a cube has a volume of 1728 cu mm. If the allowable error in the edge of a cube is 0.04 mm,compute the allowable error in the volume of the cube?
The edge of a cube with a volume of 1728mm^3 = 12mm
So....the allowable error, E, would be
(12 - .04)^3 < E < (12 + .04)^3
1710.77mm^3 < E < 1745.34mm^3
