What region in 3D space is the inequality (1<=x^2+y^2+z^2<=4 ) ?
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Answer:
Sphere is correct. The surface of a sphere in 3D space is ax^2 + by^2 + cz^2 = r^2 where a=b=c. x^2 = r^2 -> Parabola x^2 + y^2 = r^2 -> Circle x^2 + y^2 + z^2 = r^2 -> Sphere x^2 + y^2 + z^2 + w^2 = r^2 -> Hyper-Sphere (4D space) v(0)^2 + v(1)^2 + v(2)^2 + ... + v(n)^2 = r^2 -> n-Sphere (nD+1 space)
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Other answers
Sphere is correct. The surface of a sphere in 3D space is ax^2 + by^2 + cz^2 = r^2 where a=b=c. x^2 = r^2 -> Parabola x^2 + y^2 = r^2 -> Circle x^2 + y^2 + z^2 = r^2 -> Sphere x^2 + y^2 + z^2 + w^2 = r^2 -> Hyper-Sphere (4D space) v(0)^2 + v(1)^2 + v(2)^2 + ... + v(n)^2 = r^2 -> n-Sphere (nD+1 space)
Brian
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