SOLUTION: In the summer a large pool evaporates water at 15% per day. Given 25,700 gallons of water, which function models the pool’s loss?
f(x) = 25700(0.15)^x
f(x) = 25700(0.85)^x
f(x)
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-> SOLUTION: In the summer a large pool evaporates water at 15% per day. Given 25,700 gallons of water, which function models the pool’s loss?
f(x) = 25700(0.15)^x
f(x) = 25700(0.85)^x
f(x)
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Question 965332: In the summer a large pool evaporates water at 15% per day. Given 25,700 gallons of water, which function models the pool’s loss?
f(x) = 25700(0.15)^x
f(x) = 25700(0.85)^x
f(x) =25700(1.85)^x
f(x) =25700(1.15)^x
You can put this solution on YOUR website! In the summer a large pool evaporates water at 15% per day. Given 25,700 gallons of water, which function models the pool’s loss?
f(x) = 25700(0.15)^x
f(x) = 25700(0.85)^x
f(x) =25700(1.85)^x
f(x) =25700(1.15)^x
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The 1st one, f(x) = 25700(0.15)^x models the pool's loss for the 1st day only.
None of the others model its loss.
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The 2nd one, f(x) = 25700(0.85)^x models the amount remaining in the pool - not the loss.
