The formula is: The distance of a point in 3D space from point (0,0,0) is the hypotenuse of a right triangle where the adjacent sides have lengths of z, and square root of (x² + y²).
What is the formula?
Calculate it with this and I guarantee the 10 points.
A Sphere: (2, -5, 8). Find the distance to the center.
What is the formula?
Calculate it with this and I guarantee the 10 points.
A Sphere: (2, -5, 8). Find the distance to the center.
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Here,
origin O(0, 0, 0),
and center of sphere C(2, -5, 8),
distance, OC = sqrt[(2-0)^2 +(-5-0)^2 +(8-0)^2],
OC = sqrt[4+25+64] = sqrt(93) = 9.644 >==========< ANSWER
origin O(0, 0, 0),
and center of sphere C(2, -5, 8),
distance, OC = sqrt[(2-0)^2 +(-5-0)^2 +(8-0)^2],
OC = sqrt[4+25+64] = sqrt(93) = 9.644 >==========< ANSWER
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Your assumption is wrong!. The "other guy" did not give anyone a "thumbs down", lotosoya.
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D(x,y,z) = √(x²+y²+z²)
D(2, -5, 8) = √(4+25+64) = √93 ≈ 9.644
If the point (2, -5, 8) is on a sphere centered on the origin, the radius of the sphere is √93 or about 9.644.
You've already said that the distance is the hypotenuse of a right triangle with sides equal to z and √(x²+y²). From the Pythagorean Theorem, the length of the hypotenuse is the square root of the square of the sides:
D = √{z² + [√(x² + y²)]²} = √[z² + (x² + y²)] = √(x²+y²+z²)
D(2, -5, 8) = √(4+25+64) = √93 ≈ 9.644
If the point (2, -5, 8) is on a sphere centered on the origin, the radius of the sphere is √93 or about 9.644.
You've already said that the distance is the hypotenuse of a right triangle with sides equal to z and √(x²+y²). From the Pythagorean Theorem, the length of the hypotenuse is the square root of the square of the sides:
D = √{z² + [√(x² + y²)]²} = √[z² + (x² + y²)] = √(x²+y²+z²)