(-3, 0) and (4, 0)
2007-04-21 08:37:33 · 10 answers · asked by Dave 4 in Science & Mathematics ➔ Mathematics
Distance formula
d = √(x₂- x₁)² + (y₂- y₁)²
Ordered Pair
(- 3, 0)(4, 0)
d = √[4 - (- 3)] ² + (0 - 0)²
d = √[ 4 + 3]² + 0
d = √[7]²
d = √49
d = 7
- - - - - - s-
2007-04-21 09:02:44 · answer #1 · answered by SAMUEL D 7 · 0⤊ 0⤋
7
2007-04-21 08:44:25 · answer #2 · answered by the man 2 · 0⤊ 0⤋
7
2007-04-21 08:42:12 · answer #3 · answered by Anonymous · 0⤊ 0⤋
7/0
2007-04-21 08:40:56 · answer #4 · answered by ~LOZ~ 6 · 0⤊ 0⤋
These points lie on the x axis and question becomes that of finding the distance , d , between - 3 and 4. This leads to:-
d = 7
2007-04-21 08:41:40 · answer #5 · answered by Como 7 · 1⤊ 0⤋
You could use the distance formula...but its kinda pointless in this case. Just by looking, you can see that there is no change in y-position - only in x-position.
|-3| + |4| = 3 + 4 = 7 units away
The points are 7 units away from each other.
2007-04-21 08:41:44 · answer #6 · answered by Bhajun Singh 4 · 1⤊ 0⤋
Use the formula for distance,
d = sqr[(x1-x2)^2 + (y1-y2)^2]
Here x1 = -3, x2 = 4
y1 = 0, y2 = 0
d = sqr[(-3 -4)^2 + (0-0)^2] = sqr[49] = 7
2007-04-21 08:41:50 · answer #7 · answered by kellenraid 6 · 1⤊ 0⤋
we are able to apply the pythagorean theorem to discover a distance formula. a^2 + b^2 = c^2 a is the version interior the x path and b is the version interior the y, than c is the area ( x1 - x2)^2 + (y1 - y2)^2 = c^2 sqrt(( x1 - x2)^2 + (y1 - y2)^2) = c than plug interior the numbers sqrt(( 10- 13)^2 + (20- sixteen)^2) = c sqrt(( -3)^2 + (4)^2) = c sqrt(9 +sixteen) = c sqrt(25) = c 5 = c
2016-10-13 03:13:42 · answer #8 · answered by ? 4 · 0⤊ 0⤋
distance can not be negative
3+4=7
2007-04-21 08:49:19 · answer #9 · answered by iyiogrenci 6 · 0⤊ 0⤋
it is really really easy
to find the difference you simply have x2 - x1
so 4 - -3
which would be 4+3
so the answer is 7
2007-04-21 08:41:27 · answer #10 · answered by l0uislegr0s 3 · 0⤊ 0⤋