How do you evaluate #f^-2g# for #f=3# and #g=27#?

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Answer 1 · 2017-03-17T04:42:14

#f^(-2)g=3#

To evaluate #f^(-2)g# for #f=3# and #g=27#

we have to put given values of #f# and #g# in to the function #f^(-2)g#.

As such #f^(-2)g=3^(-2)xx27#

As #3^(-2)=1/3^2#

#f^(-2)g=1/3^2xx27=(3xx3xx3)/(3xx3)#

= #(3xxcancel3xxcancel3)/(cancel3xxcancel3)#

= #3#