3. Using the replacement set, find the solutions for the equation.

y = 0.5x + 10 Replacement Set: {(-2, 9.5), (0, 10),(3, 11.5),(4, 13)}(Points : 1)

{(-2, 9.5), (0, 10),(3, 11.5),(4, 13)}

{(-2, 9.5), (0, 10),(3, 11.5)}

{(0, 10),(3, 11.5),(4, 13)}

{ (0, 10),(3, 11.5)}

4. Find the solution, given the replacement set.

5x - 2y = 6 {(-2, -8), (0, -3), (1, -1), (2, 4), (4, 7)}(Points : 1)

{(-2, -8), (2, 4), (4, 7)}

{(-2, -8), (0, -3), (4, 7)}

{(-2, -8), (0, -3), (1, -1)}

{(1, -1), (2, 4), (4, 7)}

5. Solve the equation, using the domain of x = {-1, 0, 2}. Write your solutions as a set of ordered pairs.

2x + 3y = 7(Points : 1)

6. Give the range of x - 2y = 0 for the domain {-1, 0, 4}.(Points : 1)

7. The interest on a loan at an interest rate of 9% over a year is I = (0.09)P, where P is the principle (loan amount) and I is the interest rate. Use a domain in the table to find the interest of these amounts for one year. Write the range only in your solution.

Principle 10 100 1000 10,000 100,000 1,000,000

Interest -- -- -- -- -- --

(Points : 1)

8. What is the independent variable and what is the dependent variable of this relation?

Principle 10 100 1,000 10,000

Interest 2 20 200 2,000

(Points : 1)

y = 0.5x + 10 Replacement Set: {(-2, 9.5), (0, 10),(3, 11.5),(4, 13)}(Points : 1)

{(-2, 9.5), (0, 10),(3, 11.5),(4, 13)}

{(-2, 9.5), (0, 10),(3, 11.5)}

{(0, 10),(3, 11.5),(4, 13)}

{ (0, 10),(3, 11.5)}

4. Find the solution, given the replacement set.

5x - 2y = 6 {(-2, -8), (0, -3), (1, -1), (2, 4), (4, 7)}(Points : 1)

{(-2, -8), (2, 4), (4, 7)}

{(-2, -8), (0, -3), (4, 7)}

{(-2, -8), (0, -3), (1, -1)}

{(1, -1), (2, 4), (4, 7)}

5. Solve the equation, using the domain of x = {-1, 0, 2}. Write your solutions as a set of ordered pairs.

2x + 3y = 7(Points : 1)

6. Give the range of x - 2y = 0 for the domain {-1, 0, 4}.(Points : 1)

7. The interest on a loan at an interest rate of 9% over a year is I = (0.09)P, where P is the principle (loan amount) and I is the interest rate. Use a domain in the table to find the interest of these amounts for one year. Write the range only in your solution.

Principle 10 100 1000 10,000 100,000 1,000,000

Interest -- -- -- -- -- --

(Points : 1)

8. What is the independent variable and what is the dependent variable of this relation?

Principle 10 100 1,000 10,000

Interest 2 20 200 2,000

(Points : 1)

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