A conical container can hold 120π cubic centimeters of water. The diameter of the base of the container is 12 centimeters. The height of the container is
centimeters. If its diameter and height were both doubled, the container's capacity would be
times its original capacity

cone volume=(1/3)hpir^2

d/2=r

d=12

d/2=12/2=6=r

V=120pi

120pi=(1/3)hpi6^2

120pi=(1/3)hpi36

120pi=12hpi

divide both sides by pi

120=12h

divide 12

10=h cm

if height and diameter is doubled aka height and radius are doubled

V=(1/3)(2h)pi(2r^2)

we want to see how it compares to V=(1/3)hpir^2

V=(1/3)(2h)pi(2r^2)

V=2(1/3)(h)pi4r^2

V=8((1/3)hpir^2)

the volume is increased by 8 times

blanks are

10

8

Step-by-step explanation:

well the 1 and -1 are always after the 0 and the 0 is inthe middle. s thats what i would guess. but i dont know what a slope of 2 is sorry

step-by-step explanation: