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Sliva [168]
3 years ago
12

The radius of a sphere is increasing at a rate of 2 mm/s. How fast is the volume increasing (in mm3/s) when the diameter is 40 m

m
Mathematics
1 answer:
daser333 [38]3 years ago
6 0

Answer:

\dfrac{dV}{dt}=502.65\ mm^3/s

Step-by-step explanation:

The volume of a sphere is given by :

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

The rate of change of volume means,

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

We have, \dfrac{dr}{dt}=2\ mm/s\ and\ r=40\ mm

So,

\dfrac{dV}{dt}=\dfrac{4}{3}\pi \times 3\times 20\times 2\\\\=502.65\ mm^3/s

So, the volume is increasing at the rate of 502.65\ mm^3/s.

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Simplify √(x^2-10x+25) if -5≤x<5
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\qquad\qquad\huge\underline{{\sf Answer}}

Let's simplify ~

\qquad \sf  \dashrightarrow \: \sqrt{ {x}^{2} - 10x + 25 }

\qquad \sf  \dashrightarrow \: \sqrt{  {x}^{2} - 5x - 5x   + 25  }

\qquad \sf  \dashrightarrow \: \sqrt{  {x}^{} (x- 5) - 5(x    - 5)  }

\qquad \sf  \dashrightarrow \: \sqrt{   (x- 5) (x    - 5)  }

\qquad \sf  \dashrightarrow \: \sqrt{   (x- 5)  {}^{2}  }

\qquad \sf  \dashrightarrow \: (x- 5)

value x lies between :

\qquad \sf  \dashrightarrow \: - 5 \leqslant x \leqslant 5

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\qquad \sf  \dashrightarrow \: - 10

if value of x is taken as 5

\qquad \sf  \dashrightarrow \:5 - 5

\qquad \sf  \dashrightarrow \:0

So, the possible values of the required expression lies between ~

\qquad \sf  \dashrightarrow \: - 10 \leqslant  \sqrt{ {x}^{2} - 10x + 25 }  < 0

I hope you understood the whole procedure. let me know if you have any doubts in given steps ~

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