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The table shows the cost of a game from 2000 to 2004, which has been increasing in a quadratic fashion. Let x = 0 in 2000, and find the best-fit quadratic equation. What will game cost in 2010? A) $417 Eliminate B) $746 C) $960 D) $1,586

Respuesta :

Answer: Here we are going to use the equation y=15x2+3x+56.

x=0 in 2,000 so x=10 in 2010.  

Substitute 10 for x.

After this we have the answer: D. $1,586

Hope this helps

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