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Smades
@Smades
January 2021
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Bonsoir. Aidez moi svp c'est pour lundi, urgent !
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mina151
CB^2=AC^2+AB^2
=169+196
=365
CB =19,10
EB^2=BD^2+DE^2
=400+324
=724
EB= 26,9
CE^2=CB^2+EB^2
=365+725=1089
CE=33
alors CB ET BE sont perpendiculaires
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Bonsoir, aidez-moi svp, merci. Cordialement Smades
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Lista de comentários
=169+196
=365
CB =19,10
EB^2=BD^2+DE^2
=400+324
=724
EB= 26,9
CE^2=CB^2+EB^2
=365+725=1089
CE=33
alors CB ET BE sont perpendiculaires