Respuesta :
Set the function to equal zero:
[tex]-x^2 - 5x + 50 = 0[/tex]
To find the zeros, or solutions, to this problem, we will use the quadratic formula:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Plug the values into the equation:
[tex]a = -1, b = -5, c = 50[/tex]
[tex]x=\frac{5\pm\sqrt{-5^2-4(-1)(50)}}{2(-1)}[/tex]
[tex]x=\frac{5\pm\sqrt{25+200}}{-2}[/tex]
[tex]x=\frac{5\pm15}{-2}[/tex]
Solve for both plus and minus:
[tex] \frac{5 + 15}{-2} = \frac{20}{-2} = -10 [/tex]
[tex]\frac{5 - 15}{-2} = \frac{-10}{-2} = 5[/tex]
[tex]x = -10, x = 5[/tex]
The question cannot use a negative value for x, as the diver cannot dive a negative distance from the board.
The answer is 'x = 5; the diver hits the water 5 feet away horizontally from the board', because x represents the horizontal distance away from the board.
[tex]-x^2 - 5x + 50 = 0[/tex]
To find the zeros, or solutions, to this problem, we will use the quadratic formula:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Plug the values into the equation:
[tex]a = -1, b = -5, c = 50[/tex]
[tex]x=\frac{5\pm\sqrt{-5^2-4(-1)(50)}}{2(-1)}[/tex]
[tex]x=\frac{5\pm\sqrt{25+200}}{-2}[/tex]
[tex]x=\frac{5\pm15}{-2}[/tex]
Solve for both plus and minus:
[tex] \frac{5 + 15}{-2} = \frac{20}{-2} = -10 [/tex]
[tex]\frac{5 - 15}{-2} = \frac{-10}{-2} = 5[/tex]
[tex]x = -10, x = 5[/tex]
The question cannot use a negative value for x, as the diver cannot dive a negative distance from the board.
The answer is 'x = 5; the diver hits the water 5 feet away horizontally from the board', because x represents the horizontal distance away from the board.
Answer:
x = 5; the diver hits the water 5 feet away horizontally from the board
Step-by-step explanation: