Answer :
To determine the slope of the line that represents the volume of water in the pool over time, we use the provided data points:
[tex]\[ \begin{tabular}{|c|c|} \hline Time (min) & Water in Pool (gal) \\ \hline 0 & 50 \\ \hline 1 & 44 \\ \hline 2 & 38 \\ \hline 3 & 32 \\ \hline 4 & 26 \\ \hline 5 & 20 \\ \hline \end{tabular} \][/tex]
We can use the formula for the slope [tex]\( m \)[/tex], which is the rate of change of the dependent variable (water in pool) with respect to the independent variable (time):
[tex]\[ m = \frac{\Delta y}{\Delta x} = \frac{\text{change in water}}{\text{change in time}} \][/tex]
To calculate the slope, we take any two points from the table. For example, let's use the first two points: [tex]\((0, 50)\)[/tex] and [tex]\((1, 44)\)[/tex].
[tex]\[ \Delta y = 44 - 50 = -6 \][/tex]
[tex]\[ \Delta x = 1 - 0 = 1 \][/tex]
Hence, the slope is:
[tex]\[ m = \frac{\Delta y}{\Delta x} = \frac{-6}{1} = -6 \][/tex]
We can validate the consistency of this slope calculation by checking other consecutive data points:
[tex]\[ \begin{aligned} (1, 44) \text{ and } (2, 38): \quad & \Delta y = 38 - 44 = -6 \\ & \Delta x = 2 - 1 = 1 \\ & \frac{\Delta y}{\Delta x} = \frac{-6}{1} = -6 \end{aligned} \][/tex]
[tex]\[ \begin{aligned} (2, 38) \text{ and } (3, 32): \quad & \Delta y = 32 - 38 = -6 \\ & \Delta x = 3 - 2 = 1 \\ & \frac{\Delta y}{\Delta x} = \frac{-6}{1} = -6 \end{aligned} \][/tex]
[tex]\[ \begin{aligned} (3, 32) \text{ and } (4, 26): \quad & \Delta y = 26 - 32 = -6 \\ & \Delta x = 4 - 3 = 1 \\ & \frac{\Delta y}{\Delta x} = \frac{-6}{1} = -6 \end{aligned} \][/tex]
[tex]\[ \begin{aligned} (4, 26) \text{ and } (5, 20): \quad & \Delta y = 20 - 26 = -6 \\ & \Delta x = 5 - 4 = 1 \\ & \frac{\Delta y}{\Delta x} = \frac{-6}{1} = -6 \end{aligned} \][/tex]
The slope remains consistent at [tex]\( -6 \)[/tex] for each pair of points.
The term that describes the slope of the line is "negative" because as time increases, the volume of water in the pool decreases, giving a negative rate of change. Therefore, the slope of the line graphed to represent the volume of water in a pool over time is negative.
[tex]\[ \begin{tabular}{|c|c|} \hline Time (min) & Water in Pool (gal) \\ \hline 0 & 50 \\ \hline 1 & 44 \\ \hline 2 & 38 \\ \hline 3 & 32 \\ \hline 4 & 26 \\ \hline 5 & 20 \\ \hline \end{tabular} \][/tex]
We can use the formula for the slope [tex]\( m \)[/tex], which is the rate of change of the dependent variable (water in pool) with respect to the independent variable (time):
[tex]\[ m = \frac{\Delta y}{\Delta x} = \frac{\text{change in water}}{\text{change in time}} \][/tex]
To calculate the slope, we take any two points from the table. For example, let's use the first two points: [tex]\((0, 50)\)[/tex] and [tex]\((1, 44)\)[/tex].
[tex]\[ \Delta y = 44 - 50 = -6 \][/tex]
[tex]\[ \Delta x = 1 - 0 = 1 \][/tex]
Hence, the slope is:
[tex]\[ m = \frac{\Delta y}{\Delta x} = \frac{-6}{1} = -6 \][/tex]
We can validate the consistency of this slope calculation by checking other consecutive data points:
[tex]\[ \begin{aligned} (1, 44) \text{ and } (2, 38): \quad & \Delta y = 38 - 44 = -6 \\ & \Delta x = 2 - 1 = 1 \\ & \frac{\Delta y}{\Delta x} = \frac{-6}{1} = -6 \end{aligned} \][/tex]
[tex]\[ \begin{aligned} (2, 38) \text{ and } (3, 32): \quad & \Delta y = 32 - 38 = -6 \\ & \Delta x = 3 - 2 = 1 \\ & \frac{\Delta y}{\Delta x} = \frac{-6}{1} = -6 \end{aligned} \][/tex]
[tex]\[ \begin{aligned} (3, 32) \text{ and } (4, 26): \quad & \Delta y = 26 - 32 = -6 \\ & \Delta x = 4 - 3 = 1 \\ & \frac{\Delta y}{\Delta x} = \frac{-6}{1} = -6 \end{aligned} \][/tex]
[tex]\[ \begin{aligned} (4, 26) \text{ and } (5, 20): \quad & \Delta y = 20 - 26 = -6 \\ & \Delta x = 5 - 4 = 1 \\ & \frac{\Delta y}{\Delta x} = \frac{-6}{1} = -6 \end{aligned} \][/tex]
The slope remains consistent at [tex]\( -6 \)[/tex] for each pair of points.
The term that describes the slope of the line is "negative" because as time increases, the volume of water in the pool decreases, giving a negative rate of change. Therefore, the slope of the line graphed to represent the volume of water in a pool over time is negative.
