I need help asap!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

madreJ [45]2 years ago

8 0

Answer:

# 2 is correct :)

Step-by-step explanation:

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At 8:00 am, here's what we know about two airplanes: Airplane #1 has an elevation of 80870 ft and is decreasing at the rate of 4

wel

Let's begin by listing out the information given to us:

8 am

airplane #1: x = 80870 ft, v = -450 ft/ min

airplane #2: x = 5000 ft, v = 900ft/min

We must note that the airplanes are moving at a constant speed. The equation for the airplanes is given by:

$\begin{gathered} E=x_1+vt----1 \\ E=x_2+vt----2 \\ where\colon E=elevation,ft;x=InitialElevation,ft; \\ v=velocity,ft\text{/}min;t=time,min \\ x_1=80,870ft,v=-450ft\text{/}min \\ E=80870-450t----1 \\ x_2=5,000ft,v=900ft\text{/}min \\ E=5000+900t----2 \end{gathered}$

We equate equations 1 & 2 to get the time both airlanes will be at the same elevation. We have:

$\begin{gathered} 5000+900t=80870-450t \\ \text{Add 450t to both sides, we have:} \\ 900t+450t+5000=80870-450t+450t \\ 1350t+5000=80870 \\ \text{Subtract 5000 from both sides, we have:} \\ 1350t+5000-5000=80870-5000 \\ 1350t=75870 \\ \text{Divide both sides by 1350, we have:} \\ \frac{1350t}{1350}=\frac{75870}{1350} \\ t=56.2min \\ \\ \text{After }56.2\text{ minutes, both airplanes will be at the same elevation} \end{gathered}$

The elevation at that time (when the elevations of the two airplanes are the same) is given by substituting the value of time into equations 1 & 2. We have:

$\begin{gathered} E_1=80870-450t \\ E_1=80870-450(56.2) \\ E_1=80870-25290 \\ E_1=55580ft \\ \\ E_2=5000+900t \\ E_2=5000+900(56.2) \\ E_2=5000+50580 \\ E_2=55580ft \\ \\ \therefore E_1\equiv E_2=55580ft \end{gathered}$

6 0

8 months ago

HELP!! PLEASE!!! QUICK!!! 30 PTS!!!

FromTheMoon [43]

X + 2/3 = -2(4/5x-7)
Is equivalent to -2(x+2/3)=4/5x-7
So the answer would be A

8 0

3 years ago

Simplify using the order of operations.<br> 3/8-1/9

svetoff [14.1K]

Answer:

19 /72 ≈0.264

Step-by-step explanation: