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Algebra 17 Online
OpenStudy (anonymous):

Jimmy's Chocolate shop sells two different fudge bars. The dark chocolate fudge bar sells for $6.00 per pound and the milk chocolate fudge bar sells for $4.00 per pound. You purchased a combination of these bars for all of your family and friends for the holidays and your bill came to $120. This scenario can be modeled by the equation 6d + 4m = 120. Which of the following is not a possible combination of dark chocolate bars and milk chocolate bars?

OpenStudy (anonymous):

Helpp

OpenStudy (johnweldon1993):

Okay...."which of the following" So what are your answer choices?

OpenStudy (anonymous):

A.10 dark bars and 15 milk bars B. 12 dark bars and 12 milk bars C.14 dark bars and 18 milk bars D, 6 dark bars and 21 milk bars

OpenStudy (johnweldon1993):

Okay...your equation \[6d + 4m = 120\] What you do is take your answer choices...and plug them into your equation *I'll do 1 example for A* 10 dark bars and 15 milk bars \[6d + 4m = 120\] 'd' gets replaced by 10...and 'm' gets replaced with 15 \[6(10) + 4(15) = 120\] Now we simplify \[60 + 60 = 120\] \[120 = 120\] So this one works out...and is NOT your answer..but this is how you would solve this problem...just test the other choices

OpenStudy (anonymous):

ahh ok iget it now thanks

OpenStudy (johnweldon1993):

No problem!

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