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jeka57 [31]
3 years ago
6

What is the domain of y= log_4(x+3)? all real numbers less than -3 all real numbers greater than –3 all real numbers less than 3

all real numbers greater than 3
Mathematics
1 answer:
antiseptic1488 [7]3 years ago
7 0

Step-by-step answer:

The domain of log functions (any legitimate base) requires that the argument evaluates to a positive real number.

For example, the domain of log(4x) will remain positive when x>0.

The domain of log_4(x+3) requires that x+3 >0, i.e. x>-3.

Finally, the domain of log_2(x-3) is such that x-3>0, or x>3.

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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
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Answer:

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.

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Recall that the volume of a cylinder is given by:

\displaystyle V = \pi r^ 2h

Since the radius is a constant 3 meters:

\displaystyle V = 9\pi h

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 <em>t: </em>

<em />\displaystyle \frac{d}{dt}\left[ V\right] = \frac{d}{dt}\left[ 9\pi h\right]<em />

Implicitly differentiate:

\displaystyle \frac{dV}{dt} = 9\pi \frac{dh}{dt}

The rate at which the water level is rising is represented by dh/dt. Substitute and solve:

\displaystyle \left(4 \right) = 9\pi \frac{dh}{dt}

Therefore:

\displaystyle \frac{dh}{dt} = \frac{4}{9\pi} \approx 0.1415\text{ m/min}

In conclusion, the water level is rising at a rate of approximately 0.1415 meters per minute.

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