To determine which of the given points is part of the solution set for the inequality \( y \leq 25x + 1 \), we need to substitute each point into the inequality and see if it holds true.
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For the point \( (10, 5.5) \): \[ y = 5.5 \quad \text{and} \quad x = 10 \] Substitute into the inequality: \[ 5.5 \leq 25(10) + 1 \ 5.5 \leq 250 + 1 \ 5.5 \leq 251 \quad \text{(True)} \]
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For the point \( (10, 6) \): \[ y = 6 \quad \text{and} \quad x = 10 \] Substitute into the inequality: \[ 6 \leq 25(10) + 1 \ 6 \leq 250 + 1 \ 6 \leq 251 \quad \text{(True)} \]
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For the point \( (10, 4) \): \[ y = 4 \quad \text{and} \quad x = 10 \] Substitute into the inequality: \[ 4 \leq 25(10) + 1 \ 4 \leq 250 + 1 \ 4 \leq 251 \quad \text{(True)} \]
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For the point \( (-10, 6) \): \[ y = 6 \quad \text{and} \quad x = -10 \] Substitute into the inequality: \[ 6 \leq 25(-10) + 1 \ 6 \leq -250 + 1 \ 6 \leq -249 \quad \text{(False)} \]
The points \( (10, 5.5) \), \( (10, 6) \), and \( (10, 4) \) are all part of the solution set because they satisfy the inequality. However, \( (-10, 6) \) does not satisfy the inequality.
Thus, the options \( (10, 5.5) \), \( (10, 6) \), and \( (10, 4) \) are all valid solutions.