Answer
$525_{\text{nine}}$
Work Step by Step
The place values of base nine are: $...,9^3, 9^2, 9^1, 1$.
The place values that are less than or equal to $428$ are $9^2$, $9^1$, and $1$
Divide $428$ by $9^2$ or $81$ to obtain:
$428 \div 81 = 5$ remainder $23$
Divide the remainder $23$ by $9^1$ or $9$ to obtain:
$23 \div 9 = 2$ remainder $5$
Divide the remainder $5$ by $1$ to obtain:
$5\div 1 = 5$ remainder $0$
Thus,
$428=(5 \times 9^2) + (2 \times 9^1) + (5 \times 1)
\\428=525_{\text{nine}}$