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Bryce is on a soccer pitch currently about 82 1/5 feet away from the goal. He first runs 2/3 of the way to the goal. Then he dribbles the ball another 12 1/2 feet, before shooting. How far away from the goal was Bryce when he took the shot?

Respuesta :

mayokunoke

Answer:

Bryce was 14[tex]\frac{9}{10}[/tex] feet away from goal, when he took the shot.

Step-by-step explanation:

  • Total distance from goal = 82[tex]\frac{1}{5}[/tex] = [tex]\frac{411}{5}[/tex] feet
  • Bryce runs 2/3 of this distance, that means [tex]\frac{2}{3}[/tex] × [tex]\frac{411}{5}[/tex] = [tex]\frac{274}{5}[/tex] feet
  • Bryce dribbles another 12[tex]\frac{1}{2}[/tex] feet = [tex]\frac{25}{2}[/tex] feet
  • So, Distance covered by Bryce = Distance run + Distance dribbled

So, the distance of Bryce from the goal at the time of shooting can be obtained by subtracting the distance covered by Bryce, from the total distance from goal, such that;

Distance away from goal =  [tex]\frac{411}{5}[/tex] - ([tex]\frac{274}{5}[/tex] + [tex]\frac{25}{2}[/tex]) = [tex]\frac{149}{10}[/tex] feet

Converting the answer to mixed fraction,

Bryce was 14[tex]\frac{9}{10}[/tex] feet away from goal, when he took the shot.

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