₦airaland Forum

Karmanaut:
Substituted so much I felt like a coach.

aahh boss mi. integral of 1/(1-w^2) = arctan(w) ?

hey guys try this asap..

integrate dx/√(1+sinx)

::::: Some useful mathematical results ::
(α+в+¢)²= α²+в²+¢²+2(αв+в¢+¢α) 1. (α+в)²= α²+2αв+в² 2. (α+в)²= (α-в)²+4αв b 3. (α-в)²= α²-2αв+в² 4. (α-в)²= (α+в)²-4αв 5. α² + в²= (α+в)² - 2αв. 6. α² + в²= (α-в)² + 2αв. 7. α²-в² =(α + в)(α - в) 8. 2(α² + в²) = (α+ в)² + (α - в)² 9. 4αв = (α + в)² -(α-в)² 10. αв ={(α+в)/2}²-{(α-в)/2}² 11. (α + в + ¢)² = α² + в² + ¢² + 2(αв + в¢ + ¢α) 12. (α + в)³ = α³ + 3α²в + 3αв² + в³ 13. (α + в)³ = α³ + в³ + 3αв(α + в) 14. (α-в)³=α³-3α²в+3αв²-в³ 15. α³ + в³ = (α + в) (α² -αв + в²) 16. α³ + в³ = (α+ в)³ -3αв(α+ в) 17. α³ -в³ = (α -в) (α² + αв + в²) 18. α³ -в³ = (α-в)³ + 3αв(α-в)
ѕιη0° =0 ѕιη30° = 1/2 ѕιη45° = 1/√2 ѕιη60° = √3/2 ѕιη90° = 1 ¢σѕ ιѕ σρρσѕιтє σƒ ѕιη тαη0° = 0 тαη30° = 1/√3 тαη45° = 1 тαη60° = √3 тαη90° = ∞ ¢σт ιѕ σρρσѕιтє σƒ тαη ѕє¢0° = 1 ѕє¢30° = 2/√3 ѕє¢45° = √2 ѕє¢60° = 2 ѕє¢90° = ∞ ¢σѕє¢ ιѕ σρρσѕιтє σƒ ѕє¢
2ѕιηα¢σѕв=ѕιη(α+в)+ѕιη(α-в) 2¢σѕαѕιηв=ѕιη(α+в)-ѕιη(α-в) 2¢σѕα¢σѕв=¢σѕ(α+в)+¢σѕ(α-в) 2ѕιηαѕιηв=¢σѕ(α-в)-¢σѕ(α+в)

ѕιη(α+в)=ѕιηα ¢σѕв+ ¢σѕα ѕιηв. » ¢σѕ(α+в)=¢σѕα ¢σѕв - ѕιηα ѕιηв. » ѕιη(α-в)=ѕιηα¢σѕв-¢σѕαѕιηв. » ¢σѕ(α-в)=¢σѕα¢σѕв+ѕιηαѕιηв. » тαη(α+в)= (тαηα + тαηв)/ (1−тαηαтαηв) » тαη(α−в)= (тαηα − тαηв) / (1+ тαηαтαηв) » ¢σт(α+в)= (¢σтα¢σтв −1) / (¢σтα + ¢σтв) » ¢σт(α−в)= (¢σтα¢σтв + 1) / (¢σтв− ¢σтα) » ѕιη(α+в)=ѕιηα ¢σѕв+ ¢σѕα ѕιηв. » ¢σѕ(α+в)=¢σѕα ¢σѕв +ѕιηα ѕιηв. » ѕιη(α-в)=ѕιηα¢σѕв-¢σѕαѕιηв. » ¢σѕ(α-в)=¢σѕα¢σѕв+ѕιηαѕιηв. » тαη(α+в)= (тαηα + тαηв)/ (1−тαηαтαηв) » тαη(α−в)= (тαηα − тαηв) / (1+ тαηαтαηв) » ¢σт(α+в)= (¢σтα¢σтв −1) / (¢σтα + ¢σтв) » ¢σт(α−в)= (¢σтα¢σтв + 1) / (¢σтв− ¢σтα)
α/ѕιηα = в/ѕιηв = ¢/ѕιη¢ = 2я » α = в ¢σѕ¢ + ¢ ¢σѕв » в = α ¢σѕ¢ + ¢ ¢σѕα » ¢ = α ¢σѕв + в ¢σѕα » ¢σѕα = (в² + ¢²− α²) / 2в¢ » ¢σѕв = (¢² + α²− в²) / 2¢α » ¢σѕ¢ = (α² + в²− ¢²) / 2¢α » Δ = αв¢/4я » ѕιηΘ = 0 тнєη,Θ = ηΠ » ѕιηΘ = 1 тнєη,Θ = (4η + 1)Π/2 » ѕιηΘ =−1 тнєη,Θ = (4η− 1)Π/2 » ѕιηΘ = ѕιηα тнєη,Θ = ηΠ (−1)^ηα

1. ѕιη2α = 2ѕιηα¢σѕα 2. ¢σѕ2α = ¢σѕ²α − ѕιη²α 3. ¢σѕ2α = 2¢σѕ²α − 1 4. ¢σѕ2α = 1 − ѕιη²α 5. 2ѕιη²α = 1 − ¢σѕ2α 6. 1 + ѕιη2α = (ѕιηα + ¢σѕα)² 7. 1 − ѕιη2α = (ѕιηα − ¢σѕα)² 8. тαη2α = 2тαηα / (1 − тαη²α) 9. ѕιη2α = 2тαηα / (1 + тαη²α) 10. ¢σѕ2α = (1 − тαη²α) / (1 + тαη²α) 11. 4ѕιη³α = 3ѕιηα − ѕιη3α 12. 4¢σѕ³α = 3¢σѕα + ¢σѕ3α

» ѕιη²Θ+¢σѕ²Θ=1 » ѕє¢²Θ-тαη²Θ=1 » ¢σѕє¢²Θ-¢σт²Θ=1 » ѕιηΘ=1/¢σѕє¢Θ » ¢σѕє¢Θ=1/ѕιηΘ » ¢σѕΘ=1/ѕє¢Θ » ѕє¢Θ=1/¢σѕΘ » тαηΘ=1/¢σтΘ » ¢σтΘ=1/тαηΘ » тαηΘ=ѕιηΘ/¢σѕΘ

#shalom.

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

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

we are all learning.

lopzy:
Can anyone help confirm this answers? A committee of 3 is to be chosen from 5 men and 4 women.find the probability that there will be one woman in the committee. Is the answer 10/21 or 1/21

5men
4women

Total = 9persons
forming a-3-man committee
Total ways of selection =9C3 =84 ways

now for the given condition, if one woman had to be in the committee, its means to other two gat to be men

thus,

probability that there will be one woman in the committee

5C2*4C1 /9C3= 10x4/84 = 10/21

that's it dear, hope you get?

Seamione:
Good morning bro. It's not that I find joy in arguing with you over this o as you be my oga patapata but I can't help but to point this out;
You said possible sample space

S={(s1s2s4) ,(s1s2s4),(s1s2s5),(s1s2s6),(s1s4s5),(s1s5s6), (s2s4s5), (s6s4s5),(s6s4s1),(s2s6s5) . . . }

Don't you think it's possible that the sample space could also possibly be

S={(s4s1s2) ,(s5s1s2),(s6s1s2),(s5s1s4),(s6s1s5), (s5s2s4), (s5s6s4),(s1s6s4),(s5s2s6), (s5s6s2) . . . } and in which your theory doesn't hold in this case.
Moreover, if you check your sample space, it's already up to 10 and you agreed that we could still continue, so many contradictions.
You see, my boss, whenever they use "no repetition" in this type of question, it simply means the same number can't show up twice in the same group like (s1s1s2). So, if you don't mind, you can also look this up.

Good morning to y'all
#Peace

sorry check well, dia nine, i just made small mistake in writing d first group twice. den plus one other 1 i didn't include that will make it 10.

ur still repeating same sample space , no matter how far u go
hmm. i don't think i can explain this any further/longer maybe i should 'nt av used combinatorial approach for this question, i believe u will understand better some day, i really love ur type, who challenges/questions the 'lemmas' . I should have used the method u used ('wrongly ') , but i won't do that now as that will prompt another episode of arguments .m not really feeling strong. ..


nice solving.... keep the fire burning.

#shalom.

jackpot:
Let me follow your argument to see if it holds water:
you gave your sample space as
S={(124),(125),(126),(245),(246),(145),
(156),(645),(641),(265)}.

Following your logic,
S={(124),(512),(126),(524),(246),(514),(156),(564),(164),(526)} because, for instance, 125 and 512 are the same numbers without repetition (whatever that means).

So, probability = 0/10=0.

Contradiction!

Class dismiss. Everybody go home.

hmmm,

i normally don't like arguing much, but let me try to explain this for the last time & if u guys think otherwise, that's Ok . so be it.

to explain this,

[b]lets say we have 5 students in a class ( named S1, S2, S3 , S4 & S5 ) then asked to form a group of three (probably, study group ) with no repetition allowed.(no repetition of same student(s) in the group or the sample space ) . already noted that S1 is the class rep. & S5 is the assistant class rep. , Now if a group is picked at random , whats the probability (or chance )of picking a group which has its last member as either S1 or S5 ? ( i. e either the class rep. or the assistant )

NB: In combination, we give attention to order of arrangements

total possible arrangement without repetition = 5C3
=10 ways

Now, possible sample space

S={ (s1s2s4),(s1s2s5),(s1s2s6),(s1s4s5),(s1s5s6), (s2s4s5), (s6s4s5),(s6s4s1),(s2s6s5) . . . }

the possible groups continues to about ten of them , i just stopped there cuz, I've gotten what i needed , any more will result to repetition of already exiting group which goes against the given condition

Probability of having either s1 or s5 as last member of the group , could be (s1s2s5) or (s1s4s5) or (s2s4s5) or (s6s4s1) or (s2s6s5) or (s6s4s5)

= 1/10 + 1/10 + 1/10 +1/10 +1/10 + 1/10 =6/10 =3/5 [/b]
Now you guys be the judge

am off this question, happy solving .



Good night!!!!

@personal59

www.google.com.ng/search?q=numerical+integration&client=ms-opera-mini-android&channel=new&gws_rd=cr&ei=HEXfVYzPJoX7aN_XmcgM

click on the site above n download d e copy, it really helps, u could understand well enough questions related to yours, seeing d exams is 2maro.


i don't wanna bring new formulas dat cud cause confusion,

leme advice u to just study ur notes carefully n all tests /assignments. u can't get all solutions from us here, at least my boss has done justice to some of ur problem.

this is how i could help on this


hope u get my gist..

Karmanaut:
You used the third derivative?
Isn't it the second derivative for the trapezoid rule and the fourth derivative for Simpson's rule?

no bro,

that's it, wanna see the proof. ?

personal59:
He did 1 and 3 But he said he doesnot understand question 4 especially how to go about solving it and he said question 2 does not have error bound so I said he should assume an error bound so that I will be able to solve if eventually given in exam with error bound too


So pls can u help me sove question 2 using an assumed error bound and also question 4.

Thanks

ok sir

we gat basically two types of errors when using trapezoidal rule

local error ,

e = -1/12 *h^3 f' ' '(α) : x₁<α<x₂

&
Global error ,

e= -1/12 * n f' ' ' (α) : x₁<α<x_n


guess this helps





brb leme recheck the question behind.

boss karmanaut , great works sir
kindly look into that probability question on the previous page
good day.

personal59:
Pls bro can u help out on the question I post a page before this karmanaut he said he his aving problem with question 4 and call on u to help out he also include question two but I ask him to assume the error bound value so that I will b able to sove it wen eventually given in d exam thanks in anticipation

guess he has done it already.

jackpot:
ok. Thank God you said scarcely, not completely. My time here is somehow irregular. Most of the times, I come to see new questions and solutions. If I see an unsolved one that is not too trivial and interests me, then I'll go for it. Questions like Perm and Combination, simultaneous solution of 4^x+6^y=100 and x+y=5 do not interest me.

Some of the questions that I post is for others to learn. Under normal circumstances, elementary university maths treat conic sections at the surface level. That question requires knowledge of rotation of axes. You'll agree with me that sometimes we do have an unnecessarily long solution. In other words, i'm looking for shorter solutions, that is why I seek for other people's POV.

Don't feel that you're doing me a favour by solving the question. It is not about me. Other people are learning too. Gracias.

Aright dear i get your point,


BTW: look into that probability question i solved in the previous page , your honest mathematical input is required. tnx.

jackpot:
please, kindly show full workings.

ur own na 2 just dey post problems n tag people but u scarcely help others in solving dias

i could av post solutions to some of ur problems even if i av limited ideas cud av done all possible .

but feels lazy/discouraged to do so

Seamione:
Exactly, waiting for others to come in on this

u definitely know little about probability..

hmmm.

its ok then..

Seamione:
Oga General Ben, I think the "no repetition" in the question means we can't have numbers like 111, 222, 444, 555 and 666.
Moreover, 421 and 124 are not exactly the same number, they're just numbers with the same digits so we can't consider that as repetition because if that's the case, I don't think we should also have 124 and 125 because 1 and 2 have been repeated.
I hope you understand what I'm trying to say

hmmm, i won't say much bro

let others shed more light.

Seamione:
No, I kinda disagree with your sample space, it can't just be 5C3. For instance, where Is 421, 521 and 621 in your sample space? At least, they're also of 3 digits which are odd numbers..

c'mmon bro, the question says no repetition

5C3 means total number of ways we could form the 3-digit number without repetition

421 (124) ,521(125) , 621( 126) already exist in the sample space.

donholy28:
Show us d math puzzle

i tire oo, av been longing to see it.

@Op
and you emerged best at tribalism huh?

Btw : similar thread already exist.

Seamione:
Good evening guys,
Please take a look at this question.

Three digit numbers re formed from the digits 1,2,4,5,6. If repetition is not allowed and a number is picked at random, find the probability that it's a multiple of 5 or an odd number.

From my calculations, I got 2/5 but the text says it's 3/5. Who is wrong?

total ways of selection = 5C3 = 10ways

sample space

S={(124),(125),(126),(245),(246),(145),(156),(645),(641),(265)}

check where we have an odd number or multiple of 5

thus, we get

6/10=3/5


hope you get?

worthing:
integrate In(cosx)dx

I doubt if an analytical solution to this exist.

Ooh my God!! so pathetic and unfortunate

But please!!! i know how painful it is,, the husband should think twice before sending his wife packing please, he should remember their love over the years and all they passed through, please i know its painful ......

take heart sir /ma, at least she has learnt her lessons now

Life goes on!!! , Another will come.


#Shalom

can't think of a justifiable Excuse though,

went for a night vigil (from 9:30pm -5:30am) came home so tired and sleepy, just woke up about an hour ago, i believe the service should be at climax by now .


There's always one excuse or the other we give when it comes to spiritual things ,

"Do you love me more than these "

"where there's a will, there's always a way " . God help us all. ...
happy Sunday folks,

::::1luv..

#shalom.

kidrossy:
anyone...
If the quadratic function 3x²-7x+R is a perfect square find R...anyways I got 49/12...just hoping to be correct

for a given quadratic expression to be a perfect square ,

b^2 -4ac=0

here a=3 , b=-7 c= R

thus 49 = 12R

R =49/12.


ur right

kidrossy:
Pls can anyone help me with this...
A Cone with a sloping edge is 12m has a total surface area of 160piecm^2, what is it's base radius...Thanx

A= πr(r+s)

where s=√(r² +h² )

s=sloping edge , h= perpendicular height ,r=base radius , A=total surface area

A=160πcm² , s= 12m = 1200cm


plug in the values and solve , guess you can finish up from there?

personal59:
Its still dsame answer I was given bro nd I don't knw how the step by step was gotten will u pls explain better just like teaching a lay man

hmm, i did nothing much here bro just a simple integration. , though there are different ways of solving exact DEs, but trust me i just used the simplest one i thought you should comprehend easily.

@personal59 ,

since the given D. E is exact,

by definition $Mdx + $Ndy =$0dx =C

thus, it follows that

$(1+x² y² + y)dx +$xdy = C

= x +1/3 x³ y² +2xy =C


that's it man .

hope u get?

personal59:
It remain the solution for the equation

u said prove not solve hope u know d difference?

personal59:
Oya come and solve o





Prove that (1+x^2y^2+y)dx + xdy =0 is exact

here

worthing:
argentofchange pls the working

::::::::::::::::::::Solution:::::::::::::::::


For second result,

1/sin4x =cosec4x

=> 1/4$cosecu du. (where u=4x)

on taking product & ratio with 'cosecu -cotu' ,

we get

1/4$ [cosecu (cosecu-cotu) /(cosecu-cotu)] du

=1/4 $[(cosec²u - cotu. cosecu )/(-cotu+cosecu)] du

which takes the form $f '(u) /f(u) du = ln[f(u)] + C

hence we have

1/4 In( cosec4x - cot4x) + c ......( replace. u)



for the first result

worthing:
gurus in the house help out with this question



integral dx/sin4x

1/8ln(tan2x) + c

or. 1/4 ln(csc4x -cot4x) +C