
<u>We </u><u>have</u><u>, </u>
- Line segment AB
- The coordinates of the midpoint of line segment AB is ( -8 , 8 )
- Coordinates of one of the end point of the line segment is (-2,20)
Let the coordinates of the end point of the line segment AB be ( x1 , y1 ) and (x2 , y2)
<u>Also</u><u>, </u>
Let the coordinates of midpoint of the line segment AB be ( x, y)
<u>We </u><u>know </u><u>that</u><u>, </u>
For finding the midpoints of line segment we use formula :-

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>
- The coordinates of midpoint and one of the end point of line segment AB are ( -8,8) and (-2,-20) .
<u>For </u><u>x </u><u>coordinates </u><u>:</u><u>-</u>





<h3><u>Now</u><u>, </u></h3>
<u>For </u><u>y </u><u>coordinates </u><u>:</u><u>-</u>





Thus, The coordinates of another end points of line segment AB is ( -14 , 36)
Hence, Option A is correct answer
Answer:
2. b
3. d.
4. c.
Step-by-step explanation:
Answer:
The ball can be knocked down at a horizontal distance of 3.09 feet or 102.17 feet from the marshal.
Step-by-step explanation:
We have the function that represents the height h (x) of the ball 
Where x is the horizontal distance of the ball.
We want to find the horizontal distance the ball is at (horizontal distance between the field marshal and the ball) when it is at a height of 10 feet.
To do this, we must do h (x) = 10

Now we must solve the second degree equation. For this we use the formula of the resolvent:

and

ft
and
ft
Then, the ball can be knocked down at a horizontal distance of 3.09 feet or 102.17 feet from the marshal.
Y= mx+b
(16, -7) = (x,y)
2(16) - 3y = 12
32 - 3y = 12
32 - 12 = 3y
20/3 = y
6.6 = y
2x-3(-7) = 12
2x+21 = 12
2x = 12 - 21
x = -9/2
Insert the values into y = mx + b
solve for m and then solve for b