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Blake is playing a racing game on his computer. The game tracks the locations of objects using and y coordinates (see graph below). Suddenly, Blake spots a shortcut that would take him straight to the finish line. How much shorter is the shortcut than the normal course? Round the final answer to the nearest tenth of a unit. Do not round Intermediate calculations.

Blake Is Playing A Racing Game On His Computer The Game Tracks The Locations Of Objects Using And Y Coordinates See Graph Below Suddenly Blake Spots A Shortcut class=

Sagot :

Answer: 2.2 units

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Explanation:

I'll define these point labels

  • B = Blake's starting position
  • F = finish line
  • C = the third unmarked point of the triangle

The locations of the points are

  • B = (-8,1)
  • C = (-6,-3)
  • F = (4,-2)

Use the distance formula to find the distance from B to C

[tex]B = (x_1,y_1) = (-8,1) \text{ and } C = (x_2,y_2) = (-6,-3)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-8-(-6))^2 + (1-(-3))^2}\\\\d = \sqrt{(-8+6)^2 + (1+3)^2}\\\\d = \sqrt{(-2)^2 + (4)^2}\\\\d = \sqrt{4 + 16}\\\\d = \sqrt{20} \ \text{ ... exact distance}\\\\d \approx 4.47214 \ \ \text{... approximate distance}\\\\[/tex]

Segment BC is roughly 4.47214 units long.

Following similar steps, you should find that segment CF is approximately 10.04988 units long.

If Blake doesn't take the shortcut, then he travels approximately BC+CF = 4.47214+10.04988 = 14.52202 units. This is the path from B to C to F in that order.

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Use the distance formula again to find the distance from B to F. This distance is about 12.36932 units. He travels this amount if he takes the shortcut.

Subtract this and the previous result we got

14.52202 - 12.36932 = 2.1527

That rounds to 2.2

This is the amount of distance he doesn't have to travel when he takes the shortcut.

In other words, the track is roughly 2.2 units shorter when taking the shortcut.

Side note: Replace "units" with whatever units you're working with (eg: feet or meters).

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