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Mathematics 70 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):

SOMEONE HELP

OpenStudy (anonymous):

what are the possible answers?

OpenStudy (anonymous):

we need the "possible combinations"

OpenStudy (anonymous):

I agree!

OpenStudy (anonymous):

10 dark bars and 15 milk bars 12 dark bars and 12 milk bars 14 dark bars and 18 milk bars 6 dark bars and 21 milk bars

OpenStudy (anonymous):

Jk :P

OpenStudy (anonymous):

thank you. <3

OpenStudy (anonymous):

all you gotta do is plug the numbers into the equation which will give u the answer of C

OpenStudy (anonymous):

youre wrong actually..

OpenStudy (anonymous):

14 dark bars and 18 milk bars that's he correct answer

OpenStudy (anonymous):

6(14)+4(18) DOES NOT = 120

OpenStudy (anonymous):

the*

OpenStudy (anonymous):

your right haha good job. sorry, and thanks

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