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love history [14]
3 years ago
14

Miri counted her money and found that her 25 coins which were nickels, dimes, and quarters were worth $3.20. the number of dimes

exceeded the number of nickels by 4. how many coins of each kind did she have?
Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
4 0
Nickels = x

dimes = x + 4

quarters = x

Value of coins:

nickel = 5 cents

dime = 10 cents

quarter = 25 cents

5x + 10(x + 4) + 25x = 3.20

5x + 10x + 40 + 25x = 320

40x = 320 - 40

40x = 280

x = 280/40

x = 7

There are 7 quarters, 7 nickels and 11 dimes.



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This problem tackles the place values of numbers. From the rightmost end of the number to the leftmost side, these place values are ones, tens, hundreds, thousands, ten thousands, hundred thousands, millions, ten millions, one hundred millions, and so on and so forth. My idea for the solution of this problem is to add up all like multiples. In this problem, there are 5 multiples expressed in ones, thousands, hundred thousands, tens and hundreds. Hence, you will add up 5 like terms. The solution is as follows

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2 years ago
Solve<br>4(2x - 3) = 6x + 2​
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。☆✼★ ━━━━━━━━━━━━━━  ☾  

4(2x - 3) = 6x + 2

Expand it:

8x - 12 = 6x + 2

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2500

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