igormacedorrn
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Water flows from the bottom of a storage tank at a rate of r(t) = 400 − 8t liters per minute, where 0 ≤ t ≤ 50. Find the amount of water that flows from the tank during the first 30 minutes.
______ Liters.

Answer :

Ropafadzo

Answer:

Step-by-step explanation:

Just integrate r(t) with respect to t  

Integrate r(t) dt = 400t -(4t^2)/2 + C

                      = 400t -4t^2 + C

where C is a constant

then replace the above answer with 30 and 0, then do subtraction for both answers

hence the amount of water flows during first 30 mins is

=   {400*30 - 4*(30)^2 + C} - {400*0 - 4*(0)^2 + C}

=  12000 - 3600

=   8400 liters

abiolataiwo2015

The amount of water that flows from the tank during the first 30 minutes is 8,400Liters.

Based on the information definite integral will be use to find the amount of water.

Using this formula

[tex]\int_{a}^{b} r(t) dt\\\\\\Where:\\r(t)=400-8t\\\\\\ \int_{a}^{b} r(t) dt\ = \int_{0}^{30} (400-8t) dt\\\\\\\\ \int_{a}^{b} r(t) dt\ = (400t-8t^2/2)_{0}^{30}\\\\\\\\ = (400t-4t^2)_{0}^{30}\\\\\\\\=(400\times30)-4(30)^2\\\\\\=12,000-4(900)\\\\\\=12,000-3,600\\\\\\=8,400\ liters\\[/tex]

Inconclusion the amount of water that flows from the tank during the first 30 minutes is 8,400Liters.

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