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KJourdan19
09.08.2021 •
Mathematics
Water flows into a right cylindrical shaped swimming pool with a circular base at a rate of 4 m33/min. The radius of the base is 3 m. How fast is the water level rising inside the swimming pool? The volume of a right cylinder with a circular base is V=πr2hV=πr2h, where rr is the radius of the base and hh is the height of the cylinder.
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Ответ:
The water level is rising at a rate of approximately 0.1415 meters per minute.
Step-by-step explanation:
Water is flowing into a right cylindrical-shaped swimming pool at a rate of 4 cubic meters per minute. The radius of the base is 3 meters.
And we want to determine the rate at which the water level of the pool is rising.
Recall that the volume of a cylinder is given by:
Since the radius is a constant 3 meters:
Water is flowing at a rate of 4 cubic meters per minute. In other words, dV/dt = 4 m³ / min.
Take the derivative of both sides with respect to t:
Implicitly differentiate:
The rate at which the water level is rising is represented by dh/dt. Substitute and solve:
Therefore:
In conclusion, the water level is rising at a rate of approximately 0.1415 meters per minute.
Ответ:
1.5×10^17
Step-by-step explanation:
it would be better to first multiply the two then divide the answer by 9.4×10^-3
I hope this helps