anonymous
  • anonymous
Heidi and Mike have 51 dimes and nickles. If the value of the coins is $4.10, how many coins of each type were there?
Mathematics
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anonymous
  • anonymous
Heidi and Mike have 51 dimes and nickles. If the value of the coins is $4.10, how many coins of each type were there?
Mathematics
schrodinger
  • schrodinger
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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lgbasallote
  • lgbasallote
let x = no. of nickles y = no. of dimes x + y = 51 5(cents)x + 10(cents)y = 410 cents multiply the first equation by -10.. -10x - 10y = -510 5x + 10y = 410 -5x = -100 x = 20 y = 31 There are 20 nickles and 31 dimes
anonymous
  • anonymous
thank you for being so specific! :)

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