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Jamesvidonne
@Jamesvidonne
May 2019
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Bonjours à tous pouvez vous m'aider svp pour mon dm de maths de 3eme
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Bonjour,
partie 1 :
1) DHC rectangle en C
pythagore
HD² = DC²+HC²
tu as les mesures, calcule HD
2) cosdHc = HC/HD = 3/5 =0,6
dHc = 53° arrondi
3) bHd+dHc = 180°
bHd =180-53 = 127°
partie 2 :
1) 0<x<12
2) f(x) = (HC*CD)/2 = 4*x/2 = 2x
3) A ABCD - f(x) = 48-2x
g(x) =-2x+48
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Verified answer
Bonjour,partie 1 :
1) DHC rectangle en C
pythagore
HD² = DC²+HC²
tu as les mesures, calcule HD
2) cosdHc = HC/HD = 3/5 =0,6
dHc = 53° arrondi
3) bHd+dHc = 180°
bHd =180-53 = 127°
partie 2 :
1) 0<x<12
2) f(x) = (HC*CD)/2 = 4*x/2 = 2x
3) A ABCD - f(x) = 48-2x
g(x) =-2x+48