To rearrange the equation for volume \( V = lwh \) to solve for the width \( w \), you would isolate \( w \) on one side of the equation. Here’s how you can do that:
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Start with the original equation: \[ V = lwh \]
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To solve for \( w \), divide both sides by \( lh \): \[ w = \frac{V}{lh} \]
In this rearrangement, you've expressed the width \( w \) in terms of the volume \( V \), length \( l \), and height \( h \).
When justifying the rearrangement, you may want to mention the units involved:
- The unit for volume \( V \) is typically cubic meters (m³).
- The unit for length \( l \) is meters (m).
- The unit for height \( h \) is also meters (m).
Thus, when you divide \( V \) by \( lh \), the units would be: \[ \text{Units of } w = \frac{\text{m}^3}{\text{m} \cdot \text{m}} = \frac{\text{m}^3}{\text{m}^2} = \text{m} \]
This confirms that the resulting unit for \( w \) is consistent with its physical interpretation as a width, which is measured in meters.