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elixir [45]
3 years ago
6

Please help me. Any help is greatly appreciated

Mathematics
2 answers:
timama [110]3 years ago
4 0
Let’s add the two speeds together so we can tell how far apart they are, so they move apart at 1,200 mph. At 3 PM, this is how far away they will be from each other. At 4 PM, they will be double that, which is 2,400. They still need to go 300 miles, which is a quarter of how fast they go in an hour, meaning they need to go for another quarter of an hour. That is 15 minutes, making the time 4:15 when they are 2,700 miles apart, which is answer b.
iragen [17]3 years ago
3 0

Answer:

c

Step-by-step explanation:

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Check the picture below.

so it hits the ground when y = 0, thus

\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{16}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{5}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}

\bf h(t)=-16t^2+16t+5\implies \stackrel{h(t)}{0}=-16t^2+16t+5
\\\\\\
16t^2-16t-5=0\implies (4t+1)(4t-5)=0\\\\
-------------------------------\\\\
4t+1=0\implies 4t=-1\implies t=-\cfrac{1}{4}\\\\
-------------------------------\\\\
4t-5=0\implies 4t=5\implies t=\cfrac{5}{4}\implies \boxed{t=1\frac{1}{4}}

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Answer:

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Step-by-step explanation:

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