Question 164267
((-3mn)^2*64(m^2n)^3)/(16m^2n^4)*(mn^2)^3(OVER)24(m^2n^2)^4/(3m^2n^3)^2
Let's deal with the first part first (out to the "over"):
((-3mn)^2*64(m^2n)^3)/(16m^2n^4)*(mn^2)^3=
(9m2^2n^2)*64(m^6n^3)/(16m^2n^4)(m^3n^6)=
(576m^8n^5)/(16m^5n^10)=
36m^3/n^5-------------------------------first part
Next, the part after the "over"
24(m^2n^2)^4/(3m^2n^3)^2=
(24m^8n^8)/(9m4n^6)=
8m^4n^2/3---------------------------second part

Now, we'll put the first part and second part back together

36m^3/n^5 over 8m^4n^2/3  multiply numerator and denominator by 3/8m^4n^2 ( this will make the denominator of the complex fraction equal to 1 and thus get rid of the complex fraction):

(36m^3/n^5)*(3/8m^4n^2) over 1=
(108m^3/(8m4n^7)=
27/2mn^7----------------------------ans
Easy to make a mistake on this one!!!!!!!!


Hope this helps---ptaylor