SOLUTION: Hi Please help me. 1. 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 sin

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: Hi Please help me. 1. 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 sin      Log On

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Question 1142489: Hi Please help me.
1. 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 1 cm? Round to the nearest tenth of a second.
2. A small pipe can fill a tank in 9 min more time than it takes a larger pipe to fill the same tank. Working together, the pipes can fill the tank in 6 min. How long would it take each pipe, working alone, to fill the tank?
3. A car travels 238 mi. A second car, traveling 9 mph faster than the first car, makes the same trip in 1 h less time. Find the speed of each car.
4. A homeowner hires a mason to lay a brick border around a rectangular patio that measures 8 ft by 10 ft. If the total area of the patio and border is 168 ft2, what is the width of the border?

Found 2 solutions by ikleyn, Alan3354:
Answer by ikleyn(52794) About Me  (Show Source):
You can put this solution on YOUR website!
.
1.   d = 0.0034t2 − 0.52518t + 20.


     To find " t " when the depth "d" is equal to 1 cm, you need to solve the quadratic equation

         0.0034t2 − 0.52518t + 20 = 1.

     To solve this equation, first write it in the standard form and then use the quadratic formula.


     If you don't know how to use the quadratic formula, learn it from the lessons

          - Introduction into Quadratic Equations

          - PROOF of quadratic formula by completing the square

     in this site.



2.  Let " t " be the time for the larger pipe to fill the tank working alone.

    Then the time to fill the tank for the smaller pipe is (t+9) minutes.


    Now, in one minute the larger pipe fills   1%2Ft     of the tank volume;

                       the smaller pipe fills  1%2F%28t%2B9%29 of the tank volume.


    Working together, the two pipes fill  1%2Ft + 1%2F%28t%2B9%29  of the tank volume per minute.


     From the other side,  the two pipes, working together, fill  1%2F6  of the tank volume, according to the condition.


    It gives you an equation

        1%2Ft + 1%2F%28t%2B9%29 = 1%2F6.


    To solve it, multiply both sides by 6t*(t+9). You will get

        6(t+9) + 6t = t*(t+9).


     Transform this equation to the standard form


         6t + 54 + 6t = t^2 + 9t

         t^2 - 3t - 54 = 0


     Now you can solve it by factoring

         (t-9)*(t+6) = 0.


     It has two roots t= -6  and  t= 9.

     Of these two roots, only positive t= 9 is the valid solution to the problem.


     ANSWER.  Large pipe can fill the tank in 9 minutes, working alone.

              Small pipe can fill the tank in (9+9) = 18 minutes.


     CHECK.  1%2F18 + 1%2F9 = 1%2F18+%2B+2%2F18 = 3%2F18 = 1%2F6.    ! Correct !

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Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
1. 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 = 0.0034t^2 − 0.52518t + 20, where t is the time since a stopper was removed from the hole. When will the depth be 1 cm? Round to the nearest tenth of a second.
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d = 0.0034t^2 − 0.52518t + 20 = 1
0.0034t^2 − 0.52518t + 19 = 0
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 0.0034x%5E2%2B-0.52518x%2B19+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-0.52518%29%5E2-4%2A0.0034%2A19=0.0174140324.

Discriminant d=0.0174140324 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--0.52518%2B-sqrt%28+0.0174140324+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-0.52518%29%2Bsqrt%28+0.0174140324+%29%29%2F2%5C0.0034+=+96.6385644909046
x%5B2%5D+=+%28-%28-0.52518%29-sqrt%28+0.0174140324+%29%29%2F2%5C0.0034+=+57.8261413914483

Quadratic expression 0.0034x%5E2%2B-0.52518x%2B19 can be factored:
0.0034x%5E2%2B-0.52518x%2B19+=+%28x-96.6385644909046%29%2A%28x-57.8261413914483%29
Again, the answer is: 96.6385644909046, 57.8261413914483. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+0.0034%2Ax%5E2%2B-0.52518%2Ax%2B19+%29

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t = the smaller solution.
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