## Algebra 2 Common Core

$|d-0.11885|\le0.00015$
Half-way between the given values, $0.1187\text{ and } 0.1190 ,$ is \begin{array}{l}\require{cancel} \dfrac{0.1187+0.1190}{2} \\\\= \dfrac{0.2377}{2} \\\\= 0.11885 .\end{array} The distance from $0.11885$ to $0.1187$ and $0.1190$ is $0.00015 .$ Since $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), then the given inequality, $0.1187\le d \le0.1190 ,$ is equivalent to \begin{array}{l}\require{cancel} |d-0.11885|\le0.00015 .\end{array}