Only selections B and D give a maximum height of 13 at t=3. However, both of those functions have the height be -5 at t=0, meaning the ball was served from 5 ft below ground. This does not seem like an appropriate model.
We suspect ...
• the "correct" answers are probably B and D
• whoever wrote the problem wasn't paying attention.
Answer: 2/3 * (× + 2 )
Step-by-step explanation:
((x²-4)/(3x)) ÷ ((x-2)/(2x)). ⇒ [ ( ײ - 4 ) * 2x ] ÷ [ ( × - 2 ) *3x ]
Simplifying by x [ 2 * ( ײ - 4 ) ] ÷ [ ( × - 2 ) *3 ] ⇒ (2/3)*{ [ ( x-2 )*(×+2)]÷ (×-2) }
Simplifying by ( ×+2) 2/3 * (× + 2 )