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Brums [2.3K]
3 years ago
10

A spherical balloon is inflating with helium at a rate of 64 pi StartFraction ft cubed Over min EndFraction . How fast is the​ b

alloon's radius increasing at the instant the radius is 2 ​ft?
Mathematics
1 answer:
olga2289 [7]3 years ago
5 0

Answer:

The radius of the balloon is increasing at a rate of 4 feet per minute.

Step-by-step explanation:

We are given the following in the question:

\dfrac{dV}{dt} = 64\pi \dfrac{\text{ ft}^3}{\text{min}}

Volume of sphere is given by

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

where r is the radius of the balloon.

Instant radius, r = 2 ft

Rate of change of volume =

\dfrac{dV}{dt} = \dfrac{d}{dt}(\dfrac{4}{3}\pi r^3)\\\\\dfrac{dV}{dt} =4\pi r^2\dfrac{dr}{dt}

Putting values, we get,

64\pi = 4\pi (2)^2\dfrac{dr}{dt}\\\\\Rightarrow \dfrac{dr}{dt} = 4

Thus, the radius of the balloon is increasing at a rate of 4 feet per minute.

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1) \frac{(-2)^{-5}}{(-2)^{-10}}=-32

2) 2^{-1}.2^{-4} = \frac{1}{32}

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Step-by-step explanation:

1) \frac{(-2)^{-5}}{(-2)^{-10}}

Solving using exponent rule: a^{-m}=\frac{1}{a^m}

\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32

So, \frac{(-2)^{-5}}{(-2)^{-10}}=-32

2) 2^{-1}.2^{-4}

Using the exponent rule: a^m.a^n=a^{m+n}

We have:

2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}

We also know that: a^{-m}=\frac{1}{a^m}

Using this rule:

2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}

So, 2^{-1}.2^{-4} = \frac{1}{32}

3) (-\frac{1}{2} )^3.(-\frac{1}{2} )^2

Solving:

(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}

So, (-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}

4) \frac{2}{2^{-4}}

We know that: a^{-m}=\frac{1}{a^m}

\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32

So, \frac{2}{2^{-4}} = 32

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