After painting his porch, jamil has \dfrac14 4 1 start fraction, 1, divided by, 4, end fractionof a can of paint remaining. the can has a radius of 888 cm and a height of 202020 cm. he wants to pour the remaining paint into a smaller can for storage. the smaller can has a radius of 555 cm. what does the height of the smaller can need to be to hold all of the paint?

The volume of the remaining paint is:
 V1 = (1/4) * (pi) * (r ^ 2) * (h)
 Where,
 r: radio
 h: height
 Substituting values we have:
 V1 = (1/4) * (3.14) * (8 ^ 2) * (20)
 V1 = 1004.8 cm ^ 3
 The smallest can volume is:
 V2 = pi * r ^ 2 * h
 Substituting values:
 V2 = (3.14) * 5 ^ 2 * h
 Matching the volumes:
 V2 = V1
 (3.14) * 5 ^ 2 * h = 1004.8
 Clearing h:
 h = (1004.8) / ((3.14) * 5 ^ 2)
 h = 12.8 cm
 
 The height of the smallest part of the paint is:
 h = 12.8 cm

Y’all got to do it all over