Points: (-5, -2) and (3, 13)Distance works out as 17 units
3.61 units
A. 8.54 units B. 7.28 units C. 3.61 units D. 12.04 units
The distance is the square root of 125 (which is about 11.18) 1 to -9 = 10 units -2 to 3 = 5 units distance = sqrt (102 + 52) = sqrt (125)
Points: (2, 4) and (5, 0) Distance: 5
Points: (-5, -2) and (3, 13)Distance works out as 17 units
Points: (1, -2) and (1, -5) Distance: 3 units by using the distance formula
Points: (-6, 1) and (-2, -2) Distance: 5 units
If you mean points of (-5, -3) and (7, 7) then by using the distance formula it is 120 units
Using Pythagoras: 5 squared + 2 squared = √21 which is about 4.58.
10 units
3.61 units
A. 8.54 units B. 7.28 units C. 3.61 units D. 12.04 units
To find the distance between two points (x0, y0) and (x1, y1) use Pythagoras: distance = √(change_in_x² + change_in_y²) → distance = √((x1 - x0)² + (y1 - y0)²) → distance = √((8 - -7)² + (-5 - -13)²) → distance = √(15² + 8²) → distance = √289 → distance = 17 units.
If you mean points of (-2, 4) and (5, 4) then using the distance formula it is 7
The run, combined with the rise (the distance in units up) creates the slope of a line. In the slope 5/3 , 5 is the rise and 3 is the run, meaning that to find the next point on the line you would first move up five units, then go to the right 3 units.
The distance is the square root of 125 (which is about 11.18) 1 to -9 = 10 units -2 to 3 = 5 units distance = sqrt (102 + 52) = sqrt (125)