Math Problem Solver

$$180 - 132=48$$

$$\begin{aligned} \mathtt{1}\overset{\mathtt{{\scriptscriptstyle 7}}}{\cancel{\mathtt{8}}}\overset{\mathtt{{\scriptscriptstyle 10}}}{\cancel{\mathtt{0}}}\\ \mathtt{-\phantom{0}}\mathtt{1}\mathtt{3}\mathtt{2}\\ \hline \mathtt{0}\mathtt{4}\mathtt{8} \end{aligned}$$

Borrow $10^{1}$, resulting in $7$ in the $10^{1}$ place, and $10$ in the $10^{0}$ place.

$8$ is the digit in the $10^{0}$ place. $10 \times 10^{0} - 2 \times 10^{0} = 8 \times 10^{0}$.

$4$ is the digit in the $10^{1}$ place. $7 \times 10^{1} - 3 \times 10^{1} = 4 \times 10^{1}$.

$0$ is the digit in the $10^{2}$ place. $1 \times 10^{2} - 1 \times 10^{2} = 0 \times 10^{2}$.