Can someone explain how knowing a certain amount of ft (36ft^3) = volume 240in^3 then your able to find volume in^2?
I struggle with geometric formulas not familiar with understanding what I should already know
Sorry, but volume is always measured in (units)^3, such as in ft^3, and never in (units)^2, or ft^2. Please reread your question and then write it more clearly. Thank you.
Re-read the original question first, if it's available. Volume is the product of length times width times height if you're discussing a rectangular right prism (such as a shoebox).
ok say the information true but your given say 10 in^2 find the volume of in^2
Sorry, but again I just do not understand your phrasing. Given that the volume is 10 in^2, then the volume is 10 in^2. You see, this doesn't make sense.
Are you sure you're not talking about area as well as volume? And what is the shape whose volume you want?
would that make sense say area is 10 in^2 so find the volume?
Only if you also provide the SHAPE and IDENTITY of the figure in question. Is it a cube? Is it a right rectangular prism? Is it a cone? You see, I need more info.
Example. A cube has a surface area of 10 square inches. Find its volume. 1. Find the area of ONE side of the cube. It is (10/6) square inches. 2. Find the volume of the cube. It is equal to the CUBE of the length of one side, or [(10/6) square inches]^3. End of problem.
this is because a cube has SIX sides, all of the same dimensions. Find the length of one side; then cube that to obtain the volume of the cube.
You can't change the units from cubed to square
The dimension of a straight line is simply x. The area of a square is x^2. The volume of a cube of side length x is x^3. So, given the SURFACE AREA of a cube, you CAN find the surface area of ONE of its six sides. From this area, you can find the length of the length of the side of the square. From the length of the side of the square, you can find the volume of the cube using a different formula: The volume is V = x^3. I'm sorry, but I have to get off the 'Net. If you're stillw orking on this problem in 2 or 3 hours, I'll continue working with you then.
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