Question 202432
I'll do the first four to get you started


1)


{{{3*4^0=3*1=3}}}



So {{{3*4^0=3}}}


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2) 


{{{1/(4^(-2))=1/(1/4^2)=1(4^2/1)=4^2=16}}}



So {{{1/(4^(-2))=16}}}



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3) 


{{{(1/4)^(-1/2)=(4/1)^(1/2)=(4)^(1/2)=sqrt(4)=2}}}



So {{{(1/4)^(-1/2)=2}}}



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4) 



{{{(bm)^(-3)=(1/(bm))^3=(1^3)/(bm)^3=1/(b^3*m^3)}}}



So {{{(bm)^(-3)=1/(b^3*m^3)}}}