" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
25-9c² > 0
-9c²+25 > ²0
a = -9
c = 25
Δ = b²-4ac
Δ = -4*(-9)*25
Δ = 36*25
Δ = 900
√Δ = 30
Δ > 0 czyli 2 pierwiastki
x1 = -b-√Δ /2a
x1 = -30/2*(-9)
x1 = -30/-18
x1 = 5/3
x2 = -b+√Δ/2a
x2 = 30/-18
x2 = -5/3
a < 0 to ramiona paraboli w dół
25-9c² > 0 dla x∈ (-5/3 ; 5/3)
(n-5)kwadrat-(4-n)kwadrat<bądź równe 0 błagam bo na prace kontrolną na jutro potrzebuje
(n-5)²-(4-n)² ≤ 0
n²-10n+25-(16-8n+n²) ≤ 0
n²-10n+25-16+8n-n² ≤ 0
-2n+9 ≤ 0
-2n ≤ -9 |*(-1)
2n ≥ 9 |:2
n ≥ 9/2
n ≥ 4,5
n ∈ < 4,5 ; +∞)