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raj tested his new flashlight by shinning it on his bedroom wall the beam of the light can be described by the equation x^2 y^2-2x-4y-31=0 how many inches wide is the beam of light on the wall
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May 15, 2017
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raj tested his new flashlight by shinning it on his bedroom wall the beam of the light can be described by the equation x^2 y^2-2x-4y-31=0 how many inches wide is the beam of light on the wall
Mathematics
high-school
Gary Kerr
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Answer:
12 inches.
Explanation:
Did the question, got it right.
Frank Conry
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May 17, 2017
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Frank Conry
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I can get it into standard form of that certain conic section by completing the square
(x^2-2x)+(y^2-4y)-31=0
(x^2-2x+1-1)+(y^2-4y+4-4)-31=0
(x-1)^2-1+(y-2)^2-4-31=0
(x-1)^2+(y-2)^2-36=0
(x-1)^2+(y-2)^2=36
(x-1)^2+(y-2)^2=6^2
in form
(x-h)^2+(y-k)^2=r^2
r=radius
so how wide it is is the diameter which is 2 times of radius
6=radius
6 times 2=12
12 inches I assume if the units in equation is in inches
Clam
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May 21, 2017
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Clam
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