1 over 12
Distance = (9-5)2+(-6-1)2 = 65 and the square root of this is the distance between the points which is about 8.062257748
the answer is 3/16 and there more click here 1/6 = 2/12 2/6 = 4/12 Halfway in between is 3/12 = 1/4
For (2,1) & ( 14,6) Use Pythagoras. d^2 = (2 - 14)^2 + ( 1 - 6)^2 d^2 = (-12)^2 + (-5)^2 d^2 = 144 + 25 d^2 = 169 d = sqrt(169) d = 13
4
1 over 12
If you mean points of (-4, 2) and (1, 2) then the distance works out as 5
It is [(1 - x1)^2 + (2 - y1)^2]^0.25
12 and 1/2 miles
Area = 1/2 x sum_of_lengths_of_parallel_sides x distance_between_themIf 14 in and 12 in are the parallel sides and 8 in is the distance between them:Area = 1/2 x (14 in + 12 in) x 8 in = 104 in2If 14 in and 8 in are the parallel sides and 12 in is the distance between them:Area = 1/2 x (14 in + 8 in) x 12 in = 132 in2If 12 in and 8 in are the parallel sides and 14 in is the distance between them:Area = 1/2 x (12 in + 8 in) x 14 in = 140 in2
What is the distance between (4, -2) and (-1,6)?
20 km, about 12 and 1/2 miles
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 There is a distance of 12.
What is the distance between (4, -2) and (-1,6)?
(12, 4) and (12, - 10)Distance = sqrt[(Y2 - Y1)2 + (X2 - X1)2]Distance = sqrt[(- 10 - 12)2 + (12 - 12)2]Distance = sqrt(- 22)2Distance = 22==============Could be...(0, 22) as I am not sure here.
Points: (-6, 1) and (-2, -2) Distance: 5 units
Use Pythagoras to find the distance between two points (x0,.y0) and (x1, y1): distance = √(change_in_x² + change_in_y²) → distance = √((x1 - x0)² + (y1 - y0)²) → distance = √((4 - 1)² + (-1 -2)²) → distance = √(3² + (-2)²) → distance = √(9 + 9) → distance = √18 = 3 √2