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kvv77 [185]
3 years ago
5

Convert 6 miles to feet. hint: 1 mile is 5280 feet

Mathematics
2 answers:
seropon [69]3 years ago
8 0

Answer:

the answers got deleted because they didn't give an explanation. so, the explanation is: you multiply 5,280 by 6 to get 31,680 ft

Morgarella [4.7K]3 years ago
7 0

Answer:

31680 feet

Step-by-step explanation:

You multiply 5280 nd  and you get 31680...

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Write the equation in standard form.<br><br> 3x−9=7y
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1. (5 points) At 6pm, ghost A is 5 kilometers due west of ghost B. Ghost A is flying west
zaharov [31]

Answer:

The distance between the ghost changes at 10 pm approximately at a rate of 24.981 kilometers per hour.

Step-by-step explanation:

At first we assume that north and east directions both represent positive quantities. Let suppose that \vec r_{A,o} = (0\,km,0\,km) and \vec r_{B,o} = (5\,km, 0\,km). If both ghosts moves at constant velocity such that \vec v_{A} = \left(-15\,\frac{km}{h}, 0\,\frac{km}{h} \right) and \vec v_{B} = \left(0\,\frac{km}{h},20\,\frac{km}{h}  \right), then the final positions of both ghosts are, respectively:

Ghost A

\vec r_{A} = \vec r_{A,o}+t\cdot \vec v_{A} (Eq. 1)

Ghost B

\vec r_{B} = \vec r_{B,o}+t\cdot \vec v_{B} (Eq. 2)

Where t is the time, measured in hours.

Then, the equations of motion of each ghost are, respectively:

Ghost A

\vec r_{A} = (0\,km,0\,km)+t\cdot \left(-15\,\frac{km}{h}, 0\,\frac{km}{h}  \right)

\vec r_{A} = \left(-15\cdot t, 0)\,\,\,\left[km \right]

Ghost B

\vec r_{B} = (5\,km, 0\,km)+t\cdot \left(0\,\frac{km}{h}, 20\,\frac{km}{h}  \right)

\vec r_{B} = (5, 20\cdot t)\,\,\,\left[km\right]

Then, the distance between both ghosts is:

\vec r_{B/A} = (5,20\cdot t)-(-15\cdot t, 0)\,\,\,[km]

\vec r_{B/A} =(5+15\cdot t, 20\cdot t)\,\,\,[km] (Eq. 3)

The magnitude of the relative is represented by the following Pythagorean identity:

r^{2}_{B/A} = (5+15\cdot t)^{2}+(20\cdot t)^{2}

Then, we find the rate of change of the relative distance (\dot r_{B/A}), measured in kilometers per hour, by implicit differentiation:

2\cdot r_{B/A}\cdot \dot r_{B/A} = 2\cdot (5+15\cdot t)\cdot 15+2\cdot (20\cdot t)\cdot 20

r_{B/A}\cdot \dot r_{B/A} = 15\cdot (5+15\cdot t)+20\cdot (20\cdot t)

\dot r_{B/A} = \frac{75+625\cdot t}{r_{B/A}}

\dot r_{B/A} = \frac{75+625\cdot t}{\sqrt{(5+15\cdot t)^{2}+(20\cdot t)^{2}}} (Eq. 4)

If we know that t = 4\,h, then the rate of change of the relative distance at 10 PM is:

\dot r_{B/A} = \frac{75+625\cdot (4)}{\sqrt{[5+15\cdot (4)]^{2}+[20\cdot (4)]^{2}}}

\dot r_{B/A} \approx 24.981\,\frac{km}{h}

The distance between the ghost changes at 10 pm approximately at a rate of 24.981 kilometers per hour.

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The answer will be D
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