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borishaifa [10]
3 years ago
9

Jireh flew his crop duster from the ground to an altitude of 3,500 feet. He continued to fly at that height for 20 minutes until

he descended to 2,000 feet. He then flew back to the ground and landed his plane.
Which part of the scenario is best represented by a linear increasing interval?


Jireh flew his crop duster from the ground to an altitude of 3,500 feet.

Jireh flew at 3,500 feet for 20 minutes.

Jireh descended to 2,000 feet.

Jireh landed his plane.
Mathematics
2 answers:
miss Akunina [59]3 years ago
7 0

That is the first choice. Answer

Rus_ich [418]3 years ago
3 0

Answer:

A

Step-by-step explanation:

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IgorLugansk [536]
Now, we know that he's charging $12 per session to each of his students, and he has 14 students currently, so his revenue is just 14 * 12 or 168 bucks.

now, let's take a peek as the session price goes up in jumps of 2, from 12, to 14, 16, 18 and so on, as each jumps happen, the students drop by 1, from 14, to 13, to 12 and so on.

\bf \begin{array}{ccccllll}
\stackrel{increase}{x}&\stackrel{new}{price}&students&\stackrel{revenue}{y}\\
------&------&------&------\\
0&12&14&168\\
1&14&13&182\\
2&16&12&192\\
3&18&11&198\\
\boxed{4}&20&10&\boxed{200}\\
5&22&9&198\\
6&24&8&192\\
7&26&7&182\\
8&28&6&168\\
..&..&..&..
\end{array}

notice, the revenue starts off at 168, goes up up, reaches 200 bucks and then starts to drop back down.

thus, that means the U-turn or vertex of that revenue function is at 4,200, namely h = 4, and k = 200

\bf \qquad \textit{parabola vertex form}\\\\
\begin{array}{llll}
\boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\
x=a(y-{{ k}})^2+{{ h}}
\end{array} \qquad\qquad  vertex\ ({{ h}},{{ k}})\\\\
-------------------------------\\\\
\stackrel{c(x)}{y}=a(x-\stackrel{h}{4})\stackrel{k}{+200}
\\\\\\
\textit{now, we also know that }
\begin{cases}
x=2\\
y=192
\end{cases}\implies 192=a(2-4)^2+200
\\\\\\
-8=a(-2)^2\implies -8=4a\implies \cfrac{-8}{4}=a\implies -2=a
\\\\\\


\bf therefore\qquad \stackrel{ c(x)}{y}=-2(x-4)^2+200\impliedby \textit{let's expand that}
\\\\\\
y=-2(x^2-8x+16)+200\implies y=-2x^2+16x-32+200
\\\\\\
\stackrel{y}{c(x)}=-2x^2+16x+168
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