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mikailah0988
06.10.2019 •
Mathematics
Solve by using a matrix.
a collection of nickels, dimes and quarters totals $6.00. if there are 52 coins altogether and twice as many dimes as nickels, how many of each kind of coin are there?
a.
q = 5
d = 60
n = 30
c.
q = 15
d = 28
n = 6
b.
q = 35
d = 10
n = 5
d.
q = 10
d = 28
n = 14
Solved
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Ответ:
Option D
q = 10
d = 28
n = 14
Step-by-step explanation:
Given : A collection of nickels, dimes and quarters totals $6.00. If there are 52 coins altogether and twice as many dimes as nickels.
To find : How many of each kind of coin are there?
Solution :
Let n be the nickels
Let d be the dims
Let q be the quarters.
According to question,
There are twice as many dimes as nickels -![d=2n](/tpl/images/0295/1974/f0e7d.png)
Total number of coins = 52
So,![n+d+q=52](/tpl/images/0295/1974/174a2.png)
Substitute![d=2n](/tpl/images/0295/1974/f0e7d.png)
Coins have a total value of $6 which is equal to 600 cents.
We know,
one nickel is worth $0.05= 5 cent,
one dimes are worth $0.10=10 cent
one quarters are worth $0.25=25 cent.
So,![25q+10d+5n=600](/tpl/images/0295/1974/35890.png)
Now, put d=2n and q=52-3n
Substitute n in q and d,
There are 10 quarters, 28 dimes, and 14 nickels
Therefore, Option D is correct.
Ответ:
im unsure. sorry
Step-by-step explanation: