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If \(z = \frac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }\)

 

, find \(\lfloor z \rfloor\)

 Apr 20, 2018
 #1
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We can simplify this as : 

 

z  =  3 - 2(2)                - 1

      _______    ≈     ______  ≈  0.9121

       √3 - √8              -1.096

 

So....the floor  of  this    is    0

 

 

cool cool cool

 Apr 20, 2018

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