Seudónimo Seudónimo
  • 23-10-2018
  • Mathematics
contestada

plz help me on this question

plz help me on this question class=

Respuesta :

frika
frika frika
  • 01-11-2018

Let number [tex]c=\sqrt{45}.[/tex]

Square this number:

[tex]c^2=45.[/tex]

Since

[tex]36<45<49,[/tex]

you can state that

[tex]\sqrt{36}<\sqrt{45}<\sqrt{49},\\ \\6<\sqrt{45}<7.[/tex]

Also [tex]\sqrt{45}\approx 6.708[/tex] and [tex]6.5<x_C<7[/tex], then point C is the best approximation of [tex]\sqrt{45}.[/tex]

Answer: option C.

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