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lora16 [44]
4 years ago
13

s:Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar and ^ to indicate

an exponent. The formula for the volume of a cylinder is V = πr2h. The volume of a cylinder is three times the volume of a cone with the same radius and height. If the volume of a cone with the same height as a cylinder equals the volume of the cylinder, the equation for the radius of cone R in terms of the radius of cylinder r is
Mathematics
2 answers:
anzhelika [568]4 years ago
8 0
Vco = 1/3×πr2h
Vcy = Vco --> πr^2×h = 1/3πR^2×h
h's cancel, as well as π's
r^2 = 1/3R^2 --> R^2 = 3r^2 -->
SR (R^2) = SR (3r^2)
R = SR(3)×r
Feliz [49]4 years ago
4 0

Answer:

R=r* \sqrt{3}

Step-by-step explanation:

The formula for the volume of a cylinder is V = \pi r^{2} h

The formula for the volume is = \frac{1}{3}\pi r^{2} h

The volume of a cylinder is three times the volume of a cone with the same radius and height.

If the volume of a cone with the same height as a cylinder equals the volume of the cylinder, then the equation for the radius of cone R in terms of the radius of cylinder r is :

Equating both radii:

r^{2} =\frac{1}{3} R^{2}

so, R^{2} =3r^{2}

R=\sqrt{3r^{2} }

R=r* \sqrt{3}

This is the final answer.

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Let <em>s(t)</em> be the amount of salt in the tank at time <em>t</em>. Then <em>s(0)</em> = 50 lb.

Salt flows into the tank at a rate of

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and flows out at a rate of

(2 gal/min) (<em>s(t)</em>/300 lb/gal) = <em>s(t)</em>/150 lb/min

so that the net rate at which salt is exchanged through the tank is

d<em>s(t)</em>/d<em>t</em> = 12 - <em>s(t)</em>/150 … (lb/min)

Solve for <em>s(t)</em>. The DE is separable, so we have

d<em>s</em>/d<em>t</em> = 12 - <em>s</em>/150

150 d<em>s</em>/d<em>t</em> = 1800 - <em>s</em>

150/(1800 - <em>s</em>) d<em>s</em> = d<em>t</em>

Integrate both sides to get

-150 ln|1800 - <em>s</em>| = <em>t</em> + <em>C</em>

Solve for <em>s</em> :

ln|1800 - <em>s</em>| = -<em>t</em>/150 + <em>C</em>

1800 - <em>s</em> = exp(-<em>t</em>/150 + <em>C </em>)

1800 - <em>s</em> = <em>C</em> exp(-<em>t</em>/150)

<em>s</em> = 1800 - <em>C</em> exp(-<em>t</em>/150)

Now given that <em>s(0)</em> = 50, we solve for <em>C</em> :

50 = 1800 - <em>C</em> exp(-0/150)

50 = 1800 - <em>C</em>

<em>C</em> = 1750

Then the amount of salt in the tank at any time <em>t</em> ≥ 0 is

<em>s(t)</em> = 1800 - 1750 exp(-<em>t</em>/150)

To find the time it takes for the tank to hold 100 lb of salt, solve for <em>t</em> in

100 = 1800 - 1750 exp(-<em>t</em>/150)

1700 = 1750 exp(-<em>t</em>/150)

34/35 = exp(-<em>t</em>/150)

ln(34/35) = -<em>t</em>/150

<em>t</em> = -150 ln(34/35) ≈ 4.348

So it would take approximately 4.348 minutes.

By the way, we didn't have to solve for <em>s(t)</em>, we could have instead stopped with

-150 ln|1800 - <em>s</em>| = <em>t</em> + <em>C</em>

Solve for <em>C</em> - this <em>C</em> <u>is not</u> the same as the one we found using the other method. <em>s(0)</em> = 50, so

-150 ln|1800 - 50| = 0 + <em>C</em>

<em>C</em> = -150 ln|1750|

==>   <em>t</em> = 150 ln(1750) - 150 ln|1800 - <em>s</em>|

Then <em>s(t)</em> = 100 lb when

<em>t</em> = 150 ln(1750) - 150 ln(1700)

<em>t</em> = 150 ln(1750/1700)

<em>t</em> = 150 ln(35/34)

<em>t</em> ≈ 4.348

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