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

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Answer 1 · 2018-03-29T11:24:01

$\frac{2}{3 x^{4}}$ $(2)/(3x^4)$

First $y^{0} = 1$ $y^0=1$ as anything to the power of 0 is 1

So it looks more like $\frac{2 x}{3 x^{5}}$ $(2x)/(3x^5)$

When we divide exponets they subtract so $\frac{x}{x^{5}} = x^{1 - 5} = x^{- 4} = \frac{1}{x^{4}}$ $x/x^5=x^(1-5)=x^-4=1/x^4$

So it is merely $\frac{2}{3 x^{4}}$ $(2)/(3x^4)$

Answer 2 · 2018-03-29T11:29:35

$\frac{2 x y^{0}}{3 x^{5}} = \frac{2}{3 x^{4}}$ $(2xy^0)/(3x^5)=color(blue)(2/(3x^4)$

Simplify:

$\frac{2 x y^{0}}{3 x^{5}}$ $(2xy^0)/(3x^5)$

Apply the zero exponent rule: $a^{0} = 1$ $a^0=1$

Simplify $y^{0}$ $y^0$ to $1$ $1$ .

$\frac{2 x \times 1}{3 x^{5}}$ $(2x xx1)/(3x^5)$

$\frac{2 x}{3 x^{5}}$ $(2x)/(3x^5)$

Apply quotient exponent rule: $\frac{a^{m}}{a^{n}} = a^{m - n}$ $a^m/a^n=a^(m-n)$

$\frac{2 x^{1 - 5}}{3}$ $(2x^(1-5))/3$

Simplify.

$\frac{2 x^{- 4}}{3}$ $(2x^(-4))/3$

Apply negative exponent rule: $a^{- m} = \frac{1}{a^{m}}$ $a^(-m)=1/(a^m)$

$\frac{2}{3 x^{4}}$ $2/(3x^4)$

How do you simplify 2xy03x5\frac { 2x y ^ { 0} } { 3x ^ { 5} }?