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Re: Nairaland Mathematics Clinic by Oxide65: 1:07pm On Aug 14, 2014
Arithmetic: Question:
Solve:dy/dx+2ytanx=sinx. By 1Book , from Calculus Mental Challenge(CMC).

:::SOLUTION:::.
Integrating both sides,
$(dy/dx+2ytanx)dx=$sinxdx.
$dy/dx.dx+2y$tanxdx=$sinxdx.
$dy+2y(-lncosx)=-cosx+c
NB: y is taken as a constant in $2ytanxdx.
y-2ylncosx=-cosx+c.
y(1-2lncosx)=-cosx+c.
.'.y=(-cosx+c)/(1-2lncosx). Which is the required solution.
Re: Nairaland Mathematics Clinic by Oxide65: 1:09pm On Aug 14, 2014
Arithmetic: Question:
Solve:dy/dx+2ytanx=sinx. By 1Book , from Calculus Mental Challenge(CMC).

:::SOLUTION:::.
Integrating both sides,
$(dy/dx+2ytanx)dx=$sinxdx.
$dy/dx.dx+2y$tanxdx=$sinxdx.
$dy+2y(-lncosx)=-cosx+c
NB: y is taken as a constant in $2ytanxdx.
y-2ylncosx=-cosx+c.
y(1-2lncosx)=-cosx+c.
.'.y=(-cosx+c)/(1-2lncosx). Which is the required solution.


I'm trying to send d solution But It's not appearing
Re: Nairaland Mathematics Clinic by Oxide65: 1:13pm On Aug 14, 2014
Arithmetic: Question:
Solve:dy/dx+2ytanx=sinx. By 1Book , from Calculus Mental Challenge(CMC).

:::SOLUTION:::.
Integrating both sides,
$(dy/dx+2ytanx)dx=$sinxdx.
$dy/dx.dx+2y$tanxdx=$sinxdx.
$dy+2y(-lncosx)=-cosx+c
NB: y is taken as a constant in $2ytanxdx.
y-2ylncosx=-cosx+c.
y(1-2lncosx)=-cosx+c.
.'.y=(-cosx+c)/(1-2lncosx). Which is the required solution.

Re: Nairaland Mathematics Clinic by Nobody: 1:13pm On Aug 14, 2014
benbuks: Reasons why some questions on this forum might not be /are not solved includes

a) the poster never try to help others on their questions , but wants others to help on his ( help is reciprocal . don't be inconsiderate / selfish )

b) the poster always posts too-long questions which is hard to type out solutions

c) most of gurus / those with more competence hands to handle the question are offline /busy

d) the notion of " how much will they even pay me in sacrificing my time ,Mb & energy in solving some one else's problem .

e) otherwise ...


so guys lets take note of the above points & do something about it .

For me, helping out with solution to questions on this thread became problematic the moment I changed to android. Unlike my former qwerty phone, typing out solution to problems with this one takes too much time. sad
Re: Nairaland Mathematics Clinic by 1Book(m): 2:46pm On Aug 14, 2014
[quote author=Oxide65][/quote]
...Just upload the final answer where y=? so that it will be clear.
Re: Nairaland Mathematics Clinic by Oxide65: 3:17pm On Aug 14, 2014
1Book:
...Just upload the final answer where y=? so that it will be clear.



y=cosx (1+cosx )
Re: Nairaland Mathematics Clinic by Oxide65: 3:17pm On Aug 14, 2014
1Book:
...Just upload the final answer where y=? so that it will be clear.



y=cosx (1+cosx ) That's what I found After solving it.
Re: Nairaland Mathematics Clinic by Oxide65: 3:27pm On Aug 14, 2014
1Book:
...Just upload the final answer where y=? so that it will be clear.





y=cosx (1+Acosx )
Re: Nairaland Mathematics Clinic by Nobody: 11:49am On Aug 15, 2014
hmmmm.
Re: Nairaland Mathematics Clinic by Nobody: 11:50am On Aug 15, 2014
Tolzeal: Help Gurus At Home:


Solve this..
◻➕◻➕◻➕◻➕◻=30
Fill the boxes using (1, 3, 5, 7, 9, 11, 13, 15)
U can also repeat the numbers..This question came in UPSC Final Exams held in Dec '13 !!!



hmm

seems...like just four numbers can do d job .

.leme try sha...
Re: Nairaland Mathematics Clinic by Nobody: 11:54am On Aug 15, 2014
Arithmetic: x is a very small value which ranges from 0 to infinity. Hence x approaches infinity.

x=9/4


prove by solution if u can bro...

i think newton-raphson cud b d guy 2 call ..

what thinketh thou .?
Re: Nairaland Mathematics Clinic by Nobody: 12:09pm On Aug 15, 2014
smurfy:

For me, helping out with solution to questions on this thread became problematic the moment I changed to android. Unlike my former qwerty phone, typing out solution to problems with this one takes too much time. sad
.
ok bro u can also try when chanced ...m also using Android ...


greetings sir..
Re: Nairaland Mathematics Clinic by Arithmetic(m): 3:38pm On Aug 15, 2014
benbuks:

x=9/4


prove by solution if u can bro...

i think newton-raphson cud b d guy 2 call ..

what thinketh thou .?
Oh!, I'm very sorry sir, I made a silly mistake along the way.
:::SOLUTION:::
18¥x=729¥x-1.
Let a=¥x, we have; NB:¥ rep. square root.
18a=729a-1,
=>32.2a=36(a-1).
=>2a=3{6(a-1)-2}.
=>2a=3(6a-6-2).
=>3{6a-8}-2a=0.
Using Binomial Expansion, we have;
(1+2)6a-8-2a=0.
=>1+2{6a-8}+22.{6a-8}(6a-9)/2+...-2a=0.
=>1+12a-16+2(36a2-102a+72)-2a=0.
=>1+12a-16+72a2-204a+144-2a=0.
On simplifying, we have;
=>72a2-194a+129=0.
Solving the affected quadratic eqn.
=>a=1.5 or 1.194444.
=>a=3/2 or 43/36.
We take the value of a=3/2, which satisfies the equation.
Recall a=¥x,
=>x=a2.
...x=(3/2)2=9/4.
Newton Raphon's iterative method will also do.#stillyourboy
Re: Nairaland Mathematics Clinic by Drniyi4u(m): 4:03pm On Aug 15, 2014
Tolzeal: Help Gurus At Home:


Solve this..
◻➕◻➕◻➕◻➕◻=30
Fill the boxes using (1, 3, 5, 7, 9, 11, 13, 15)
U can also repeat the numbers..This question came in UPSC Final Exams held in Dec '13 !!!


I guess u mean we should take 9 numbers(some may be repeated) from d set of numbers that can sum up to 30, m I right?
Re: Nairaland Mathematics Clinic by dejt4u(m): 4:14pm On Aug 15, 2014
Tolzeal: Help Gurus At Home:


Solve this..
◻➕◻➕◻➕◻➕◻=30
Fill the boxes using (1, 3, 5, 7, 9, 11, 13, 15)
U can also repeat the numbers..This question came in UPSC Final Exams held in Dec '13 !!!


after spending minutes on this question, i concluded within myself dat it is not possible.. There is no way you will add odd numbers together in odd numbers of way and you will get even number..it is jst nt possible QED
Re: Nairaland Mathematics Clinic by Soneh(m): 6:05pm On Aug 16, 2014
if|x y z |
| a b c | =5
| 1 2 3 |
use properties of determinants to find
|1 2 9. |
|a b 3c |
|x y 3z |
pls I need the solution with explanations
Re: Nairaland Mathematics Clinic by efficiencie(m): 9:13pm On Aug 16, 2014
Soneh: if|x y z |
| a b c | =5
| 1 2 3 |
use properties of determinants to find
|1 2 9. |
|a b 3c |
|x y 3z |
pls I need the solution with explanations

Let the matrix A with element a(ij) be defined as:
A=[row(1)(a(11) a(12) a(13) ...a(1m), row(2)(a(21) a(22) a(23) ...a(2m), ... row(n)(a(n1) a(n2) a(n3) ...a(nm)]

Hence if d determinant:
|row(1)(x y z), row(2)(a b c), row(3)(1 2 3)|=5
Then
|row(1)(1 2 9), row(2)(a b 3c), row(3)(x y 3z)|
Multiple of a column
=3. |row(1)(1 2 3), row(2)(a b c), row(3)(x y z)|
Interchange of rows
=-3. |row(1)(x y z), row(2)(a b c), row(3)(1 2 3)|
=-3(5)
=-15

hence
|row(1)(1 2 9), row(2)(a b 3c), row(3)(x y 3z)|=-15

I hope i'm ryt!!!
Re: Nairaland Mathematics Clinic by Soneh(m): 12:14pm On Aug 17, 2014
efficiencie:

Let the matrix A with element a(ij) be defined as:
A=[row(1)(a(11) a(12) a(13) ...a(1m), row(2)(a(21) a(22) a(23) ...a(2m), ... row(n)(a(n1) a(n2) a(n3) ...a(nm)]

Hence if d determinant:
|row(1)(x y z), row(2)(a b c), row(3)(1 2 3)|=5
Then
|row(1)(1 2 9), row(2)(a b 3c), row(3)(x y 3z)|
Multiple of a column
=3. |row(1)(1 2 3), row(2)(a b c), row(3)(x y z)|
Interchange of rows
=-3. |row(1)(x y z), row(2)(a b c), row(3)(1 2 3)|
=-3(5)
=-15

hence
|row(1)(1 2 9), row(2)(a b 3c), row(3)(x y 3z)|=-15

I hope i'm ryt!!!
pls explain better, i'm preparing for an exam
Re: Nairaland Mathematics Clinic by efficiencie(m): 12:28pm On Aug 17, 2014
Soneh:
pls explain better, i'm preparing for an exam

the determinant of a matrix A with row elements r(.j)=k×a(.j) or column elements c(i.)=k×a(i.) equals the product 'k' and the matrix with elements a(.j) or a(i.) respectively. hence
|1 2 9|
|a b 3c|
|x y 3z|
equals 3 multiplied by d determinant
|1 2 3|
|a b c|
|x y z|
it's lik factoring 3 out of d 3rd column

The second move involves interchanging d 1st n 3rd rows and d@ renders the determinant negative.
Re: Nairaland Mathematics Clinic by Nobody: 11:00am On Aug 18, 2014
dejt4u:
after spending minutes on this question, i concluded within myself dat it is not possible.. There is no way you will add odd numbers together in odd numbers of way and you will get even number..it is jst nt possible QED

yea bro.....this very enigma had made me do some crazy things with numbers & algebra . @ a glance of it , one could conclude that , its mathematically "impossible " to sum odd integers , odd number of times to give an even number & vice versa .

however , for some infinitesimal or infinitude reasons , i'm kinda reluctant to also accept the notion . (maybe am not crazy enough yet ..lolz. ) ..

its well.
Re: Nairaland Mathematics Clinic by Nobody: 3:39pm On Aug 18, 2014
Two coins are biased in such a way so that
when they are both tossed once: i) the
probability of getting two heads is the same as
the probability of getting two tails ii)
the probability of getting one head and one tail is
5/8 For each coin, what is the probability of
getting a head?
Re: Nairaland Mathematics Clinic by Soneh(m): 9:49pm On Aug 19, 2014
pls gurus in the house should help with this:
write in standard form, (a)3i^105+5i^73-8i^50-2i^119
(b) (3i-5)^4 -(2-I)^4
PLS I will appreciate if the solutions are accompanied with explanation
Re: Nairaland Mathematics Clinic by Oxide65: 12:48pm On Aug 20, 2014
Guys Please I'm trying to prove that lim(SQROOT(n^2 +n)-n)=1/2 But the estimates I'm making aren't moving well. Please can someone help?.Even After derationalizing I still got stuck. I need help Please.

Re: Nairaland Mathematics Clinic by Nobody: 1:28pm On Aug 20, 2014
Oxide65: Guys Please I'm trying to prove that lim(SQROOT(n^2 +n)-n)=1/2 But the estimates I'm making aren't moving well. Please can someone help?.Even After derationalizing I still got stuck. I need help Please.

as n --> ?...
Re: Nairaland Mathematics Clinic by Nobody: 2:08pm On Aug 20, 2014
Soneh: pls gurus in the house should help with this:
write in standard form, (a)3i^105+5i^73-8i^50-2i^119
(b) (3i-5)^4 -(2-I)^4
PLS I will appreciate if the solutions are accompanied with explanation

--------SOLUTION--------------
(a)

= 3(i104 )i + 5(i72 )i -8(i2 )25 -2(i118 ) i

=>3i(1) +5i(1) - 8(-1) -2i(-1)

= 10i +8

(b)

by difference of two squares.
we have
[(3i-5)2 ]2 - [(2-i)2 ]2

=>(9i2 -30i+25)2 -(4-4i+i2)2

=>(16-30i)2 - (3-4i)2

=> (256-960i +900i2 )-(9-24i+16i2 )
=>-544-960i +7 +24i

hence we have ,

-537 -936i
Re: Nairaland Mathematics Clinic by Oxide65: 2:50pm On Aug 20, 2014
benbuks:

as n --> ?...


as n--->.infinity. Thanks for your time.
Re: Nairaland Mathematics Clinic by Soneh(m): 3:01pm On Aug 20, 2014
benbuks:

--------SOLUTION--------------
(a)

= 3(i104 )i + 5(i72 )i -8(i2 )25 -2(i118 ) i

=>3i(1) +5i(1) - 8(-1) -2i(-1)

= 10i +8

(b)

by difference of two squares.
we have
[(3i-5)2 ]2 - [(2-i)2 ]2

=>(9i2 -30i+25)2 -(4-4i+i2)2

=>(16-30i)2 - (3-4i)2

=> (256-960i +900i2 )-(9-24i+16i2 )
=>-544-960i +7 +24i

hence we have ,

-537 -36i
pls I don't get the (a) part, what happened to the powers.
thanks
Re: Nairaland Mathematics Clinic by Nobody: 3:57pm On Aug 20, 2014
Oxide65:


as n--->.infinity. Thanks for your time.

ok

"""""""""SoLuTiOn"""""""""""""""

take product by the conjugate expression √(n2+ n ) + n at the numerator & denominator

=> Lim n-->infinity =[[ √(n2 +n) - n ] [[√(n2 + n) + n ] ] / [√(n2 + n ) + n ]

=> Lim n-->infinity ( n2 +n - n2 ) / [ √[n 2(1 + 1/n) ] + n ]

=> simplifying & dividing throughout by 'n' we have

1/√( 1 + 1/n ) + 1 since n>0 ,as n--> infinity

we thus have 1/[ √(1+0) +1 ]

=1/2 ......... (proved )
Re: Nairaland Mathematics Clinic by Oxide65: 5:14pm On Aug 20, 2014
benbuks:

ok

"""""""""SoLuTiOn"""""""""""""""

take product by the conjugate expression √(n2+ n ) + n at the numerator & denominator

=> Lim n-->infinity =[[ √(n2 +n) - n ] [[√(n2 + n) + n ] ] / [√(n2 + n ) + n ]

=> Lim n-->infinity ( n2 +n - n2 ) / [ √[n 2(1 + 1/n) ] + n ]

=> simplifying & dividing throughout by 'n' we have

n/√( 1 + 1/n ) + 1 since n>0 ,as n--> infinity

we thus have 1/[ √(1+0) +1 ]

=1/2 ......... (proved )







Very well. Thanks alot. Im very sorry i could have been more specific. Could you try using the €-N argument? I'm sure you must be familiar with it because from what I've seen, I must confess all my mathematical problems died when I found this link.
Re: Nairaland Mathematics Clinic by Oxide65: 5:16pm On Aug 20, 2014
benbuks:

ok

"""""""""SoLuTiOn"""""""""""""""

take product by the conjugate expression √(n2+ n ) + n at the numerator & denominator

=> Lim n-->infinity =[[ √(n2 +n) - n ] [[√(n2 + n) + n ] ] / [√(n2 + n ) + n ]

=> Lim n-->infinity ( n2 +n - n2 ) / [ √[n 2(1 + 1/n) ] + n ]

=> simplifying & dividing throughout by 'n' we have

n/√( 1 + 1/n ) + 1 since n>0 ,as n--> infinity

we thus have 1/[ √(1+0) +1 ]

=1/2 ......... (proved )



Very well. Thanks alot. Im very sorry i could have been more specific. Could you try using the €-N argument? I'm sure you must be familiar with it because from what I've seen, I must confess all my mathematical problems died when I found this link.
Re: Nairaland Mathematics Clinic by Oxide65: 5:25pm On Aug 20, 2014
Soneh:
pls I don't get the (a) part, what happened to the powers.
thanks


Use the concept of indices.

Re: Nairaland Mathematics Clinic by Nobody: 10:40pm On Aug 20, 2014
Soneh:
pls I don't get the (a) part, what happened to the powers.
thanks

ok...you"ll have to go back to your basics of complex numbers

rationally , negative roots are invalid e.g ± √-1 , ±√-2 ,±√-3 e.t c.

solving x2 + 1 =0 ..........(*)

produces complex roots , however mathematicians have worked severally on such irrational roots , which later discovered & agreed that solution to say (*) =±√-1

we now use ' i ' or ' j ' = ±√-1 ..........(**)

hence solution of (*) ,x = ± i

having believed that x=± i

=>i2 =-1 .......( from (**) )

i3 = i(i2) =i(-1) =-i

i4 = (i2 ) (i2 ) = (-1)(-1)=1

i5 = (i4 ) i = (1)i = i

we could observe that powers of ' i ' in .

2 ,6 , 10 ,14 . . . =-1

3 ,7 , 11 ,15 . . . =-i

4 , 8 , 12 , 16 . . . = 1

5 , 9, 13 , 17 , . . . =i

& their multiples in that order , gives same result .
.

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