How do you simplify #\frac { 2x y ^ { 0} } { 3x ^ { 5} }#?

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Answer 1 · 2018-03-29T11:24:01

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

First #y^0=1# as anything to the power of 0 is 1

So it looks more like #(2x)/(3x^5)#

When we divide exponets they subtract so #x/x^5=x^(1-5)=x^-4=1/x^4#

So it is merely #(2)/(3x^4)#

Answer 2 · 2018-03-29T11:29:35

#(2xy^0)/(3x^5)=color(blue)(2/(3x^4)#

Simplify:

#(2xy^0)/(3x^5)#

Apply the zero exponent rule: #a^0=1#

Simplify #y^0# to #1#.

#(2x xx1)/(3x^5)#

#(2x)/(3x^5)#

Apply quotient exponent rule: #a^m/a^n=a^(m-n)#

#(2x^(1-5))/3#

Simplify.

#(2x^(-4))/3#

Apply negative exponent rule: #a^(-m)=1/(a^m)#

#2/(3x^4)#