Question

class11 math sets 8(c)​

Answer 1

Answer:

(-10,5)∆(20,30)

Step-by-step explanation:

as the set A' will=【R-{(-5,20)}】

and it is interested by c then the answer will came.

thanks

Answer 2

$(a)\ A\cup B=[-5,\ 20]\cup[5,\ 100)=[-5,\ 100)$

Combine the closed -5 and the open 100 to get A∪B, because -5 and 100 are the least and the greatest among them respectively.

$(b)\ B-C=[5,\ 100)-(-10,\ 30)=[30,\ 100)$

Both these intervals contain the interval $[5,\ 30).$ Or we can say this is B∩C.

This is removed from B. We know  $B-C=B-(B\cap C)$

$\begin{aligned}(c)\ \ A'\cap C=\ \ &((-\infty,\ -5)\cup(20,\ \infty))\cap(-10,\ 30)\\\\\implies\ \ &((-\infty,\ -5)\cap(-10,\ 30))\cup((20,\ \infty)\cap(-10,\ 30))\\\\\implies\ \ &(-10,\ -5)\cup(20,\ 30)\end{aligned}$

$\begin{aligned}(d)\ A\cap B\cap C=\ \ &[-5,\ 20]\cap[5,\ 100)\cap(-10,\ 30)\\\\\implies\ \ &[-5,\ 20]\cap[5,\ 30)\\\\\implies\ \ &[5,\ 20]\end{aligned}$

$\begin{aligned}(e)\ B'\cup C'=\ \ &(B\cap C)'\\\\\implies\ \ &[5,\ 30)'\\\\\implies\ \ &(-\infty,\ 5)\cup[30,\ \infty)\end{aligned}$

And finally,

$\begin{aligned}(f)\ (A'\cup B)\cap C=\ \ &(A'\cap C)\cup(B\cap C)\\\\\implies\ \ &(-10,\ -5)\cup(20,\ 30)\cup[5,\ 30)\\\\\implies\ \ &(-10,\ -5)\cup[5,\ 30)\end{aligned}$