To solve the problem, let's analyze the situation using the information given.
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Vertices:
- Assume the square baseball field has its bases positioned at the vertices. Let:
- A (1st Base) be at (0, 0)
- B (2nd Base) be at (90, 0)
- C (3rd Base) be at (90, 90)
- D (Home Plate) be at (0, 90)
- Assume the square baseball field has its bases positioned at the vertices. Let:
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Position of the Shortstop:
- The shortstop is in the base path between second base (B) and third base (C) and is 30 feet from second base.
- Therefore, her position can be calculated as:
- 30 feet towards third base means she is positioned at (90, 0) + (0, 30) = (90, 30).
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First Base Position:
- First base is at (0, 0) as assumed above.
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Calculating the distance:
- To find the distance from the shortstop’s position at (90, 30) to first base (0, 0), we use the distance formula:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] where \((x_1, y_1) = (0, 0)\) and \((x_2, y_2) = (90, 30)\).
Plugging in the values, we get: \[ \text{Distance} = \sqrt{(90 - 0)^2 + (30 - 0)^2} \] \[ = \sqrt{90^2 + 30^2} \] \[ = \sqrt{8100 + 900} \] \[ = \sqrt{9000} \] \[ = 30\sqrt{10} \]
Approximating \(\sqrt{10} \approx 3.162\): \[ 30 \times 3.162 \approx 94.86 \]
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Rounding:
- Rounding to the nearest tenth, the distance the shortstop needs to throw the ball is approximately 94.9 feet.
Therefore, the final answer is: 94.9 feet.