Question

Given the formula for the volume of a box, $V=lwh$ which equation best represents the
remaining units when Justifying your rearrangement to solve for the width?
$\frac {in^{3}}{in^{2}}=in.$
$in.=in.$
$\frac {in^{3}}{in^{2}}=\frac {in^{3}}{in^{2}}$
$in.^{3}=in.^{3}$

Answer 1

To justify the rearrangement of the formula for the volume of a box \( V = lwh \) in order to solve for width \( w \), we can start by rearranging the equation to isolate \( w \):

\[
w = \frac{V}{lh}
\]

Since the volume \( V \) is measured in cubic inches (in³), length \( l \) and height \( h \) are measured in square inches (in²) or linear inches (in) depending on how you express the dimensions. To clarify the units:

1. Volume \( V \) has the units of cubic inches: \( V \) is in \( in³ \).
2. The product \( lh \) has the units of area, which is square inches: \( lh \) is in \( in² \).

When you divide the volume by the area \( lh \), the remaining units will be:

\[
\frac{in³}{in²} = in
\]

This shows that when you calculate the width \( w \), it is in linear inches (in).

Therefore, the equation that best represents the remaining units when justifying your rearrangement to solve for width is:

\[
\frac{in^{3}}{in^{2}} = in
\]

So, the correct answer is:

\(\frac{in^{3}}{in^{2}} = in.\)