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r-ruslan [8.4K]
3 years ago
14

Water is pumped out of a holding tank at a rate of 6-6e^-0.13t liters/minute, where t is in minutes since the pump is started. I

f the holding tank contains 1000 liters of water when the pump is started, how much water does it hold one hour later?
Mathematics
1 answer:
astraxan [27]3 years ago
3 0
Procedure:

1) Integrate the function, from t =0 to t = 60 minutues to obtain the number of liters pumped out in the entire interval, and

2) Substract the result from the initial content of the tank (1000 liters).

Hands on:

Integral of (6 - 6e^-0.13t) dt  ]from t =0 to t = 60 min =

= 6t + 6 e^-0.13t / 0.13 = 6t + 46.1538 e^-0.13t ] from t =0 to t = 60 min =

6*60 + 46.1538 e^(-0.13*60) - 0 - 46.1538 = 360 + 0.01891 - 46.1538 = 313.865 liters

2) 1000 liters - 313.865 liters = 613.135 liters

Answer: 613.135 liters



 

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Find the dimensions of an open rectangular box with a square base that holds 2000 cubic cm and is constructed with the least bui
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<h3>The dimensions of the given rectangular box are:</h3><h3>L  =   15.874 cm  , B  =  15.874 cm   , H = 7.8937 cm</h3>

Step-by-step explanation:

Let us assume that the dimension of the square base = S x S

Let us assume the height of the rectangular base = H

So, the total area of the open rectangular box  

= Area of the base +  4 x ( Area of the adjacent faces)

=  S x S  +  4 ( S x H)   = S² +  4 SH   ..... (1)

Also, Area of the box  = S x S x H  =  S²H

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Substituting the value of H in (1), we get:

A = S^2 + 4 SH =  S^2 + 4 S(\frac{2000}{S^2}) =  S^2 + (\frac{8000}{S})\\\implies A  =  S^2 + (\frac{8000}{S})

Now, to minimize the area put :

(\frac{dA}{dS} ) = 0 \implies 2S  - \frac{8000}{S^2}  = 0\\\implies S^3 = 4000\\\implies S  = 15.874 \approx 16 cm

Putting the value of S  = 15.874 cm in the value of H , we get:

\implies H = \frac{2000}{S^2}  =  \frac{2000}{(15.874)^2} = 7.8937 cm

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