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Use the rule of Pythagoras - calculate the distance as squareroot(deltax2 + deltay2), where deltax and deltay are the differences in the x and y coordinates, respectively.
Use the rule of Pythagoras - calculate the distance as squareroot(deltax2 + deltay2), where deltax and deltay are the differences in the x and y coordinates, respectively.
Use the rule of Pythagoras - calculate the distance as squareroot(deltax2 + deltay2), where deltax and deltay are the differences in the x and y coordinates, respectively.
Use the rule of Pythagoras - calculate the distance as squareroot(deltax2 + deltay2), where deltax and deltay are the differences in the x and y coordinates, respectively.
What else can I help you with?
Find the distance between the points -5 -3 and 7 6?
15
Find the distance between the points -4 15 and 6 -9?
26
What the distance between two points 6 5 and 30 15?
Points: (6, 5) and (30, 15) Distance: 26 by using the distance formula
What is the distance between the points (37) and (1516) on a coordinate plane?
To find the distance between the points (3, 7) and (15, 16) on a coordinate plane, you can use the distance formula: ( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ). Plugging in the values, ( d = \sqrt{(15 - 3)^2 + (16 - 7)^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15 ). Therefore, the distance between the points is 15 units.
What is the distance between (-63) and (6-6)?
If you mean points of (-6, 3) and (6, -6) then the distance works out as 15
Find the distance between the points -4 15 and 6 -9?
26
Find the distance between the points -5 -3 and 7 6?
15
Find the distance between the points A 9 3 and B 15 11?
10
What is the distance between the points 3 7 and 15 16 on a coordinate plane?
Distance between the points of (3, 7) and (15, 16) is 15 units
What the distance between two points 6 5 and 30 15?
Points: (6, 5) and (30, 15) Distance: 26 by using the distance formula
What is the distance between (15 -17) and (-7 -17) in the coordinate?
The distance works out as 22 between the points of (15, -17) and (-7, -17)
What is the distance between the points (37) and (1516) on a coordinate plane?
To find the distance between the points (3, 7) and (15, 16) on a coordinate plane, you can use the distance formula: ( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ). Plugging in the values, ( d = \sqrt{(15 - 3)^2 + (16 - 7)^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15 ). Therefore, the distance between the points is 15 units.
Explain how to find the distance between the points (28-17) and (-15-17) on a coordinate plane?
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
What is the distance between the points (-7 -13) and (8 -5)?
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.
What is the distance between (-63) and (6-6)?
If you mean points of (-6, 3) and (6, -6) then the distance works out as 15
What is the distance between the points -4 17 and 11 17 in the xy-plane?
It is the square root of (-4-11)2+(17-17)2 which works out as 15.
Explain how to find the distance between the points (28 -17) and (-15-17) on a coordinate plane?
To find the distance between any two points on the Cartesian plane use Pythagoras: The distance between (x0, y0) and (x1, y1) is given by: distance = √((x1 - x0)² + (y1 - y0)²) → distance between (28, -17) and (-15, -17) is: distance = √((x1 - x0)² + (y1 - y0)²) = √((-15 - 28)² + (-17 - -17)²) = √((-43)² + (0)) = √1849 = 43 ------------------------ In this case, the y-coordinates are the same (y0 = y1 = -17), so this becomes: distance = √((x1 - x0)² + (y0 - y0)²) = √((x1 - x0)² + 0²) = √((x1 - x0)²) = |x1 - x0| The vertical bars around the expression mean the absolute value of the expression, which is the numerical value of the expression ignoring the sign. distance = |x1 - x0| = |-15 - 28| = |-43| = 43.
