Answer:
See explanation
Step-by-step explanation:
An equation that represents the path of a diver jumping off a diving board is
[tex]y = -7x^2 + 5x + 16.[/tex]
The diver jumped to the water when x = 0.
Then
[tex]y=-7\cdot 0^2+5\cdot 0+16=16[/tex]
The jumper reached the water when y = 0, then
[tex]-7x^2+5x+16=0\\ \\D=5^2-4\cdot (-7)\cdot 16=25+448=473\\ \\x_{1,2}=\dfrac{-5\pm\sqrt{473}}{-14}[/tex]
So, point where the jumper reached the water is
[tex]\left(\dfrac{5+\sqrt{473}}{14},0\right)\approx (1.93,0)[/tex]
