Roger has 3.10 consisting of quarters dimes and nickles. The number of nickles was 3 less than the number of dimes. The number of dimes was 5 more than the number of quarters. How many coins of each type did he use? Please help me, thank you very much!! <3

5 dimes 2 nickels 10 quarters

Step-by-step explanation:

Let n = number of nickels

      d = number of dimes

      q = number of quarters

n = d - 3 (The number of nickles was 3 less than the number of dimes)

d= q - 5 (The number of dimes was 5 more than the number of quarters)

0.05n + 0.1d + 0.25q = 3.1

Use substitution:

n=q-8

0.05n + 0.1d + 0.25q = 3.1

0.05(q-8) + 0.1(q-5) + 0.25q = 3.1

0.05q-.4 + 0.1q -.5 +0.25q = 3.1

0.4q-0.9=3.1

0.4q=4

q = 10

n =q-8    n = 10-8     n=2

d=q - 5     d = 10-5     d = 5

c

step-by-step explanation:

c

not enough information

step-by-step explanation:

we cannot know this, because the other students could have handed out 10 flyers, or they could have handed out 100. one way, the mean would go up, and another way, the mean would go down.