Given the formula for the volume of a box, V=lwh , which equation best represents the remaining units when justifying your rearrangement to solve for the width?

To rearrange the equation \( V = lwh \) to solve for the width \( w \), you would divide both sides of the equation by \( lh \):

\[
w = \frac{V}{lh}
\]

Now, considering the units:

- Volume \( V \) is measured in cubic inches (\( \text{in}^3 \)).
- Length \( l \) and height \( h \) both have units of square inches (\( \text{in}^2 \)) since they are linear measurements.

Thus, when you compute \( \frac{V}{lh} \), you get:

\[
w = \frac{\text{in}^3}{\text{in}^2} = \text{in}
\]

This shows that the units of width \( w \) are linear inches (\( \text{in} \)).

Now, let's analyze the given options:

A) \( \frac{\text{in}^3}{\text{in}^2} = \frac{\text{in}^3}{\text{in}^2} \) - Correct but doesn't directly represent the final unit.

B) \( \frac{\text{in}^3}{\text{in}^2} = \text{in} \) - This directly shows what \( w \) equals, which represents the unit of width clearly.

C) \( \text{in}^3 = \text{in}^3 \) - This is true but doesn't help in showing the rearrangement for \( w \).

D) \( \text{in} = \text{in} \) - This is also true but does not convey the entire equation.

The option that best represents the remaining units when justifying your rearrangement to solve for the width is:

**B) \( \frac{\text{in}^3}{\text{in}^2} = \text{in} \)**.