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PLEASE ANSWER ASAP: Danielle's wallet had $2.65 in change:all quarters,dimes and nickels. She had 3 times as many quarters as dimes and one more dime than nickels. How many…
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PLEASE ANSWER ASAP: Danielle's wallet had $2.65 in change:all quarters,dimes and nickels. She had 3 times as many quarters as dimes and one more dime than nickels. How many…
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Sep 20, 2016
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PLEASE ANSWER ASAP: Danielle's wallet had $2.65 in change:all quarters,dimes and nickels. She had 3 times as many quarters as dimes and one more dime than nickels. How many of each coin did she have?
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Hinterbu
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2.65=3x(25)+x(10)+(x-1)(5)
9 quarters 3 dimes 2 nickels
9*.25+2.25
3*.10=.30
2*.5=10
2.25+.30+.10=2.65
Jonathan Gleason
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Sep 23, 2016
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We can write an equation in terms of n, the amount of nickels:
25(3n+3)+10(n+1)+5n=265.
Thus:
90n+85=265
Subtracting 85, we have:
90n=180
Therefore, n=2. As such, Danielle has 2 nickels, 3 dimes, and 9 quarters.
Sponge Bob
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Sep 26, 2016
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