Question

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 determine which of the points is a solution to the linear inequality \( y > 19x - 29 \), we can substitute the x-coordinate of each point into the inequality and see if the y-coordinate satisfies it.

The inequality we are working with is \( y > 19x - 29 \).

Since all the given points have \( x = 0 \), we can simplify the inequality by substituting \( x = 0 \):

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

Now we will check each point to see if it satisfies \( y > -29 \):

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, because it is not greater)

Now, the points that satisfy the inequality \( y > -29 \) are:

• (0, -13)
• (0, -\(\frac{1}{3}\))
• (0, -19)
• (0, -\(\frac{1}{9}\))

Thus, any of the following points are solutions to the inequality:

• (0, -13)
• (0, -\(\frac{1}{3}\))
• (0, -19)
• (0, -\(\frac{1}{9}\))

The only point that does not satisfy the inequality is (0, -49) and (0, -29).

So, valid responses to the inequality include (0, -13), (0,-\(\frac{1}{3}\)), (0,-19), and (0,-\(\frac{1}{9}\)).

Answer 2

Show which of the following points is part of the solution set for the inequality y≤25x+1.(1 point)
Responses

(10,5.5)
left parenthesis 10 comma 5.5 right parenthesis

(−10,6)
left parenthesis negative 10 comma 6 right parenthesis

(10,6)
left parenthesis 10 comma 6 right parenthesis

(10,4)

Answer 3

To determine which of the given points is part of the solution set for the inequality \( y \leq 25x + 1 \), we can substitute the \( x \) values of each point into the inequality and check if the corresponding \( y \) value satisfies it.

The inequality we are working with is:

\[ y \leq 25x + 1 \]

We will test each point:

1. Point (10, 5.5):
   Substitute \( x = 10 \) into the inequality.
   \[ y \leq 25(10) + 1 \ y \leq 250 + 1 \ y \leq 251 \] Since \( 5.5 \leq 251 \) (True), this point is part of the solution set.

2. Point (−10, 6):
   Substitute \( x = -10 \) into the inequality.
   \[ y \leq 25(-10) + 1 \ y \leq -250 + 1 \ y \leq -249 \] Since \( 6 \leq -249 \) (False), this point is NOT part of the solution set.

3. Point (10, 6):
   Substitute \( x = 10 \) into the inequality.
   \[ y \leq 25(10) + 1 \ y \leq 250 + 1 \ y \leq 251 \] Since \( 6 \leq 251 \) (True), this point is part of the solution set.

4. Point (10, 4):
   Substitute \( x = 10 \) into the inequality.
   \[ y \leq 25(10) + 1 \ y \leq 250 + 1 \ y \leq 251 \] Since \( 4 \leq 251 \) (True), this point is part of the solution set.

Summary of Results:

• (10, 5.5): True (part of the solution set)
• (−10, 6): False (not part of the solution set)
• (10, 6): True (part of the solution set)
• (10, 4): True (part of the solution set)

Thus, the points (10, 5.5), (10, 6), and (10, 4) are part of the solution set for the inequality \( y \leq 25x + 1 \).

Answer 4

Kiani is shopping at a bookstore with a budget of $60. Hardbacks cost $12 and paperbacks cost $5. Write a two-variable inequality and determine the constraints of the problem. Which of the following points represents a viable solution?(1 point)
Responses

(1,4.5)
left parenthesis 1 comma 4.5 right parenthesis

(−2,6)
left parenthesis negative 2 comma 6 right parenthesis

(6,6)
left parenthesis 6 comma 6 right parenthesis

(2,4)

Answer 5

Let's define the variables for the problem:

• Let \( x \) represent the number of hardbacks purchased.
• Let \( y \) represent the number of paperbacks purchased.

The prices are as follows:

• Hardbacks cost $12 each.
• Paperbacks cost $5 each.

Step 1: Write the two-variable inequality

Since Kiani has a budget of $60, the inequality representing this constraint is:

\[ 12x + 5y \leq 60 \]

Step 2: Determine the constraints of the problem

1. Non-negativity: Kiani cannot purchase a negative number of books, so: \[ x \geq 0 \ y \geq 0 \]

Step 3: Identify viable solutions

Now we can test the proposed points to see if they satisfy the inequality \( 12x + 5y \leq 60 \) and the non-negativity constraints.

1. Point (1, 4.5):
   \[ 12(1) + 5(4.5) = 12 + 22.5 = 34.5 \ 34.5 \leq 60 \quad \text{(True)} \] Both \( x \geq 0 \) and \( y \geq 0 \) are satisfied.

2. Point (−2, 6):
   \[ 12(-2) + 5(6) = -24 + 30 = 6 \ 6 \leq 60 \quad \text{(True)} \] However, \( x = -2 \) does not satisfy \( x \geq 0 \) (False).

3. Point (6, 6):
   \[ 12(6) + 5(6) = 72 + 30 = 102 \ 102 \leq 60 \quad \text{(False)} \] Both \( x \geq 0 \) and \( y \geq 0 \) are satisfied.

4. Point (2, 4):
   \[ 12(2) + 5(4) = 24 + 20 = 44 \ 44 \leq 60 \quad \text{(True)} \] Both \( x \geq 0 \) and \( y \geq 0 \) are satisfied.

Conclusion

The viable solutions among the proposed points are:

• (1, 4.5): True
• (−2, 6): False (not valid as \( x \) cannot be negative)
• (6, 6): False
• (2, 4): True

Final Answer: The viable solutions are (1, 4.5) and (2, 4).

Answer 6

Use the image to answer the question.

The x- and y- axes of a coordinate plane range from 0 to 30 and 0 to 20 in increments of 5 and are labeled 'Lawns' and 'Dogs,' respectively. A line connects left parenthesis 0 comma 10 right parenthesis and left parenthesis 8 comma 0 right parenthesis.

This graph shows the inequality 25x+15y≥200, and represents Lina’s earnings for lawn mowing and dog walking where x represents the number of lawns mowed and y represents the number of dogs walked. Which of the labeled points is a viable solution in this context?

(1 point)
Responses

(10,10)
left parenthesis 10 comma 10 right parenthesis

(20,11.5)
left parenthesis 20 comma 11.5 right parenthesis

None of the points are a viable solution.
None of the points are a viable solution.

(2,3)

Answer 7

To determine which of the labeled points is a viable solution for the inequality \( 25x + 15y \geq 200 \), we can substitute each of the proposed points into the inequality and see if the inequality is satisfied.

Given inequality:

\[ 25x + 15y \geq 200 \]

Testing the points:

1. Point (10, 10): \[ 25(10) + 15(10) = 250 + 150 = 400 \ 400 \geq 200 \quad \text{(True)} \]

2. Point (20, 11.5): \[ 25(20) + 15(11.5) = 500 + 172.5 = 672.5 \ 672.5 \geq 200 \quad \text{(True)} \]

3. Point (2, 3): \[ 25(2) + 15(3) = 50 + 45 = 95 \ 95 \geq 200 \quad \text{(False)} \]

Summary of Results:

• (10, 10): Satisfies the inequality (True).
• (20, 11.5): Satisfies the inequality (True).
• (2, 3): Does NOT satisfy the inequality (False).

Conclusion:

Both (10, 10) and (20, 11.5) are viable solutions to the inequality \( 25x + 15y \geq 200 \).

Given the responses:

• (10,10)
• (20,11.5)
• None of the points are a viable solution.
• (2,3)

Since at least one point is a viable solution, the correct response would be both (10,10) and (20,11.5). However, if the question allows only one selection, you can choose either (10, 10) or (20, 11.5) as a viable solution. If instructed to select just one, I would recommend selecting (10, 10) as it appears first. If you must choose only one option, please confirm if the interface restricts you from choosing multiple points.