Question

write the given equation in standard form: x^2+y^2-4x-6y-13=0

Answer 1

How about (x-2)^2 + (y-3)^2 = 26

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Answer 2

do you know how to 'complete the square"?

Answer 3

first rewrite as
\[x^2-4x+y^2-6x=13\] then half of -4 is -2 and half of -6 is -3 so write as
\[(x-2)^2+(y-3)^2=13+2^2+3^2=13+4+9=26

Answer 4

\[(x-2)^2+(y-3)^2=13+2^2+3^2=13+4+9=26\]

Answer 5

\((x-2)^2+(y-3)^2=13+4+9\)
\((x-2)^2+(y-2)^2=30\)
\(\frac{(x-2)^2}{30}+\frac{(y-3)^2}{30}=1\)

Answer 6

lol you were correct satellite

Answer 7

hi watchman

Answer 8

yeah arithmetic is tough this time of night!

Answer 9

still energetic answering questions :)?

Answer 10

so is typing "watchmath" instead of "watchman" sorry

Answer 11

thanks guys, very helpful! I appreciate it!

Answer 12

bb911 is it clear how to complete a square?

Answer 13

yes, I understand now, thanks!