I know I take the inverse of Kc for the reverse reaction. But, it is double so I

Question

I know I take the inverse of Kc for the reverse reaction. But, it is double so I square the answer to 13?  

Same scenario logic for 14) I know it is something to do with the coefficients but I don't know what to do with those and the initial Kc values.  The logic behind both of these  are the same so it must be a related solution.

13) The equilibrium constant is given for one of the reactions below.  Determine the value of the missing equilibrium constant.

            H2(g) + Br2(g)  ?  2 HBr(g)   Kc = 3.8

Explanation / Answer

13) Kc for doubled reverse reaction is Kc = ( 1/Kc^2) = ( 1/3.8 x10^-4)^2 = 6.9 x 10^-10

hence D is answer

14) Kc for doubled reverse reaction Kc = ( 1/Kc)^2 = ( 1/0.28)^2 = 13

hence D is answer

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