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#1 Re: Help Me ! » Integrate ∫(1+4y^2)^(1/2) dy from 0 to 1 » Yesterday 18:13:22

Yes, sorry it was an error. I edited. Anyway, Im going to go sleep.

#2 Help Me ! » Integrate ∫(1+4y^2)^(1/2) dy from 0 to 1 » Yesterday 12:03:43

Replies: 3

I have the following integral to solve : … rom+0+to+1

I was able to integrate and I got : (y/4)+((e^(2y)-e^(-2y))/16)

I don't what to do after that. I know I can replace ((e2y-e-2y)/2) with sinh(2y)... but after that I'm lost. Help please ? Thanks

I'm using the following substitution : y=sinhθ/2

#3 Help Me ! » integral of 1/(x^2+2) dx » 2015-03-15 12:40:09

Replies: 1

So, I have no idea how I'm supposed to evaluate the following integral : 1/(x^2+2) I know that it has the form of the derivative of arctan but I'm not sure how to arrive at this result.

Thank you

#4 Re: Help Me ! » Evaluating arctan(−10) » 2015-01-03 04:51:15

OK, I thought that we could always represent them in algebraic expressions....anywya, thank you!

#5 Re: Help Me ! » Evaluating arctan(−10) » 2015-01-03 03:45:50

Ok, but how am I supposed to know if there is an algebraic expression or not ? Whether we're talking about arctan, arcsin, arccos, etc.

#6 Re: Help Me ! » Evaluating arctan(−10) » 2015-01-02 15:55:20

I didn't see taylor series... and i want a closed form for arctan(-10)...

#7 Help Me ! » Evaluating arctan(−10) » 2015-01-02 14:04:36

Replies: 12

I was wondering, how could I find the exact value of arctan(−10) ? We know that an approximation of the exact value would be arctan(−10)≈−1.47112 rad but if we wanted the exact value in radians, How would I find it ?

Thank you!

#8 Help Me ! » Quick question on tangent » 2014-10-18 11:43:18

Replies: 1


Knowing that the tangent line to a curve of f(x) at the point A(4,3) passes through the point (0,2) find f(4) and f^'(4) (derivative at x=4)

I'm not sure but is it f(4)=3
and f^'(4)=1/4


Thank you!

#9 Help Me ! » scalene triangle question » 2014-08-15 08:45:24

Replies: 5

Hi, if we have a scalene triangle with : 112,28 and 40 degrees. and that we have a side of lenght 4 between 28 and 40. Can we find the height of the triangle without using the sine law ?

Thank you!

#10 Help Me ! » Transcendental numbers » 2014-08-03 12:06:27

Replies: 27


I was wondering, what are Transcendental numbers ? I've never heard of such type of numbers. Thank you!

#11 Help Me ! » Small problem of logic. » 2014-08-01 11:57:33

Replies: 1

Hi, there's just one thing which I'm not sure with this problem :

There are 24 pounds of nails in a sack. Can you measure out 9 pounds
of nails using only a balance with two pans?

I answered yes and here's the reason. Consider a number X of nails which weights 24 pounds. This number X has atleast a factor of 2^3 (It can be higher.) in its factors. Why ? Because we need to divide evenly the number of nails so we can have equal weight on each side of balance.

So :



x/2=12 pounds x/4=6 pounds x/8=3 pounds

So we have the following groups (after successive divisons.):


Clearly, x/4+x/8 nails would give 9 pounds

Now, the only thing bugging me is that my solution is founded on the assumption that all nails must have equal weights. The question isn't clear if the nails do have all the same weight or not...

What do you think ???

THank you!

#13 Help Me ! » Proportion » 2014-07-25 05:57:20

Replies: 1

All the proportions which we have deduced from a:b::c:d may be represented generally as follows :

ma+nb:pa+qb::mc+nd: pc + qd

For the product of the extreme terms is mpac+npbc+mqad +nqbd; which, since ad = bc, becomes mpac+npbc+mqad +nqbd; also the product of the mean terms is mpac+mqbc+npad+nqbd; or, since ad = bc, it is mpac+npbc+mqad +nqbd: so that the two products are equal.

Here are some of the proportions deduced from a:b::c:d   :


and some others (I'm not putting all of them.)

I don't understand the part : "All the proportions which we have deduced from a:b::c:d may be represented generally as follows :

ma+nb:pa+qb::mc+nd: pc + qd"

What does it mean ??? I'm not following for a certain reason....

Thank you!

#14 Re: Help Me ! » Limits at infinity » 2014-07-14 04:19:12

Yeah, that would be an idea!

#15 Re: Help Me ! » Limits at infinity » 2014-07-14 04:08:44

bobbym wrote:

That would be 9h. Think about it, if x goes to 0 how does that affect 9h. It does not.

Well, at first i thought the same thing but wasn't too sure... ok thanks smile

#16 Re: Help Me ! » Limits at infinity » 2014-07-14 03:59:04

Well, I have another question.

If had something like

lim  9h

What do you do in these occasions ? Is there a definition that tells us the actions needed ? Thank you (I'm referring to the different letters.)

#17 Re: Help Me ! » Limits at infinity » 2014-07-11 09:20:33

bobbym wrote:

Also, there is a theorem for when there is one polynomial over another. I put it into post #13.

Yes, I saw it tongue

#18 Re: Help Me ! » Limits at infinity » 2014-07-11 09:13:39

Ah ok, thanks for the web page.

#19 Re: Help Me ! » Limits at infinity » 2014-07-11 09:02:56

But... how do you know that - ∞ / 5 is -∞ ???

#20 Re: Help Me ! » Limits at infinity » 2014-07-11 08:26:54

I didn't learn differentiation...

#21 Re: Help Me ! » Limits at infinity » 2014-07-11 08:04:17

Btw, if you want to see the problem, go her ;

it's example 4 at the end

#22 Re: Help Me ! » Limits at infinity » 2014-07-11 08:02:36

I didn't learn that...  yet.

#23 Re: Help Me ! » Finding the second and third root of a square and cube. » 2014-07-11 08:00:41

bobbym wrote:


Did you understand what happened? I can do more examples if you need them.

Nah it's okay !

#24 Help Me ! » Limits at infinity » 2014-07-11 07:24:08

Replies: 21

Hi, basic question here:

I have

lim 4z^2+z^6

x->positive infinity

Now, I know that this reduces to :

________=         0+infinity
1/z^3 -5

which gives infinity over -5. Now, how do I evaluate infinity over a constant ??? Is there a fact or definition that tells us how to deal with this ?

Thank you

#25 Re: Help Me ! » Finding the second and third root of a square and cube. » 2014-07-11 07:18:26

bobbym wrote:

According to the template we now just need to solve

we get

Now we have n1,n2 and n3 so

we have found the cube root of the RHS!

Ah ok! Well, thank you for taking your time.

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