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artcher [175]
3 years ago
6

I'm clueless.

Mathematics
1 answer:
Alex Ar [27]3 years ago
6 0
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Step-by-step explanation:

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Preview Activity 3.5.1. A spherical balloon is being inflated at a constant rate of 20 cubic inches per second. How fast is the
ArbitrLikvidat [17]

Answer:

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

Because \frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi } the radius is changing more rapidly when the diameter is 12 inches.

Step-by-step explanation:

Let r be the radius, d the diameter, and V the volume of the spherical balloon.

We know \frac{dV}{dt}=20 \:{in^3/s} and we want to find \frac{dr}{dt}

The volume of a spherical balloon is given by

V=\frac{4}{3} \pi r^3

Taking the derivative with respect of time of both sides gives

\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}

We now substitute the values we know and we solve for \frac{dr}{dt}:

d=2r\\\\r=\frac{d}{2}

r=\frac{12}{2}=6

\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\\\\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^2} \\\\\frac{dr}{dt}=\frac{20}{4\pi(6)^2 } =\frac{5}{36\pi }\approx 0.044

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

When d = 16, r = 8 and \frac{dr}{dt} is:

\frac{dr}{dt}=\frac{20}{4\pi(8)^2}=\frac{5}{64\pi }\approx 0.025

The radius is increasing at a rate of approximately 0.025 in/s when the diameter is 16 inches.

Because \frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi } the radius is changing more rapidly when the diameter is 12 inches.

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Just divide 7225 by 8. :) I put it in fraction form but decimal form is just 903.125. I hope this helps you out. :)

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Find the median, first quartile, third quartile, and interquartile range of the data
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