− − 5t 6 = 0 t 2 only of the graph of is 1. x-intercept f (t (t − 6) + 1) = 0 85. = = 6 or = −1 A( x) t t 143 + = 2) x( x Discard the negative solution since the time 143 of flight must be positive. The ball will + 2x − 143 = x2 strike the ground after 6 seconds. 0 ( + 13)( − x x b. When the ball passes the top of the building, 11) = 0 = x −13 it will be 96 feet from the ground. Therefore, or = 11 x Discard the negative solution width cannot we solve be negative. The width of the rectangular = 96 s the length is 13 feet. + − = 96 80t 16t 2 = 86. 306 A( x) 96 + = 1) 306 x( x −16t + 80t = 2 + − 306 = 0 x2 x 0 is 11 − = 5t 0 t feet 2 (t and − 5) = 0 t ( + 18)( −17) x x = 0 = 0 or = 5 t t = x = The ball is at the top of the building at time or 17 −18 x Discard the negative solution since width cannot = 0 seconds when it is thrown. It will pass t be negative. The width of the rectangular the top of the building on the way down window is 17 cm and the length is 18 cm. after 5 seconds.
t 2 − 5t − 6 = 0 only x-intercept of the graph of f is 1. (t − 6) (t + 1) = 0 85. A( x) = 143 t = 6 or t = −1 x( x + 2) = 143 x2 + 2x − 143 = 0 ( x + 13)( x − 11) = 0 Discard the negative solution since the time of flight must be positive. The ball will strike the ground after 6 seconds. b. When the ball passes the top of the building, x = −13 or x = 11 it will be 96 feet from the ground. Therefore, Discard the negative solution width cannot be negative. The width of the rectangular we solve s = 96 86. is 11 feet and the length is 13 feet. A( x) = 306 x( x + 1) = 306 x2 + x − 306 = 0 96 + 80t − 16t 2 = 96 −16t 2 + 80t = 0 t 2 − 5t = 0 t (t − 5) = 0 ( x + 18)( x −17) = 0 t = 0 or t = 5 x = −18 or x = 17 The ball is at the top of the building at time Discard the negative solution since width cannot be negative. The width of the rectangular window is 17 cm and the length is 18 cm. t = 0 seconds when it is thrown. It will pass the top of the building on the way down after 5 seconds.