Answer:
Step-by-step explanation:
We can use the distance formula derived from the Pythagorean theorem
D = ![\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28x_2-x_1%29%5E2%20%2B%28y_2-y_1%29%5E2%7D)
the two points given are
(0, 3) and (-2, -3)
![(x_2-x_1) = (0-(-2)) = 2\\(y_2-y_1) = (3-(-3)) = 6\\D = \sqrt{(2)^2 + (6)^2} \\D = \sqrt{4 + 36} \\D = \sqrt{40} = 6.324](https://tex.z-dn.net/?f=%28x_2-x_1%29%20%3D%20%280-%28-2%29%29%20%3D%202%5C%5C%28y_2-y_1%29%20%3D%20%283-%28-3%29%29%20%3D%206%5C%5CD%20%3D%20%5Csqrt%7B%282%29%5E2%20%2B%20%286%29%5E2%7D%20%5C%5CD%20%3D%20%5Csqrt%7B4%20%2B%2036%7D%20%5C%5CD%20%3D%20%5Csqrt%7B40%7D%20%3D%206.324)
Answer:
B
Step-by-step explanation:
1. Plot the points on a graph (reflected across x-axis)
Graphs of linear functions
Since we are graphing a line, if we have two points we can plot the whole line
We evaluate the line in two points in order to locate two places where it passes through. We say y = h(x)
1. If x = 0 then
h(x) = - 5(0) +3
= 0 + 3
= 3
Then y = 3
2. If x = 2 then
h(x) = - 5(2) +3
= -10 + 3
= -7
Then y = -7
Now, we have two points: (0, 3) and (2, -7)
We locate them and then we plot the only straight line that passes through both points