Math Problem Solver

$$50 + 90=140$$

$$\begin{aligned} \overset{\mathtt{1}}{\phantom{0}}\mathtt{5}\mathtt{0}\\ \mathtt{+\phantom{0}}\phantom{0}\mathtt{9}\mathtt{0}\\ \hline \mathtt{1}\mathtt{4}\mathtt{0} \end{aligned}$$

$0$ is the digit in the $10^{0}$ place. $0 \times 10^{0} + 0 \times 10^{0} = 0 \times 10^{0}$.

$4$ is the digit in the $10^{1}$ place. $5 \times 10^{1} + 9 \times 10^{1} = 14 \times 10^{1} = 1 \times 10^{2} + 4 \times 10^{1}$.$1 \times 10^{2}$ carries to the $10^{2}$ place.

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