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exercice 1
Si l'arbre est perpendiculaire au sol, on peut appliquer le théorème de Pythagore.
On a donc : 10² = 7²+h²
100 = 49+h²
h² = 100-49
h = 51
donc h = √51 ≈ 7,1 m
exercice 2
a = -4 b = 3 c = 2 d = -6
(-b) x c | a^2 | (-c) x b ?
d^2 | a x b | c^2
a x (-c) | b^2 | -a x d
(-b) x c = (-3) x (-6) = 18
a^2 = (-4)^2 = 16
(-c) x b = -2 x 3 = -6
d^2 = (-6)^2 = 36
a x b = -4 x 3 = -12
c^2 = 2^2 = 4
a x (-c) = -4 x -2 = 8
b^2 = 3^2 = 9
-a x d = 4 x -6 = -24
Total : -1728
bonne soirée
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bonsoir
exercice 1
Si l'arbre est perpendiculaire au sol, on peut appliquer le théorème de Pythagore.
On a donc : 10² = 7²+h²
100 = 49+h²
h² = 100-49
h = 51
donc h = √51 ≈ 7,1 m
exercice 2
a = -4 b = 3 c = 2 d = -6
(-b) x c | a^2 | (-c) x b ?
d^2 | a x b | c^2
a x (-c) | b^2 | -a x d
(-b) x c = (-3) x (-6) = 18
a^2 = (-4)^2 = 16
(-c) x b = -2 x 3 = -6
d^2 = (-6)^2 = 36
a x b = -4 x 3 = -12
c^2 = 2^2 = 4
a x (-c) = -4 x -2 = 8
b^2 = 3^2 = 9
-a x d = 4 x -6 = -24
Total : -1728
bonne soirée