Use Pythagoras:
distance = √(difference_in_x^2 + difference_in_y^2)
= √((6 - -3)^2 + (2 - -2)^2)
= √(9^2 + 4^2)
= √(81 + 16)
= √97 ≈ 9.85 units
(-3-(-6))2 + (7-4)2 = 18 and the square root of this is the distance between the two points
3 and 1/2 miles
The distance between the points of (4, 3) and (0, 3) is 4 units
-3
The distance between points: (9, 4) and (3, 4) is 6
The distance is the square root of [(6--3)squared+(2--2)squared] = 9.849 rounded to 3 decimal places
(-3-(-6))2 + (7-4)2 = 18 and the square root of this is the distance between the two points
3 and 1/2 miles
3 and 1/2 miles
If you mean points of (6, -2) and (3, -9) then it is the square root of 58 using the distance formula
Using Pythagoras: distance = √(difference_in_x^2 + difference_in_y^2) = √((6 - 2)^2 + (3 - 4)^2) = √(16 + 1) = √17 ≈ 4.12
The distance between the points of (4, 3) and (0, 3) is 4 units
6?! 1-2 2-3 3-4 4-5 5-6 6-7
To find the distance between the points (7, 6) and (-5, 3), you can use the distance formula: (d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}). Plugging in the values, we get (d = \sqrt{(-5 - 7)^2 + (3 - 6)^2} = \sqrt{(-12)^2 + (-3)^2} = \sqrt{144 + 9} = \sqrt{153}). The distance is approximately 12.25 units.
If you mean: (4, 6) and (7, -3) then it is:- Distance is the square root of (4-7)^2+(6--3)^2 = 9.487 rounded to 3 decimal places
Points: (6, -2) and (6, 2)Using the distance formula: 4
What is the distance between (4, -2) and (-1,6)?