To determine which point is part of the solution set for the inequality \( y \leq \frac{2}{5} x + 1 \), we first need to substitute each point into the inequality and see if it holds true.
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Point A: (10, 6) \[ y = 6, \quad x = 10 \quad \Rightarrow \quad 6 \leq \frac{2}{5} \cdot 10 + 1 \] \[ 6 \leq \frac{20}{5} + 1 = 4 + 1 = 5 \quad \text{(False)} \]
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Point B: (10, 5.5) \[ y = 5.5, \quad x = 10 \quad \Rightarrow \quad 5.5 \leq \frac{2}{5} \cdot 10 + 1 \] \[ 5.5 \leq 5 \quad \text{(False)} \]
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Point C: (-10, 6) \[ y = 6, \quad x = -10 \quad \Rightarrow \quad 6 \leq \frac{2}{5} \cdot (-10) + 1 \] \[ 6 \leq -4 + 1 = -3 \quad \text{(False)} \]
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Point D: (10, 4) \[ y = 4, \quad x = 10 \quad \Rightarrow \quad 4 \leq \frac{2}{5} \cdot 10 + 1 \] \[ 4 \leq 5 \quad \text{(True)} \]
Only Point D, \((10, 4)\), satisfies the inequality. Therefore, the solution is:
D. (10, 4)