Witam, mam drobnym problem z tym zadaniem :
Wyznacz promień i środek okręgu .
a) (x - 2/3) do kwadratu +(y-1) do kwadratu = 144/9
b) (x- pierwiastek z dwóch ) do kwadratu + (y + pierwiastek z trzech) do kwadratu=2
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2025 KUDO.TIPS - All rights reserved.
S=(a,b)
a)
b)
a)S(2/3,1) srodek. r=12/3
b)s(pierwaistek 2,-pierwiastek z 3) r=pierwiastek z 2
bo ogolny wzor to taki (x-x1)+(y-y1)=r kwadrat
gdzie s(x1,y1)