To determine which of the given points is part of the solution set for the inequality \( y \leq 25x + 1 \), we will substitute each point into the inequality and check if it holds true.
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Point (10, 5.5):
- Substitute \( x = 10 \) and \( y = 5.5 \): \[ 5.5 \leq 25(10) + 1 \] \[ 5.5 \leq 250 + 1 \] \[ 5.5 \leq 251 \quad \text{(True)} \]
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Point (10, 4):
- Substitute \( x = 10 \) and \( y = 4 \): \[ 4 \leq 25(10) + 1 \] \[ 4 \leq 250 + 1 \] \[ 4 \leq 251 \quad \text{(True)} \]
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Point (-10, 6):
- Substitute \( x = -10 \) and \( y = 6 \): \[ 6 \leq 25(-10) + 1 \] \[ 6 \leq -250 + 1 \] \[ 6 \leq -249 \quad \text{(False)} \]
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Point (10, 6):
- Substitute \( x = 10 \) and \( y = 6 \): \[ 6 \leq 25(10) + 1 \] \[ 6 \leq 250 + 1 \] \[ 6 \leq 251 \quad \text{(True)} \]
Conclusion: The points (10, 5.5), (10, 4), and (10, 6) are part of the solution set, while (-10, 6) is not. Therefore, the points (10, 5.5), (10, 4), and (10, 6) are all valid responses. If you need only one of them as part of the solution set, all three can be valid.