Mathematics, 03.05.2020 13:38 majesticfart7736

Identify the graph of the equation. Then find 0 to the nearest degree.
2x² + 4xy + 4y2 = 0

Answers: 1

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Answer from: genyjoannerubiera

It is a type of conic section, but there is only one point .

This represent a single point (0, 0) in the real plane.

This is a complex graph x = i*y - y, there are imaginary numbers.

Step-by-step explanation:

2 xx + 4xy + 4yy = 0

...xx + 2xy + y*y + yy = 0

(x + y)^2 + yy = 0

if x is a constant

then the cross sections would be parabolas in the zy-plane

... 2yy + 4xy - 2xy + xx = 0

(2yy + 4xy + 2xx -2xx) - 2xy + xx = 0

2(y + x)^2 - 2xx - 2xy + xx = 0

2(y + x)^2 - xx - 2xy = 0

2(y + x)^2 - xx - 2xy +yy - yy = 0

...

(x + y)^2 + yy = 0

x = i*y - y

if x = r cos a

y = r sin a

cos a )^2 + cos a sina + sin a)^2 + sin a)^2 = 0

cos a sin a + sin a)^2 = -1

Answer from: Quest

answer: the answer is 23

step-by-step explanation:

Answer from: Quest

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Identify the graph of the equation. Then find 0 to the nearest degree.
2x² + 4xy + 4y2 = 0...