Answer
a) 12 in
b) 50 mi/h
c) 3 mi
d) 0.2 in
Work Step by Step
We need to use table 1.5, as the author mentioned.
a) $\rm 30\;cm=\rm 30\;cm\cdot \left(\dfrac{4\;in}{10\;cm}\right)$
the unit of $\rm cm$ cancels and the unit of $\rm in$ remains.
Hence,
$$\boxed{30\;\rm cm=12\;in}$$
b) $\rm 25\;m/s=\rm 25\;m/s\cdot \left(\dfrac{2\;mi/h}{1\;m/s}\right)$
the unit of $\rm m/s$ cancels and the unit of $\rm mi/h$ remains.
Hence,
$$\boxed{25\;\rm m/s=50\;mi/h}$$
c) $\rm 5\;km= 5\;km\cdot \left(\dfrac{0.6\;mi}{1\;km}\right)$
the unit of $\rm km$ cancels and the unit of $\rm mi$ remains.
Hence,
$$\boxed{5\;\rm km=3\;mi}$$
d) $\rm 0.5\;cm=\rm 0.5\;cm\cdot \left(\dfrac{4\;in}{10\;cm}\right)$
the unit of $\rm cm$ cancels and the unit of $\rm in$ remains.
Hence,
$$\boxed{0.5\;\rm cm=0.2\;in}$$
