SOLUTION: The depth d of a liquid in a bottle with a hole of area 0.5 cm2 in its side can be approximated by
d = 0.0034t2 − 0.52518t + 20,
where t is the time since a stopper was r
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-> SOLUTION: The depth d of a liquid in a bottle with a hole of area 0.5 cm2 in its side can be approximated by
d = 0.0034t2 − 0.52518t + 20,
where t is the time since a stopper was r
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Question 1003409: The depth d of a liquid in a bottle with a hole of area 0.5 cm2 in its side can be approximated by
d = 0.0034t2 − 0.52518t + 20,
where t is the time since a stopper was removed from the hole. When will the depth be 18 cm? Round to the nearest tenth of a second. Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! The depth d of a liquid in a bottle with a hole of area 0.5 cm2 in its side can be approximated by
d = 0.0034t2 - 0.52518t + 20,
where t is the time since a stopper was removed from the hole. When will the depth be 18 cm? Round to the nearest tenth of a second.
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Use ^ (Shift 6) for exponents.
d = 0.0034t^2 - 0.52518t + 20
d = 0.0034t^2 - 0.52518t + 20 = 18
Solve for t
