How do you evaluate #6\div 6- ( 1+ 2- 2)#?
1 Answer
Oct 6, 2017
Explanation:
#"following the order of operations as set out in PEMDAS"#
#["Parenthesis (brackets),Exponents (powers"),#
#"Multiplication ,Division ,Addition ,Subtraction"]#
#rArr6-:6-(color(red)1)larrcolor(blue)" brackets"#
#=1-1larrcolor(blue)" division"#
#=0#
