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How do you simplify #\frac { 4y ^ { 2} } { x ^ { 0} y ^ { 3} }#?

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Karl W. Share
Mar 15, 2018

Answer:

It becomes #4/y#

Explanation:

First of all, anything raised to the zero power is going to be 1, so that #x^0# just is one, and won't affect the problem. Second of all, you can cancel out exponents when working with fractions, so the #y^2/y^3# will become #1/y^1# and you can simplify that to just #1/y#. We can't forget the 4 either, so in all actuality, your expression simplifies to #(4*1)/y# or just #4/y#

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Write your answer here...
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Then teach the underlying concepts
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Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

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1
Mar 15, 2018

Answer:

See a solution process below:

Explanation:

First, use this rule of exponents to simplify the #x# term:

#a^color(red)(0) = 1#

#(4y^2)/(x^color(red)(0)y^3) => (4y^2)/(1 * y^3) => (4y^2)/y^3#

Next, use these rule of exponents to simplify the #y# term:

#x^color(red)(a)/x^color(blue)(b) = 1/x^(color(blue)(b)-color(red)(a))# and #a^color(red)(1) = a#

#(4y^color(red)(2))/y^color(blue)(3) => 4/y^(color(blue)(3)-color(red)(2)) => 4/y^color(red)(1) => 4/y#

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