Somebody this is my last question

Step-by-step explanation:

Each group of 1 dime and 2 quarters has a value of $0.60. Thus there must be ...

  $7.20/$0.60 = 12

groups of coins. That is, there must be 12 dimes and 24 quarters.

Comment on the problem statement

The problem is over-specified. Any one of the numbers could be left out and you could still determine the number of quarters and dimes.

You may be expected to solve this using an equation or system of equations.

Let d and q represent the numbers of dimes and quarters, respectively. The problem statement gives rise to three equations relating these two variables.

You can solve the system by choosing any two of these equations. For example, using the first two, we can substitute for d to get ...

  q/2 + q = 36

  3q = 72 . . . . . . . multiply by 2 to eliminate fractions

  72/3 = q = 24 . . . . divide by 3

  d = 24/2 = 12 . . . . find d from the second equation

17.08

correct me if i am wrong.

sweetheart, there isn't a cube below.