SOLUTION: A tank contains gallons of water. Each day one-third of the water in the tank is removed and not replaced. How much water remains in the tank at the end of 6 days?
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-> SOLUTION: A tank contains gallons of water. Each day one-third of the water in the tank is removed and not replaced. How much water remains in the tank at the end of 6 days?
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Question 37792: A tank contains gallons of water. Each day one-third of the water in the tank is removed and not replaced. How much water remains in the tank at the end of 6 days?
i tried doing this a few times and im not getting the right answer. I think you write it like this: a=5832(1/3)^(6-1) right? Answer by fractalier(6550) (Show Source):
You can put this solution on YOUR website! Well, you're close.
If 1/3 is removed, 2/3 is what's left.
Therefore, if 5832 gallons were originally there,
x0 = 5832
x1 = 5832(2/3) (after 1 day)
x2 = 5832(2/3)(2/3) (after 2 days)
.
.
.
x6 = 5832(2/3)^6 (after 6 days) =
512
