To solve for the width \( w \) in the formula for the volume of a box \( V = lwh \), we can rearrange the equation to isolate \( w \):
-
Start with the volume formula: \[ V = lwh \]
-
To solve for \( w \), divide both sides by \( lh \): \[ w = \frac{V}{lh} \]
Now, let's analyze the units involved:
- The volume \( V \) is measured in cubic inches (in³).
- The length \( l \) and the height \( h \) are measured in square inches (in²) when you consider the area they contribute.
So, the units for \( V \) and \( lh \) in the equation for \( w \) are: \[ \text{Unit of } w = \frac{\text{in}^3}{\text{in}^2} \]
When you simplify \(\frac{\text{in}^3}{\text{in}^2}\):
- Subtract the exponents of the same base (the base here is inches): \[ \text{in}^{3-2} = \text{in}^1 = \text{in} \]
Thus, the width \( w \) will be in inches.
Therefore, the best representation of the units in this context for the equation that justifies the rearrangement to solve for width is: \[ \frac{\text{in}^3}{\text{in}^2} = \text{in} \]
This shows that when you divide cubic inches by square inches, you are left with inches, which is the appropriate unit for width.