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aev [14]
2 years ago
10

Area of a triangle using sine rule: workout area of ABCD please show working out

Mathematics
1 answer:
goldfiish [28.3K]2 years ago
5 0
Answer:
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A spherical balloon is being inflated at a rate of 3 cubic inches per second. Determine the change in the rate of the radius.How
Novay_Z [31]

Answer:

The rate rate of change of radius is \frac{1}{48\pi} inches per second when the diameter is 12 inches.

The radius is changing more rapidly when the diameter is 12 inches.

Step-by-step explanation:

Consider the provided information.

A spherical balloon is being inflated at a rate of 3 cubic inches per second.

The volume of sphere is V=\frac{4}{3}\pi r^3

Differentiate the above formula with respect to time.

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

Substitute the respective values in the above formula,

3=4\pi 6^2\frac{dr}{dt}

\frac{1}{48\pi}=\frac{dr}{dt}

The rate rate of change of radius is \frac{1}{48\pi} inches per second when the diameter is 12 inches.

When d=16

3=4\pi 8^2\frac{dr}{dt}

\frac{3}{256\pi}=\frac{dr}{dt}

Thus, the radius is changing more rapidly when the diameter is 12 inches.

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I think its 14 quarters and 50 pennies
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Please help me with 2 and 3.
Marianna [84]

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