How do you evaluate #6\div 6- ( 1+ 2- 2)#?

1 Answer
Jim G.
Oct 6, 2017

#0#

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#

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