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alyssababy7
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Complete the square and write the equation in standard form. Then give the center and radius of the circle.
x^2 + y^2  10x  8y + 29 = 0
A. (x  5)^2 +(y  4)^2 = 12 (5, 4), r = 12
B. (x  5)^2 +(y  4)^2 = 12 (5, 4), r = 2 sq rt 3
C. (x + 5)^2 +(y + 4)^2 = 12 (5, 4), r = 2 sq rt 3
 one year ago
 one year ago
alyssababy7 Group Title
Complete the square and write the equation in standard form. Then give the center and radius of the circle. x^2 + y^2  10x  8y + 29 = 0 A. (x  5)^2 +(y  4)^2 = 12 (5, 4), r = 12 B. (x  5)^2 +(y  4)^2 = 12 (5, 4), r = 2 sq rt 3 C. (x + 5)^2 +(y + 4)^2 = 12 (5, 4), r = 2 sq rt 3
 one year ago
 one year ago

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Reaper534 Group TitleBest ResponseYou've already chosen the best response.0
is this calculis
 one year ago

alyssababy7 Group TitleBest ResponseYou've already chosen the best response.0
Yes precalc
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
start with \[x^2 + y^2  10x  8y + 29 = 0\] then group the terms together as \[x^210x+y^28y=29\] to complete the square, take half the coefficient of the \(x\) and \(y\) term and write \[(x5)^2+(y8)^2=29+5^2+4^2\] then compute the number on the right
 one year ago

Reaper534 Group TitleBest ResponseYou've already chosen the best response.0
ooh thats y im in geometry srry
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
you get \[(x5)^2+(y4)^2=12\] so the circle has center at \((5,4)\) and radius \(\sqrt{12}\) or \(2\sqrt{3}\) if you prefer
 one year ago

alyssababy7 Group TitleBest ResponseYou've already chosen the best response.0
thank you
 one year ago

Reaper534 Group TitleBest ResponseYou've already chosen the best response.0
see u marrow alyssa :) bye
 one year ago
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