The function f(x) is the height of a model rocket x seconds after launch.

Mathematics

1 answer:

rusak2 [61]2 years ago

3 0

Answer:

The practical domain for the function f(x) will be [0,5].

Step-by-step explanation:

A model rocket is launched from the ground and its height from the ground is given by f(x), t seconds after its launch.

It is given that the rocket reaches maximum height at t = 2.5 seconds and comes back to the ground after t = 5 seconds.

So, at t = 0 seconds, the rocket was at the ground and hence, t can not be negative and after t = 5 seconds the rocket comes back to the ground and hence, t can not be greater than 5 seconds.

Therefore, the practical domain for the function f(x) i.e. the values of t lies in the interval [0,5]. (Answer)

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PLEASE HELP!!!!!!!!!!!!! DUE SOON

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$\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ h=-16t^2+\stackrel{\stackrel{v_o}{\downarrow }}{65}t$

now, take a look at the picture below, so for 2) and 3) is the vertex of this quadratic equation, 2) is the y-coordinate and 3) the x-coordinate.

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+65}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{65}{2(-16)}~~,~~0-\cfrac{65^2}{4(-16)} \right) \implies \left( \cfrac{65}{32}~,~0- \cfrac{4225}{-64}\right)$

$\bf \left( \cfrac{65}{32}~,~0+ \cfrac{4225}{64}\right)\implies \left( \stackrel{seconds}{2\frac{1}{32}}~~,~~ \stackrel{feet~hight}{66\frac{1}{64}}\right)$