(-3,0) and (4,0) Thank you
2007-08-08 12:35:08 · 11 answers · asked by vadataborn 2 in Science & Mathematics ➔ Mathematics
7
2007-08-08 12:37:37 · answer #1 · answered by Keith P 7 · 0⤊ 0⤋
distance d between (-3,0 )& (4,0)
= sqrt[(4+3)^2+(0-0)^2]
=sqrt[]49+0]
=7AN.
Note: both points are on the X-axis. easily verified
2007-08-08 19:42:26 · answer #2 · answered by Anonymous · 0⤊ 0⤋
okay, if you have a graph that your trying to find it on, then just use math (well obviously lol) anyways -3,0 plus 3 would be 0,0 then plus 4 more. so the distance is 7. (if i'm thinking of the same thing as your talking about!!!)
2007-08-08 19:38:18 · answer #3 · answered by klumsy-kitty 2 · 0⤊ 0⤋
It's 3 on the negative side + 4 on the positive side = 7.
Don't exactly know how the other person got 5 lol.
2007-08-08 19:42:08 · answer #4 · answered by alomi_revolution 4 · 0⤊ 0⤋
7. You have no change in y, so all you have to do is subtract negative 3 from 4.
Subtracting from a negative is the same as adding, because two negative make a postive: i.e. Nobody doesn't like Sara Lee.
So 4 plus 3 is 7.
You're welcome.
2007-08-08 19:37:55 · answer #5 · answered by Anonymous · 1⤊ 0⤋
Distance between two points defined by (x1, y1) and (x2, y2) is equal to [(x2-x1)^2 + (y2-y1)^2]^0.5
In your case -
Distance = [(4- (-3))^2 + (0-0)^2]^0.5 = [(7)^2]^0.5 = [49]^0.5 = 7
2007-08-08 19:40:23 · answer #6 · answered by MedievalLoser 2 · 0⤊ 0⤋
You can use the formula of
square root of distance = [(y2-y1)square + (x2-x1)square]
2007-08-08 19:39:46 · answer #7 · answered by michael2003c2003 5 · 0⤊ 0⤋
d = sqrt((x2-x1)^2 +(y2-y1)^2)
d = sqrt((4-(-3))^2+(0-0)^2)
d = sqrt(7^2) = 7
2007-08-08 19:42:27 · answer #8 · answered by ironduke8159 7 · 0⤊ 0⤋
YAY THE DISTANCE FORMULA! or not because they are on the x-axis so u can just count.
2007-08-08 19:40:53 · answer #9 · answered by Anonymous · 0⤊ 0⤋
=SQRT( (x1-x2)^2 + (y1-y2)^2)
= SQRT ( (-3-4)^2 + (0-0)^2)
= 7
2007-08-08 19:40:17 · answer #10 · answered by pksk212 3 · 0⤊ 0⤋