kimberlyocallaghan
kimberlyocallaghan kimberlyocallaghan
  • 25-02-2020
  • Mathematics
contestada

Rearranging literal equations for z

Rearranging literal equations for z class=

Respuesta :

yasmitadp yasmitadp
  • 26-02-2020

Answer:

[tex]z = \frac{3n - 3}{6 - 2n}[/tex]

Step-by-step explanation:

[tex]\frac{1}{(n-1)^{\frac{1}{3} } } = (\frac{2z + 3}{4z} )^{\frac{1}{3} }[/tex]

cubing on both sides we get

[tex]\frac{1}{n-1} = \frac{2z + 3}{4z}[/tex]

4z = (2z+3)(n-1)

4z = 2nz + 3n - 2z - 3

4z - 2nz + 2z = 3n - 3

z(6 - 2n) = 3n - 3

[tex]z = \frac{3n - 3}{6 - 2n}[/tex]

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