To find the y-coordinate of the center of the round mirror defined by the equation \( x^2 + y^2 + 4x - 6y = 0 \), we will first rewrite the equation in standard form by completing the square for both \( x \) and \( y \).
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Reorganize the equation:
\[ x^2 + 4x + y^2 - 6y = 0 \]
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Complete the square for \( x \):
- Take the coefficient of \( x \) (which is 4), halve it (2), and then square it (which gives 4).
- Add and subtract 4 inside the equation.
\[ (x^2 + 4x + 4) + y^2 - 6y = 4 \]
Which simplifies to:
\[ (x + 2)^2 + y^2 - 6y = 4 \]
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Complete the square for \( y \):
- Take the coefficient of \( y \) (which is -6), halve it (-3), and then square it (which gives 9).
- Add and subtract 9 inside the equation.
\[ (x + 2)^2 + (y^2 - 6y + 9) = 4 + 9 \]
Which simplifies to:
\[ (x + 2)^2 + (y - 3)^2 = 13 \]
Now, the equation \((x + 2)^2 + (y - 3)^2 = 13\) is in the standard form of a circle, where the center of the circle is \((-2, 3)\) and the radius is \(\sqrt{13}\).
Therefore, the y-coordinate of the mirror's center is:
\[ \boxed{3} \]