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15 = sq rt of (81+144)
Distance formula between two points (in the plane): d = √[(x2 - x1)^2 + (y2 - y1)^2]
Let A be the point with cordinates (x1, y1), and B the point with coordinates (x2, y2).
Substitute the given values into that formula:
d = √[(x2 - x1)^2 + (y2 - y1)^2]
d = √[(3 - -6)^2 + (-8 - 4)^2]
d = √[9^2 + (-12)^2]
d = √[81 +144)
d = √225
d = 15
Thus, the distance from point A to point B is 15.
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∙ 14y ago
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What is the distance between point -3 5 and point 4 -6 in the coordinate plane?
The answer depends on the metric used. The Euclidean distance is sqrt[(-3-4)2 + (5+6)2] = sqrt[72 + 112] =sqrt(49 + 121) = sqrt(170) = 13.0384 (to 6 sf). The Minkowsky distance, on the other hand, is |-3-4| + |5+6| = 7 + 11 = 18. There are other metrics.
What is the distance between point negative 6 1 and point 4 3?
It is the square root of (-6-4)2+(1-3)2 = 2 times sq rt of 26 or about 10.198 to 3 decimal places
What is the distance from the point of 6 5 that meets the line of 4 -6 and 2 -3 at right angles on the Cartesian plane giving answer to an appropriate degree of accuracy?
Points: (4, -6) and (2, -3) Slope: -3/2 Equation: 2y = -3x Perpendicular slope: 2/3 Perpendicular equation: 3y = 2x+3 Both equations meet at: (-6/13, 9/13) Distance from (6, 5) to (-6/13, 9/13) is 7.766 units rounded to 3 decimal places
The distance from the green point on the parabola to the parabola's focus is 6 What is the distance from the green point to the directrix?
answer is 6
What is the slope of the line that passes through point 6 8 4 -3?
Points: (6, 8) and (4, -3) Slope: (-3-8)/(4-6) = 11/2 or 5.5
What is the distance between point -3 5 and point 4 -6 in the coordinate plane?
The answer depends on the metric used. The Euclidean distance is sqrt[(-3-4)2 + (5+6)2] = sqrt[72 + 112] =sqrt(49 + 121) = sqrt(170) = 13.0384 (to 6 sf). The Minkowsky distance, on the other hand, is |-3-4| + |5+6| = 7 + 11 = 18. There are other metrics.
What is the distance between point negative 6 1 and point 4 3?
It is the square root of (-6-4)2+(1-3)2 = 2 times sq rt of 26 or about 10.198 to 3 decimal places
Find the distance between the points 4 6 and 4 6?
Since they are the same point, the distance between them is 0.
What is the distance between points (9 4) and (3 4) on a coordinate plane?
The distance between points: (9, 4) and (3, 4) is 6
What is the transformation of B(2 4) when dilated with a scale factor of ½ using the point (4 6) as the center of dilation?
When doing enlargements through a centre, the new position of any point is the distance of that point from the centre multiplied by the scale factor; it is easiest to treat the x- and y- coordinates separately.To enlarge (2, 4) by a scale factor of ½ with (4, 6) as the centre of enlargement:x: distance is (4 - 2) = 2 → new distance is 2 × ½ = 1 → new x is 2 + 1 = 3y: distance is (6 - 4) = 2 → new distance is 2 × ½ = 1 → new y is 4 + 1 = 5→ (2, 4) when enlarged by a scale factor of ½ with a centre of (4, 6) transforms to (3, 5).
Find the distance between (-4 2) and (-7 -6). in units?
The distance between the points of (4, 3) and (0, 3) is 4 units
What is the distance between two points -3 7 and -6 4?
(-3-(-6))2 + (7-4)2 = 18 and the square root of this is the distance between the two points
What is the distance from the point of 6 5 that meets the line of 4 -6 and 2 -3 at right angles on the Cartesian plane giving answer to an appropriate degree of accuracy?
Points: (4, -6) and (2, -3) Slope: -3/2 Equation: 2y = -3x Perpendicular slope: 2/3 Perpendicular equation: 3y = 2x+3 Both equations meet at: (-6/13, 9/13) Distance from (6, 5) to (-6/13, 9/13) is 7.766 units rounded to 3 decimal places
What is the distance between point -4 -6 and point 4 2?
sqrt[(-4 - 4)2 + (-6 - 2)2] = sqrt[82 + 82] = sqrt(64 + 64] = sqrt(128) = 11.31 approx
What is the distance between the numbers 1 and 7?
6?! 1-2 2-3 3-4 4-5 5-6 6-7
Find the co-ordinate of points on the x-axis which are at a distance of 5 units from the point (6,-3)?
To find the coordinates of points on the x-axis that are 5 units away from the point (6, -3), we can use the distance formula. The distance formula is: Distance = √((x2 - x1)^2 + (y2 - y1)^2) In this case, we know that x1 = 6, y1 = -3, and distance = 5. We also know that the points are on the x-axis, so the y-coordinate is 0. So we can plug these values into the distance formula and solve for x2: 5 = √((x2 - 6)^2 + (0 - (-3))^2) 5 = √(x2 - 6)^2 + 9 25 = (x2 - 6)^2 x2 = √25 + 6 = √16 + 6 = 4 + 6 = 10 Therefore, the coordinates of the point on the x-axis that is 5 units away from (6, -3) in the positive direction of x-axis are (10, 0) and the point on the x-axis that is 5 units away from (6, -3) in the negative direction of x-axis is (2,0).
The distance from the green point on the parabola to the parabolas focus is 6 What is the distance from the green point to the directrix?
6
