How do you find the product of #(3ad)/(4c^4)*(8c^2)/(6d)#?

Question

1557 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-22T21:42:27

#a/c^2#

#(3ad)/(4c^4) * (8c^2)/(6d)#

cross-cancel:

#(3ad)/(3d)=a#

#(6d)/(3d) = 2#

#(8c^2)/(4c^2)=2#

#(4c^4)/(4c^2)=c^2#

simplified equation:

#a/c^2 * 2/2#

#=(2a)/(2c^2)#

#=a/c^2#

Answer 2 · 2016-12-22T22:21:44

#=a/c^2#

Product means the answer to a mulitplication.

#(3ad)/(4c^4) xx (8c^2)/(6d)" "# cancel any like factors

#(cancel3ad)/(cancel4c^4) xx (cancel8^2c^2)/(cancel6^2d)" "larr(8div4=2)/(6div3=2)#

Simplify into one numerator and one denominator

#= (cancel2ac^2canceld)/(cancel2c^4canceld)#

Simplify by subtracting indices of like bases

#=a/c^2#