Mathematics, 26.08.2019 16:20 avasteinhebel10

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

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Answer from: vlactawhalm29

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

Cameron, arthur, and jamie are playing soccer. their locations are recorded by a motion tracking sys

Answer from: Quest

The internet will you

Answer from: Quest

Ifound this on google “that means that the graph of a function with direct variation has these features: it is a straight line: because the slope is the ratio y/x = k. it passes through the origin (0,0): because y = kx = k (0) = 0.” i hope this you for what your looking for

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