Let's break down the problem step-by-step:
1. Identify the Rate of Water Emptying:
We start by determining the rate at which the water is emptying from the tank. We are given the following data points:
Time [tex]\( t \)[/tex] (in minutes) | Quarts of water [tex]\( w \)[/tex]
--------------------------|------------------------
0 | 50
2 | 45
4 | 40
From these points, we can see that for every increase in 2 minutes, the water in the tank decreases by 5 quarts:
[tex]\[ \text{Rate} = \frac{(50 - 45)}{(2 - 0)} = \frac{5 \text{ quarts}}{2 \text{ minutes}} = 2.5 \text{ quarts per minute} \][/tex]
2. Formulate the Equation:
The water decreases at a constant rate. We can represent the amount of water left [tex]\( w \)[/tex] in the tank as a function of time [tex]\( t \)[/tex]. We know:
At [tex]\( t = 0 \)[/tex], [tex]\( w = 50 \)[/tex]
The rate of water emptying is 2.5 quarts per minute.
This gives us a linear equation:
[tex]\[ w = 50 - 2.5t \][/tex]
3. Verify the Equation:
Let’s verify the equation with another data point:
At [tex]\( t = 2 \)[/tex]:
[tex]\[ w = 50 - 2.5 \times 2 \][/tex]
[tex]\[ w = 50 - 5 = 45 \][/tex]
At [tex]\( t = 4 \)[/tex]:
[tex]\[ w = 50 - 2.5 \times 4 \][/tex]
[tex]\[ w = 50 - 10 = 40 \][/tex]
Both points match the given data, confirming our equation is correct.
4. Check the Solution for [tex]\( t = 30 \)[/tex] Minutes:
Now, we need to determine if there is a viable solution when [tex]\( t = 30 \)[/tex] minutes:
[tex]\[ w = 50 - 2.5 \times 30 \][/tex]
[tex]\[ w = 50 - 75 \][/tex]
[tex]\[ w = -25 \][/tex]
Since the amount of water cannot be negative, this means there is no viable solution when [tex]\( t = 30 \)[/tex] minutes.
Summary:
- The equation modeling the relationship is:
[tex]\[ w = 50 - 2.5t \][/tex]
- There is no viable solution when the time is 30 minutes.
