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DaniilM [7]
3 years ago
8

A coin bank has 250 coins dimes and quarters worth 39.25 how many of each type of coins are there?

Mathematics
2 answers:
Maurinko [17]3 years ago
8 0
Set up a system of equations.

0.10d + 0.25q = 39.25
d + q = 250

Where 'd' represents the number of dimes, and 'q' represents the number of quarters.

d + q = 250

Subtract 'q' to both sides:

d = -q + 250

Plug in '-q + 250' for 'd' in the 1st equation:

0.10(-q + 250) + 0.25q = 39.25

Distribute 0.10:

-0.10q + 25 + 0.25q = 39.25

Combine like terms:

0.15q + 25 = 39.25

Subtract 25 to both sides:

0.15q = 14.25

Divide 0.15 to both sides:

q = 95

Now plug this into any of the two equations to find 'd':

d + q = 250

d + 95 = 250

Subtract 95 to both sides:

d = 155

So there are 95 quarters and 155 dimes.
bija089 [108]3 years ago
4 0
Coin bank= 250 coins
quarters worth=39.25
250-39.25=210.75 of coins there are
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Explanation-

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The last term, "the constant", is <span> -2 </span>

Step-1 : Multiply the coefficient of the first term by the constant <span> <span> 1</span> • -2 = -2</span> 

Step-2 : Find two factors of  -2  whose sum equals the coefficient of the middle term, which is  <span> 1 </span>.

<span><span>     -2   +   1   =   -1</span><span>     -1   +   2   =   1   That's it</span></span>


Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -1  and  2 
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Step-4 : Add up the first 2 terms, pulling out like factors :
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Answer:

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Step-by-step explanation:

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