a science class designed a container to protect an egg. each prototype is launched from a machine at 9.4…

a science class designed a container to protect an egg. each prototype is launched from a machine at 9.4 meters per second from a roof 40 meters tall. the function $f(t)=-4.9t^{2}+9.4t + 40$ represents the path from the building to the ground on the t - axis, where t is the amount of time since the launch. how do the mathematical range and reasonable range compare? mathematical: $yleq44.5$ reasonable: $ygeq0$ mathematical: $yleq44.5$ reasonable: $ygeq40$ mathematical: $yleq44.5$ reasonable: $0leq yleq44.5$ mathematical: $yleq44.5$ reasonable: $40leq yleq44.5$

a science class designed a container to protect an egg. each prototype is launched from a machine at 9.4 meters per second from a roof 40 meters tall. the function $f(t)=-4.9t^{2}+9.4t + 40$ represents the path from the building to the ground on the t - axis, where t is the amount of time since the launch. how do the mathematical range and reasonable range compare? mathematical: $yleq44.5$ reasonable: $ygeq0$ mathematical: $yleq44.5$ reasonable: $ygeq40$ mathematical: $yleq44.5$ reasonable: $0leq yleq44.5$ mathematical: $yleq44.5$ reasonable: $40leq yleq44.5$

Answer

Answer:

mathematical: $y\leq44.5$ reasonable: $0\leq y\leq44.5$

Explanation:

Step1: Encontrar el valor máximo de la función

La función $f(t)=-4.9t^{2}+9.4t + 40$ es una parábola de la forma $y = ax^{2}+bx + c$ con $a=-4.9$, $b = 9.4$ y $c = 40$. El tiempo $t$ en el vértice se da por $t=-\frac{b}{2a}=-\frac{9.4}{2\times(-4.9)}=\frac{9.4}{9.8}\approx0.96$. Sustituyendo $t = 0.96$ en $f(t)$: $f(0.96)=-4.9\times(0.96)^{2}+9.4\times0.96 + 40\approx44.5$. Así, el rango matemático es $y\leq44.5$ ya que es una parábola abierta hacia abajo.

Step2: Determinar el rango razonable

El objeto comienza en una altura de 40 metros y cae al suelo (altura $y = 0$). Entonces, el rango razonable para la altura $y$ está entre 0 y 44.5 metros, es decir $0\leq y\leq44.5$.