x2/t2+y2/t'2 = 1 T.theodor 2022.8.20
y = ax3+bx2+cx+d y =−b+2√2b2−6ac/3a x=2 ax3+bx2+cx+d−−b+2√2b2−6ac/3a= 0 (x−−b+2√2b2−6ac/3a)2(x−α) = 0 ax3+bx2+cx+d−−b−2√2b2−6ac/3a= 0 (x−−b−2√2b2−6ac/3a)2(x−β) = 0 x=1 y = ax3+bx2+cx+d y=m ax3+bx2+cx+d−m = 0 (x−γ)(a'x2+b'x2+c') = 0 b'2−4a'…
Quote saved.
Login to quote this blog
Failed to save quote. Please try again later.
You cannot quote because this article is private.