The x distance is (-2-(-8)) = 6
The y distance is (-1-(-7)) = 6
The point to point distance is from Pythagorean theorem square root of (x squared +y squared) = 8.485
You can use the Pythagorean Theorem to calculate the distance between two points. The distance is the square root of ((difference in x-coordinates)2 + (difference in y-coordinates)2). ________________________________________________________________________________________ The distance is the diagonal of a right angled triangle with equal two side and each equal 6. So, the distance is: (36 + 36)1/2 = 6 x 1.4142 = 8.4853
The distance between (3, 7) and (-3, -1) is sqrt{[(3 - (-3)]2 + [7 - (-1)]2} = sqrt{[3 + 3]2 + [7 + 1]2} =sqrt{[6]2 + [8]2} = sqrt{36 + 64} = sqrt{100} = 10 units.
6?! 1-2 2-3 3-4 4-5 5-6 6-7
Points: (7, -1) and (-2, -4) Slope: 1/3
The answer is the difference between the lengths, which is 11 yards (or 33 feet).To find the answerIt is easier to first express the first distance in feet and eliminate the fraction.7 1/3 yards x 3 = 22 feetTo find the second distance, multiply by 2 1/2. (22 x 2 1/2 = 55 feet)Subtract to find the difference 55-22 = 33 feet, which is 33/3 or 11 yards.To find the answer by multiplying fractions (which is the point of the problem)The second distance, the distance Carl threw the ball, is 7 1/3 x 2 1/2.To multiply mixed fractions (like 7 1/3), it's a good idea to convert them to improper fractions first. 7 1/3 = 22/3 and 2 1/2 = 5/222/3 x 5/2 = 110/6 = 18 2/6 yards = 18 1/3 yards (55 feet).Subtracting Devin's distance, 18 1/3 - 7 1/3 = 11 yards (33 feet).
djk = [ ( xj - xk )^2 + ( yj - yk )^2 ]1/2djk = [ ( 2 - -4 )^2 + ( 1 - -7 )^2 ]^1/2djk = [ ( 6 )^2 + ( 8 )^2 ]^1/2djk = [ 36 + 64 ]^1/2 = [ 100 ]^1/2 = 10
You can use the Pythagorean Theorem to calculate the distance between two points. The distance is the square root of ((difference in x-coordinates)2 + (difference in y-coordinates)2). ________________________________________________________________________________________ The distance is the diagonal of a right angled triangle with equal two side and each equal 6. So, the distance is: (36 + 36)1/2 = 6 x 1.4142 = 8.4853
|AB| = sqrt[(5 - 2)2 + (7 - 4)2] =sqrt[9 + 9] = 3*sqrt(2)
58 * * * * * No. Distance squared = (-2 - 3)2 + (1 - 7)2 = 52 + 62 = 25 + 36 = 61 So distance = sqrt(61) = 7.8 approx.
The distance between two points on a coordinate plane is calculated using the distance formula: Distance = √((x2 - x1)^2 + (y2 - y1)^2) In this case, the coordinates of the two points are (7, 1) and (7, 3). Since the x-coordinates are the same, we only need to calculate the difference in the y-coordinates, which is (3 - 1) = 2. Plugging this into the distance formula gives us: Distance = √((0)^2 + (2)^2) = √4 = 2. Therefore, the distance between the two points is 2 units.
Straight line equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Point: (7, 5) Equations intersect: (4, 1) Perpendicular distance: square root of [(7-4)2+(5-1)2] = 5
The distance between (3, 7) and (-3, -1) is sqrt{[(3 - (-3)]2 + [7 - (-1)]2} = sqrt{[3 + 3]2 + [7 + 1]2} =sqrt{[6]2 + [8]2} = sqrt{36 + 64} = sqrt{100} = 10 units.
58
Straight line equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Point of intersection: (4, 1) Distance: (7-4)2+(5-1)2 = 25 and the square root of this is the perpendicular distance which is 5 units of measurement
(8-7)2+(0-0)2 = 1 and the square root of 1 is 1 Answer: 1
The Pythagorean distance is sqrt[(4 - 7)2 + (7 - 8)2] = sqrt[9 + 1] = sqrt(10) = 3.162 approx.
A 1 -7 B -2 5 C 2 7 D 7 2 E 7 -1