IDNLearn.com: Your one-stop destination for finding reliable answers. Our Q&A platform offers reliable and thorough answers to ensure you have the information you need to succeed in any situation.
Sagot :
Certainly! Let's go through each part of the problem step-by-step.
### Part A: Description of the Graph
We have the following system of inequalities:
[tex]\[ \begin{array}{l} 2x + 3y \leq 15 \\ x + y \geq 3 \end{array} \][/tex]
1. Graphing the Inequalities:
- For the inequality [tex]\(2x + 3y \leq 15\)[/tex]:
- The boundary line is [tex]\(2x + 3y = 15\)[/tex].
- To graph this, find the intercepts:
- When [tex]\(x = 0\)[/tex]: [tex]\(3y = 15 \Rightarrow y = 5\)[/tex]. This gives the point [tex]\((0, 5)\)[/tex].
- When [tex]\(y = 0\)[/tex]: [tex]\(2x = 15 \Rightarrow x = 7.5\)[/tex]. This gives the point [tex]\((7.5, 0)\)[/tex].
- Plot these points and draw the line. Since the inequality is [tex]\(\leq\)[/tex], shade the area below and including the line.
- For the inequality [tex]\(x + y \geq 3\)[/tex]:
- The boundary line is [tex]\(x + y = 3\)[/tex].
- To graph this, find the intercepts:
- When [tex]\(x = 0\)[/tex]: [tex]\(y = 3\)[/tex]. This gives the point [tex]\((0, 3)\)[/tex].
- When [tex]\(y = 0\)[/tex]: [tex]\(x = 3\)[/tex]. This gives the point [tex]\((3, 0)\)[/tex].
- Plot these points and draw the line. Since the inequality is [tex]\(\geq\)[/tex], shade the area above and including the line.
2. Solution Set:
- The solution set is the region where the shadings of the two inequalities overlap.
- The lines are solid because the inequalities are [tex]\(\leq\)[/tex] and [tex]\(\geq\)[/tex], which include the boundaries.
### Part B: Checking the Point (5,1)
Given the point [tex]\((5,1)\)[/tex]:
1. Check the inequality [tex]\(2x + 3y \leq 15\)[/tex]:
[tex]\[ 2(5) + 3(1) = 10 + 3 = 13 \leq 15 \][/tex]
This condition is satisfied.
2. Check the inequality [tex]\(x + y \geq 3\)[/tex]:
[tex]\[ 5 + 1 = 6 \geq 3 \][/tex]
This condition is also satisfied.
Since both conditions are satisfied, the point [tex]\((5,1)\)[/tex] is included in the solution area.
### Part C: Choosing a Different Point
Let's choose the point [tex]\((3,1)\)[/tex]:
1. Check the inequality [tex]\(2x + 3y \leq 15\)[/tex]:
[tex]\[ 2(3) + 3(1) = 6 + 3 = 9 \leq 15 \][/tex]
This condition is satisfied.
2. Check the inequality [tex]\(x + y \geq 3\)[/tex]:
[tex]\[ 3 + 1 = 4 \geq 3 \][/tex]
This condition is also satisfied.
Since both conditions are satisfied, the point [tex]\((3,1)\)[/tex] is in the solution set.
Interpretation:
- If Michael buys 3 cupcakes and 1 piece of fudge:
- The total cost would be [tex]\(2 \times 3 + 3 \times 1 = 6 + 3 = 9\)[/tex] dollars, which is within his budget of \$15.
- This combination would feed [tex]\(3 + 1 = 4\)[/tex] siblings, which meets the requirement of feeding at least 3 siblings.
Thus, the chosen point [tex]\((3,1)\)[/tex] provides a feasible and valid combination of cupcakes and fudge within the given constraints.
### Part A: Description of the Graph
We have the following system of inequalities:
[tex]\[ \begin{array}{l} 2x + 3y \leq 15 \\ x + y \geq 3 \end{array} \][/tex]
1. Graphing the Inequalities:
- For the inequality [tex]\(2x + 3y \leq 15\)[/tex]:
- The boundary line is [tex]\(2x + 3y = 15\)[/tex].
- To graph this, find the intercepts:
- When [tex]\(x = 0\)[/tex]: [tex]\(3y = 15 \Rightarrow y = 5\)[/tex]. This gives the point [tex]\((0, 5)\)[/tex].
- When [tex]\(y = 0\)[/tex]: [tex]\(2x = 15 \Rightarrow x = 7.5\)[/tex]. This gives the point [tex]\((7.5, 0)\)[/tex].
- Plot these points and draw the line. Since the inequality is [tex]\(\leq\)[/tex], shade the area below and including the line.
- For the inequality [tex]\(x + y \geq 3\)[/tex]:
- The boundary line is [tex]\(x + y = 3\)[/tex].
- To graph this, find the intercepts:
- When [tex]\(x = 0\)[/tex]: [tex]\(y = 3\)[/tex]. This gives the point [tex]\((0, 3)\)[/tex].
- When [tex]\(y = 0\)[/tex]: [tex]\(x = 3\)[/tex]. This gives the point [tex]\((3, 0)\)[/tex].
- Plot these points and draw the line. Since the inequality is [tex]\(\geq\)[/tex], shade the area above and including the line.
2. Solution Set:
- The solution set is the region where the shadings of the two inequalities overlap.
- The lines are solid because the inequalities are [tex]\(\leq\)[/tex] and [tex]\(\geq\)[/tex], which include the boundaries.
### Part B: Checking the Point (5,1)
Given the point [tex]\((5,1)\)[/tex]:
1. Check the inequality [tex]\(2x + 3y \leq 15\)[/tex]:
[tex]\[ 2(5) + 3(1) = 10 + 3 = 13 \leq 15 \][/tex]
This condition is satisfied.
2. Check the inequality [tex]\(x + y \geq 3\)[/tex]:
[tex]\[ 5 + 1 = 6 \geq 3 \][/tex]
This condition is also satisfied.
Since both conditions are satisfied, the point [tex]\((5,1)\)[/tex] is included in the solution area.
### Part C: Choosing a Different Point
Let's choose the point [tex]\((3,1)\)[/tex]:
1. Check the inequality [tex]\(2x + 3y \leq 15\)[/tex]:
[tex]\[ 2(3) + 3(1) = 6 + 3 = 9 \leq 15 \][/tex]
This condition is satisfied.
2. Check the inequality [tex]\(x + y \geq 3\)[/tex]:
[tex]\[ 3 + 1 = 4 \geq 3 \][/tex]
This condition is also satisfied.
Since both conditions are satisfied, the point [tex]\((3,1)\)[/tex] is in the solution set.
Interpretation:
- If Michael buys 3 cupcakes and 1 piece of fudge:
- The total cost would be [tex]\(2 \times 3 + 3 \times 1 = 6 + 3 = 9\)[/tex] dollars, which is within his budget of \$15.
- This combination would feed [tex]\(3 + 1 = 4\)[/tex] siblings, which meets the requirement of feeding at least 3 siblings.
Thus, the chosen point [tex]\((3,1)\)[/tex] provides a feasible and valid combination of cupcakes and fudge within the given constraints.
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. IDNLearn.com is your go-to source for accurate answers. Thanks for stopping by, and come back for more helpful information.
