Distance = (9-5)2+(-6-1)2 = 65 and the square root of this is the distance between the points which is about 8.062257748
Pythagoras can be used to find the distance between any two points (x0, y0) and (x1, y1): Distance = √((x1 - x0)² + (y1 - y0)²) → distance = √((5 - 5)² + (13 - 9)²) = √(0 + 4²) = √(4²) = 4.
It is sqrt{[(-5 - (-2)]2 + (1 - 3)2} = sqrt(9 + 4) = sqrt(13) = 3.6056 approx.
Halfway between 2/9 and 5/9 is 3.5/9, more correctly expressed as 7/18.
In order to find the distance between two coordinates, you first need to find the difference between the x and y coordinates. In this case, the difference between the x coordinates is 3-(-2) = 5. The difference between the y coordinates is -4-5 = -9. To find the distance you add up the squares of these differences then find the square root. 52 = 25. -92 = 81. 25+81 = 106. Thus the distance is the square root of 106, or approximately 10.296
It is: 2
Distance = (9-5)2+(-6-1)2 = 65 and the square root of this is the distance between the points which is about 8.062257748
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Points: (-1, -9) and (4, -2) Distance: (-1-4)2+(-9--2)2 = 74 and the square root of this is the distance which is about 8.602 to 3 decimal places
square root of (5-9)^2+(1+6)^2
Sqrt[(5-4)2 + (5-4)2] = sqrt[1 + 1] = sqrt(2)
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.
square root of (5-9)^2+(1+6)^2
Distance = sqrt[(Y2 - Y1)2 + (X2 - X1)2]Distance = sqrt[(9 - - 3)2 + (0 - 5)2]Distance = sqrt[(12)2 + (- 5)2]Distance = sqrt(144 + 25)Distance = sqrt(169)Distance = 13 units================
|AB| = sqrt[(5 - 2)2 + (7 - 4)2] =sqrt[9 + 9] = 3*sqrt(2)
If you mean points of (-3, 5) ans (9, -2) then the distance works out as the square root of 193 which is about 14 rounded to the nearest integer
Dist2 = [(5 - 0)2 + (3 - 9)2] = [52 + 62] = 25 + 36 = 61 So distance = sqrt(61) = 7.81 (approx).