1.Mary is selling her craft to earn money she sells her bracelets for $6 and her necklaces for $10 her goal is to make at least $120 in sales which of the following represents three possible solutions to the problem
A.6x+10y<120
(5,10),(10,5),and(15,0)are three possible solutions.
B.6x+10y>=120
(5,9),(10,6),and(15,3)are three possible solutions.
C.6x-10y<120
(5,5),(10,10),and(15,15)are three possible solutions.
D.7x+5y<=140
(7,21),(14,14),and(21,7)are three possible solutions.
I think it's A.6x+10y<120
(5,10),(10,5),and(15,0)are three possible solutions. please correct me if I'm wrong.
disagree. She wants 6x+10y to be BIGGER or equal to 120
It said she wants to "make at least $120"
To me that means that your relation should contain > 120 or ≥ 120
This automatically ruled out 2 of the 4 statements, and in D she limits her sales to $140
A quick check will show that the 3 points given for
6x + 10y ≥ 120 all work
So the answer to this would be B or D??? if I have another choice i'd say D. Am I correct.
You are correct! The correct option is A.
To check if this option represents the three possible solutions, we can substitute the values in the inequalities.
Let's take the first solution (5, 10):
6(5) + 10(10) = 30 + 100 = 130
Since 130 is greater than 120, the inequality 6x + 10y < 120 is true for this solution.
Similarly, for the second solution (10, 5):
6(10) + 10(5) = 60 + 50 = 110
Since 110 is less than 120, the inequality 6x + 10y < 120 is true for this solution as well.
For the third solution (15, 0):
6(15) + 10(0) = 90 + 0 = 90
Since 90 is less than 120, the inequality 6x + 10y < 120 is true for this solution too.
Therefore, all three solutions satisfy the inequality 6x + 10y < 120, making option A the correct choice.