Did I do this right?
A cage with a volume of 27 cubic meters, if the sides are in whole meters only and the width is 3 meters, what are the possible dimensions for the length and height?
27=3X3X3
or
27=9X1X3
Yes. Your answers are correct.
To solve this problem, you correctly started by expressing the volume of the cage as a product of its dimensions, using the equation:
27 = width x length x height
Given that the width is 3 meters, you then substituted 3 for the width in the equation:
27 = 3 x length x height
Next, you need to find the possible values for the length and height. Since the dimensions must be whole numbers, the length and height should be factors of 27 that result in a whole number when multiplied by 3.
You found two possible sets of dimensions:
1) length = 3 meters, height = 9 meters
This gives: 27 = 3 x 3 x 9
2) length = 9 meters, height = 1 meter
This gives: 27 = 3 x 9 x 1
Both sets of dimensions are valid solutions for a cage with a volume of 27 cubic meters, where the width is 3 meters.