2007-09-12 15:14:14 · 10 answers · asked by hanna 1 in Science & Mathematics ➔ Mathematics
d ² = (3 - 4) ² + (- 2 - 1) ²
d ² = (-1) ² + (- 3) ²
d ² = 1 + 9
d ² = 10
d = √10
2007-09-14 19:39:50 · answer #1 · answered by Como 7 · 0⤊ 0⤋
use Pythagorean's theorem:
hypotenuse = sqrt [ a^2 + b^2 ]
a = 4 - 3 = 1
b = 1 - -2 = 1 + 2 = 3
hypotenuse = distance between the points = sqrt (1+9)
= sqrt(10)
2007-09-12 22:18:45 · answer #2 · answered by Pakyuol 7 · 2⤊ 0⤋
Distance formula!
distance = square root of [ (x2-x1)^2 + (y2 - y1)^2 ]
so (3 - 4)^2 + (-2 - 1)^2
(-1)^2 + (-3)^2
1+9
10
Take the square root, and your answer is about 3.16
2007-09-12 22:20:51 · answer #3 · answered by Regina 1 · 0⤊ 0⤋
The points are 1 apart on the x-axis, and 3 on the y
distance = sqrt(1 + 9) = sqrt (10)
2007-09-12 22:20:14 · answer #4 · answered by Beardo 7 · 0⤊ 0⤋
use the distance formula given that x1=4 and y1=1; x2=3 and y2= - 2
2007-09-12 22:19:57 · answer #5 · answered by jiggins 2 · 0⤊ 0⤋
distance is equal to square root of (4-3)^2+(1+2)^2, that will be square root of 10... (sorry I dont have a calculator here in the office... lol...)
2007-09-12 22:22:27 · answer #6 · answered by criselda 3 · 0⤊ 0⤋
distance formula
d = sqrt of ( (x2 - x1)^2 + (y2 - y1)^2 )
= sqrt of ( (1 - -2)^2 + (4 - 3)^2 )
= sqrt of (9 + 1)
= sqrt of 10
2007-09-12 22:16:24 · answer #7 · answered by cjcourt 4 · 2⤊ 0⤋
sqrt[(4-3)^2 + (1-(-2))^2]
sqrt[1^2 + 3^2]
sqrt[1 + 9]
sqrt[10]
2007-09-12 22:17:54 · answer #8 · answered by Runa 7 · 1⤊ 0⤋
The square root of ten
2007-09-12 22:18:57 · answer #9 · answered by Steve B 6 · 1⤊ 0⤋
you could graph it and do pythagorean theorem. or you could use the formula radical[(x1-x2)^2+(y1-y2)^2] the answer ends up being
2007-09-12 22:20:20 · answer #10 · answered by crosiegiveadvice 1 · 0⤊ 0⤋