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Study Island

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Generation Date: 02/22/2015

Generated By: Robert Dilliplane

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1. Jake is building a fence around his property. He wants the perimeter to be no more than 100 feet. He also wants the length to be at least 10 feet longer than the width. If he builds his fence according to these limits, which would be the maximum possible width of the fence?

|[pic|A. |30 feet |

|] | | |

|[pic|B. |10 feet |

|] | | |

|[pic|C. |50 feet |

|] | | |

|[pic|D. |20 feet |

|] | | |

[pic]

2. Which graph represents the following system of inequalities?

[pic]

|[pic] |[pic] |

|W. |X. |

|[pic] |[pic] |

|Y. |Z. |

|[pic|A. |W |

|] | | |

|[pic|B. |Y |

|] | | |

|[pic|C. |X |

|] | | |

|[pic|D. |Z |

|] | | |

[pic]

3. Choose the system of inequalities that best matches the graph below.

[pic]

|[pic|A. |[pic] |

|] | | |

|[pic|B. |[pic] |

|] | | |

|[pic|C. |[pic] |

|] | | |

|[pic|D. |[pic] |

|] | | |

[pic]

4. Graph the following system of inequalities.

[pic]

|[pic] |[pic] |

|W. |X. |

|[pic] |[pic] |

|Y. |Z. |

|[pic|A. |X |

|] | | |

|[pic|B. |Z |

|] | | |

|[pic|C. |W |

|] | | |

|[pic|D. |Y |

|] | | |

[pic]

5. Choose the system of inequalities that best matches the graph below.

[pic]

|[pic|A. |[pic] |

|] | | |

|[pic|B. |[pic] |

|] | | |

|[pic|C. |[pic] |

|] | | |

|[pic|D. |[pic] |

|] | | |

[pic]

6. Choose the system of inequalities that best matches the graph below.

[pic]

|[pic|A. |[pic] |

|] | | |

|[pic|B. |[pic] |

|] | | |

|[pic|C. |[pic] |

|] | | |

|[pic|D. |[pic] |

|] | | |

[pic]

7.

[pic]

Which quadrants contain the solution to this system of inequalities?

|[pic|A. |quadrants I and II |

|] | | |

|[pic|B. |quadrants I, II, and IV |

|] | | |

|[pic|C. |quadrants I and IV |

|] | | |

|[pic|D. |quadrants III and IV |

|] | | |

[pic]

8.

The drill team is selling candy at the football games for a fundraiser. Students can buy the candy for $2 and everyone else can buy the candy for $3. They only have 200 bags of candy and they need to make at least $500.

If the solution region is the amount of candy sold to students and everyone else, which graph's shaded area represents the solution region?

|[pic] |[pic] |

|W. |X. |

|[pic] |[pic] |

|Y. |Z. |

|[pic|A. |Y |

|] | | |

|[pic|B. |Z |

|] | | |

|[pic|C. |W |

|] | | |

|[pic|D. |X |

|] | | |

[pic]

9. Which system of inequalities is represented by the graph below?

[pic]

|[pic|A. |[pic] |

|] | | |

|[pic|B. |[pic] |

|] | | |

|[pic|C. |[pic] |

|] | | |

|[pic|D. |[pic] |

|] | | |

[pic]

10.

Julia works for a company selling light bulbs. She sells fluorescent light bulbs for $5 a piece and incandescent light bulbs for $2. To meet her quota, she needs to sell over 100 light bulbs and make at least $350.

If the solution region is the amount of light bulbs Julia needs to sell to meet her quota, which graph's shaded area represents the solution region?

|[pic] |[pic] |

|W. |X. |

|[pic] |[pic] |

|Y. |Z. |

|[pic|A. |X |

|] | | |

|[pic|B. |Z |

|] | | |

|[pic|C. |Y |

|] | | |

|[pic|D. |W |

|] | | |

[pic]

11. Graph the following system of inequalities.

[pic]

|[pic] |[pic] |

|W. |X. |

|[pic] |[pic] |

|Y. |Z. |

|[pic|A. |Z |

|] | | |

|[pic|B. |Y |

|] | | |

|[pic|C. |W |

|] | | |

|[pic|D. |X |

|] | | |

[pic]

12. Graph the following system of inequalities.

[pic]

|[pic] |[pic] |

|W. |X. |

|[pic] |[pic] |

|Y. |Z. |

|[pic|A. |W |

|] | | |

|[pic|B. |X |

|] | | |

|[pic|C. |Z |

|] | | |

|[pic|D. |Y |

|] | | |

[pic]

Answers

1. D

2. D

3. A

4. A

5. A

6. C

7. A

8. C

9. A

10. C

11. A

12. A

Explanations

1. Set up and graph a system of inequalities to solve the problem.

Letting x represent the width and y represent the length of the fenced-in area produces the inequality below when the perimeter restrictions are considered.

|2x + 2y |< |100 |

|2y |< |-2x + 100 |

|y |< |-x + 50 |

Furthermore, the length must be at least 10 feet longer than the width.

|y |> |x + 10 |

Graph both inequalities to determine the solution.

|[pic] |

The greatest possible width is 20 feet.

2. The first inequality has a y-intercept of 2 and a slope of [pic].

The inequality symbol is less than or equal to, so the line will be a solid line, and the solution region will be shaded below this line.

The second inequality has a y-intercept of 5 and a slope of -1.

The inequality symbol is greater than, so the line will be a dashed line, and the solution region will be shaded above this line.

Therefore, the correct graph is graph Z.

3. The pink line crosses the y-axis at (0,2). Since b represents the y-intercept of the equation, b = 2. The line also rises -2 and runs 3 to each point. Since the slope of a line is rise over run, and m represents the slope of the line, m = -[pic].

Thus, the equation of the line is y = -[pic]x + 2.

Now, plug in a point from the shaded area, such as (-3,0).

[pic]

Since the line is solid, the less than or equal to symbol should be used to get the inequality y < -[pic]x + 2.

The blue line crosses the y-axis at (0,0). Since b represents the y-intercept of the equation, b = 0. The line also rises -1 and runs 4 to each point. Since the slope of a line is rise over run, and m represents the slope of the line, m = -[pic].

Thus, the equation of the line is y = -[pic]x.

Now, plug in a point from the shaded area, such as (-4,0).

[pic]

Since the line is dotted, then the less than symbol should be used to get the inequality y < -[pic]x.

The only system of the answer choices that matches both inequalities is y < -[pic]x + 2. and y < -[pic]x.

4. The first inequality, y > [pic]x - 2, has a y-intercept of b = -2 and a slope of m = [pic]. The inequality symbol says greater than, so this line should be dotted with the common shaded area above the line.

The second inequality, y < -[pic]x, has a y-intercept of b = 0 and a slope of m = -[pic]. The inequality symbol says less than, so this line should be dotted with the common shaded area below the line.

The only graph that fits every description is X.

5. The red line crosses the y-axis at (0,1). Since b represents the y-intercept of the equation, b = 1. The line also rises 2 and runs 1 to each point. Since the slope of a line is rise over run, and m represents the slope of the line, m = 2.

Thus, the equation of the line is y = 2x + 1.

Now, plug in a point from the shaded area, such as (-1,1).

[pic]

Since the line is solid, then the greater than or equal to symbol should be used to get the inequality y > 2x + 1.

The blue line is a vertical line with no slope that crosses at -2 on the x-axis.

Thus, the equation of the line is x = -2.

Notice that the shading is to the right of the solid line, so the greater than or equal to symbol should be used to get the inequality x > -2.

The only system of the answer choices that matches both inequalities is x > -2 and y > 2x + 1.

6. The purple line is a horizontal line with a slope of m = 0 that crosses at -1 on the y-axis.

Thus, the equation of the line is y = -1.

Notice that the shading is below the dotted line, so the less than symbol should be used to get the inequality y < -1.

The pink line is a vertical line with undefined slope that crosses at 1 on the x-axis.

Thus, the equation of the line is x = 1.

Notice that the shading is to the left of the solid line, so the less than or equal to symbol should be used to get the inequality x < 1.

The only system of the answer choices that matches both inequalities is y < -1 and x < 1.

7.

| |

To determine which quadrants contain the solution, graph each inequality separately.

|The first inequality, [pic], |[pic] |

|has a y-intercept of [pic], a slope of [pic], is a solid line, and is shaded above. | |

|The second inequality, [pic], |[pic] |

|has a y-intercept of [pic], a slope of [pic], is a dotted line, and is shaded above. | |

|The solution is in the overlapping sections. |[pic] |

| | |

|Therefore, the quadrants that contain the solution are quadrants I and II. | |

8. First, form the inequalities needed for the graph using x for the amount of candy sold to everyone else and y for the amount of candy sold to students.

Since they only have 200 bags of candy, use the following inequality.

x + y < 200

Since candy is sold to students for $2, everyone else for $3, and they need to make at least $500, use the following inequality.

3x + 2y > 500

Thus, the solution region can be found using this system of inequalities.

|x + y < |200 |

|3x + 2y > |500 |

In order to graph the inequalities, convert each equation into slope-intercept form.

|x + y |< |200 | |3x + 2y |> |500 |

|x - x + y |< |200 - x | |3x - 3x + 2y |> |500 - 3x |

|y |< |-x + 200 | |2y |> |-3x + 500 |

| | | | |[pic] | |[pic] |

| | | | |2  | |2    |

| | | | |y |> |- |

| | | | | | |3 |

| | | | | | |x + 250 |

| | | | | | | |

| | | | | | | |

| | | | | | |[pic] |

| | | | | | | |

| | | | | | | |

| | | | | | | |

| | | | | | |2 |

| | | | | | | |

| | | | | | | |

Since both equations are allowed to be equal, graph them using solid lines.

The equation on the left shows y is less than or equal to the right side, so the solution region will be below this line.

The equation on the right shows y is greater than or equal to the right side, so the solution region will be above this line.

The region that is below the first line and above the second line forms the solution region, which is shaded as shown below.

[pic]

9. The dashed line, l1, has a y-intercept of 3 and a slope of [pic].

Since the shaded region is below the dashed line, the inequality is less than.

[pic]

The dashed line, l2, has a y-intercept of 0 and a slope of [pic].

Since the shaded region is below the dotted line, the inequality is less than.

[pic]

Therefore, the graph represents the following system of inequalities.

[pic]

10. First, form the inequalities needed for the graph using x for the amount of incandescent light bulbs sold and y for the amount of fluorescent light bulbs sold.

Since Julia needs to sell more than 100 light bulbs, use the following inequality.

x + y > 100

Since each fluorescent light bulb sells for $5, each incandescent light bulb sells for $2, and Julia needs to make at least $350, use the following inequality.

2x + 5y > 350

Thus, the solution region can be found using this system of inequalities.

|x + y > |100 |

|2x + 5y > |350 |

In order to graph the inequalities, convert each equation into slope-intercept form.

|x + y |> |100 | |2x + 5y |> |350 |

|x - x + y |> |100 - x | |2x - 2x + 5y |> |350 - 2x |

|y |> |-x + 100 | |5y |> |-2x + 350 |

| | | | |[pic] | |[pic] |

| | | | |5  | |5  |

| | | | |y |> |- |

| | | | | | |2 |

| | | | | | |x + 70 |

| | | | | | | |

| | | | | | | |

| | | | | | |[pic] |

| | | | | | | |

| | | | | | | |

| | | | | | | |

| | | | | | |5 |

| | | | | | | |

| | | | | | | |

Since the equation on the left is not allowed to be equal, graph it using a dotted line.

Since the equation on the right is allowed to be equal, graph it using a solid line.

The equation on the left shows y is greater than the right side, so the solution region will be above this line.

The equation on the right shows y is greater than or equal to the right side, so the solution region will be above this line.

The region that is above the first line and above the second line forms the solution region, which is shaded as shown below.

[pic]

11. The first inequality, x > 1, only contains one variable, x, so it makes a vertical line with no slope that crosses at 1 on the x-axis. The inequality symbol says greater than or equal to, so this vertical line should be solid with the common shaded area to the right of the line.

The second inequality, y < -x + 3, has a y-intercept of b = 3 and a slope of m = -1. The inequality symbol says less than or equal to, so this line should be solid with the common shaded area below the line.

The only graph that fits every description is Z.

12. The first inequality, y < x + 3, has a y-intercept of b = 3 and a slope of m = 1. The inequality symbol says less than, so this line should be dotted with the common shaded area below the line.

The second inequality, y > x - 3, has a y-intercept of b = -3 and a slope of m = 1. The inequality symbol says greater than, so this line should be dotted with the common shaded area above the line.

The only graph that fits every description is W.

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