How do you simplify #(27r^5s^2t^4)/(3r^7s^7t^2)#"?

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Answer 1 · 2016-07-13T17:56:29

#(9t^2)/(r^2s^5)#

Some rules to follow:

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#(27r^5s^2t^4)/(3r^7s^7t^2)#

#27/3# can be simplified to #9#

#(9r^5s^2t^4)/(r^7s^7t^2)#

Following the first rule from above, simplify #r^5/r^7# to #r^-2/1#

#(9r^-2s^2t^4)/(s^7t^2)#

Use the same rule to simplify #s^2/s^7# to #s^-5/1#

#(9r^-2s^-5t^4)/t^2#

Let's use the rule one last time to simplify #t^4/t^2# to #t^2/1#

#9r^-2s^-5t^2#

Now we must use the second rule from above to simplify #r^-2# to #1/r^2#

#(9s^-5t^2)/r^2#

Use the second rule another time to make #s^-5# equivalent to #1/s^5#

#(9t^2)/(r^2s^5)#