SOLUTION: Where is the center of 25(x+5)^2 + 4(y-3)^2 = 100 ?
Algebra
->
Sequences-and-series
-> SOLUTION: Where is the center of 25(x+5)^2 + 4(y-3)^2 = 100 ?
Log On
Algebra: Sequences of numbers, series and how to sum them
Section
Solvers
Solvers
Lessons
Lessons
Answers archive
Answers
Click here to see ALL problems on Sequences-and-series
Question 6524
:
Where is the center of 25(x+5)^2 + 4(y-3)^2 = 100 ?
Answer by
rapaljer(4671)
(
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