Triangle ABC has vertices at A (1,-3), B (3,0)x and C (5,-3). What is the distance from vertex B to the midpoint of AC?

The distance from vertex B to the midpoint of AC is 3 units

Step-by-step explanation:

we are given a triangle with vertices  A (1,-3), B (3,0) and C (5,-3).

Mid point of of AC  (x,y) =  ( 1+5/2 , -3 -3/2)

                                     = ( 6/2 , -6/2)

                                     = (3 , -3)

Distance formula =

                              =

                               =

                                = 3 units

use google you idiot omg you're so annoying

step-by-step explanation:

-2x + 7 = 7

subtract 7 from both sides.

-2x = 0

divide both sides by -2.

x = 0