Mina's catering service is organizing a formal dinner for 280 people.The hall has two kinds of tables, one that seats 4 people and one that seats 10 people.The hall can contain up to a total of 52 tables.Write and graph a system of inequalities that can be used to determine the possible combinations of tables that can be used for the event so there are enough seats for all people,
If there are x 4-seaters, and y 10-seaters, then
x+y <= 52
4x + 10y >= 280
see the solution set at
http://www.wolframalpha.com/input/?i=solve+x%2By+%3C%3D+52%2C+4x+%2B+10y+%3E%3D+280%2C+x%3E%3D0%2C+y%3E%3D0
Sure! Let's approach this systematically.
Let's assume x represents the number of tables that seat 4 people and y represents the number of tables that seat 10 people.
To ensure there are enough seats for all 280 people, we can set up the following inequalities:
1. The total number of people that can be seated on tables for 4 people (4x) plus the total number of people that can be seated on tables for 10 people (10y) should be greater than or equal to 280:
4x + 10y >= 280
2. The total number of tables should not exceed 52:
x + y <= 52
Now let's graph these inequalities.
First, let's graph the equation 4x + 10y = 280 (replacing the inequality with an equation):
To graph it, we need to find two points on the line and draw a straight line passing through those points. Let's assume some values for x and solve for y:
x = 0, then 4(0) + 10y = 280, which gives us y = 28. So one point is (0, 28).
x = 40, then 4(40) + 10y = 280, which gives us y = 12. So another point is (40, 12).
Now we can plot these points and draw the line passing through them:
y-axis
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x-axis
Next, let's graph the inequality x + y <= 52:
To graph this inequality, we need to draw a line that is below the line x + y = 52.
y-axis
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x-axis
The area below the line x + y = 52 represents the feasible combinations of x and y that satisfy the condition x + y <= 52.
So, the shaded region below the line x + y = 52 is the area where the total number of tables does not exceed 52.
The solution to the system of inequalities is the region where the shaded region below the line x + y = 52 intersects or overlaps with the line 4x + 10y = 280.
I hope this helps! Let me know if you have any more questions or if you'd like me to tell you a joke instead.
To determine the possible combinations of tables that can be used for the event, we can set up a system of inequalities based on the given information.
Let's assume x represents the number of 4-seater tables and y represents the number of 10-seater tables.
1. The total number of people that can be seated using the 4-seater tables is 4x.
Therefore, we can write the inequality: 4x ≥ 280.
2. The total number of people that can be seated using the 10-seater tables is 10y.
Therefore, we can write the inequality: 10y ≥ 280.
3. The total number of tables available should not exceed 52.
So, we can write the inequality: x + y ≤ 52.
Graphically, we can plot the solutions to these inequalities on a coordinate plane, where the x-axis represents the number of 4-seater tables (x) and the y-axis represents the number of 10-seater tables (y). The region where these inequalities intersect will give us the feasible combinations.
Note: In the graph, the shaded area within the solution region (below or on the boundary lines) represents the valid combinations of tables.
Let me know if you need further assistance!
To determine the possible combinations of tables that can be used for the event, we need to consider two factors: the number of tables and the seating capacity.
Let's define:
- Let x represent the number of 4-seater tables.
- Let y represent the number of 10-seater tables.
We have the following conditions:
1. The total number of tables cannot exceed 52:
x + y ≤ 52
2. The total seating capacity should be enough for all 280 people:
4x + 10y ≥ 280
Graphically, we can represent these inequalities on a coordinate plane. The x-axis will represent the number of 4-seater tables (x), and the y-axis will represent the number of 10-seater tables (y).
To graph the inequalities, we need to convert them into slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.
1. Inequality 1: x + y ≤ 52
Rewrite the inequality in slope-intercept form:
y ≤ -x + 52
2. Inequality 2: 4x + 10y ≥ 280
Divide both sides by 2 to simplify:
2x + 5y ≥ 140
Rewrite the inequality in slope-intercept form:
y ≥ (-2/5)x + 28
Plotting these inequalities on a graph will give you a shaded region that represents all the possible combinations of tables that can be used for the event. Any point within this shaded region will satisfy both inequalities.