Chapter 7 - Polynomial Equations and Factoring - 7.3 - Special Products of Polynomials - Exercises - Page 375: 30

$99\frac{8}{9}\approx99.89$

$10\frac{1}{3}\cdot9\frac{2}{3}=(10+\frac{1}{3})(10-\frac{1}{3})$ Using sum and difference product pattern $(a+b)(a-b)=a^{2}-b^{2}$ where $a=10$ and $b=\frac{1}{3}$, we have $(10+\frac{1}{3})(10-\frac{1}{3})=10^{2}-(\frac{1}{3})^{2}$ $=100-\frac{1}{9}=99\frac{8}{9}\approx99.89$