The volume of a box (or rectangular prism) can be expressed with the formula:
\[ V = l \times w \times h \]
Where:
- \( V \) is the volume,
- \( l \) is the length,
- \( w \) is the width, and
- \( h \) is the height.
In your case, it seems you want to solve for the width \( w \). To do that, you can rearrange the formula for volume:
\[ w = \frac{V}{l \times h} \]
If you have a volume given in cubic inches (in³) and lengths or heights in square inches (in²), you want to ensure the units cancel correctly to get the width in inches.
Starting from the rearranged formula:
\[ w = \frac{V}{l \times h} \]
Substituting \( V \) in cubic inches and \( l \) and \( h \) in square inches gives:
\[ w = \frac{\text{in}^3}{\text{in}^2} = \text{in} \]
This equation shows that the units for width resolve correctly when you plug in the volumes and areas.
To put it concisely based on your wording, the statement:
\[ \frac{\text{in}^3}{\text{in}^2} = \text{in} \]
is the equation that best represents the justification for solving for the width \( w \).
So the final statement would be:
\[ w = \frac{V}{l \times h} \quad \text{(units: } \text{in} = \frac{\text{in}^3}{\text{in}^2}\text{)} \]