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Amanda [17]
3 years ago
14

Can someone please help me with this I'm on a time limit

Mathematics
2 answers:
PilotLPTM [1.2K]3 years ago
5 0
2/7(2s + 3) =
(2/7)2s + (2/7)3 =

4/7s + 6/7
Elan Coil [88]3 years ago
4 0
2/7(2s + 3)

2/7*2/1s + 2/7*3/1

4/7s + 6/7

Final Answer: 4/7s + 6/7
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a spherical balloon is deflated at a rate of 256pi/3 cm^3/sec. at what rate is the radius of the balloon changing when the radiu
Hunter-Best [27]
What you need to look for is \frac { dr }{ dt } when r=8.

Now:

Volume\quad of\quad a\quad sphere:\\ \\ V=\frac { 4 }{ 3 } \pi { r }^{ 3 }

\therefore \quad \frac { dV }{ dr } =\frac { 4 }{ 3 } \pi \cdot 3{ r }^{ 2 }=4\pi { r }^{ 2 }\\ \\ \therefore \quad \frac { dr }{ dV } =\frac { 1 }{ \frac { dV }{ dr }  } =\frac { 1 }{ 4\pi { r }^{ 2 } }

And:\\ \\ \frac { dV }{ dt } =-\frac { 256\pi  }{ 3 } \\ \\ Therefore:\\ \\ \frac { dr }{ dt } =\frac { dr }{ dV } \cdot \frac { dV }{ dt }

\\ \\ =\frac { 1 }{ 4\pi { r }^{ 2 } } \cdot -\frac { 256\pi  }{ 3 } \\ \\ =-\frac { 256 }{ 12 } \cdot \frac { \pi  }{ \pi  } \cdot \frac { 1 }{ { r }^{ 2 } }

\\ \\ =-\frac { 64 }{ 3{ r }^{ 2 } } \\ \\ When\quad r=8,\\ \\ \frac { dr }{ dt } =-\frac { 64 }{ 3\cdot { 8 }^{ 2 } } =-\frac { 1 }{ 3 } \\ \\ Answer:\quad -\frac { 1 }{ 3 } \quad cm/sec
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3 years ago
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Vesna [10]

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See I don't know the Answer but I Need points to ask question so Sorry

Step-by-step explanation:

3 0
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GalinKa [24]

Answer:

  1. 1 1/14 seconds
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Step-by-step explanation:

We are being asked for times, so we can express the rate in terms of seconds per meter.

  (15 seconds)/(14 meters) = (15/14) seconds per meter = 1 1/14 seconds/meter

__

1) It will take Robot A 1 1/14 seconds to go 1 meter.

2) it will take Robot A 4 2/7 seconds to go 4 meters. (4 times as long)

5 0
2 years ago
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