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.
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
The distance is the square root of [(6--3)squared+(2--2)squared] = 9.849 rounded to 3 decimal places
(-2-3)2+(7-7)2 = 25 and the square root of this is the distance which is 5
Points: (2, 3) and (2, 7) Distance works out as: 4 units
58
Just subtract the lowest number from the greatest number. For example, the distance between 3 and 8, is 8 - 3 = 5 units, the distance between -2 and 3, is 3 - (-2) = 3 + 2 = 5 units, the distance between -4 and -2, is -2 - (-4) = -2 + 4 = 2 units.
3 because 5- 2 is 3.
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
(3-1)2 + (5-8)2 = 13 and the square root of this is the distance between the points
The distance is the square root of [(6--3)squared+(2--2)squared] = 9.849 rounded to 3 decimal places
Points: (-5, -2) and (3, 13)Distance works out as 17 units
(-2-3)2+(7-7)2 = 25 and the square root of this is the distance which is 5
Points: (2, 3) and (2, 7) Distance works out as: 4 units
3 and 1/2 miles
3 and 1/2 miles
The distance between the points of (4, 3) and (0, 3) is 4 units
It is: 2