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anonymous

  • one year ago

I really need help!! Please?? The FEA club will charge each competitor $10 to enter the card stacking contest.Write an equation and find how many competitors it will take to get to the breakeven. I know the equation is 425p+6460=10p. What and how do I get the answer??

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  1. anonymous
    • one year ago
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    Can you explain where the 425 and 6460 came from please? Is it in the book?

  2. anonymous
    • one year ago
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    Yes, its in a book.the 425 is the cost (times by 10, origanly 4.25) per competitor, 6460 is the cost (originally 64.60) of extra expensive. And I need the break even for when the FEA charges the competitor $10 to enter . @rheaanderson341

  3. anonymous
    • one year ago
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    Okay.. I'm a little confused but we'll try and solve this together. okay, so you multiplied the $4.25 and the $64.60 with the 10. How come?

  4. anonymous
    • one year ago
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    Does p represent competitors?

  5. anonymous
    • one year ago
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    My teacher has us do that because in our class gets confused when working with decimals, yes p is competitors.

  6. anonymous
    • one year ago
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    okay, i think i get this now. hold on let me write it and explain it. :)

  7. anonymous
    • one year ago
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    OK thanks 😊😊

  8. anonymous
    • one year ago
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    Okay, so I couldn't get myself to understand the multiplying part so I did it with the decimals kept in place. Sorry for awful tablet writing, lol! To explain though... To "breakeven" basically means equal in economics/finance/etc.. So you simply get x on one side and the other number on the other side. Then you solve to get x = ???. There can't be 11.43 people, so it's either 11 or 12. Does this make any sense?

  9. anonymous
    • one year ago
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    OMG THANK YOU SOOO MUCH!!!! yes it makes since. And you're fine with the writing. I understand. ☺☺ you are a life saver. You literly kept me from pulling my hair out. Again thank you sooo much!! And thank you for giving me a clearer explanation of break even!

  10. anonymous
    • one year ago
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    Always, always welcome. :)

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