To determine which points are part of the solution set for the inequality \( y \leq \frac{2}{5}x + 1 \), we will evaluate each point by substituting the x-value into the equation and checking if the resulting y-value satisfies the inequality.
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Point A: (10, 5.5)
- Substitute \( x = 10 \): \[ y \leq \frac{2}{5}(10) + 1 = \frac{20}{5} + 1 = 4 + 1 = 5 \]
- Check: \( 5.5 \leq 5 \) (FALSE)
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Point B: (5, 3)
- Substitute \( x = 5 \): \[ y \leq \frac{2}{5}(5) + 1 = \frac{10}{5} + 1 = 2 + 1 = 3 \]
- Check: \( 3 \leq 3 \) (TRUE)
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Point C: (10, 6)
- Substitute \( x = 10 \): \[ y \leq \frac{2}{5}(10) + 1 = 4 + 1 = 5 \]
- Check: \( 6 \leq 5 \) (FALSE)
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Point D: (10, 4)
- Substitute \( x = 10 \): \[ y \leq \frac{2}{5}(10) + 1 = 4 + 1 = 5 \]
- Check: \( 4 \leq 5 \) (TRUE)
Thus, the points that are part of the solution set for the inequality \( y \leq \frac{2}{5}x + 1 \) are:
- B. (5, 3)
- D. (10, 4)