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:
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Start with the original formula: \[ V = lwh \]
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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: \[ \text{Units of } w = \frac{\text{Units of } V}{\text{Units of } l \times \text{Units of } h} \]
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