x=y4 /2
y4.
(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
Assuming the vertex is 0,0 and the directrix is y=4 x^2=0
y4
x=y4 /2
y4.
(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
Assuming the vertex is 0,0 and the directrix is y=4 x^2=0
y4
y + y4 = y5 , possibly. Except that conventionally, the number (or coefficient) would be written first. y + y4 = y*(1 + y3) = y(1 + y)*(1 - y + y2)
Assuming the domain and range are suitably defined, then yes. If not, then no.
(y^2 + 8)(y^2 + 2)
x2 + y4 + x4 +y2 = x6 + y6unless you know what x and y are.* * * * *x2 + y4 + x4 + y2 ??I don't believe that this expression can be factorised or otherwise simplified.It certainly does not equal x6 + y6,for all x and all y:for example, if x = y = 1, thenx2 + y4 + x4 + y2 = 4, whilstx6 + y6 = 2;thus, they are two manifestly unequal quantities.
If x = 3 and y = 4 then the answer is 2
The GCF is y4
-3/4 if the question is 3x+4y