2007-02-22 03:49:01 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The distance formula:
y=√(y2-y1)^2+(x2-x1)^2
y=√(0-0)^2-(4-3)^2
y=√1=1
2007-02-22 04:23:16 · answer #1 · answered by Anonymous · 4⤊ 0⤋
The distance is 1. (3-4)^2 + (0-0)^2 = 1^2 = 1
2007-02-22 03:55:19 · answer #2 · answered by George V 1 · 1⤊ 0⤋
Distance formula
d = √(x₂- x₁)² + (y₂- y₁)²
d = √(4 - 3)² + (0 - 0)²
d = √(1)² + (0)²
d = 1
- - - - - - - - - -s-
2007-02-22 07:16:06 · answer #3 · answered by SAMUEL D 7 · 1⤊ 0⤋
distance = sqrt ((4-3)^2 + (0-0)^2) = sqrt(1) = 1
2007-02-22 03:54:01 · answer #4 · answered by ironduke8159 7 · 1⤊ 0⤋
once you place the factors on a graph and combine the factors, it is going to become a triangle. so with the aid of using the gap formulation, the gap from: F to E= 2 E to D= ?10 D to F=5?2 you could basically use the Pythagorean theorem on a spectacular triangle so do no longer understand the thank you to define the gap between the factors in this triangle cuz that is no longer a spectacular triangle.
2016-11-24 23:56:13 · answer #5 · answered by rasavong 4 · 0⤊ 0⤋
distance = 1
you stay on the "X" line
2007-02-22 03:54:35 · answer #6 · answered by danrouthier 2 · 1⤊ 0⤋
d=sqrt(((y2-y1)^2)+((x2-x1)^2))
d=sqrt(0+1)
d=1
2007-02-22 03:55:31 · answer #7 · answered by Al3xa 2 · 1⤊ 0⤋