Matemáticas, pregunta formulada por Catelul, hace 1 año

On a movie set, an archway is modeled by the equation y = -0.5x2 + 3x, where y is the height in feet and x is the horizontal distance in feet. A laser is
directed at the archway at an angle modeled by the equation -0.5x + 2.42y = 7.65 such that the beam crosses the archway at points A and B. At what height from the ground are the points A and B?

Help me!Please!


AnonimousD: It can be in Spanish?
Catelul: I doesn't know Spanish.

Respuestas a la pregunta

Contestado por Piscis04
2
y=-0.5x^2+3x \to archway  \\  \\ -0.5x + 2.42y = 7.65 \\  \\ y= (7.65+0.5x): 2.42\to y= 0.21x +3.16\to laser \\  \\   \left \{ {y=-0.5x^2+3x \atop { y= 0.21x +3.16}} \right.  \\  \\ y=y  \\  \\ -0.5x^2+3x= 0.21x +3.16 \\  \\ -0.5x^2+3x- 0.21x -3.16= 0 \\  \\ -0.5x^2+2.79x -3.16= 0 \\  \\ x_ {1\ y\ 2} =  \frac{-b\pm \sqrt{b^2-4ac}}{2a}\qquad a=-0.5\quad b=2.79\quad c= -3.16 \\  \\  x_ {1\ y\ 2} =  \frac{-2.79\pm \sqrt{(2.79)^2-4(-0.5)(-3.16)}}{2(-0.5)}

 x_ {1\ y\ 2} =  \frac{-2.79\pm \sqrt{7.78-6.32}}{-1} \\  \\ x_ {1\ y\ 2} =  \frac{-2.79\pm \sqrt{1.46}}{-1} \\  \\  x_ {1\ y\ 2} =  \frac{-2.79\pm 1.21}{-1} \\  \\ x_1= 2.79+1.21\to  \boxed{x_1=4}  \\  \\ x_2= 2.79-1.21\to  \boxed{x_2=1.58}  \\  \\  Points \ A \ y\ B \to  \\  \\ A=(x, y) \to  \boxed{A= (4,4)} \ height = 4\\  \\ B=(x, y) \to  \boxed{B= (1.58\ ,3.49)} \ height = 3.49


Hope this can help you!

Espero que te sirva, salu2!!!!
Otras preguntas