A collection of dimes and quarters has a total value of two dollars and contains 28 coins how many dimes are there in the collection.

I have modified the problem to make it solvable.

The collection has 20 dimes and 8 quarters

Step-by-step explanation:

Let's call:

x=number of dimes

y=number of quarters

There are 28 coins in total, thus:

The total value of the collection is $2, or 200 cents. Since each dime is worth 10 cents and each quarter is worth 25 cents:

10x+25y=200

Dividing by 5:

From [1]:

x=28-y

Substituting in [2]

2(28-y)+5y=40

Operating:

56-2y+5y=40

Simplifying:

3y=40-56=-16

This equation yields a negative value of y, which is unacceptable as a number of coins.

The problem does not have a valid solution.

To make the problem coherent I'll correct the second condition. Instead of $2, the collection is worth $4 or 400 cents. The equation would be now:

10x+25y=400

Dividing by 5:

2x+5y=80\qquad\qquad [2]

Now:

56-2y+5y=80

Simplifying:

3y=80-56=24

Solving:

y=24/3=8

And:

x=28-y=20

The collection has 20 dimes and 8 quarters

location: point 0

step-by-step explanation:

well, -8 to 10 on a number line is 18 spaces.

1/9 of 18 is 2.

so, 4/9 of 18 must be 8.

8 spaces from -8 (a) is 0.

so, 4/9 of the way from -8 to 10 is 0!