Decide which of the ordered pairs are solutions for the equation y=4x+2

Question

.(4 points)
Solution Not A Solution
(−2, −6)

(−1, 0)

(−2,6)

(0, 2)

Answer 1

To determine which of the ordered pairs are solutions for the equation \( y = 4x + 2 \), we will substitute the x-value of each ordered pair into the equation and see if the resulting y-value matches the given y-value in the ordered pair.

1. For the ordered pair \((-2, -6)\): \[ y = 4(-2) + 2 = -8 + 2 = -6 \] The y-value matches, so \((-2, -6)\) is a solution.

2. For the ordered pair \((-1, 0)\): \[ y = 4(-1) + 2 = -4 + 2 = -2 \] The y-value does not match (0 ≠ -2), so \((-1, 0)\) is not a solution.

3. For the ordered pair \((-2, 6)\): \[ y = 4(-2) + 2 = -8 + 2 = -6 \] The y-value does not match (6 ≠ -6), so \((-2, 6)\) is not a solution.

4. For the ordered pair \((0, 2)\): \[ y = 4(0) + 2 = 0 + 2 = 2 \] The y-value matches, so \((0, 2)\) is a solution.

Summary:

• Solutions: \((-2, -6)\), \((0, 2)\)
• Not Solutions: \((-1, 0)\), \((-2, 6)\)

Answer 2

Ms. Mac asked her students to determine a solution for 2x−5y=9
.

Bella believes (-2, 1) is a solution to the equation.

Marco believes (11, 2) is a solution to the equation.

Liam believes (2, -1)is a solution to the equation.

(1 point)
Response area named a point that is a solution that lies on the graph of the equation. Response area named a point that is not a solution and lies above the graph of the equation. Response area named a point that is not a solution and lies below the graph of the equation.

Answer 3

To determine if the points are solutions for the equation \( 2x - 5y = 9 \), we will substitute each ordered pair into the equation and see if they satisfy it.

1. For the point \((-2, 1)\): \[ 2(-2) - 5(1) = -4 - 5 = -9 \] Since \(-9 \neq 9\), \((-2, 1)\) is not a solution and lies below the graph of the equation.

2. For the point \((11, 2)\): \[ 2(11) - 5(2) = 22 - 10 = 12 \] Since \(12 \neq 9\), \((11, 2)\) is not a solution and lies above the graph of the equation.

3. For the point \((2, -1)\): \[ 2(2) - 5(-1) = 4 + 5 = 9 \] Since \(9 = 9\), \((2, -1)\) is a solution to the equation.

Summary:

• Point that is a solution: \((2, -1)\)
• Point that is not a solution and lies above the graph: \((11, 2)\)
• Point that is not a solution and lies below the graph: \((-2, 1)\)

Answer 4

Mrs. Wilbanks is taking a group of students to see a play. The cost of the bus rental to transport her class is $500 and the cost per ticket to see the play is $20. Create an equation with two variables that models this situation, then find the total cost for the event if 60 students & teachers plan to attend.(1 point)
Responses

cap c is equal to 520 t; $31,200

cap c is equal to 500 t plus 20 t; $1,700

cap c is equal to 500 t plus 20; $30,020

C=500+20t ; $1,700

Answer 5

To model the situation, we let:

• \( C \) represent the total cost of the trip.
• \( t \) represent the number of tickets sold (or attendees).

The fixed cost of the bus rental is $500, and the cost per ticket for the play is $20. Therefore, the equation modeling the total cost \( C \) can be expressed as:

\[ C = 500 + 20t \]

Now, if 60 students and teachers are attending, we substitute \( t = 60 \) into the equation:

\[ C = 500 + 20 \times 60 \] \[ C = 500 + 1200 \] \[ C = 1700 \]

So, the total cost for the event if 60 students and teachers plan to attend is $1,700.

Thus the correct response is:

C = 500 + 20t ; $1,700

Answer 6

Sophia is selecting a planning committee for her school's annual talent show. The committee must consist of 12 students, with members from both the freshman and sophomore classes. At least two students must be from each class. Which of the following is a viable solution for this situation?(1 point)
Responses

0 freshmen and 12 sophomores

1 freshman and 11 sophomores

5 freshmen and 7 sophomores

6 freshmen and 4 sophomores

Answer 7

To satisfy the requirement for the planning committee, we need a total of 12 students with at least 2 students from each class (freshmen and sophomores).

Let's analyze the responses:

1. 0 freshmen and 12 sophomores: This does not satisfy the requirement of at least 2 freshmen. Not a viable solution.

2. 1 freshman and 11 sophomores: This does not satisfy the requirement of at least 2 freshmen. Not a viable solution.

3. 5 freshmen and 7 sophomores: This meets the requirement of at least 2 freshmen and at least 2 sophomores. Viable solution.

4. 6 freshmen and 4 sophomores: This also meets the requirement of at least 2 freshmen and at least 2 sophomores. Viable solution.

From the options, both 5 freshmen and 7 sophomores and 6 freshmen and 4 sophomores are viable solutions for the situation.

If the question is asking for a single viable solution, then 5 freshmen and 7 sophomores is a clear option.

Answer 8

Alex won a gift card worth $96.00 for winning a design competition. She decided to spend her winnings on sketchbooks and paint sets. Each sketchbook (x) costs $12.00, while each paint set (y) costs $15.00.

Which of the following combinations of sketchbooks and paint sets can Alex purchase, assuming she spends her entire gift card?

(1 point)
Responses

2 sketchbooks and 5 paint sets

1 sketchbook and 4 paint sets

3 sketchbooks and 4 paint sets

​​​​ 4 sketchbooks and 3 paint sets

Answer 9

To determine which combinations of sketchbooks (x) and paint sets (y) Alex can purchase with her $96 gift card, we need to set up an equation based on the costs of the items:

The cost of the sketchbooks is \(12x\) (where \(x\) is the number of sketchbooks), and the cost of the paint sets is \(15y\) (where \(y\) is the number of paint sets). The total cost must equal $96:

\[ 12x + 15y = 96 \]

Now we'll evaluate each of the provided combinations to see if they satisfy this equation:

1. 2 sketchbooks and 5 paint sets: \[ 12(2) + 15(5) = 24 + 75 = 99 \] This exceeds $96. Not a viable option.

2. 1 sketchbook and 4 paint sets: \[ 12(1) + 15(4) = 12 + 60 = 72 \] This is less than $96. Not a viable option.

3. 3 sketchbooks and 4 paint sets: \[ 12(3) + 15(4) = 36 + 60 = 96 \] This equals $96. Viable option.

4. 4 sketchbooks and 3 paint sets: \[ 12(4) + 15(3) = 48 + 45 = 93 \] This is less than $96. Not a viable option.

Based on the calculations, the only combination that allows Alex to use the entire gift card is:

3 sketchbooks and 4 paint sets.

Answer 10

Is the point (0,−3) a solution to the inequality y > 4/5 x − 3, as shown on the graph?

(1 point)
Responses

No, because (0,−3) falls within the shaded region.

No, because (0,−3) falls on the dashed boundary line.

Yes, because (0,−3) falls within the shaded region.

Yes, because (0,−3) falls on the boundary line.

Answer 11

To determine whether the point \((0, -3)\) is a solution to the inequality \(y > \frac{4}{5}x - 3\), we can follow these steps:

1. Substitute the point into the inequality:

   • The point is \((x, y) = (0, -3)\).
   • Substitute \(x = 0\) and \(y = -3\) into the inequality: \[ -3 > \frac{4}{5}(0) - 3 \] Simplifying, we get: \[ -3 > -3 \] This statement is not true; \(-3\) is not greater than \(-3\).

2. Analyze the graph:

   • The inequality \(y > \frac{4}{5}x - 3\) means that the region above the line \(y = \frac{4}{5}x - 3\) is the solution area.
   • The boundary line corresponding to the equation \(y = \frac{4}{5}x - 3\) is dashed because the inequality is strict (greater than, not greater than or equal to).
   • Since \(-3\) is equal to the value of the function at \(x = 0\) (where the line crosses the y-axis), the point \((0, -3)\) falls on the dashed line.

3. Conclusion:

   • Since the point is on the boundary line (which is dashed), it is not included in the solution set of the inequality.

Based on the analysis, the correct response is:

No, because (0,−3) falls on the dashed boundary line.