Question

Cameron, Arthur, and Jamie are playing soccer. Their locations are recorded by a motion tracking system. The distance between grids is 5 meters. SEE EXAMPLES 3 AND 4 x 50 60 70 40 30 20 10 O 10 20 30 40 50 60 70 80 90 y Arthur Cameron Jamie 20. How far apart are Arthur and Jamie? Round to the nearest tenth of a meter. 21. Who is closer to Cameron? Explain. 22. The soccer ball is located at the point (35, 60). Who is closest to the soccer ball

Answer 1

Answer:

20. Arthur and Jamie are 39.1 meters are apart

21. Arthur is closer to Cameron

22. Jamie is the closest to the soccer ball

Step-by-step explanation:

* "I have added screenshot of the complete question as well as

  the diagram)

20.

∵ The location of Arthur is (20 , 30)

∵ The location of Jamie is (45 , 60)

- Lets use the rule of the distance to find how they are far apart

∵ $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

- Let the location of Arthur is $(x_{1},y_{1})$ and the location

 of Jamie is $(x_{2},y_{2})$

∴ $d=\sqrt{(45-20)^{2}+(60-30)^{2}}=\sqrt{(25)^{2}+(30)^{2}}=39.1$

∴ d = 39.1 meters

* Arthur and Jamie are 39.1 meters are apart

21.

- To find who is closest to Cameron, find the distance between

  Cameron and each one of them

∵ The location of Cameron is (70 , 10)

- By using the rule of the distance find the distance between

 Cameron and Arthur d1 and the distance between Cameron

 and Jamie d2

∵ $d1=\sqrt{(70-20)^{2}+(10-30)^{2}}=\sqrt{(50)^{2}+(-20)^{2}}=53.9$

∴ d1 = 53.9 meters

∵ $d2=\sqrt{(70-45)^{2}+(10-60)^{2}}=\sqrt{(25)^{2}+(-50)^{2}}=55.9$

∴ d2 = 55.9 meters

∵ The distance between the Cameron and Arthur is 53.9 meters

∵ The distance between the Cameron and Jamie is 55.9 meters

∵ The distance between Arthur and Cameron < the distance

  between Jamie and Cameron

∴ Arthur is closer to Cameron

22.

- To know who is closest to the soccer ball find the distance

 between it and them

∵ The location of the soccer ball is (35 , 60)

∵ d1 is the distance between Cameron and it

∵ d2 is the distance between Arthur and it

∵ d3 is the distance between Jamie and it

∴ $d1=\sqrt{(70-35)^{2}+(10-60)^{2}}=\sqrt{(35)^{2}+(-50)^{2}}=61.0$

∴ d1 = 61.0 meters

∴ $d2=\sqrt{(20-35)^{2}+(30-60)^{2}}=\sqrt{(-15)^{2}+(-30)^{2}}=33.5$

∴ d2 = 33.5 meters

∴ $d3=\sqrt{(45-35)^{2}+(60-60)^{2}}=\sqrt{(10)^{2}+(0)^{2}}=10$

∴ d3 = 10 meters

∵ d3 is the shortest distance

∴ Jamie is the closest to the soccer ball

- You can find this answer by easiest way if you put the location of the soccer ball on the picture of the field