Question

Rational Equations

Question is picture in comments

Answer 1

@Nnesha @theDeviliscoming @ Just about anyone...

Answer 2

I'm confused on how to do this

Answer 3

@ultrilliam

Answer 4

BBL. I need Jay to do some things first.

Answer 5

Oh um, thank you!

Answer 6

Okay, so to set this problem up, we will use apply the D = rt formula.

According to the given information, it takes longer to go Eastward then Westward therefore:

the time for Eastward will be
\(\dfrac{2400}{(450 - w)}\)

The time for Westward will be
\(\dfrac{2400}{(450 + w)} = t\)

The total time it takes is eleven hours, so Time Eastward + Time Westward = 11hrs:
\(\dfrac{2400}{450 - w} + \dfrac{2400}{450 + w} = 11\)

Answer 7

Okay so Table is this

Answer 8

I have to admit, I don't really feel like doing the rest, but you can at least re-arrange the equation so that you have:
\(\dfrac{1}{450 - w} + \dfrac{1}{450 + w} = \dfrac{11}{2400}\)

This equation is very annoying and I don't like it, but you should be able to solve for \(w\) the speed of the wind

Answer 9

LaTex... on a computer that can't display it. So... Ignore the d-frac?

1/450-w +1/450+w=11/2400?

Answer 10

And then I cross multiply?

Answer 11

Come to think of it, let's just try to finish it. We would multiply the 1st fraction top and bottom by \(450 + w\) and multiply the 2nd fraction top and bottom by \(450 - w\) to get

\(\dfrac{450 + w}{450^2 - w^2} + \dfrac{450 - w}{450^2 - w^2} = \dfrac{11}{2400}\)

Which simplifes to \(\dfrac{900}{450^2 - w^2} = \dfrac{11}{2400}\)

Definitely not doing the rest.

Answer 12

Okay, so to finish this, you could continue to isolate \(w\) as follows:
\(\dfrac{(900)(2400)}{11} = 450^2 - w^2\)

Then re-write the above as:
\(w^2 = 450^2 - \dfrac{(900)(2400)}{11}\)

Then take the square root of both sides to get \(w\)
\(w = \sqrt{450^2 - \dfrac{(900)(2400)}{11}}\)

Answer 13

can't see... at all

Answer 14

Right click to copy image address if you can't see it.

Answer 15

Paste the image address in your URL BAR

Answer 16

still can't

Answer 17

https://assets.questioncove.com/attachments/1495900200-5928d3ef364c21edf141b400-RationalEquations.png

Answer 18

I can see that. Thank you Hero. Sorry about this

Answer 19

Speed of the wind seems kind of high though.

Answer 20

high?

Answer 21

Yes, I think the max wind speed in mph for extreme conditions for commercial airlines is 57mph. In this particular case, it's almost 80mph.

Answer 22

Holy crapbaskets... I know math can do whatever it wants but could they at least try being realistic?

Answer 23

Would you mind posting the rest of the problem? I only see parts I and II.

Answer 24

Part 3 was to solve and find the wind speed

"Solve your equation to find the speed of the wind. Round your answer to the nearest mile per hour."

Answer 25

K

Answer 26

Thank you Hero!