# Algebra

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.

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1. disagree. She wants 6x+10y to be BIGGER or equal to 120

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posted by Damon
2. 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

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posted by Reiny
3. So the answer to this would be B or D??? if I have another choice i'd say D. Am I correct.

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