To determine which of the given points satisfies the inequality \(y \leq \frac{2}{5}x + 1\), we can substitute the \(x\) and \(y\) values of each point into the inequality and check whether it holds true.
The inequality is \(y \leq \frac{2}{5}x + 1\).
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For the point \((10, 6)\): \[ y = 6 \quad \text{and} \quad x = 10 \] Substitute \(x\) into the inequality: \[ \frac{2}{5}(10) + 1 = 4 + 1 = 5 \] Check the inequality: \[ 6 \leq 5 \quad \text{(false)} \]
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For the point \((-10, 6)\): \[ y = 6 \quad \text{and} \quad x = -10 \] Substitute \(x\) into the inequality: \[ \frac{2}{5}(-10) + 1 = -4 + 1 = -3 \] Check the inequality: \[ 6 \leq -3 \quad \text{(false)} \]
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For the point \((10, 5.5)\): \[ y = 5.5 \quad \text{and} \quad x = 10 \] Substitute \(x\) into the inequality: \[ \frac{2}{5}(10) + 1 = 4 + 1 = 5 \] Check the inequality: \[ 5.5 \leq 5 \quad \text{(false)} \]
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For the point \((10, 4)\): \[ y = 4 \quad \text{and} \quad x = 10 \] Substitute \(x\) into the inequality: \[ \frac{2}{5}(10) + 1 = 4 + 1 = 5 \] Check the inequality: \[ 4 \leq 5 \quad \text{(true)} \]
Therefore, the point that is part of the solution set for the inequality \(y \leq \frac{2}{5}x + 1\) is \((10, 4)\).