(8-2)2+(-4-15)2 = 397 and the square root of this is the distance which is
19.925 rounded to 3 decimal places
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.
The shortest distance between 2 parallel lines is a perpendicular drawn between 2 parallel lines the diagram shows it clearly 1 parallel line ------------------------------------|-------------------------------------------------------------------- | | | the vertical line is the shortest distance | | ------------------------------------|------------------------------------------------------------------- 2nd parallel line
The sq.root of 122+162=20
If d is the distance between them, then d2 = (-6 -10)2 + (1 - (-8))2 = (-16)2 + 92 =256 + 81 = 337 so d = sqrt(337) = 18.36
Points: (-6, 1) and (-2, -2) Distance: 5 units
13
the first point is x = 28 and y = -17. The second point is x = -15 and y = -17. Since both points have the same y coordinate then the points are on a straight horizontal line and distance is the difference of the x coordinates, or 28 - (-15) = 43
5
0 0
13 The distance between 2 points in a cordinate space is SQRT ((x2 -x1)2 + (y2- y1)2) Hence the answer to your question would be square root of (1-(-4))2 + (3-15)2 which would be square root of (25 + 144) or square root of (169) which is nothing but 13!!!
To find the distance between the points (-2, 5) and (-2, 0), we can use the distance formula. Since both points have the same x-coordinate (-2), the distance is simply the difference in their y-coordinates: |5 - 0| = 5. Therefore, the distance between the two points is 5 units.
10
8.54
You can get distance and hours and directions from Mapquest.com
You can get distance and hours and directions from Mapquest.com
To find the distance between two points in 3D space (x1, y1, z1) and (x2, y2, z2), use the distance formula: Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2) Substitute the coordinates into the formula to find the distance.
2 and -6