HELP!

Jerrek said:
Luis, he spent one minute looking it over. Now leave it be ok? I don't give a flying fuck.

ooOOOoOoOoOoOoOOoOooOOOooOoOoOooOOooOoOOoooOOOOOoooOoooooOOOoooooohhh!!!!!!! :tardbang:

HANDBAG!

Green%20handbag.jpg
 
OK..Maybe you can explain this...

If 6 x 2 = a dozen, then why doesn't 6 x 2.5 = a dozen and a half?
 
Squiggy said:
OK..Maybe you can explain this...

If 6 x 2 = a dozen, then why doesn't 6 x 2.5 = a dozen and a half?

Because 0.5 is not the half of 2 ;)
 
Squiggy said:
OK..Maybe you can explain this...

If 6 x 2 = a dozen, then why doesn't 6 x 2.5 = a dozen and a half?

Trick question. It's all in the wording:

a dozen and a half = 12 + 1/2 = 12 1/2

6 x 2.5 = 18 or one and a half dozen.


:D :D :D
 
PuterTutor said:

i was thinking that when nixy asked her first q, when luis and jerllrek [;)] started up i was just glad i'd taken the time to explode before.

:nuts2:
 
Jerrek said:
*frown* you don't have to get rid of all "y"s. Whoever told you that is a moron.

ask him how would he obtain the speed (first differentiation) of the particle whose movement is described by the equation x^y = y^x

Seriously, the hard part of the problem is to get rid of all the "y"s if you can't do it, just say it, instead of making lame remarks when you are not giving an answer to the problem i posted, and somehow trying to say that the problem i posted is "incorrect" and "not needed".

Now, if you are so cool at calculus and algebra, just get rid of all the damn y, and if you don't, just give up and accept that your knowledge is not that cool.

Edited: just so YOU can get it, i will rephrase the problem, differentiate
x^y = y^x and express the result in the form of y' = something expressed in terms of x, i don't want anything else, just x's and numbers.
 
Now i got tired of this......

1st, The value of the derivative at any point of a curve is the slope of the tangent in such point.

Now given that, obtain the slope of the curve expressed by x^y = x^y, when x=5


2nd part, if we have the movement in time of a particle described by a formula, the first derivative will be the speed, and the second will be the acceleration.

Obtain the speed when the time x=5 of the particle whose movement is described by x^y=y^x


Getting rid of all the ys is not "explicitly mandatory" in calculus, however, if you don't need to get rid of the y in order to obtain the value of y' when x=5 (both problems), then whoever told me that it is not needed to get rid of all the ys is indeed a moron.
 
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