Answer:
Janet has coins worth 252.5, 101, and 50.5.
Step-by-step explanation:
Let us call the three coins [tex]a,b[/tex], and [tex]c[/tex].
The first coin [tex]a[/tex] is 5 times another coin (let's say [tex]b[/tex]); therefore,
[tex]a = 5b[/tex],
and the third coin is 2 times the value of another coin (it's coin [tex]b[/tex]):
[tex]c =2b[/tex].
The total worth of all coins is 404; therefore,
[tex]a+b+c =404[/tex]
or
[tex]5b+b+2b =404[/tex]
[tex]8b =404[/tex]
[tex]\boxed{b = 50.5}[/tex]
with the value of [tex]b[/tex] in hand, we find [tex]a[/tex] and [tex]c[/tex]:
[tex]a = 5b = 5(50.5)[/tex]
[tex]\boxed{a = 252.5}[/tex]
and for [tex]c:[/tex]
[tex]c = 2 b = 2 (50.5)[/tex]
[tex]\boxed{c = 101}[/tex]
Thus, Janet has 3 coins worth 252.5, 101, and 50.5.
