How do you simplify $\frac{- 9 x^{3} y z^{2}}{27 x^{4} y z}$ $(-9x^3yz^2)/(27x^4yz)$ ?

Question

1578 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-12T02:47:04

$\frac{- 9 x^{3} y z^{2}}{27 x^{4} y z} = - \frac{z}{3 x}$ $(-9x^(3)yz^(2))/(27x^(4)yz)=color(blue)(-z/(3x)$

Simplify:

$\frac{- 9 x^{3} y z^{2}}{27 x^{4} y z}$ $(-9x^(3)yz^(2))/(27x^(4)yz)$

Simplify $- \frac{9}{27}$ $-9/27$ to $- \frac{1}{3}$ $-1/3$ .

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

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

$- \frac{x^{3 - 4} y^{1 - 1} z^{2 - 1}}{3}$ $-(x^(3-4)y^(1-1)z^(2-1))/3$

Simplify.

$- \frac{x^{- 1} y^{0} z^{1}}{3}$ $-(x^(-1)y^(0)z^(1))/3$

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

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

Simplify.

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

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

$- \frac{z}{3 x}$ $-z/(3x)$

How do you simplify (-9x^3yz^2)/(27x^4yz)−9x3yz227x4yz(-9x^3yz^2)/(27x^4yz)?