What is the volume of a box that will hold exactly 567 cubes with 1/3 inch sides? I am somewhat confused on how to solve this. Please help.
each cube has a volume of (1/3)^3 = 1/27 in^3
so, the total volume is 567/27 = 21 in^3
21
it's 21
21 in^3
idk but i need help in iready
To find the volume of a box that can hold a certain number of smaller cubes, you need to know the size of the smaller cubes.
In this case, the cubes have sides of 1/3 inch. To find the volume of a cube, you cube the length of one side. So the volume of each cube is (1/3)^3 = 1/27 cubic inches.
To find the volume of the box, you multiply the volume of each cube by the number of cubes it can hold. In this case, the box can hold 567 cubes.
Therefore, the volume of the box is (1/27 cubic inches/cube) × (567 cubes) = 567/27 cubic inches.
To simplify the calculation, we can write 567/27 as a mixed number: 567 ÷ 27 = 21 with a remainder of 0. So the volume of the box is 21 cubic inches.
Therefore, a box that can hold exactly 567 cubes with 1/3 inch sides has a volume of 21 cubic inches.