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The Height Y (In Feet)of a Ball That You Throw y=1180x2+x+6y = - \frac { 1 } { 180 } x ^ { 2 } + x + 6

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The height y (in feet) of a ball that you throw is given by y=1180x2+x+6y = - \frac { 1 } { 180 } x ^ { 2 } + x + 6 where x is the horizontal distance (in feet) from where you release the ball. How high is the ball when it reaches its maximum height?

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