To solve for the width \( w \) in the formula for the volume of a box \( V = lwh \), you can rearrange the equation as follows:
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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 with regard to units, you can analyze the units involved in the equation.
- Volume \( V \) is measured in cubic units (e.g., cubic meters, \( m^3 \)).
- Length \( l \) and height \( h \) are measured in linear units (e.g., meters, \( m \)).
Thus, the units on both sides of the original equation \( V = lwh \) can be represented as: \[ [\text{Volume}] = [\text{Length}] \times [\text{Width}] \times [\text{Height}] \] This means: \[ [m^3] = [m] \times [w] \times [m] \]
When you rearrange to solve for width \( w \): \[ [w] = \frac{[m^3]}{[m] \times [m]} = \frac{[m^3]}{[m^2]} = [m] \]
So, the equation that best represents the remaining units after rearranging to solve for the width is: \[ [w] = \frac{[V]}{[l] \times [h]} \]
This shows that the width \( w \) has the proper unit of length.