₦airaland Forum

yemstok:
Thanks bro Q5 is integration

OK bro...

will post solutions asap..


but wait oo..how much you go pay me for. burning my MB....? lolz. ( just kidding ..never mind )

bolkay47:
no 1::: [b]3/2[/b]ln(2x+1)+2ln(x-2)+K

yea man you try


just a li2 correction (the bolded its 3/2. not 3 )

nice 1 try post full solutions next time

1luv ..happy solving.

yemstok:
Here

1. Integral 7x-4/2x^2-3x-2 dx

2. Integral x^3/x+1 dx

3. Integral cosx/7+sinx dx = In (7+sinx)+C

4. Integrate with respect to X
A. Sinxcosx
B. Sin6x
C. X^2 e^3

5. Y= sin^-1 (2x-1)
b. Y=tan^-1 (1+x/1-x)

am not with pen/paper now.
Q3 put. u=7+sinx

du=cosxdx
dx= du/sinx

thus => $cosx/u *du/cosx

= $du/u = ln (u) +C

hence we have ln(7+sinx) +C . ...(as expected)
Q4.
A $sinxcosxdx

set u= sinx
dx=du/cosx

=> $u*sinx*du/sinx

= >u^2 /2 + C
= 0.5sin^2 x + C .

B. similarly

let t= 6x
dt= 6dx
dx= dt/6

=>1/6 $sint .dt
= -1/6 cost + C
= -1/6 cos6x + C .

C) guess the question is like this

x^2 e^(3x)

let's use integration by parts (I.B.P)

put u=x^2 ; du =2xdx

dv=e^(3x) dx ; v=e^(3x) /3

by $udv =u.v -$vdu. , we have

x^2 e^(3x) /3 -2/3 $ x e^(3x) dx..............(*)

again let's compute $xe^(3x)dx from. (*) above

set p= x. ; dp = dx

dq=e^(3x)dx ; q=e^(3x) /3

by $pdq = p.q -$qdp ,

we have xe^(3x) /3 - 1/3 $ e^(3x) dx

=>xe^(3x) /3. -e^(3x) /9 ..........(**)

putting (**) in (*) to get the final ans..

=>x^2 e^(3x) /3 -2/3 [ xe^(3x) /3 -e^(3x) /9 ]. + C

simply. further..
rest solution coming later... m not with pen/paper ...will use partial fraction for 1 &2 .

check Q5. are we to integrate or differentiate .?

here's a lit2 problem that is becoming a problem to the problem solvers ...


try out

1. Find the value of x,y and z if x^2-yz=1,
y^2-xz=4, and z^2-xy=9.



anticipating.

hmmmm see mathematicians a beg....


@sir laplacian & prof. emmyeuler.


I greet thee in CAPITAL letters...

efficiencie:
∫(e^x/x)dx
=∫(1/x+1+x/2!+x^2/3!+x^3/4!...) dx
=lnx+x+x^2/2.2!+x^3/3.3!+x^4/4.4!+...+x^n/
n.n!+...

seems you're on track.

series's cool.

This is non linear , non homogeneous & irrational differential equation of the nth-degree /order (where n >infinity ) which drills out the madness in so many Ingenious "mathemaholics." Myself inclusive yet an unseen & divine solution exist beyond the realms of the terrestrial cosmos ,

Guess what , its call the Almighty equation of LOVE ...irrespective of my infinitesimal faithfulness , his infinitude love , mercy & grace exponents all odds ...I thought I could find love , peace & joy in girls , friends or family , only to realize that the formula is long out- dated .the equation of love can't be solved by analytical means , its complex yet seem simple to handle , ...it rules the heart, controls the mind & possibly determines the destiny of a man .

Oooh lord I need the formula !!! .


" only my imagination" .


Happy val. In advance. Fellas .

#shalom.

makasimatics:
Who can prove the almighty formula using this magic formula?

F'(x)=±sqrt(f'(x)^2 -4f(x))

This formula derives the quadratic formula.

Do you know? @ Richez

derive it let's see asap....

Profmaojo:
Limit n tend to infinity square root of n^2 +n _n =1/2
note _n is not in d square root

snap & post ..not clear

yemstok:
Thanks for helping out the other day bro. I need ur assistance once more. Should I bring iit on?

OK..let's see if I can ...

or others can help out ..

Thorby:
Thanks bro. pls u can still help o.

here

Thorby:
Indices

=2

ok thanks man but not tooo clear oo

still expecting part2...make it clear plz.

Oyasub11:
Part 2

Oyasub11:
How do I go about it,please put me through.

take a photo of the solution ...then save with a name u will easily remember.... now when posting , b4 u click 'summit'. scroll down a bit where you see three spaces . ('choose" now click there then locate the directory you saved the image , then post

that's it bro...try it..

[quote author=Oyasub11 post=30060049]X=56,Y=60,Z=44.The workings is too long don't know how to post it.[/quot

try snap & post abeg

greet thee all

try this plz.


Q1)Use Lagrange multiplier to find the values of x,y,z that minimises the objective function f(x,y,z) =11xy+14yz+15xz, subject to the constrain xyz=147840

Q2) A rectangular box open at the top is to have a volume of 32cm-cube , what must be the dimension of the box so that the total surface is maximum .

Danny4show:
All my effort to solve this two questions out of the assignments given to us was fruitless..pls I need your help sir's and ma's..
1) IN AN EXAMINATION THE AVERAGE GRADE WAS 74 AND THE STANDARD DEVIATION WAS 7. IF 12% OF THE CLASS ARE ALL GIVEN A'S AND THE GRADES ARE CURVED TO FOLLOW A NORMAL DISTRIBUTIONS. WHAT IS THE LOWEST POSSIBLE A'S AND THE HIGHEST POSSIBLE B'S?

2) A LOT CONTAIN 5 ITEMS OF WHICH 4 ARE EFFECTIVE, 3 ITEMS ARE DRAWN AT RANDOM FROM THE LOT ONE AFTER THE ONE WITHOUT REPLACEMENT, FIND THE PROBABILITY THAT ALL THE 3 ARE NOT EFFECTIVE?

GOD BLESS YOU AS YOU HELP A BROTHER....

*****Solution *****

Q2 given 4 effective , it means. 1 not effective

now total ways of selection = 5C3

p(not effective) = 1- p(effective)

=> 1 -[ 4C3/5C3 ] = 1-( 4/10)


hence probability that all three selected are not effective

I.e p(not effective) = 6/10 =0.6 that's it .

hope it helps ..

brb....need to charge my phone.

badmus45:
1. The equation of motion of a forced oscillation is d2y/dt+2dy/dt
+5y=4sin5t solve the equation given that t=0 ,y=0 and dy/dt=0.
(2)determine the equation of the curve which satisfies the equation
xydy/dx=x^2-1 which passes. thru point (1,2)........abeg

*******solutions *******

Q1.

put y ' = D

we have axillary equation as D^2-2D+5=0
solving , we obtain

D=(-2±4i)/2
D=-1±2i
Yp=4sint/(-1+2D+5)
Yp=4sint/(2D+4)
Yp=2sint/(D+2)
Yp=2(D-2)sint/(D^2-2)
Yp=2cost-4sint/(-1-2)
Yp=2(cost-2sint)/-3
Y=(Acos2t+Bsin2t)e^(-t)-2(cost-2sint)/3
At t=0 y=0
0=(Acos0+Bsin0)e^(-0)-2(cos0-2sin0)/3
0=A-2/3
A=2/3
At. Y`=0
B=-1/3
Y=[(2cos2t-sin2t)e^(-t)-2cost+4sint)]/3



...(particular solution)

dejt4u:
u cn easily use crammer's method to solve this if it is for UTME..

nice job...boss.

its me. benbuks ...

#shalom.

here

help out here oo

badmus45:
i was not sure of the answer....sowi

post your maths problems here. www.nairaland.com/1147658/nairaland-mathematics-clinic/158#29703680.

nice1 boss Laplacian ..


greetings sir....

#offs my boxer 4 u..

hay guys...try this cheep 1 .

A box contains a 10-kobo piece ,a 5-kobo piece and a 1-kobo piece. Two coins are drawn , one at a time with replacement .
What is the probability that
i) the two coins are not of same value
ii )they are of same value .?



happy Sunday .

Dekins88:
Good!
You have what it takes to attend those top-tier schools.
Where are you doing your Undergraduate studies now?

here in Nigeria. fed.uni.of tech .

yemstok:
Ure the best man! Gracias bro!

God is the best !! give him praise .

happy solving

..h.

here

badmus45:
lets try these :if y=x/√(a^2 -x^2) show that (a^2-x^2)d2y/dx2=3xdy/dx and pls can i get your mobile num

hmm

yemstok:
Pls guys kindly help out with these
'Derivative of logarithm functions'

1. y=a^cosx

2. y=x^x

3. y=x^xsquare

4. y=e^1+x/1-x

5. y=2^inx

**Solution **

1.)

put t=cosx ; dt/dx =-sinx
=>y=a^t dy/dt =a^t *lna
dy/dx =dt/dx* dy/dt

=>dy/dx =-a^t *sinx * lna

thus , dy/dx =-a^cosx sinx *lna


2.)

taking Napierian logarithm of both sides , we have

lny=xlnx

put F(x,y)=lny-xlnx

f_x=-lnx -1

f_y=1/y

now dy/dx=-f_x /f_y

=>dy/dx =-(-lnx-1)/(1/y)
dy/dx= y(lnx+1)

hence dy/dx =x^x (lnx +1)

3.) similarly

take natural logarithm of both sides

lny = x² lnx

y'/y = 2xlnx + x

y'=dy/dx = y(2xlnx +x)

replace 'y'


4.) y= e^{[ 1+ x/(1-x)]}
put
u =[1 + x/(1-x)] =1/(1-x) ; du/dx =(1-x)^-2

now y' = dy/dx = u'e^u (where u '= du/dx )

dy/dx =(1-x)^-2 * e ^{[1 + x/(1-x) ]}

………(replace u)

5.) y=2^lnx

put u = lnx ; du/dx = 1/x

=> y= 2^u dy/du = 2^u *ln2

now y' = 2^u*ln2 * 1/x

hence dy /dx = 2^lnxln2 /x



that's it man