What is the value of #(2a^-1 + (a^-1)/2) / a # when #a=1/2#?

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Answer 1 · 2017-03-19T07:36:41

#(2a^(-1)+a^(-1)/2)/a=10#

As #a=1/2#, #a^(-1)=1/a=2#

Hence #(2a^(-1)+a^(-1)/2)/a#

= #(2xx2+2/2)/(1/2)#

= #(4+1)xx2/1#

= #10#

Answer 2 · 2017-03-19T17:15:07

#10#

#(2a^-1+a^-1/2)/a#

#:.=((2 xx 1/a)+(1/a)/2)/a#

#:.=((2/1 xx 1/a)+(1/a xx 1/2))/a#

#:.=(2/a+1/(2a))/a#

#:.=((4+1)/(2a))/a#

#:.=(5/(2a))/a#

#:.=5/(2a) xx 1/a#

#:.=5/(2a^2)#

substitute #color(red)(a=1/2#

#:.=5/(2 xx color(red) ((1/2))^2)#

#:.=5/(2 xx 1/4)#

#:.=5/(1/2)#

#:.=5/1 xx 2/1#

#:.=10#