How do you evaluate #2+6(9-3^2)-2#?

1 Answer
John D.
Mar 25, 2017

The answer is #0#.

Explanation:

Use the order of operations #("PEMDAS")#.

#color(red)"P"# = parentheses
#color(blue)"E"# = exponents
#color(orange)"MD"# = multiplying and dividing
#color(violet)"AS"# = adding and subtracting

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#2+6 color(red)((9-3^2)) -2 color(white)"XXX"# Evaluate the #color(red)"P"#arentheses

within parentheses, we start over with #"PEMDAS"#

#(9-3^2) color(white)"XXXXXXX"# No #color(red)"P"#arentheses
#(9-color(blue)(3^2)) color(white)"XXxxxxxxx"# Evaluate the #color(blue)"E"#xponents
#(9-9) color(white)"xxXXxxxxxx"# No #color(orange)"M"#ultiplication or #color(orange)"D"#ivision
#(color(violet)(9-9))color(white)"xxXXxxxxxx"# Evaluate the #color(violet)"A"#ddition and #color(violet)"S"#ubtraction
#=(0)#

#2 + 6(0) - 2 color(white)"XXXXXX"# No #color(blue)"E"#xponents

#2 + color(orange)(6(0)) - 2 color(white)"XXXXXX"# Evaluate the #color(orange)"M"#ultiplication and #color(orange)"D"#ivision

#color(violet)(2+0-2) color(white)"XXXXXXX/"# Evaluate the #color(violet)"A"#ddition and #color(violet)"S"#ubtraction

#=0#

Final Answer

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