Question

What is the distance of AB rounded to the nearest tenth and why?

Answer 1

Answer:

Distance is 2.83 units

Step-by-step explanation:

We are given a straight line on graph and asked to find its length using Distance Formula. The Distance formula is given as under

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Where

$(x_1,y_1) = (-3,1))$

$(x_2,y_2) = (-1,-1)$

Putting these values in Distance Formula we get

$D=\sqrt{(1-(-1))^2+(-3-(-1))^2}$

$D=\sqrt{(2)^2+(-2)^2}$

$D=\sqrt{4+4}$

$D=\sqrt{8}$

$D=2\sqrt{2}$

$D=2 \time 1.414$

$D=2.828$

Answer 2

Answer:

The  distance of AB rounded to the nearest tenth is:

                                   2.8 units

Step-by-step explanation:

We know that the distance between the two points :

  $(x_1,y_1)\ and\ (x_2,y_2)$ is calculated using the distance formula as:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Here point A lie at (-3,1) and point B lie at (-1,-1)

Hence, the distance between A and B is given by:

$d=\sqrt{(-3-(-1))^2+(1-(-1))^2}\\\\d=\sqrt{(-2)^2+2^2}\\\\d=\sqrt{4+4}\\\\d=\sqrt{8}\\\\d=2.8284$

which to the nearest tenth is:

$d=2.8\ units$