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I think they got rid of it for all of the inconvenience from whenever you wanted to post an answer. You would have to do the captcha EVERY TIME.

Hi, Guest!

I agree, there's nobody on.

I also agree that this is a math site.

However, when no questions are asked, the members like to make social posts for fun.

It looks like Shane has neither questions to ask nor questions to answer.

I do not see how there being nobody active indicate Shane is looking for a date.

If you need help, though, feel free to ask.

Wouldn't it actually be 5[2pi-1]?

*anymore

well duh this is not a dating site so get over it and only post if you have a mathe problem that some one can help you with or u are just beond helping.

118097

Is it just me, or do you not have to do the Recaptcha confirmation yet?

Go online here and learn about "e" as much as you can:

http://www.mathsisfun.com/numbers/e-eulers-number.html 

Well, yeah, she joined a while before you, but it still was pretty recent....

But if you check some of the wraps, she was here before me.

No, she was actually a Guest known as Victoria when Hayley1 (I miss her...) convinced her to join!

Actually, I think there was a glitch with her join date. She was here WAAAAAY before me.

Naw, it's fine.

You actually did go up really quickly!

I was surprised the other day, I saw ColdPlay had like 2.5k points and she joined not too long ago.

People like me, sort of? :| _{Please don't get mad at me for mentioning that...}

Haha, after the first year I totally forgot the forum at about 450 points, then I started going on and soon enough, I got tons of points.

There are people that get 50-100 points a day, I'm sure you can get tons in no time!

I assume this is

f(x)= (-5x)/ [ sin(x)+cos(x)  ]   .....if so.....we can write it as

(-5x) [ sin (x) + cos(x)]^(-1)    ......so.....using the product and chain rules, we have.....

f ' (x) =  -5 [ sin (x) + cos(x)]^(-1)   +  (5x)[sin (x) + cos(x)]^(-2)  (cos(x) - sin(x) ]

So    f '  (2pi) =

-5 [ sin (2pi) + cos(2pi)] ^(- 1)  + (5*2pi) [ sin(2pi) + cos(2pi)]^(-2) ( cos(2pi)  - sin (2pi) ]  =

-5  [ 0 + 1]^(-1)  + (5*2pi) [ 0 + 1^(-2) ( 1 - 0)  =

-5 + 5*2pi  =

5 [2pi - 1]  =  about 26.416

Thanks to the guest for pointing out my earlier error....!!!!

I'm still at 500...

:(

Yes, I'm pretty sure stars = points.

I'm not sure, you've actually been on a lot longer than me, too.

Keep rising up! You're awesome!

Congrats on 1000 points! Keep it up! 

No ur not im the retard of this group chat bree

No ur not im the retard of this group chat bree

Yea, sorry gtg bye