A fish tank with a rectangular base has a volume of 3,360 cubic inches. The length and width of the tank are 14 inches and 12 inches, respectivily. Find the height, in inches, of the tank?
V=LxWxH
3360=14x12xH
3360=168xH
H=3360/168
H=2O
14x12=168 3360/168=20 which would be the height-- basic volume is lxhxw but you are missing h (height) so you work backwards
wAIT, so is 20 the height then?
Was I right?
yes.
yes, and much more eloquent
Eloquent?
Were you typing that about me?
I'm confused...I know what eloquent means, but it doesn't make sense how you used it.
Unless you were attempting some sort of joke. I don't know, but I'm confused, so if you clear that up for me I would appreciate it.
brie
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
What? I am confused
You are on the right track! Let's break down the problem further.
To find the possible dimensions for the length and height of the cage, we have to consider the volume equation:
V = length x width x height
In this case, the volume is given as 27 cubic meters, and the width is 3 meters.
So, we have the equation:
27 = length x 3 x height
Now, we need to find the possible dimensions for the length and height, given that they must be whole meters.
One possible solution is if we let length = 3 meters and height = 3 meters. Using these values in the equation, we get:
27 = 3 x 3 x 3
27 = 27
This solution satisfies the equation and the given conditions, so it is valid.
Another possible solution is if we let length = 9 meters and height = 1 meter. Using these values in the equation, we get:
27 = 9 x 3 x 1
27 = 27
Again, this solution satisfies the equation and the given conditions, so it is also valid.
Therefore, the possible dimensions for the length and height of the cage are:
1) Length = 3 meters, Height = 3 meters
2) Length = 9 meters, Height = 1 meter