What is #(4xx4(6+4)-3(8+1)+2)-:3+(5+2)(5+2)-2xx2+(42-:6)(42-:6)+1# ?
1 Answer
Jan 15, 2017
Explanation:
Use PEMDAS to help with the order of operations:
P for Parentheses
E for Exponents
MD for Multiplication and Division (left to right)
AS for Addition and Subtraction (left to right)
So we find:
#(4xx4(6+4)-3(8+1)+2)-:3+(5+2)(5+2)-2xx2+(42-:6)(42-:6)+1#
#=(4xx4(10)-3(9)+2)-:3+(7)(7)-2xx2+(7)(7)+1#
#=(4xx40-27+2)-:3+49-2xx2+49+1#
#=(160-27+2)-:3+49-2xx2+49+1#
#=(133+2)-:3+49-2xx2+49+1#
#=135-:3+49-2xx2+49+1#
#=45+49-4+49+1#
#=94-4+49+1#
#=90+49+1#
#=139+1#
#=140#
