SOLUTION: Using Torrecelli's Principle, it can be shown that the depth d of a liquid in a bottle with a hole of area 0.5 cm^2 in its side can be approximated by d=0.0034t^2-0.52518t + 20, wh

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Using Torrecelli's Principle, it can be shown that the depth d of a liquid in a bottle with a hole of area 0.5 cm^2 in its side can be approximated by d=0.0034t^2-0.52518t + 20, wh      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 626697: Using Torrecelli's Principle, it can be shown that the depth d of a liquid in a bottle with a hole of area 0.5 cm^2 in its side can be approximated by d=0.0034t^2-0.52518t + 20, where t is the time since a stopper was removed from the hole. When will the depth be 10cm? Round to the nearest tenth of a second.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Using Torrecelli's Principle, it can be shown that the depth d of a liquid
in a bottle with a hole of area 0.5 cm^2 in its side can be approximated by
d=0.0034t^2-0.52518t + 20,
where t is the time since a stopper was removed from the hole.
When will the depth be 10cm? Round to the nearest tenth of a second.
:
d = 10 cm find t
0.0034t^2-0.52518t + 20 = 10
:
0.0034t^2-0.52518t + 20 - 10 = 0
a quadratic equation
0.0034t^2-0.52518t + 10 = 0
use the quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
in this equation: x=t; a=.0034; b=-.52518; c=10
:
After a lot of tedious math, two solutions t=22.2 sec, t=132.2 sec