Karmanaut:

arctan (x)
It follows the same procedure.
Start from y =arctan (x)
Therefore x = tan (y)(Call this equation 1)
Differentiating x wrt y gives sec^2 (y)
Also from trigonometry sec^2 (x) =1 + tan^2 (x)
Therefore dx/dy = 1 + tan^2 (y) (call this equation 2)
Recall that x =tan (y) from equation 1
Substitute it into equation 2
which guves:
dx/dy=(1+x^2)
Dy/dx Which is required of us is the inverse of this. Turning the equation upside down gives:
dy/dx = 1/(1+x^2)
Q. E. D
I worked backwards from the answer since it is a standard. Do Not Try This At Home.

hahaha...

guy u cunning ooo, mathematically.


i never agree with ur solution finish , still proof-reading.

agentofchange1:


hahaha...

guy u cunning ooo, mathematically.

" The great advances in mathematics have not been made by logic but by creative imagination."
- George Frederick James Temple, 100 Years of Mathematics: a Personal Viewpoint (1981)

Karmanaut:

Written your exam yet?
How was it?

Done tnx very much is an afternoon paper not morning iit was great atleast A is sure by god grace

personal59:


Done tnx very much is an afternoon paper not morning iit was great atleast A is sure by god grace

Good luck, mate.

masperano:


*Smiles* see who is talking because of the small small sec school/undergrad maths wey u dey solve na im dey shark you abi?

yes sir/ma. I do know it means throwing curved balls.

PS: when replying me next time be more civil

by your reply, you're the most civil. Miss/Mr Civility 2015

There are gays in here. I wonder why a sane man will register as a 'female' on a forum like this. Your sex is your identity.be real!

hey guys, who could derive the moment generating funtion of hypergeometric distribution. note that it is discrete random variable.

lexiconkabir:
hey guys, who could derive the moment generating funtion of hypergeometric distribution. note that it is discrete random variable.

Check here
Or this one instead:
[img]http://amstat.tandfonline.com/na101/home/literatum/publisher/tandf/journals/content/utas20/1973/utas20.v027.i03/00031305.1973.10479008/production/00031305.1973.10479008.fp.png_v03[/img]

Karmanaut:

Check here
Or this one instead:
[img]http://amstat.tandfonline.com/na101/home/literatum/publisher/tandf/journals/content/utas20/1973/utas20.v027.i03/00031305.1973.10479008/production/00031305.1973.10479008.fp.png_v03[/img]

i think i now see why that statistics lecturer skipped it. hypergeometric distribution has no M.G.F

lexiconkabir:
i think i now see why that statistics lecturer skipped it. hypergeometric distribution has no M.G.F

Your lecturer told you to derive the MGF?

Karmanaut:

Your lecturer told you to derive the MGF?

actually he did not, but out of curiousity i wanted to, because he never asked us to derive that of normal distribution, and i derived it myself!

lexiconkabir:
actually he did not, but out of curiousity i wanted to, because he never asked us to derive that of normal distribution, and i derived it myself!

Ok.

Karmanaut:

Ok.

thanks! can you please integrate exponential raised to the power of 5x all divided by 5x?

lexiconkabir:
thanks! can you please integrate exponential raised to the power of 5x all divided by 5x?

The answer is ...
(Ei(5x))/5 + C
Where Ei(x) is the Exponential Integral given as
or the derivative of (e^x)/x.
Cheers.

Karmanaut:


This does not have an anti-derivative; There is no closed form expression for the indefinite integral.
The primitives of ln(cos(x)) cannot be expressed as a finite combination of usual functions.
The analytic expression is complicated and includes a special function (polylogarithm).

integral log(cos(x)+sec(x)) dx = 1/2 (-i Li_2(-e^(2 i x))+i (Li_2((-3+2 sqrt(2)) e^(2 i x))+Li_2(-(3+2 sqrt(2)) e^(2 i x)))+i x^2+2 x log(1+e^(2 i x))-2 log(1+(3-2 sqrt(2)) e^(2 i x)) (x+sin^(-1)(sqrt(2)))-2 log(1+(3+2 sqrt(2)) e^(2 i x)) (x-sin^(-1)(sqrt(2)))-4 i sin^(-1)(sqrt(2)) tan^(-1)((tan(x))/sqrt(2))+2 x log(cos(x)+sec(x)))+constant



Where Li_n(x) is the Polylogarithmic integral function which is given by the integral of dt/ln (t) from 0 to x for all positive real numbers where x=! 1.
the integral for x > 1 has to be interpreted as a Cauchy Principal value

The polylog in this case is specifically the 'dilogarithm' because of the power of n is 2 in the denominator of the sum.
cos (x) = (e^ix + e^-ix)/2

jackpot, richiez, agentofchange1 ………… should i believe him? is he right?

Porthos:


jackpot, richiez, agentofchange1 ………… should i believe him? is he right?

Cheers.

Porthos:


jackpot, richiez, agentofchange1 ………… should i believe him? is he right?

you see someone using a powerful terabyte computer (probably, Wolfram Alpha) to solve maths and you're asking whether to believe him.

Who am I, abeg? I trust that computer more. Believe him, FFS!

jackpot:
you see someone using a powerful terabyte computer (probably, Wolfram Alpha) to solve maths and you're asking whether to believe him.

Who am I, abeg? I trust that computer more. Believe him, FFS!

lol……… u dey make me laff. u say na computer dey solve am? haha… na karmanaut ooo

hmmmm.

Porthos:


lol……… u dey make me laff. u say na computer dey solve am? haha… na karmanaut ooo

fact: The guy is good no doubt..............

But maybe you need to learn how to distinguish between problems solved with computer apps and those solved manually,as many seen here are solved with the aid of system software.

Mathematics is now made easy with recent mth app. I would have posted some earlier, but that is like contributing to...............*you know the rest*...........................

I may be posting guides/hints tho to solving some(if my hussle/sku decide to give me chance to).
But sincerly it takes a mathbrain to effectively use and interprete results from these apps....................#backtosidelines

Karmanaut:


The answer is ...
(Ei(5x))/5 + C
Where Ei(x) is the Exponential Integral given as
or the derivative of (e^x)/x.
Cheers.

guy geme d app u dey use take solve all those questions na. ..tnx.

agentofchange1:


guy geme d app u dey use take solve all those questions na. ..tnx.

Cheers
Costs like #600

Karmanaut:

Cheers
Costs like #600

ok. thanks man.

pls, help me look into this;
1) A boy of mass 40kg sits from the pivot of a bench while his 30kg friend sits on the other side. Calculate the distance of the two friends when the bench is at equilibrium.
2) A uniform beam is 12m long and has a mass of 50kg, and masses of 30kg and 40kg are suspended from its ends. At what point must the beam be supported so that it may rest horizontally?

Mayour11:
pls,
2) A uniform beam is 12m long and has a mass of 50kg, and masses of 30kg and 40kg are suspended from its ends. At what point must the beam be supported so that it may rest horizontally?

6.5, if my assumptions are correct

Mayour11:
pls, help me look into this;
1) A boy of mass 40kg sits from the pivot of a bench while his 30kg friend sits on the other side. Calculate the distance of the two friends when the bench is at equilibrium.

info nt complete.. Check again pls

2) A uniform beam is 12m long and has a mass of 50kg, and masses of 30kg and 40kg are suspended from its ends. At what point must the beam be supported so that it may rest horizontally?

30(6+x) + 50x = 40(6-x);
120x = 60;
x = 0.5

The answer is 0.5m to the midpoint..
=6+0.5 = 6.5m

nB: sum of clockwise moment=sum of anticlockwise moment

If a prime number and a[a-1!+1] is divisible by 2a. Then a^a is ?

Seamione thanks very much, dejt4u thanks bro, but that was all the lecturer gave, maybe he's trying something fishy.

Help me check this out
1) A uniform half-meter bar AB is balanced horizontally across a knife-edge placed 15cm from point A. A mass of 30kg is hang at the end of A. What is the weight of the bar?
I got 450N, pls help me verify it.

2) A student want to compare the performance of an old spring-scale with a new one. He placed the scales at the end of a 2m long uniform horizontal bar of mass 4kg. If both scales are in good working condition, what should be their readings?

Thanks in anticipation.

Seamione:
I don't know how to say this without sounding arrogant, but I strongly believe I perfectly understand what I need to in this context. It is you rather, who are refusing to change your perspective about the question, which is all good though.

Take a look at the question again If you don't mind
The term "repetition is not allowed" here simply means when forming a 3-digit number (just one), no digit should appear more than once. So 124, 412, 214, 142, 421 and 241 are all in conformity with the rule guiding the formation of the numbers (i. 3-digit number from the digits 1,2,4,5,6 ii. No repetition of digits)
and are also totally different numbers such cannot be regarded as repeated numbers. Therefore, they all will make the sample space.
At this point if you still feel what about 3 of us here on this thread has said on this question is wrong, then I give up. Hoping that you change your perspective about it one day.

Mind you, it's not a matter of challenging/questioning your knowledge on this, but about not misleading lots of young ones who are gaining one or two things from this forum.

Happy weekend to y'all
#Peace

agentofchange1:



yea bro, i get, its ok now lets forget the whole thing... thanks anyways....

we are all learning.

Seamione is actually right and you are wrong. Your sample space is incorrect. This is not a combination problem, but a permutation problem.

Selecting a multiple of 5 and an odd number from 1,2,4,5,6.
Will give 4x3x1 + 4x3x1.
To form a 3-digit number will give a space of 5x4x3
Probability = (2x4x3)/(5x4x3) = 2/5

Antoinne:




Seamione is actually right and you are wrong. Your sample space is incorrect. This is not a combination problem, but a permutation problem.

Selecting a multiple of 5 and an odd number from 1,2,4,5,6.
Will give 4x3x1 + 4x3x1.
To form a 3-digit number will give a space of 5x4x3
Probability = (2x4x3)/(5x4x3) = 2/5

yea bro, we've settled this a long ago... stop trying rise the death ,

happy returning....

#shalom.

Antoinne:




Seamione is actually right and you are wrong. Your sample space is incorrect. This is not a combination problem, but a permutation problem.

Selecting a multiple of 5 and an odd number from 1,2,4,5,6.
Will give 4x3x1 + 4x3x1.
To form a 3-digit number will give a space of 5x4x3
Probability = (2x4x3)/(5x4x3) = 2/5

Thanks bro, the case has long been settled though.