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lesantik [10]
3 years ago
5

How to solve n²-10n+22=-2

Mathematics
1 answer:
Aliun [14]3 years ago
7 0
It's almost solved. Just move the -2 to the other side.
n^2-10n+22=-2
n^2-10n+24 done!
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Answer: \frac{(x-4)^2}{196} +\frac{(y-1)^1}{169} =1

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You first need to determine if the ellipse is horizontal or vertical, which can be found by the length of the major axis.

The vertices are at (18,1) and (-10,1). The distance is 28 units.  (horiz)

The co-vertices are at (4,14) and (4,-12) The distance 26 units (vert)

The major axis is horizontal, so the standard form of the ellipse is

\frac{(x-h)^2}{a^2} +\frac{(y-k)^2}{b^2} =1 where (h,k) are the coordinates of the center.

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The center is at (4,1)

Since the major axis is 2a, a would be 28 divided by 2=14 squared : 196

The minor axis is 2b, so b would e 26/2=13 squared: 169

Inserting the center and a,b values into the standard form eq:

\frac{(x-4)^2}{196} +\frac{(y-1)^1}{169} =1

hope this helps.

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