To solve for the width \( w \) in the formula \( V = lwh \), we can rearrange the equation as follows:
\[ w = \frac{V}{lh} \]
When justifying this rearrangement in terms of units, we start with the volume \( V \) which is in cubic inches \( \text{in}^3 \). The base area \( l \times h \) is in square inches \( \text{in}^2 \) (since both \( l \) and \( h \) have units of inches).
Thus, when we divide the volume by the area \( l \times h \), we have:
\[ w = \frac{\text{in}^3}{\text{in}^2} = \text{in} \]
Now we need to match this to the options provided:
- A. in. = in. (not applicable since it does not represent a ratio)
- B. in.^3 over in.^2 = in.^3 over in.^2 (too complex and does not clearly illustrate the simplification)
- C. in.^3 = in.^3 (not relevant for showing the units after rearrangement)
- D. in.^3 over in.^2 = in. (this correctly captures the process of dividing cubic inches by square inches resulting in inches)
Therefore, the best answer is:
D. in.^3 over in.^2 = in.