SOLUTION: Suppose that the water depth in a leaking water tank changes at a constant rate. The water in the tank was 8 feet deep after leaking for 1 hour and was 6 feet deep after leaking fo

Algebra ->  Volume -> SOLUTION: Suppose that the water depth in a leaking water tank changes at a constant rate. The water in the tank was 8 feet deep after leaking for 1 hour and was 6 feet deep after leaking fo      Log On


   

Question 930346: Suppose that the water depth in a leaking water tank changes at a constant rate. The water in the tank was 8 feet deep after leaking for 1 hour and was 6 feet deep after leaking for 3 hours.
How many feet deep was the water when the tank first began to leak?
Found 2 solutions by TimothyLamb, lezalonzo:
Answer by TimothyLamb(4379) About Me  (Show Source):
You can put this solution on YOUR website!
leak rate equation starting at 1-hour after leak began:
x = hours
y = feet
y = mx + b
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leak rate = m = slope = dy/dx
leak rate = (8-6)/(3-1)
leak rate = (2)/(2)
leak rate = 1 ft/hour
m = 1 ft/hour
---
y-intercept = b = water level 1-hour after leak began
b = 8 ft
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leak rate equation starting at 1-hour after leak began:
y = x + 8
---
water level when the tank first began to leak:
y = x + 8
y = (0 - 1) + 8
y = -1 + 8
y = 7 ft
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Answer by lezalonzo(2) About Me  (Show Source):
You can put this solution on YOUR website!
You can see that in 2 hrs we lost 2 feet. Between 1 hr. and 3 hrs it went from 8 feet to 6 feet, meaning in two hours the level decreased 2 feet, or 1 ft per hour.