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Lanuldeesmath
@Lanuldeesmath
January 2021
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bonjour pouvez vous m'aider pour
l'exercice 3 ? merci d'avance
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A- nombre de feuille achetées en 2000
nf = 2000/8 * 1000 = 250 *1000 = 250 000 feuilles
b- m(t) = 2000 / (8+ 8*t/100) = 2000/8(1+t/100) = 250/(1+t/100)
c- le sens de variation de m(t) sur [0,100]
t 0 100
m(t) 250 125 diminition
d- diminution de 8% de la consommation rames de 1000 feuille
250-250*8/100 = 250-20 = 230 rames
le prix de rames devient alors 2000/230 = 8.70 euro
augmentation en % du prix maximum est (8.7-8.) / 8 = 8.75 %
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nf = 2000/8 * 1000 = 250 *1000 = 250 000 feuilles
b- m(t) = 2000 / (8+ 8*t/100) = 2000/8(1+t/100) = 250/(1+t/100)
c- le sens de variation de m(t) sur [0,100]
t 0 100
m(t) 250 125 diminition
d- diminution de 8% de la consommation rames de 1000 feuille
250-250*8/100 = 250-20 = 230 rames
le prix de rames devient alors 2000/230 = 8.70 euro
augmentation en % du prix maximum est (8.7-8.) / 8 = 8.75 %