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The golfer hits the golf ball. The quadratic y=-8*x^2 +48*x gives the time x seconds when the golf ball is at height 0 feet. In total, how long is the golf ball in the air?

Sagot :

[tex]y=-8\cdot x^2 +48\cdot x =-8x(x-6)\\ \\-8x(x-6)=0\ \ \ \Leftrightarrow\ \ \ -8x=0\ \ \ \vee\ \ \ x-6=0\\ \\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ x=6\\ \\ Ans.\ the\ golf\ ball\ is\ in\ the\ air\ 6\ seconds.[/tex]