To determine if a point is part of the solution set for the inequality \( y \leq 25x + 1 \), we can substitute the coordinates of each point into the inequality and see if it holds true.
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For the point \( (10, 4) \): \[ y = 4 \quad \text{and} \quad 25x + 1 = 25(10) + 1 = 250 + 1 = 251 \] \[ 4 \leq 251 \quad \text{(True)} \]
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For the point \( (10, 5.5) \): \[ y = 5.5 \quad \text{and} \quad 25x + 1 = 25(10) + 1 = 251 \] \[ 5.5 \leq 251 \quad \text{(True)} \]
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For the point \( (10, 6) \): \[ y = 6 \quad \text{and} \quad 25x + 1 = 25(10) + 1 = 251 \] \[ 6 \leq 251 \quad \text{(True)} \]
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For the point \( (-10, 6) \): \[ y = 6 \quad \text{and} \quad 25x + 1 = 25(-10) + 1 = -250 + 1 = -249 \] \[ 6 \leq -249 \quad \text{(False)} \]
Based on the evaluations:
- \( (10, 4) \): True
- \( (10, 5.5) \): True
- \( (10, 6) \): True
- \( (-10, 6) \): False
Every point except \( (-10, 6) \) satisfies the inequality \( y \leq 25x + 1 \). Thus, the points \( (10, 4) \), \( (10, 5.5) \), and \( (10, 6) \) are part of the solution set.