How do you simplify #5(2^3+4)/sqrt36 times 7#? Algebra Expressions, Equations, and Functions PEMDAS 1 Answer P dilip_k Apr 15, 2016 #(5 (2^3+4)/sqrt36)xx7=(5 12/6)xx7=(5+2)xx7=49# OR if there exists a dot #(5. (2^3+4)/sqrt36)xx7=(5 xx12/6)xx7=(5xx2)xx7=70# Please comment 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 1514 views around the world You can reuse this answer Creative Commons License