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almond37 [142]
3 years ago
8

Solvef(x)=2x^2+6, fund the value if f(x+3)​

Mathematics
1 answer:
zavuch27 [327]3 years ago
6 0

Answer:

F=2x^2+24

Step-by-step explanation:

F (x+3)=2x^2+6

F=2(x+3)^2+6

F=2(X^2+9)+6

F=2x^2+18+6

F=2x^2+24

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