To find the length of segment AB, we can use the distance formula. Given points A (-1, 3) and B (11, -8), the length of AB is calculated as follows:
[ AB = \sqrt{(11 - (-1))^2 + (-8 - 3)^2} = \sqrt{(11 + 1)^2 + (-11)^2} = \sqrt{12^2 + (-11)^2} = \sqrt{144 + 121} = \sqrt{265} \approx 16.28. ]
Therefore, the length of AB is approximately 16.28 units.
<ab> = |a|*|b|*cos(x) where |a| is the length of the vector a, |b| is the length of the vector b, and x is the angle between them.
To find the length of segment AB between points A(-1, -3) and B(11, -8), we can use the distance formula: [ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ] Substituting the coordinates, we have: [ d = \sqrt{(11 - (-1))^2 + (-8 - (-3))^2} = \sqrt{(12)^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13. ] Thus, the length of AB is 13 units.
Area of Triangle= 1/2(ab) You must multiply length b and length a together and then half it.
To find the length of the line segment AB, you can use the distance formula: ( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ). For points A(0, 0) and B(6, 3), the calculation is ( AB = \sqrt{(6 - 0)^2 + (3 - 0)^2} = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} ). Therefore, the length of AB is ( 3\sqrt{5} ).
b*ab = ab2 Suppose b*ab = ab + b2. Assume a and b are non-zero integers. Then ab2 = ab + b2 b = 1 + b/a would have to be true for all b. Counter-example: b = 2; a = 3 b(ab) = 2(3)(2) = 12 = ab2 = (4)(3) ab + b2 = (2)(3) + (2) = 10 but 10 does not = 12. Contradiction. So it cannot be the case that b = 1 + b/a is true for all b and, therefore, b*ab does not = ab + b2
Length AB is 17 units
The length of ab can be found by using the Pythagorean theorem. The length of ab is equal to the square root of (0-8)^2 + (0-2)^2 which is equal to the square root of 68. Therefore, the length of ab is equal to 8.24.
Endpoints: A (-2, -4) and B (-8, 4) Length of AB: 10 units
Using the distance formula the length of ab is 5 units
Using the distance formula the length of ab is 5 units
If you mean endpoints of (-1, -3) and (11, -8) then the length works out as 13
6.71
<ab> = |a|*|b|*cos(x) where |a| is the length of the vector a, |b| is the length of the vector b, and x is the angle between them.
If you mean endpoints (-1, -3) and (11, -8) then by using the distance formula the length between the points is 13 units
The length is 3*sqrt(5) = 6.7082, approx.
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End points: (-2, -4) and (-8, 4) Length of line AB: 10