To complete the statements regarding the scales for the axes of the function \( A(d) = d(250 - 2d) \), let's analyze the function a bit more:
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For the horizontal axis (d):
- The variable \( d \) represents the width, and it must be a positive value. Since Peter has 250 feet of fencing, the maximum width \( d \) can be is 125 feet (as \( 250 - 2d \) must still be positive). A reasonable scale for the horizontal axis would allow us to visualize a range from 0 to 125 feet effectively.
- Considering the options: 1 ft, 2 ft, 10 ft, or 75 ft, a scale of 10 ft would be reasonable, allowing for clear visualization of the width increments without making the graph too crowded.
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For the vertical axis (A(d)):
- The area \( A(d) \) can be calculated as follows: \[ A(d) = d(250 - 2d) \]
- To find the maximum area, we can analyze the function. The area is a quadratic function, and its maximum value occurs at the vertex. The vertex \( d \) value can be calculated as: \[ d = -\frac{b}{2a} = -\frac{250}{2(-2)} = 62.5 \]
- Substituting back into the area function: \[ A(62.5) = 62.5(250 - 2(62.5)) = 62.5(125) = 7812.5 \text{ square feet} \]
- The area goes to 0 when \( d = 0 \) or \( d = 125 \). Thus, the range for the area is from 0 to approximately 7812.5 square feet.
- Considering the options for the vertical axis: 75 sq ft, 100 sq ft, 500 sq ft, or 2000 sq ft, a scale of 500 sq ft would be appropriate and manageable based on the maximum area.
Combining these observations, we can complete the statements:
A reasonable scale for the horizontal axis of the function \( A(d) = d(250 - 2d) \) is 10 ft, and a reasonable scale for the vertical axis of the function \( A(d) = d(250 - 2d) \) is 500 sq ft.
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