To rearrange the equation \( V = lwh \) to solve for the width \( w \), you would divide both sides of the equation by \( lh \):
\[ w = \frac{V}{lh} \]
Now, considering the units:
- Volume \( V \) is measured in cubic inches (\( \text{in}^3 \)).
- Length \( l \) and height \( h \) both have units of square inches (\( \text{in}^2 \)) since they are linear measurements.
Thus, when you compute \( \frac{V}{lh} \), you get:
\[ w = \frac{\text{in}^3}{\text{in}^2} = \text{in} \]
This shows that the units of width \( w \) are linear inches (\( \text{in} \)).
Now, let's analyze the given options:
A) \( \frac{\text{in}^3}{\text{in}^2} = \frac{\text{in}^3}{\text{in}^2} \) - Correct but doesn't directly represent the final unit.
B) \( \frac{\text{in}^3}{\text{in}^2} = \text{in} \) - This directly shows what \( w \) equals, which represents the unit of width clearly.
C) \( \text{in}^3 = \text{in}^3 \) - This is true but doesn't help in showing the rearrangement for \( w \).
D) \( \text{in} = \text{in} \) - This is also true but does not convey the entire equation.
The option that best represents the remaining units when justifying your rearrangement to solve for the width is:
B) \( \frac{\text{in}^3}{\text{in}^2} = \text{in} \).
1 answer