2007-01-26 12:17:08 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The formula for the distance is:
sqrt((x2-x1)^2+(y2-y1)^2)
sqrt(((-1-(-2))^2+((-5-(-3))^2)
sqrt(1^2+(-2)^2)
sqrt(1+4)
sqrt(5)
~ 2.236
2007-01-26 12:23:44 · answer #1 · answered by Mark P 5 · 0⤊ 0⤋
isn't it something like
y2-y1 / x2-x1
so (x,y) =
-3--5 / -2--1
-3 + 5 / -2 +1
2/-1
=
-2
I could be wrong it has been awhile so check in your book!
2007-01-26 20:27:09 · answer #2 · answered by crackermelons 3 · 0⤊ 0⤋
(-1,-5) and (-2,-3)
D = sqrt((-2 - (-1))^2 + (-3 - (-5))^2)
D = sqrt((-2 + 1)^2 + (-3 + 5)^2)
D = sqrt((-1)^2 + 2^2)
D = sqrt(1 + 4)
D = sqrt(5)
ANS : about 2.2361
2007-01-26 22:44:04 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
the answer is the square root of 13 .. but i don't have a calculator with the square root function because i left school a long time ago... i guess its about 3.6052
ps why are all these people complicating the formula.. it's simple pythagorus..... draw a grid ... it's easier that way...
2007-01-26 20:32:36 · answer #4 · answered by StuCee 1 · 0⤊ 0⤋
Not very much. Maybe about an 8th of an inch. I stand corrected that's about a forth of an inch.
2007-01-26 20:22:42 · answer #5 · answered by pollywollydoda 3 · 0⤊ 0⤋
d=(x2-x1)^2 + (y2-y1)^2
(-2-(-1))^2 + (-3-(-2)^2
(-1)^2+(-1)^2
(1)+(1)
sqrt of 2 and D will equal 1.41
2007-01-26 20:21:31 · answer #6 · answered by ~Zaiyonna's Mommy~ 3 · 0⤊ 0⤋
sqrt(5)
2007-01-26 20:24:04 · answer #7 · answered by 1988_Escort 3 · 0⤊ 0⤋