Question

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

Answer 1

To rearrange the formula for the volume of a box, given by \( V = lwh \), to solve for the width \( w \), we can follow these steps:

1. Start with the original formula:
\[
V = lwh
\]

2. To isolate \( w \), divide both sides by \( lh \):
\[
w = \frac{V}{lh}
\]

Now, when justifying this rearrangement in terms of units, you need to consider the units involved:

- Let the units of volume \( V \) be in cubic units (e.g., cubic meters, \( m^3 \)).
- Let the length \( l \) and height \( h \) be in linear units (e.g., meters, \( m \)).

When you divide \( V \) by \( l \) and \( h \), the units remaining are:

- The unit for volume \( (length^3) \) divided by the unit for area \( (length^2) \):
\[
\text{Units of } w = \frac{\text{units of } V}{\text{units of } l \cdot \text{units of } h} = \frac{length^3}{length^2} = length
\]

Thus, the units for width \( w \) will be in linear units, justifying that \( w \) retains proper dimensional consistency after rearrangement.

So, the best equation representing the remaining units when justifying the rearrangement to solve for the width would be:
\[
\text{Units of } w = \frac{\text{Units of } V}{\text{Units of } l \times \text{Units of } h}
\]