What is the inverse of (2z^{-1})/(1-(1/3)(z^{-1))} ?

The inverse of (2z^{-1})/(1-(1/3)(z^{-1))} is (6+z)/(3z)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of (2z^{-1})/(1-(1/3)(z^{-1))}

inverse

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

\frac{6 + z}{3 z}

Solution steps

y = \frac{2 z ^{- 1}}{1 - \frac{1}{3} z ^{- 1}}

Replace

z

with

y

z = \frac{2 y ^{- 1}}{1 - \frac{1}{3} y ^{- 1}}

Solve for

y, z = \frac{2 y ^{- 1}}{1 - \frac{1}{3} y ^{- 1}}

\frac{6 + z}{3 z}

5^{x}

f (x) = 2 tan^{2} (t) + 4 tan (t) - 3 = 0

\frac{5}{x + 10}

What is the inverse of (2z^{-1})/(1-(1/3)(z^{-1))} ?
The inverse of (2z^{-1})/(1-(1/3)(z^{-1))} is (6+z)/(3z)