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The Euclidean distance is sqrt[(-2 - 3)2+ (2 - -2)2] = sqrt[52+ 42] = sqrt[25 + 16] = sqrt(41) = 6.40 approx.
The Euclidean distance is sqrt[(-2 - 3)2+ (2 - -2)2] = sqrt[52+ 42] = sqrt[25 + 16] = sqrt(41) = 6.40 approx.
The Euclidean distance is sqrt[(-2 - 3)2+ (2 - -2)2] = sqrt[52+ 42] = sqrt[25 + 16] = sqrt(41) = 6.40 approx.
The Euclidean distance is sqrt[(-2 - 3)2+ (2 - -2)2] = sqrt[52+ 42] = sqrt[25 + 16] = sqrt(41) = 6.40 approx.
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What is the distance between points 1 2 and 4 -1?
Use Pythagoras to find the distance between two points (x0,.y0) and (x1, y1): distance = √(change_in_x² + change_in_y²) → distance = √((x1 - x0)² + (y1 - y0)²) → distance = √((4 - 1)² + (-1 -2)²) → distance = √(3² + (-2)²) → distance = √(9 + 9) → distance = √18 = 3 √2
What is the distance between (6 2) and (-3 -2)?
The distance is the square root of [(6--3)squared+(2--2)squared] = 9.849 rounded to 3 decimal places
What is the distance between two points -2 7 and 3 7?
(-2-3)2+(7-7)2 = 25 and the square root of this is the distance which is 5
What is the distance between the points named by (2 3) and (2 7) on the coordinate plane?
Points: (2, 3) and (2, 7) Distance works out as: 4 units
What is the distance between -2 1 and 3 7?
58
