SOLUTION: Let the function f be defined by F(x)=x^2+28. If f(3y)=2f(y), what is the one possible value of y? A)-1 B)1 C)2 D)-3

SOLUTION: Let the function f be defined by F(x)=x^2+28. If f(3y)=2f(y), what is the one possible value of y? A)-1 B)1 C)2 D)-3

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Question 1088766: Let the function f be defined by
F(x)=x^2+28. If f(3y)=2f(y), what is the one possible value of y?
A)-1
B)1
C)2
D)-3
Answer by MathLover1(20849) (Show Source): You can put this solution on YOUR website!

or
so, the one possible value of is: C)
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