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pesqueremma
@pesqueremma
June 2021
2
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arrivé devant cette boite, Mathix arrivera t-il a poster sa lettre rectangulaire sans la plier ?
Sachant que la lettre fait ___________
! !
! ! 30,3 cm
! ___________!
50 cm
et la boite au lettre 30 cm de large et 5 cm ou on fait passer la lettre
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winner123
Verified answer
Diagonale ² de la lettre
30.3 ² + 50 ² = 3418.09
diagonale de la lettre = √3418.09 = 58.46 cm
diagonale ² de la boite
30² + 5²
900 +25 = 925
diagonale boîte = √ 925 = 30.41 cm
donc la lettre ne passera pas sans la plier
3 votes
Thanks 11
jgalereenmath
Ou sinon tu la fait passé dans la largeur de la lettre, mais pour les 3 mm restant tu force un peu
0 votes
Thanks 0
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pesqueremma
June 2021 | 0 Respostas
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Verified answer
Diagonale ² de la lettre30.3 ² + 50 ² = 3418.09
diagonale de la lettre = √3418.09 = 58.46 cm
diagonale ² de la boite
30² + 5²
900 +25 = 925
diagonale boîte = √ 925 = 30.41 cm
donc la lettre ne passera pas sans la plier