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What else can I help you with?
How can you find the distance between two numbers on a number line?
Just subtract the lowest number from the greatest number. For example, the distance between 3 and 8, is 8 - 3 = 5 units, the distance between -2 and 3, is 3 - (-2) = 3 + 2 = 5 units, the distance between -4 and -2, is -2 - (-4) = -2 + 4 = 2 units.
How do you find the shortest distance between two parallel lines?
The shortest distance between 2 parallel lines is a perpendicular drawn between 2 parallel lines the diagram shows it clearly 1 parallel line ------------------------------------|-------------------------------------------------------------------- | | | the vertical line is the shortest distance | | ------------------------------------|------------------------------------------------------------------- 2nd parallel line
Find the distance between the points -2 -10 and 10 6?
The sq.root of 122+162=20
Find the distance between points -6 1 and 10 -8?
If d is the distance between them, then d2 = (-6 -10)2 + (1 - (-8))2 = (-16)2 + 92 =256 + 81 = 337 so d = sqrt(337) = 18.36
What is linear scale?
A scale where the distance between 1 and 2 is the same as the distance between 2 and 3; and so on. By way of contrast, in a logarithmic scale, the distance between 1 and 10 would be the same as the distance between 10 and 100, or 100 and 1000. There are many other scales.
What is the distance between the points (3 7) and (15 16) on a coordinate plane?
To find the distance between the points (3, 7) and (15, 16) on a coordinate plane, we can use the distance formula: (d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}). Substituting the coordinates, we get (d = \sqrt{(15 - 3)^2 + (16 - 7)^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15). Therefore, the distance between the two points is 15 units.
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 15 and negative 2?
13
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
Find the distance between (6, -2) and (1, -2)?
5
Find the distance between the points: (-4, 15) and (1, 3)?
13 The distance between 2 points in a cordinate space is SQRT ((x2 -x1)2 + (y2- y1)2) Hence the answer to your question would be square root of (1-(-4))2 + (3-15)2 which would be square root of (25 + 144) or square root of (169) which is nothing but 13!!!
Find the distance between the points -2 2 and -2 2?
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Find the distance between the points with the given coordinates (-2 5) and (-2 0).?
To find the distance between the points (-2, 5) and (-2, 0), we can use the distance formula. Since both points have the same x-coordinate (-2), the distance is simply the difference in their y-coordinates: |5 - 0| = 5. Therefore, the distance between the two points is 5 units.
Find the distance between the points-2 -10 and -8 -2?
10
Find the distance between the points 2 2 and -1 -6?
8.54
What is the distance between -14 -6 and 1 -14?
To find the distance between the points (-14, -6) and (1, -14), you can use the distance formula: ( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ). Substituting the values, ( d = \sqrt{(1 - (-14))^2 + (-14 - (-6))^2} = \sqrt{(15)^2 + (-8)^2} = \sqrt{225 + 64} = \sqrt{289} = 17 ). Therefore, the distance between the two points is 17 units.
Find the Distance between 2 2 5 -1?
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.
