Odpowiedź:
Z tw. Pitagorasa
[tex]p^{2} =d^2+(\sqrt{52} )^2\\p^{2} =(2\sqrt{3} )^2+52\\p^{2}=12+52=64/\sqrt{} \\p=8[/tex]
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Odpowiedź:
Z tw. Pitagorasa
[tex]p^{2} =d^2+(\sqrt{52} )^2\\p^{2} =(2\sqrt{3} )^2+52\\p^{2}=12+52=64/\sqrt{} \\p=8[/tex]