solve this x÷2x^2+3xy+y^2
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Answer:
answer is 0 plz mark me brainiest
Answered by
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ANSWER
=
2y−3x+2
3y−4x−1
Step-by-step explanation:
2x
2
−3xy+y
2
+x+2y−8=0
So
dx
dy
4x−3xy
′
−3y+2y.y
′
+1+2y
′
=0
y
′
(−3x+2y+2)+(4x−3y+1)=0
y
′
=
−3x+2y+2
−(4x−3y+1)
y
′
=
2y−3x+2
−4x+3y−1
y
′
=
2y−3x+2
3y−4x−1
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