Question

terle (lms)
homework (lms graded)
cameron, arthur, and jamie are playing soccer. their locations are recorded by a motion tracking system. the grid shows distances in meters. who is closer to cameron? explain.
click the icon to view the soccer field.
arthur is about \boxed{} meters away from cameron, and jamie is about \boxed{} meters away from cameron, so \boxed{} is closer to cameron.
(round to the nearest whole number as needed.)
help me solve this view an example get more help
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Explanation:

To solve this problem, we need the coordinates of Cameron, Arthur, and Jamie from the soccer field grid (which is not fully visible here). Assuming we have their coordinates \((x_1,y_1)\) for Cameron, \((x_2,y_2)\) for Arthur, and \((x_3,y_3)\) for Jamie, we use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) for Arthur's distance from Cameron and \(d=\sqrt{(x_3 - x_1)^2+(y_3 - y_1)^2}\) for Jamie's distance from Cameron. Then we compare the two distances to see who is closer. Since the grid details (coordinates) are missing, we can't calculate the exact values, but here's the general process:

Step 1: Identify Coordinates

Let Cameron's coordinates be \((x_C,y_C)\), Arthur's be \((x_A,y_A)\), and Jamie's be \((x_J,y_J)\) from the grid.

Step 2: Calculate Arthur's Distance

Use the distance formula:
\(d_{Arthur}=\sqrt{(x_A - x_C)^2+(y_A - y_C)^2}\)

Step 3: Calculate Jamie's Distance

Use the distance formula:
\(d_{Jamie}=\sqrt{(x_J - x_C)^2+(y_J - y_C)^2}\)

Step 4: Compare Distances

If \(d_{Arthur}

For example, if Cameron is at \((0,0)\), Arthur at \((3,4)\), and Jamie at \((5,0)\):

• \(d_{Arthur}=\sqrt{(3 - 0)^2+(4 - 0)^2}=\sqrt{9 + 16}=\sqrt{25}=5\)
• \(d_{Jamie}=\sqrt{(5 - 0)^2+(0 - 0)^2}=\sqrt{25 + 0}=5\) (tie) or if Jamie is at \((6,0)\), \(d_{Jamie}=6\), so Arthur is closer.

Since the grid is not provided, please share the coordinates (or the grid details) to get the exact answer.

Answer:

To solve this problem, we need the coordinates of Cameron, Arthur, and Jamie from the soccer field grid (which is not fully visible here). Assuming we have their coordinates \((x_1,y_1)\) for Cameron, \((x_2,y_2)\) for Arthur, and \((x_3,y_3)\) for Jamie, we use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) for Arthur's distance from Cameron and \(d=\sqrt{(x_3 - x_1)^2+(y_3 - y_1)^2}\) for Jamie's distance from Cameron. Then we compare the two distances to see who is closer. Since the grid details (coordinates) are missing, we can't calculate the exact values, but here's the general process:

Step 1: Identify Coordinates

Let Cameron's coordinates be \((x_C,y_C)\), Arthur's be \((x_A,y_A)\), and Jamie's be \((x_J,y_J)\) from the grid.

Step 2: Calculate Arthur's Distance

Use the distance formula:
\(d_{Arthur}=\sqrt{(x_A - x_C)^2+(y_A - y_C)^2}\)

Step 3: Calculate Jamie's Distance

Use the distance formula:
\(d_{Jamie}=\sqrt{(x_J - x_C)^2+(y_J - y_C)^2}\)

Step 4: Compare Distances

If \(d_{Arthur}

For example, if Cameron is at \((0,0)\), Arthur at \((3,4)\), and Jamie at \((5,0)\):

• \(d_{Arthur}=\sqrt{(3 - 0)^2+(4 - 0)^2}=\sqrt{9 + 16}=\sqrt{25}=5\)
• \(d_{Jamie}=\sqrt{(5 - 0)^2+(0 - 0)^2}=\sqrt{25 + 0}=5\) (tie) or if Jamie is at \((6,0)\), \(d_{Jamie}=6\), so Arthur is closer.

Since the grid is not provided, please share the coordinates (or the grid details) to get the exact answer.