To determine which points satisfy the inequality \( y \leq 25x + 1 \), we need to substitute the \( x \) values of each point into the inequality and check if the corresponding \( y \) value is less than or equal to \( 25x + 1 \).
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Check (−10, 6):
- Substitute \( x = -10 \): \[ y \leq 25(-10) + 1 \ y \leq -250 + 1 \ y \leq -249 \]
- Since \( 6 > -249 \), this point does not satisfy the inequality.
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Check (10, 5.5):
- Substitute \( x = 10 \): \[ y \leq 25(10) + 1 \ y \leq 250 + 1 \ y \leq 251 \]
- Since \( 5.5 < 251 \), this point satisfies the inequality.
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Check (10, 6):
- Substitute \( x = 10 \): \[ y \leq 25(10) + 1 \ y \leq 250 + 1 \ y \leq 251 \]
- Since \( 6 < 251 \), this point satisfies the inequality.
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Check (10, 4):
- Substitute \( x = 10 \): \[ y \leq 25(10) + 1 \ y \leq 250 + 1 \ y \leq 251 \]
- Since \( 4 < 251 \), this point satisfies the inequality.
Conclusion: The points (10, 5.5), (10, 6), and (10, 4) are part of the solution set for the inequality \( y \leq 25x + 1 \). However, for a single point selection, any of these three points would be correct.