+380 A pool is being drained at a constant rate. The amount of water is a function of the number of minutes the pool has been draining, as shown in the table.
| Time (min) | 12 | 20 | 50 |
| Volume (gal) | 4962 | 4754 | 3974 |
Write an equation in slope intercept from that represents a function. Then find the amount of water in the pool after two and a half hours.
+129852 Using the first two values in the table, we can first find the slope to write an equation of a line, so we have
[4754 - 4962 ] / [20 - 12 ] = -26
So we have
y - 4754 = -26(x - 20)
y - 4754 = -26x + 520
y = -26x + 5274
Let's confirm that the amount after 50 minutes is correct
y = -26(50) + 5274 = 3974
So after 2 + 1/2 hrs (150 min) we have
y = -26(150) + 5274 = 1374 gallons
+129852 Using the first two values in the table, we can first find the slope to write an equation of a line, so we have
[4754 - 4962 ] / [20 - 12 ] = -26
So we have
y - 4754 = -26(x - 20)
y - 4754 = -26x + 520
y = -26x + 5274
Let's confirm that the amount after 50 minutes is correct
y = -26(50) + 5274 = 3974
So after 2 + 1/2 hrs (150 min) we have
y = -26(150) + 5274 = 1374 gallons
