Question #28a84

2 Answers
Sep 27, 2016

#(x^4)/(6y^8)#

Explanation:

I am assuming that the expression is #((x^4/3)/y^8) * 1/2#.

What we have here is #x^4/3# divided by #y^8# and this result multiplied by #1/2#

That is #(x^4/3÷y^8)1/2#

When dividing fractions we change division to multiplication and tun the second fraction' upside down'

#rArrx^4/3÷y^8/1=x^4/3xx1/y^8=x^4/(3y^8)#

Now multiply this result by #1/2#

#rArrx^4/(3y^8)xx1/2=x^4/(6y^8)#

Sep 27, 2016

Assuming that the question is supposed to read as #(x^(4/3)/y^8)^(1/2)#

we can proceed in two ways...

Recall: #x^(1/2)# is another way of writing a square root. #sqrtx#

#(x^(4/3)/y^8)^(1/2) = sqrt((x^(4/3)/y^8))#

The square root of a fraction can be split...

# = sqrt((x^(4/3)))/sqrt(y^8)#

To find the square root ... divide the index by 2.

#sqrt((x^(4/3)))/sqrt(y^8) = x^(2/3)/y^4#

Recall: #(x^m)^n = x^(mn) " "larr# multiply the indices

#(x^(4/3)/y^8)^(1/2) = x^(4/3xx1/2)/y^(8xx1/2)#

=#x^(2/3)/y^4#