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Sagot :
We are given a problem that can be solved by a system of equations. Let N be the number of nickels, D the number of dimes, and Q the number of Quarters. Since in total he has 4.2, this means mathematically:
[tex]N+D+Q=4.2,\text{ (1)}[/tex]We are told that he has 6 more dimes than nickels, this can be written like this:
[tex]D=6N,\text{ (2)}[/tex]We are told that he has three-time Quarters than nickles, this is:
[tex]Q=3N,\text{ (3)}[/tex]Now, if we replace equation (2) and (3) in equation (1), we get:
[tex]N+6N+3N=4.2[/tex]Solving for N, we get;
[tex]\begin{gathered} 10N=4.2 \\ N=\frac{4.2}{10}=0.42 \end{gathered}[/tex]Replacing the value of N in equation (2), we get:
[tex]\begin{gathered} D=6N \\ D=6(0.42)=2.52 \end{gathered}[/tex]Now we replace the value of N in equation (3):
[tex]\begin{gathered} Q=3N \\ Q=3(0.42)=1.26 \end{gathered}[/tex]Therefore, he has, 0.42 in nickels, 2.52 in dimes, and 1.26 in quarters.
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