Mike wants to fence in part of his backyard. He wants the length of the fenced-in area to be at least 20 feet long, l ≥ 20. He has 200 feet of fencing. The inequality that models the possible perimeter of the yard is 2l + 2w ≤ 200.

Question

Which are possible dimensions for Mike’s backyard? Check all that apply.

w = 50 ft; l = 10 ft
w = 10 ft; l = 50 ft
w = 20 ft; l = 60 ft
w = 90 ft; l = 30 ft
w = 50 ft; l = 40 ft

Answer 1

To determine which combinations of \( w \) (width) and \( l \) (length) are possible for Mike's fenced-in backyard, we need to check each pair against the two conditions:

1. The length must be at least 20 feet: \( l \geq 20 \)
2. The total perimeter must not exceed 200 feet: \( 2l + 2w \leq 200 \)

This can be simplified to:

\[ l + w \leq 100 \]

Now, let's check each of the given pairs:

1. \( w = 50 , \text{ft}; , l = 10 , \text{ft} \):

   • Check length: \( 10 < 20 \) (does not satisfy)
   • Not possible

2. \( w = 10 , \text{ft}; , l = 50 , \text{ft} \):

   • Check length: \( 50 \geq 20 \) (satisfies)
   • Check perimeter: \( 50 + 10 = 60 \leq 100 \) (satisfies)
   • Possible

3. \( w = 20 , \text{ft}; , l = 60 , \text{ft} \):

   • Check length: \( 60 \geq 20 \) (satisfies)
   • Check perimeter: \( 20 + 60 = 80 \leq 100 \) (satisfies)
   • Possible

4. \( w = 90 , \text{ft}; , l = 30 , \text{ft} \):

   • Check length: \( 30 \geq 20 \) (satisfies)
   • Check perimeter: \( 90 + 30 = 120 \not\leq 100 \) (does not satisfy)
   • Not possible

5. \( w = 50 , \text{ft}; , l = 40 , \text{ft} \):

   • Check length: \( 40 \geq 20 \) (satisfies)
   • Check perimeter: \( 50 + 40 = 90 \leq 100 \) (satisfies)
   • Possible

The possible dimensions for Mike’s backyard are:

• \( w = 10 , \text{ft}; , l = 50 , \text{ft} \)
• \( w = 20 , \text{ft}; , l = 60 , \text{ft} \)
• \( w = 50 , \text{ft}; , l = 40 , \text{ft} \)

Answer 2

Erin and her friends are taking a road trip. She has agreed to drive between hours 3 and 6 of the trip. If h represents the hours of the trip Erin will be driving, then h ≥ 3 and h≤ 6 represents her portion of the driving. While she is driving, her distance and time are modeled by the inequality d < 70h. What are the possible hours and distances Erin could be driving?

When graphing this scenario, let h be the horizontal variable and d be the vertical variable.

For the inequality h ≥ 3, the graph should be shaded
the boundary line.

For the inequality h ≤ 6, the graph should be shaded
the boundary line.

For the inequality d < 70h, the graph should be shaded
the boundary line.

One possible solution for Erin is
.

Answer 3

Let's analyze the scenario provided.

1. Inequalities for Erin's Driving Time:

• The first inequality is \( h \geq 3 \): This represents that Erin will be driving for at least 3 hours.
• The second inequality is \( h \leq 6 \): This indicates that Erin will be driving for no more than 6 hours.

These two inequalities define a range of hours for Erin, specifically: \[ 3 \leq h \leq 6 \]

2. Distance Inequality:

• The third inequality is \( d < 70h \): This states that the distance \( d \) traveled by Erin during the hours she drives is less than \( 70 \) times the number of hours she drives.

3. Graphing the Inequalities: We can now graph these inequalities:

• For \( h \geq 3 \):

  • The line \( h = 3 \) will be a vertical line at \( h = 3 \). The graph should be shaded to the right of this line.

• For \( h \leq 6 \):

  • The line \( h = 6 \) will be a vertical line at \( h = 6 \). The graph should be shaded to the left of this line.

• For \( d < 70h \):

  • This represents a line with a slope of 70, and it passes through the origin. The line is \( d = 70h \), and the graph should be shaded below this line, indicating that Erin's distance is less than \( 70h \).

4. Identifying Possible Solutions: To find a possible combination of \( h \) and \( d \), we must choose a value for \( h \) within the range of \( 3 \leq h \leq 6 \) and then calculate \( d \) using the inequality \( d < 70h \).

Let's choose:

• For \( h = 4 \): \[ d < 70 \times 4 = 280 \implies d < 280 \] Therefore, when \( h = 4 \), any distance \( d \) that is less than 280 would be valid.

Thus, one possible solution for Erin could be:

• \( h = 4 \) hours and \( d = 250 \) miles (as an example since \( d < 280 \)).

So, the answer would be: One possible solution for Erin is (4, 250). You can substitute different values of \( h \) between 3 and 6 and calculate respective distances to find other valid solutions.

Answer 4

When graphing this scenario, let h be the horizontal variable and d be the vertical variable.

For the inequality h ≥ 3, the graph should be shaded
✔ to the right of
the boundary line.

For the inequality h ≤ 6, the graph should be shaded
✔ to the left of
the boundary line.

For the inequality d < 70h, the graph should be shaded
✔ below
the boundary line.

One possible solution for Erin is
✔ (4.5, 200)
.

Elam is packing his room to move into a new house. A small box can hold 8 books without breaking, while a large box can hold 12 books without breaking. He has at most 160 books to pack and less than 30 boxes total. Let s represent the number of small boxes and l represent the number of large boxes.

The inequalities s ≥ 0 and l ≥ 0 are part of the system that models this scenario.

Which inequalities complete the system?

s – l < 30
8s – 12l ≤ 160
s + l < 30
8s + 12l ≤ 160
s + l > 30
8s + 12l ≤ 160
s + l < 30
8s + 12l ≥ 160

Answer 5

Let's break down the problem related to Elam packing his books. We need to determine the correct inequalities that model the situation regarding the boxes and books.

Given Information:

• A small box can hold 8 books.
• A large box can hold 12 books.
• Elam has at most 160 books to pack.
• Elam has less than 30 boxes total.
• Let \( s \) represent the number of small boxes and \( l \) represent the number of large boxes.

Inequalities Based on the Problem Statement:

1. Box Capacity:

   • Each small box holds 8 books, and each large box holds 12 books. Hence, the total number of books in small and large boxes must not exceed 160 books: \[ 8s + 12l \leq 160 \]

2. Total Boxes:

   • The total number of boxes (small and large) he uses must be less than 30: \[ s + l < 30 \]

3. Non-Negativity Constraints:

   • Since the number of boxes cannot be negative, we have: \[ s \geq 0 \quad \text{and} \quad l \geq 0 \]

Summary of the Inequalities:

So far, we have identified the following inequalities:

• \( s \geq 0 \)
• \( l \geq 0 \)
• \( 8s + 12l \leq 160 \)
• \( s + l < 30 \)

Which Inequalities Complete the System?

From the provided options, the correct inequalities that complete the system are:

• \( s + l < 30 \)
• \( 8s + 12l \leq 160 \)

The other options, such as \( s - l < 30 \) and \( s + l > 30 \), do not fit the criteria stated in the problem.

Conclusion:

The complete inequality system for Elam's scenario is:

• \( s \geq 0 \)
• \( l \geq 0 \)
• \( 8s + 12l \leq 160 \)
• \( s + l < 30 \)

Thus, the correct options to complete the system are:

• \( s + l < 30 \)
• \( 8s + 12l \leq 160 \)

Answer 6

Which graph represents the solution set for the system s + l < 30 and 8s + 12l ≤ 160? 1. On a coordinate plane, 2 straight lines are shown. The first dashed line has a positive slope and goes through (negative 30, 0) and (0, 30). Everything to the left of the line is shaded. The second solid line has a negative slope and goes through (negative 25, 30) and (negative 10, 20). Everything to the left of the line is shaded. 2. On a coordinate plane, 2 straight lines are shown. The first solid line has a negative slope and goes through (5, 10) and (20, 0). Everything below and to the left of the line is shaded. The second dashed line has a negative slope and goes through (0, 30) and (30, 0). Everything above and to the right of the line is shaded. 3. On a coordinate plane, 2 straight lines are shown. The first solid line has a negative slope and goes through (5, 10) and (20, 0). Everything above and to the right of the line is shaded. The second dashed line has a negative slope and goes through (0, 30) and (30, 0). Everything below and to the left of the line is shaded. 4. On a coordinate plane, 2 straight lines are shown. The first solid line has a negative slope and goes through (5, 10) and (20, 0). Everything below and to the left of the line is shaded. The second dashed line has a negative slope and goes through (0, 30) and (30, 0). Everything below and to the left of the line is shaded.

Answer 7

To determine which graph represents the solution set for the system of inequalities \( s + l < 30 \) and \( 8s + 12l \leq 160 \), we need to analyze both inequalities.

Analyzing the Inequalities

1. First Inequality: \( s + l < 30 \):

   • This inequality represents a line where \( s + l = 30 \).
   • The line will have a negative slope (as \( l \) decreases with an increase in \( s \)).
   • It has intercepts at (30, 0) and (0, 30).
   • Since the inequality is strict (<), the line will be dashed and the area below and to the left of this line will be shaded.

2. Second Inequality: \( 8s + 12l \leq 160 \):

   • This can be rewritten in slope-intercept form: \[ 12l \leq 160 - 8s \quad \Rightarrow \quad l \leq \frac{160}{12} - \frac{2}{3}s \approx 13.33 - 0.67s \]
   • This line has a negative slope and intercepts the axes (when \( s = 0\), \( l \approx 13.33 \); when \( l = 0\), \( s = 20 \)).
   • Since the inequality is non-strict (≤), this line will be solid, and the area below (including the line) will be shaded.

Finding the Correct Graph

Now, we will look for a graph that:

• Has a dashed line representing \( s + l < 30 \) with intercepts at (30, 0) and (0, 30), shaded below and to the left of it.
• Has a solid line representing \( 8s + 12l \leq 160 \) with intercepts at approximately (20, 0) and (0, 13.33), shaded below it.

Review of the Provided Graphs

1. Graph 1:

   • Dashed line (potentially \( s + l = 30 \)).
   • Solid line (potentially \( 8s + 12l = 160 \)).
   • Shading seems appropriate, but check the intercepts.

2. Graph 2:

   • Solid line, suggests \( 8s + 12l \) inequality but is defined incorrectly in the direction of shading (shading above and to the right, which is not correct).

3. Graph 3:

   • Solid line similar to Graph 2, also with incorrect shading above for one line.

4. Graph 4:

   • Solid line that might represent the book capacity inequality, shaded below. Dashed line consistent with smaller box inequality, but they are shaded the same way here.

Conclusion

Given the explanations:

• Graph 1 most closely represents the solution set because it has the correct configuration, with a dashed line and shading for \( s + l < 30 \) and a solid line with downward shading for \( 8s + 12l \leq 160 \).

Thus, Graph 1 is the correct representation of the solution set for the system \( s + l < 30 \) and \( 8s + 12l \leq 160 \).

Answer 8

On a coordinate plane, 2 straight lines are shown. The first solid line has a negative slope and goes through (5, 10) and (20, 0). Everything below and to the left of the line is shaded. The second dashed line has a negative slope and goes through (0, 30) and (30, 0). Everything below and to the left of the line is shaded.

there are 3 graphs where the lines go the same right? well the top color Is white middle pink then bottom greyish blue?

Joshua wants to burn at least 400 calories per day, but no more than 600. He does this by walking and playing basketball. Assuming he burns 4 calories per minute walking, w, and 5 calories per minute spent playing basketball, b, the situation can be modeled using these inequalities:

4w + 5b ≥ 400

4w + 5b ≤ 600

Which are possible solutions for the number of minutes Joshua can participate in each activity? Check all that apply.

40 minutes walking, 40 minutes basketball
60 minutes walking, 20 minutes basketball
20 minutes walking, 60 minutes basketball
50 minutes walking, 50 minutes basketball
60 minutes walking, 80 minutes basketball
70 minutes walking, 60 minutes basketball

Answer 9

To determine which combinations of minutes walking and playing basketball are valid solutions for Joshua's calorie burning goals, we need to evaluate each given pair against the two inequalities:

1. Inequalities:
   • \( 4w + 5b \geq 400 \) (burn at least 400 calories)
   • \( 4w + 5b \leq 600 \) (burn no more than 600 calories)

Analyzing Each Option

Let's evaluate each scenario to see if they satisfy both inequalities.

1. 40 minutes walking, 40 minutes basketball: \[ 4(40) + 5(40) = 160 + 200 = 360 \]

   • \( 360 \geq 400 \)? No
   • \( 360 \leq 600 \)? Yes
   • Not a valid solution.

2. 60 minutes walking, 20 minutes basketball: \[ 4(60) + 5(20) = 240 + 100 = 340 \]

   • \( 340 \geq 400 \)? No
   • \( 340 \leq 600 \)? Yes
   • Not a valid solution.

3. 20 minutes walking, 60 minutes basketball: \[ 4(20) + 5(60) = 80 + 300 = 380 \]

   • \( 380 \geq 400 \)? No
   • \( 380 \leq 600 \)? Yes
   • Not a valid solution.

4. 50 minutes walking, 50 minutes basketball: \[ 4(50) + 5(50) = 200 + 250 = 450 \]

   • \( 450 \geq 400 \)? Yes
   • \( 450 \leq 600 \)? Yes
   • Valid solution!

5. 60 minutes walking, 80 minutes basketball: \[ 4(60) + 5(80) = 240 + 400 = 640 \]

   • \( 640 \geq 400 \)? Yes
   • \( 640 \leq 600 \)? No
   • Not a valid solution.

6. 70 minutes walking, 60 minutes basketball: \[ 4(70) + 5(60) = 280 + 300 = 580 \]

   • \( 580 \geq 400 \)? Yes
   • \( 580 \leq 600 \)? Yes
   • Valid solution!

Summary of Valid Solutions

After checking all combinations, the valid solutions that fit both inequalities are:

• 50 minutes walking, 50 minutes basketball
• 70 minutes walking, 60 minutes basketball

Thus, the solutions for the number of minutes Joshua can participate in each activity are:

• 50 minutes walking, 50 minutes basketball
• 70 minutes walking, 60 minutes basketball

Answer 10

Compost costs $10 per cubic yard and topsoil costs $2 per cubic yard. Maria needs more than 10 cubic yards of topsoil. She can spend no more than $50. The graph below represents the amount of each she can buy.On a coordinate plane, 2 straight lines are shown. The x-axis shows cubic yards of compost, from 0 to 10, and the y-axis shows cubic yards of topsoil, from 0 to 16. A horizontal dashed line is at y = 10. Everything above the line is shaded. A solid straight line has a negative slope and goes through (3, 10) and (5, 0). Everything to the left of the line is shaded.Which amounts of materials satisfy the relationship?2 cubic yards of compost, 8 cubic yards of topsoil4 cubic yards of compost, 5 cubic yards of topsoil1 cubic yard of compost, 12 cubic yards of topsoil3 cubic yards of compost, 12 cubic yards of topsoil

Answer 11

To determine which combinations of compost and topsoil satisfy Maria's requirements, we first need to outline the inequalities that represent the constraints on her purchases:

Given Information:

• Cost of Compost: $10 per cubic yard.
• Cost of Topsoil: $2 per cubic yard.
• Requirement: More than 10 cubic yards of topsoil.
• Budget Constraint: She can spend no more than $50.

Setting Up the Inequalities:

1. Topsoil Requirement: \[ y > 10 \quad (\text{where } y \text{ is cubic yards of topsoil}) \]

2. Budget Constraint (where \( x \) is cubic yards of compost): \[ 10x + 2y \leq 50 \]

Simplifying the Budget Constraint:

We can divide the entire budget inequality by 2 for easier calculations: \[ 5x + y \leq 25 \]

Combined Inequalities:

So, the system of inequalities becomes:

1. \( y > 10 \)
2. \( 5x + y \leq 25 \)

Analyzing the Graph:

• The dashed horizontal line at \( y = 10 \) indicates the minimum amount of topsoil needed (everything above this line represents a valid increase in topsoil).
• The solid line with a negative slope (going through the points (3, 10) and (5, 0)) represents the budget constraint \( 5x + y = 25 \). Everything below this line is shaded.

Checking Each Option:

Now, let’s check each of the given combinations:

1. 2 cubic yards of compost, 8 cubic yards of topsoil: \[ x = 2, \quad y = 8 \implies 5(2) + 8 = 10 + 8 = 18 \leq 25 \quad \text{(Within budget)} \]

   • But \( y = 8 \) does not satisfy \( y > 10 \).
   • Not a valid solution.

2. 4 cubic yards of compost, 5 cubic yards of topsoil: \[ x = 4, \quad y = 5 \implies 5(4) + 5 = 20 + 5 = 25 \leq 25 \quad \text{(Within budget)} \]

   • But \( y = 5 \) does not satisfy \( y > 10 \).
   • Not a valid solution.

3. 1 cubic yard of compost, 12 cubic yards of topsoil: \[ x = 1, \quad y = 12 \implies 5(1) + 12 = 5 + 12 = 17 \leq 25 \quad \text{(Within budget)} \]

   • Also, \( y = 12 \) satisfies \( y > 10 \).
   • Valid solution!

4. 3 cubic yards of compost, 12 cubic yards of topsoil: \[ x = 3, \quad y = 12 \implies 5(3) + 12 = 15 + 12 = 27 \nleq 25 \quad \text{(Exceeds budget)} \]

   • Not a valid solution.

Summary of Valid Solutions:

The only combination that satisfies both conditions (more than 10 cubic yards of topsoil and within budget) is:

• 1 cubic yard of compost, 12 cubic yards of topsoil.

Therefore, the valid amounts of materials that satisfy the relationship are:

• 1 cubic yard of compost, 12 cubic yards of topsoil.