Simplify: #(12−5^2-:5)*4^2-:2^3+2^2−1+[(2^4-:2^3)^3+4^3-:4+2^5]-:7#? Prealgebra Fractions Equivalent Fractions and Simplifying 1 Answer Shwetank Mauria May 2, 2016 #(12-5^2-:5)*4^2-:2^3+2^2-1+[(2^4-:2^3)^3+4^3-:4+2^5]-:7=25# Explanation: #(12-5^2-:5)*4^2-:2^3+2^2-1+[(2^4-:2^3)^3+4^3-:4+2^5]-:7# = #(12-25-:5)*4^2-:2^3+2^2-1+[(2)^3+4^2+2^5]-:7# = #(12-5)*(2^2)^2-:2^3+2^2-1+[8+16+32]-:7# = #7*2^4-:2^3+2^2-1+56-:7# = #7*2+4-1+8# = #14+4-1+8# = #25# Answer link Related questions What is 0.098 divided by 7? How do you find the fraction notation and simplify 16.6%? How do you convert #0.bar(45)# (meaning the #45# is being repeated) to a fraction? How do you convert 0.40 (40 repeating) to a fraction? How do you convert 8/15 to a decimal? How do you express #3/4, 7/16,# and #5/8# with the #LCD#? How do you simplify fractions? What is the simplified form of #30/42#? How do you subtract and simplify #9-5 1/3#? How do you add and simplify #5/7 + 3/4#? See all questions in Equivalent Fractions and Simplifying Impact of this question 3874 views around the world You can reuse this answer Creative Commons License