SOLUTION: Where is the center of 25(x+5)^2 + 4(y-3)^2 = 100 ?

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Question 6524: Where is the center of 25(x+5)^2 + 4(y-3)^2 = 100 ?
Answer by rapaljer(4551) About Me  (Show Source):
You can put this solution on YOUR website!
Zero out the (x+5), which gives you x=-5, and zero out the (y-3), which gives you y=3. The center of this (it is an ellipse) is at the point (-5,3).

R^2 at SCC