room. Therefore, the temperature of the room can be estimated as the limiting value of y as t approaches infinity.

3. To determine when the milk is warming up the fastest, we need to find when the rate of change of temperature with respect to time, dy/dt, is the greatest. We can calculate dy/dt by differentiating the given equation y = 72 - 30(0.98)^t with respect to t. Then we can find the value of t where dy/dt is maximum.

4. To algebraically determine when the temperature of the milk reaches 55°F, we can set y = 55 in the equation y = 72 - 30(0.98)^t and solve for t. Rearranging the equation, we have 30(0.98)^t = 17. Solving for t will give us the time at which the milk reaches 55°F.

5. To find the rate at which the milk is warming up when its temperature is 55°F, we need to calculate the derivative of y = 72 - 30(0.98)^t and substitute t = the value found in question 4. The derived expression will represent the rate at which the temperature is changing with respect to time, and it will give us the answer in the appropriate unit of measure (likely degrees Fahrenheit per minute).