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 remov

Question 1139972: 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 14 cm? Round to the nearest tenth of a second.
Answer by ankor@dixie-net.com(22740) (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 14 cm? Round to the nearest tenth of a second.
:
0.0034t^2 − 0.52518t + 20 = 14
0.0034t^2 − 0.52518t + 20 - 14 = 0
0.0034t^2 − 0.52518t + 6 = 0
use the quadratic equation:

a = .0034; b = -.52518; c=6

Did math and got a reasonable solution of
x = 12.424 sec
:

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