How do you simplify #(22 - 6) |4 2 - 13| # using order of operations?

Question

1407 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-08T00:53:44

Parenthesis first ...

First parenthesis and absolute values ...

#(22-6)=16#

#abs(4xx2-13)=abs(-5)=5#

Next, multiplication ...

#16xx5=80#

hope that helped

How do you simplify #(22 - 6) *|4 * 2 - 13| # using order of operations?