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assuming that the pipes are all the same length, then you can work it out as follows:

The pipes are cylindrical, and the volume of a cylinder is equal to pi multiplied by the square of it's radius multiplied by the length of the pipe - or:

v = πr2h

we can compare the two then by saying the volume of our two smaller pipes is:

v = πr12h + πr22h

r1 = 3 and r2 = 4, therefore

v = π32h + π42h

v = π9h + π16h

v = π25h

and then we can say that the volume of our biggest pipe is:

v = πr2h

r = 5, therefore

v = π52h

v = π25h

therefore, the two small pipes carry the same amount as the large pipe

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Q: Which will carry the most water two pipes one with a 3 Cm radius and the other with a 4 Cm radius or one pipe with a 5 Cm radius?
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