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

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 -

Total number of coins = 52

So,  

Substitute

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,

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.

step-by-step explanation:

first, eliminate the parenthesis:

3

x

+

2

−

4

x

second, group like terms:

3

x

−

4

x

+

2

third, combine like terms:

(

3

−

4

)

x

+

2

→

−

1

x

+

2

→

−

x

+

2

what does x eaqual

step-by-step explanation: