SOLUTION: PLEASE HELP. A CUBE OF ICE 3 FT BY 3 FT BY 3 FT. THE ICE MELTS UNTIL IT BECOMES A CUBE WHICH IS ONE HALF AS HEAVY AS THE ORIGINAL. FIND THE DIMENSION OF A NEW CUBE (DENSITY OF ICE=
Algebra ->
Test
-> SOLUTION: PLEASE HELP. A CUBE OF ICE 3 FT BY 3 FT BY 3 FT. THE ICE MELTS UNTIL IT BECOMES A CUBE WHICH IS ONE HALF AS HEAVY AS THE ORIGINAL. FIND THE DIMENSION OF A NEW CUBE (DENSITY OF ICE=
Log On
Question 868882: PLEASE HELP. A CUBE OF ICE 3 FT BY 3 FT BY 3 FT. THE ICE MELTS UNTIL IT BECOMES A CUBE WHICH IS ONE HALF AS HEAVY AS THE ORIGINAL. FIND THE DIMENSION OF A NEW CUBE (DENSITY OF ICE=62LB/CUB.FT. THANK YOU Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! volume of cube = 3 * 3 * 3 = 27 cubic feet
ice weighs 57.4 pounds per cubic foot, therefore
our cube of ice weighs 57.4 * 27 = 1549.8 pounds
now 1/2 of that is 1549.8 / 2 = 774.9 pounds
x^3 * 57.4 = 774.9
x^3 = 774.9 / 57.4 = 13.5
x = 13.5 ^ (1/3) = 2.38 feet
