## Intermediate Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 1 - Section 1.1 - Algebraic Expressions, Real Numbers, and Interval Notation - Exercise Set: 83

#### Answer

$True$

#### Work Step by Step

$\frac{2}{5}$+$\frac{3}{5}$ =$\frac{5}{5}=1$, and 1 is an element of the natural numbers, which means the first statement is true. For the second statement, plug in 100. $9\times(100)^{2}\times(100+11)-9\times(100+11)\times(100)^{2}=0$ $90000\times111-999\times10000=0$ $9990000-9990000=0$ $0=0$, the second statement is also true. Since both of the statements are true, the answer is $True$.

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