To justify the rearrangement of the formula for the volume of a box \( V = lwh \) in order to solve for width \( w \), we can start by rearranging the equation to isolate \( w \):
\[ w = \frac{V}{lh} \]
Since the volume \( V \) is measured in cubic inches (in³), length \( l \) and height \( h \) are measured in square inches (in²) or linear inches (in) depending on how you express the dimensions. To clarify the units:
- Volume \( V \) has the units of cubic inches: \( V \) is in \( in³ \).
- The product \( lh \) has the units of area, which is square inches: \( lh \) is in \( in² \).
When you divide the volume by the area \( lh \), the remaining units will be:
\[ \frac{in³}{in²} = in \]
This shows that when you calculate the width \( w \), it is in linear inches (in).
Therefore, the equation that best represents the remaining units when justifying your rearrangement to solve for width is:
\[ \frac{in^{3}}{in^{2}} = in \]
So, the correct answer is:
\(\frac{in^{3}}{in^{2}} = in.\)
1 answer