bolkay47:
no 1::: 3ln(2x+1)+2ln(x-2)+K

here

tobillionaire:
Given that CosZ =L, where Z is an acute angle, find an
expression for (CotZ - CosecZ )/(SecZ +TanZ )
2.Find there numbers in G.P whose sum is 28 and whose
product is 512.
3. Find the derivative of Y with respect to X If
Y=√{(1+ x )/(1- x )
Where √=square root

here .

Q 2 use. x , y ,z as the g.p

which is tantamount to

a , ar , ar²


sum. : a+ ar +ar²=28 .........(i)
product (ar)³ =512. .............(ii)

from (ii) , ar =8.

=>a =8/r put in (I)

we have 8/r +8r=20 or 8r² -20r +8=0

solving we obtain r=2 or 0.5

we pick r=2
=> a=8/2 =4

hence we have 4 , 8 , 16

that's it , you can verify .

agentofchange1:


here .

Q 2 use. x , y ,z as the g.p

which is tantamount to

a , ar , ar²


sum. : a+ ar +ar²=28 .........(i)
product (ar)³ =512. .............(ii)

from (ii) , ar =8.

=>a =8/r put in (I)

we have 8/r +8r=20 or 8r² -20r +8=0

solving we obtain r=2 or 0.5

we pick r=2
=> a=8/2 =4

hence we have 4 , 8 , 16

that's it , you can verify .

pls hw do i download NL pix

agentofchange1:


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 .?

e^3x/3{x^2-2x/3+2/3}+K

Hi buddies pls i need your help on this question.Find the value of £ for which the following equation possess a solution.
X-3y+2z=4
2x+y-z=1
3x-2y+z=£
Using Gauss method.Tanx in Anticipation.

tobillionaire:
pls hw do i download NL pix

use a PC

Onase:
Hi buddies pls i need your help on this question.Find the value of £ for which the following equation possess a solution.
X-3y+2z=4
2x+y-z=1
3x-2y+z=£
Using Gauss method.Tanx in Anticipation.

£=5

express the equations in matrix form then row-reduce to echelon standard , we then deduce that £-5=0
=>£= 5 for the set of the linear equation to posses a solution I.e (x,y,z)=(1 ,-1 ,0)

that's it


hope u get .?

agentofchange1:


Q5
b

ans = xarctan[(1+x)/(1-x)] - 1/2 ln | 1+x² | + C

Good morning bro! How was your night? Pls, mail me your contact, let's discuss business.

yemstok:


Good morning bro! How was your night? Pls, mail me your contact, let's discuss business.

hmm OK ....done. check thy mail. sir.

tobillionaire:
pls hw do i download NL pix

mail me via ..

benbuks10@gmail.com ...(exclusively )

can't view the p.m you sent earlier.

..

y=§cos²@xsin²@d@

=§(cos@xsin@)²d@

=§[(1/2)xsin2@]²d@

=§1/4xsin²2@d@

=§1/8x2sin²2@d@

=§1/8x(1-cos4@)d@

=1/8x(@-sin4@/4)+C

NOTE; 1-2sin²x=cos2x

Who knows the difference between "y = f(x)" and "y = F(x)" ?

..

What are the next three numbers in this series
4,6,12,18,30,42,60,72,102,108,?,?,?

Determine the two values of c for which the line 3x+4y+c=0 is a tangent to the circle x^2+y^2-6x-2y-15=0.

The centre of the circle is Q=(3,1). If the given point is actually a tangent, then the normal thru the point @ which it is tangent, i.e the normal passes thru Q=(3,1). Now, since the slope of the tangent is -3/4, the slope of the normal must be 4/3, hence, the equation of the normal is; (y-1)/(x-3)=4/3 or y+3=4x/3, substitute this for y in the equation of the circle,
x^2+(-3+4x/3)^2-6x-2(-3+4x/3)-15=0
or
25x^2/9-38x/3=0
i.e x=0 or 114/25, correspondingly, y=-3 or 77/25,
the two points where the normal cut the circle (or the two points where the tangents touch the circle) are; A=(0,-3) and B=(114/25,77/25), substitute each cordinate in the equation of the straight line to get c.
I.e c=12 and c=-26.
(cross check, i have no calculator)

Prove that the line 3x +4y=13 is a tangent to the circle x^2+y^2-2x-3=0 and find the equation of the two tangents perpendicular to this.

If the line is a tangent, then, eliminating y should give two real but equal values of x. Hence, 16x^2+(4y)^2-32x-48=0 or
16x^2+(13-3x)^2-32x-48=0 or
25x^2-110x+121=0
if b^2=4ac, then we are done (pls verify that).
To find the two tangents perpendicular to the first, we take a line parallel to the first and passes thru the center. The center Q=(1,0) the line is 3x+4y=c but iit passes thru the center, so c=3, hence 3x+4y=3, we now find where this line cut the circle, 16x^2+(3-3x)^2-32x-48=0 or
25x^2-50x-39=0,
get the two points x1 and x2, from 3x+4y=3 get y1 and y2, so the points of intersection are A=(x1,y1) and B=(x2,y2). The slope of the perpendicular must be 4/3 so its equation must be 3y-4x=k, subtitute the cordinate of A into the line to get k, do the same B. And that gives the equation of the two perpendiculars.

agentofchange1:
What are the next three numbers in this series
4,6,12,18,30,42,60,72,102,108,?,?,?

(the bolded)
i don't have the table of primes but i 'll give u the general rule:
1.) write the pairs of twin primes in ascending order
2.) write the greater of each pair in ascending order
3.) subtract 1 (one) from each term of sequence (2.) above.
Observation; it turns quite amazing that each term (aside the first) turns out to be divisible by 3. I'll leave it to the author of this post to offer a proof if he can!!

AlphaMaximus:
Good morning, Math Clinic! Kudos to all that have been thoroughly active on this thread, its not easy. I seem to be in a fix as regards Complex Analysis: Non-Linear Transformations of the form w=(az+b)/(cz+d), which can be found in page 927 of Advanced Engineering Math By K.A Stroud......its specifically involves mapping a circle of modulus, |z|=2 onto the w- plane by a particular transformation equation of w= (z+2j)/(z+j). And the question demands that i determine the centre and radius of the resulting circle in the w-plane which normally is easy but for an impedement i encountered which i shall expatiate on.

w= (z+2j)/(z+j), w(z+j)=z+2j or

z=(2j+wj)/(w-1)=[(u+2)j-v]/[(u-1)+vj] or
taking conjugates;
z=[(u+2)j-v]*[(u-1)-vj]/[(u-1)^2+v^2]

={[(u-1)(u+2)+v^2]j+[-uv+v(u+2)]}/[(u-1)^2+v^2]
so
x=[-uv+v(u+2)]/[(u-1)^2+v^2]
and
y=[(u-1)(u+2)+v^2]/[(u-1)^2+v^2]

if u make this substitution u 'll get the centre

agentofchange1:


£=5

express the equations in matrix form then row-reduce to echelon standard , we then deduce that £-5=0
=>£= 5 for the set of the linear equation to posses a solution I.e (x,y,z)=(1 ,-1 ,0)

that's it


hope u get .?

Tanx bro.I will check and go thru.your solution.Tanx.

Laplacian:

w= (z+2j)/(z+j), w(z+j)=z+2j or

z=(2j+wj)/(w-1)=[(u+2)j-v]/[(u-1)+vj] or
taking conjugates;
z=[(u+2)j-v]*[(u-1)-vj]/[(u-1)^2+v^2]

={[(u-1)(u+2)+v^2]j+[-uv+v(u+2)]}/[(u-1)^2+v^2]
so
x=[-uv+v(u+2)]/[(u-1)^2+v^2]
and
y=[(u-1)(u+2)+v^2]/[(u-1)^2+v^2]

if u make this substitution u 'll get the centre

Thanks a lot!

Onase:
Tanx bro.I will check and go thru.your solution.Tanx.

OK BRo...UWc....

Laplacian:

(the bolded)
i don't have the table of primes but i 'll give u the general rule:
1.) write the pairs of twin primes in ascending order
2.) write the greater of each pair in ascending order
3.) subtract 1 (one) from each term of sequence (2.) above.
Observation; it turns quite amazing that each term (aside the first) turns out to be divisible by 3. I'll leave it to the author of this post to offer a proof if he can!!

OK sir...

just skip primes . that's d pattern.

tobillionaire:
pls hw do i download NL pix

If you are using OPERA MINI on your mobile phone, place your cursor to the image and press th no "1" key.
Click on save image, give it a name and an extension either .jpg or .png

If .jpg does not makes the fonts on the picture clearer, use .png. Or better still, ask the sender for the format.



Hope it helps!

Welldone, gurus in the house! I envy you guys-problem solvers.

Hello everyone, kudos and nice job.

agentofchange1:
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 .

that is the method under application to calculus(maximum and minimum problem))

Laplacian:

w= (z+2j)/(z+j), w(z+j)=z+2j or

z=(2j+wj)/(w-1)=[(u+2)j-v]/[(u-1)+vj] or
taking conjugates;
z=[(u+2)j-v]*[(u-1)-vj]/[(u-1)^2+v^2]

={[(u-1)(u+2)+v^2]j+[-uv+v(u+2)]}/[(u-1)^2+v^2]
so
x=[-uv+v(u+2)]/[(u-1)^2+v^2]
and
y=[(u-1)(u+2)+v^2]/[(u-1)^2+v^2]

if u make this substitution u 'll get the centre

this is a course on complex analysis ( use cauchy theorem)

find the next three term of the sequence,,,
1,5,9,31 53 ,......

efficiencie:


This looks clean enough!

Given
arcsina+arcsinb+arcsinc=π

Let
arcsina=A
arcsinb=B
arcsinc=C
hence:
a=sinA
b=sinB
c=sinC
Starting from the LHS of the required equation:
a√(1-a^2)+b√(1-b^2)+c√(1-c^2)
=sinA.√(1-sinA^2)+sinB√(1-B^2)+sinC√(1-sinC^2)
=sinA.cosA+sinB.cosB+sinC.cosC
=(1/2)sin2A+(1/2)sin2B+
(1/2)sin2C
=(1/2)(sin2A+sin2B+sin2C)
=(1/2)(2sin(A+B)cos(A-B) +sin2C)
but A+B+C=π
=(1/2)(2sin(π-C).cos(A-B)
+sin2C)
=(1/2)(2sin(C)cos(A-B)+sin2C)
=(1/2)(2sin(C)cos(A-B)+2sinC.cosC)
=sinC.(cos(A-B)+cosC)
=sinC.(cos(A-B)+cos(π-A-B))
=sinC.(2cos((π/2)-B)
cos(A-(π/2)))
=sinC.(2cos((π/2)-B)
cos((π/2)-A))
=sinC.(2sinB.sinA)
=2sinA.sinB.sinC
=2abc
which is the RHS of the equation.

^QED!

Thanks bro. More power to your elbow.