rozwiąż nierówność -x^2+5x>0
z przedziałami.!!
-x²+5x>0
-x(x-5)>0
x=0
x=5
x∈(0,5)
deltra = b^2 - 4ac
delta = 5^2 - 4 * (-1) * 0
delta = 25
pierwiastek z delty = 5
x1 = -b - pierw. z delty przez 2a
x1 = (-5 -5)/2*(-1)
x1 = -10/2
x1 = -5
x2 = -b + pierw. z delty przez 2a
x2 = (-5 + 5)/2*(-1)
x2 = 0/2 = 0
xe (-5;0)
w rysunku ramiona w dół
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-x²+5x>0
-x(x-5)>0
x=0
x=5
x∈(0,5)
deltra = b^2 - 4ac
delta = 5^2 - 4 * (-1) * 0
delta = 25
pierwiastek z delty = 5
x1 = -b - pierw. z delty przez 2a
x1 = (-5 -5)/2*(-1)
x1 = -10/2
x1 = -5
x2 = -b + pierw. z delty przez 2a
x2 = (-5 + 5)/2*(-1)
x2 = 0/2 = 0
xe (-5;0)
w rysunku ramiona w dół