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If a brick which is .2 inches by .05 inches by .05 inches is
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26 Feb 2019, 10:27
26 Feb 2019, 10:27
Expert Reply
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Question Stats:
30% (03:23) correct
70% (01:18) wrong based on 10 sessions
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If a brick which is .2 inches by .05 inches by .05 inches is placed inside a right circular cylinder with a radius of .05 and a height of .2, what portion of the cylinder is not taken up by the brick?
A. \(.005 \pi - .005\)
B. \(.0005 \pi - .0005\)
C. \(05 \pi - .05\)
D. \(.0005 \pi\)
E. \(0\)
A. \(.005 \pi - .005\)
B. \(.0005 \pi - .0005\)
C. \(05 \pi - .05\)
D. \(.0005 \pi\)
E. \(0\)
Re: If a brick which is .2 inches by .05 inches by .05 inches is
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26 Feb 2019, 19:54
26 Feb 2019, 19:54
Would you please explain
Re: If a brick which is .2 inches by .05 inches by .05 inches is
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10 Mar 2019, 11:05
10 Mar 2019, 11:05
Expert Reply
First we'll find the volume of the cylinder, which is the area of the base times the height. A radius of 0.05 makes an area, πr², of π(0.05)² = 0.0025π. The height is 0.2, so the volume is 0.2(0.0025)π = 0.0005π.
The volume of the brick is the length times the width times the height, so 0.2 × 0.05 × 0.05 = 0.0005.
So the volume of the cylinder not taken up by the volume of the brick is 0.0005π - 0.0005.
The volume of the brick is the length times the width times the height, so 0.2 × 0.05 × 0.05 = 0.0005.
So the volume of the cylinder not taken up by the volume of the brick is 0.0005π - 0.0005.
