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No.
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
-5xy2 + 12xy2 = (-5 + 12)xy2 = 7xy2
x3+xy-x2y2=x(x2+y-xy2)
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xy2
15x2y2-9xy3 As x2y2 = x xy2 and xy3 = xy2 y then xy2 is in both term you can first factorize 15x2y2-9xy3 = xy2(15x-9y) as 15=3x5 and 9=3x3 15x2y2-9xy3 = 3xy2(5x-3y) and that it !
No.
xy2
xy
The compound with the formula XY2 consists of one atom of element X and two atoms of element Y.
x2yz3 = 74xy2 = 75xyz = ?Since xy2 = 75, we have:x2yz3 = 74 multiply by 75 to both sides(x2yz3)(xy2) = (74)(75)x3y3z3 = 712(x3y3z3)1/3 = (712)1/3xyz = 74
xy
The GCF is xy
48
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)