What is x^3(y^{-5})(x^{-8}) ?

The solution to x^3(y^{-5})(x^{-8}) is 1/(x^5y^5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify x^3(y^{-5})(x^{-8})

simplify

x^{3} (y^{- 5}) (x^{- 8})

\frac{1}{x ^{5} y ^{5}}

Solution steps

x^{3} (y^{- 5}) (x^{- 8})

Simplify

x^{3} (y^{- 5}) (x^{- 8}) : x^{3} y^{- 5} x^{- 8}

= x^{3} y^{- 5} x^{- 8}

Simplify

x^{3} x^{- 8} : \frac{1}{x ^{5}}

= \frac{1}{x ^{5}} y^{- 5}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

y^{- 5} = \frac{1}{y ^{5}}

= \frac{1}{x ^{5}} \cdot \frac{1}{y ^{5}}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{1 \cdot 1}{x ^{5} y ^{5}}

Multiply the numbers:

1 \cdot 1 = 1

= \frac{1}{x ^{5} y ^{5}}

∣ x + 2 ∣ < ∣ 2 x ∣

6 x - 8 = 16 + 10 x

\frac{x - 1}{2 x ^{2} - x - 1} \geq 0

\frac{2 ( 3 ) ^{2} - 1}{( 3 ) ^{2}}

\frac{x + 1}{x - 1} \leq 4