x=y4 /2
y4
Assuming the domain and range are suitably defined, then yes. If not, then no.
(x + y)2 = x2 + 2xy + y2 So x2 + y2 = (x + y)2 - 2xy = a2 - 2b Then (x2 + y2)2 = x4 + 2x2y2 + y4 So x4 + y4 = (x2 + y2)2 - 2x2y2 = (a2 - 2b)2 - 2b2 = a4 - 4a2b + 4b2 - 2b2 = a4 - 4a2b + 2b2
-3/4 if the question is 3x+4y
x=y4 /2
y4
Assuming the domain and range are suitably defined, then yes. If not, then no.
The GCF is y4
7
( 5x4 y4 −−−−−−√ ) ( 6x2y−−−−√ )
4x-y4 what = 0
The simplest equation would be y4 = 0
(x + y)2 = x2 + 2xy + y2 So x2 + y2 = (x + y)2 - 2xy = a2 - 2b Then (x2 + y2)2 = x4 + 2x2y2 + y4 So x4 + y4 = (x2 + y2)2 - 2x2y2 = (a2 - 2b)2 - 2b2 = a4 - 4a2b + 4b2 - 2b2 = a4 - 4a2b + 2b2
If 2x-y = 4 then 6x -3 years =?
-3/4 if the question is 3x+4y
(x2 + y2)(x + y)(x - y) = x4 - y4.