How do you evaluate #6*(11-6)-:2#? Algebra Expressions, Equations, and Functions PEMDAS 1 Answer EZ as pi Sep 12, 2016 #15# Explanation: #6xxcolor(red)((11-6))div2# is all one term, but work out the bracket first. #6xx5div2 " "# which can also be written as #(6xx5)/2# = # 30div2 =15# OR # 3xx5 = 15# OR #6 xx 2 1/2 = 15# Answer link Related questions What is PEMDAS? How do you use PEMDAS? How do you use order of operations to simplify #3(7-2)-8#? What are common mistakes students make with PEMDAS? How do you evaluate the expression #5[8+(3-1)]-2#? How do you simplify the expression #4(30-(3+1)^2)#? How do you evaluate the expression #x^4+x# if x=2? Is it okay to add first before subtracting in #4-6+3#? How do you simplify #(-3)^2+12*5#? How do you simplify #(4-2)^3-4*8+21div3#? See all questions in PEMDAS Impact of this question 2219 views around the world You can reuse this answer Creative Commons License