Respuesta :
Recall that:
[tex]E^c=\mleft\lbrace x\in\R\colon x\notin E\mright\rbrace\text{.}[/tex]Since
[tex]E=\mleft\lbrace x\in\R\colon-4\le x\le4\mright\rbrace\text{.}[/tex]Therefore:
[tex]\begin{gathered} E^c=\mleft\lbrace x\in\R\colon x\notin E\mright\rbrace=\lbrace x\in\R\colon x<-4\text{ or x>4}\} \\ =\mleft\lbrace x\in\R\colon x<-4\mright\rbrace\cup\lbrace x\in\R\colon x>4\rbrace\text{.} \end{gathered}[/tex]Now, the cartesian product between the complementary of E and F is:
[tex]E^c\times F=(\lbrace x\in\R\colon x<-4\rbrace\cup\lbrace x\in\R\colon x>4\rbrace)\times(\mleft\lbrace x\in\R\colon\mright|x|=x\})\text{.}[/tex]Now, recall that:
[tex]\begin{gathered} |x|=x\text{ if and only if x}\ge0, \\ (A\cup B)\times C=A\times C\cup B\times C. \end{gathered}[/tex]Therefore:
[tex]E^c\times F=(\lbrace x\in\R\colon x<-4\rbrace\times\lbrace x\in\R\colon x\ge0\})\cup(\lbrace x\in\R\colon x>4\rbrace)\times\lbrace x\in\R\colon x\ge0\})\text{.}[/tex]Answer:
[tex]\begin{gathered} E^c=\lbrace x\in\R\colon x<-4\rbrace\cup\lbrace x\in\R\colon x>4\rbrace\text{.} \\ E^c\times F=(\lbrace x\in\R\colon x<-4\rbrace\times\lbrace x\in\R\colon x\ge0\})\cup(\lbrace x\in\R\colon x>4\rbrace)\times\lbrace x\in\R\colon x\ge0\})\text{.} \end{gathered}[/tex]