Hadizaali14
Hadizaali14 Hadizaali14
  • 03-09-2018
  • Mathematics
contestada

Solve for n. F = 4/3 ​(n-1) The solution is n= F+

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gmany
gmany gmany
  • 06-09-2018

[tex]F=\dfrac{4}{3}(n-1)\ \ \ \ |\cdot\dfrac{3}{4}\\\\n-1=\dfrac{3}{4}F\ \ \ \ |+1\\\\\boxed{n=\dfrac{3}{4}F+1}[/tex]

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