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nikitaa
@nikitaa
June 2021
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Aidez moi svp, il faut calculer Delta (b²-4ac)+ les solutions de -0.03x² +0.2x +133
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Steffy1
Bonjour !!!! :)
f(x)= -0,03²+0,2x+133
a=
-0,03
b=
0,2
c=
133
Δ=b² -4ac
Δ=0,2² -4×(-0,03)×133
Δ=0,04+1596
Δ=16
Delta admet 2 solutions:
Voila !!! ^^
0 votes
Thanks 0
Cabé
Δ = b² - 4ac
a =
-0,03
b =
0,2
c = 133
b2 = 0,04
-4ac = 15,96
Δ = 16
x1 = (-b-√Δ)/2a = 70
x2= (-b+√Δ)/2a = -190/3
0 votes
Thanks 0
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f(x)= -0,03²+0,2x+133
a=-0,03 b=0,2 c=133
Δ=b² -4ac
Δ=0,2² -4×(-0,03)×133
Δ=0,04+1596
Δ=16
Delta admet 2 solutions:
Voila !!! ^^
a = -0,03
b = 0,2
c = 133
b2 = 0,04
-4ac = 15,96
Δ = 16
x1 = (-b-√Δ)/2a = 70
x2= (-b+√Δ)/2a = -190/3