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 \((10, 6)\): \[ y = 6 \quad \text{and} \quad 25x + 1 = 25(10) + 1 = 250 + 1 = 251 \] Check: \( 6 \leq 251 \) (True)
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For the point \((-10, 6)\): \[ y = 6 \quad \text{and} \quad 25x + 1 = 25(-10) + 1 = -250 + 1 = -249 \] Check: \( 6 \leq -249 \) (False)
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For the point \((10, 4)\): \[ y = 4 \quad \text{and} \quad 25x + 1 = 25(10) + 1 = 250 + 1 = 251 \] Check: \( 4 \leq 251 \) (True)
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For the point \((10, 5.5)\): \[ y = 5.5 \quad \text{and} \quad 25x + 1 = 25(10) + 1 = 250 + 1 = 251 \] Check: \( 5.5 \leq 251 \) (True)
Thus, the points \((10, 6)\), \((10, 4)\), and \((10, 5.5)\) are part of the solution set, while \((-10, 6)\) is not.
Answer: \((10, 6)\), \((10, 4)\), and \((10, 5.5)\) are part of the solution set.