When Jon went running in the park, he found 9 coins worth a total of $1.80. The coins were quarters and dimes. How many of each did he find?

Answer 1

Jon found 6 quarters and 3 dimes.

First, let's call the number of dimes Jon found #d# and the number of quarters Jon found #q#
We can now write the following equation: #d + q = 9#

And, because dimes are worth 0.10andquartersareworth0.10 and quarters are worth 0.25 we can write:

#0.1d + 0.25q = 1.80#
Solving the first equation for #d# gives:
#d + q - q = 9 - q#
#d + 0 = 9 - q#
#d = 9 - q#
We can now substitute #9 - q# for #d# in the second equation and solve for #q#:
#0.1(9 - q) + 0.25q = 1.80#
#0.9 - 0.1q + 0.25q = 1.80#
#0.9 + 0.15q = 1.80#
#0.9 - 0.9 + 0.15q = 1.80 - 0.9#
#0 + 0.15q = 0.9#
#0.15q = 0.9#
#(0.15q)/0.15 = 0.9/0.15#
#q = 6#
We can now substitute #6# for #q# in solution to the first equation and calculate #d#:
#d = 9 - 6#
#d = 3#
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Answer 2

Let ( x ) represent the number of quarters and ( y ) represent the number of dimes.

Since Jon found a total of 9 coins, the first equation is:

[ x + y = 9 ]

Since the total value of the coins is $1.80, the second equation is:

[ 0.25x + 0.10y = 1.80 ]

To solve this system of equations, we can first solve the first equation for ( x ):

[ x = 9 - y ]

Substitute this expression for ( x ) into the second equation:

[ 0.25(9 - y) + 0.10y = 1.80 ]

Expand and simplify:

[ 2.25 - 0.25y + 0.10y = 1.80 ]

Combine like terms:

[ -0.15y = -0.45 ]

Divide by -0.15:

[ y = 3 ]

Substitute ( y = 3 ) back into ( x = 9 - y ):

[ x = 9 - 3 ] [ x = 6 ]

Jon found 6 quarters and 3 dimes.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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