How do you simplify #(- 1/5)^-2 + (-2)^-2#? Prealgebra Exponents, Radicals and Scientific Notation Exponents 1 Answer Shwetank Mauria Jun 6, 2016 #1/(-1/5)^2+1/(-2)^2=25 1/4# Explanation: As #a^(-n)=1/a^n#, #(-1/5)^(-2)+(-2)^(-2)# can be written as #1/(-1/5)^2+1/(-2)^2# = #1/((-1)^2/(5)^2)+1/4# = #1/(1/25)+1/4# = #1xx25/1+1/4# = #25 1/4# Answer link Related questions How do you simplify #c^3v^9c^-1c^0#? How do you simplify #(4^6)^2 #? How do you simplify #3x^(2/3) y^(3/4) (2x^(5/3) y^(1/2))^3 #? How do you simplify #4^3ยท4^5#? How do you simplify #(5^-2)^-3#? How do you simplify and write #(-5.3)^0# with positive exponents? How do you factor #12j^2k - 36j^6k^6 + 12j^2#? How do you simplify the expression #2^5/(2^3 times 2^8)#? When can I add exponents? What is the Zero Exponent Rule? See all questions in Exponents Impact of this question 4329 views around the world You can reuse this answer Creative Commons License