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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For (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 (10, 5.5): \[ y = 5.5 \quad \text{and} \quad 25x + 1 = 25(10) + 1 = 251 \] Check: \( 5.5 \leq 251 \) (True)
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For (−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 (10, 4): \[ y = 4 \quad \text{and} \quad 25x + 1 = 25(10) + 1 = 251 \] Check: \( 4 \leq 251 \) (True)
So, the points that are part of the solution set for the inequality \( y \leq 25x + 1 \) are (10, 6), (10, 5.5), and (10, 4).
The answer is:
- (10, 6)
- (10, 5.5)
- (10, 4)
Only (−10, 6) is not a solution.