Question

Which of the following points is a solution to the linear inequality y>19x−29?(1 point) Responses (0,−13) left parenthesis 0 comma negative Start Fraction 1 over 3 End Fraction right parenthesis (0,−19) left parenthesis 0 comma negative Start Fraction 1 over 9 End Fraction right parenthesis (0,−49) left parenthesis 0 comma negative Start Fraction 4 over 9 End Fraction right parenthesis (0,−29)

Answer 1

To see which points are solutions to the inequality \( y > 19x - 29 \), we can substitute \( x = 0 \) into the inequality to find the corresponding value of \( y \).

Substituting \( x = 0 \) gives:

\[ y > 19(0) - 29 \] \[ y > -29 \]

Now we need to check which of the given points has a \( y \)-value greater than \(-29\):

1. Point (0, -13): \[ -13 > -29 \] (True)

2. Point \( (0, -\frac{1}{3}) \): \[ -\frac{1}{3} > -29 \] (True)

3. Point (0, -19): \[ -19 > -29 \] (True)

4. Point \( (0, -\frac{1}{9}) \): \[ -\frac{1}{9} > -29 \] (True)

5. Point (0, -49): \[ -49 > -29 \] (False)

6. Point \( (0, -29) \): \[ -29 > -29 \] (False)

From this analysis, the points (0, -13), \( (0, -\frac{1}{3}) \), (0, -19), and \( (0, -\frac{1}{9}) \) are solutions to the inequality \( y > 19x - 29 \), while \( (0, -49) \) and \( (0, -29) \) are not.

Thus, the valid solutions are (0, -13), (0, -\(\frac{1}{3}\)), (0, -19), and (0, -\(\frac{1}{9}\)).