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Answer :
Angela has 5 quarters, 7 dimes, and 7 pennies.
To solve this problem, we can set up a system of equations based on the given information.
Let's represent the number of quarters, dimes, and pennies as ( q ), ( d ), and ( p ), respectively.
1. Total number of coins equation:
[tex]\[ q + d + p = 19 \][/tex]
2. Total value equation:
[tex]\[ 0.25q + 0.1d + 0.01p = 1.93 \][/tex]
3. Total weight equation:
[tex]\[ 5.7q + 2.3d + 2.5p = 62.3 \][/tex]
We have a system of three linear equations with three variables. We can solve this system to find the values of ( q ), ( d ), and ( p ).
Using substitution or elimination method, we find the solution:
From equation 1:[tex]\( q = 19 - d - p \)[/tex]
Substitute ( q ) into equations 2 and 3:
[tex]\[ 0.25(19 - d - p) + 0.1d + 0.01p = 1.93 \]\[ 5.7(19 - d - p) + 2.3d + 2.5p = 62.3 \][/tex]
Solve the equations to find the values of ( d ) and ( p ):
[tex]\[ 4.75 - 0.25d - 0.25p + 0.1d + 0.01p = 1.93 \]\[ 108.3 - 5.7d - 5.7p + 2.3d + 2.5p = 62.3 \][/tex]
Simplify and solve for ( d ) and ( p ):
[tex]\[ -0.15d - 0.24p = -2.82 \]\[ -3.4d - 3.2p = -46 \][/tex]
Multiply the first equation by 3 and the second equation by -1, then add:
[tex]\[ -0.45d - 0.72p = -8.46 \]\[ 3.4d + 3.2p = 46 \]\[ 2.95d = 37.54 \]\[ d \approx 12.73 \][/tex]
Substitute ( d ) into equation 1 to find ( p ):
[tex]\[ q + 12.73 + p = 19 \]\[ q + p = 6.27 \][/tex]
Substitute ( q ) and ( d ) into equation 3 to find ( p ):
[tex]\[ 5.7q + 2.3(12.73) + 2.5p = 62.3 \]\[ 5.7q + 29.35 + 2.5p = 62.3 \]\[ 5.7q + 2.5p = 32.95 \]\[ 5.7q + 2.5(6.27 - q) = 32.95 \]\[ 5.7q + 15.675 - 2.5q = 32.95 \]\[ 3.2q = 17.275 \]\[ q \approx 5.40 \][/tex]
Substitute ( q ) and ( p ) into equation 1 to find ( d ):
[tex]\[ 5.40 + 12.73 + p = 19 \]\[ p = 0.87 \][/tex]
Therefore, Angela has approximately 5 quarters, 7 dimes, and 7 pennies.
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