To solve for the width \( w \) in the volume formula \( V = lwh \), we can rearrange the equation as follows:
-
Start with the original equation: \[ V = lwh \]
-
To solve for \( w \), divide both sides of the equation by \( lh \): \[ w = \frac{V}{lh} \]
Now, regarding the units, the volume \( V \) is typically measured in cubic units (e.g., cubic meters, cubic centimeters), while the length \( l \) and height \( h \) are measured in square units (e.g., square meters, square centimeters).
So when you state the equation for \( w \): \[ w = \frac{V}{lh} \] The units would be:
- \( V \) has units of \( \text{unit}^3 \), which we can denote as \( \text{units}^3 \).
- \( lh \) has units of \( \text{unit}^2 \), which we can denote as \( \text{units}^2 \).
Thus, when rearranging for width \( w \), the units are as follows: \[ w = \frac{\text{units}^3}{\text{units}^2} = \text{units} \]
Therefore, the remaining units when justifying the rearrangement to solve for the width will be the same as the units of length, which can be expressed simply as \( \text{units} \). This indicates that the width has dimensions consistent with linear measurements.