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

Answer:

#140#

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#