To represent the situation of a pool filling at a rate of 500 gallons per hour on a graph, you would plot time (in hours) on the x-axis and the amount of water in the pool (in gallons) on the y-axis.
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Axis setup:
- The x-axis will have time ranging from 0 to 10 hours.
- The y-axis will have the amount of water in the pool ranging from 0 to 2,500 gallons.
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Y=mx+b formula: In this situation, the relationship between time and gallons can be defined by a linear equation \( y = mx + b \).
- \(m\) is the rate at which the pool is filling (500 gallons/hour).
- \(b\) is the initial amount of water (0 gallons, since the pool starts empty).
Therefore, the equation is: \[ g = 500t \] where \(g\) is the gallons of water in the pool and \(t\) is time in hours.
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Data points: You can calculate some points to plot on the graph:
- At \(t = 0\): \(g = 500 \times 0 = 0\)
- At \(t = 1\): \(g = 500 \times 1 = 500\)
- At \(t = 2\): \(g = 500 \times 2 = 1000\)
- At \(t = 3\): \(g = 500 \times 3 = 1500\)
- At \(t = 4\): \(g = 500 \times 4 = 2000\)
- At \(t = 5\): \(g = 500 \times 5 = 2500\)
- At \(t = 6\) through \(t = 10\) will yield values greater than 2500, which you might not plot if you're focusing on the 0 to 2500 range.
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Graph shape: The graph will be a straight line starting at the origin (0,0) and increasing linearly with a slope representing the fill rate of 500 gallons/hour. As \(t\) increases, the value of \(g\) will rise steadily.
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Key points: Mark the points you calculated on the graph:
- (0, 0)
- (1, 500)
- (2, 1000)
- (3, 1500)
- (4, 2000)
- (5, 2500)
Connect these points with a straight line to visualize the increase in water in the pool over time.
Overall, the graph will be a diagonal line extending upward from the origin to the point (5, 2500), stopping there unless the y-axis continues.
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