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  • 08-04-2017
  • Mathematics
contestada

how do I solve for 'k'?

how do I solve for k class=

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Аноним Аноним
  • 08-04-2017
[tex] \sqrt{2k^2+17} -7=0[/tex] 

[tex]2k^2+17= \sqrt[ \frac{1}{2} ]{7} [/tex] 

[tex]2k^2+17=7^2[/tex] 

[tex]2k^2+17=49[/tex] 

[tex]2k^2=49-17[/tex] 

[tex]2k^2=32[/tex] 

[tex]k^2 = \dfrac{32}{2} [/tex] 

[tex]k^2=16[/tex] 

[tex]k= \sqrt{16}[/tex] 

[tex]k=4[/tex]
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