₦airaland Forum

jackpot: The number too is 10x+y

the problem is
Maximize F(x,y)=xy
subject to x+y=12
0<=x,y<10.

You may use the Lagrange multiplier method to solve.

carefully solving it yields x=6, y=6.

Thus, the number is 66.

correct....ur workin plz...

...d sum of d digits of a 2-digit number is nine...find d number if it must b a maximum...

....d sum of d digits of a 2-digit number is 12, find d number if d produt of its digits must be a maximum...

Alpha Maximus: .....seriously, are you guyz kidding me(no offence intended before you guyz blast your sirens again!!) My good man Cashio already attempted the question!! So no-one else is gonna try out this simple thing? *slowly azontos away in disappointment*

none

rhydex 247: 1. Show that V=IR^2 is not a vector space over IR with respect to operation (a,b)+(c,d)=(a+c,b+d) and k(a,b)=(ka,kb).

2. Show that the vectors u=(1,2,3), v=(0,1,2) and w=(0,0,1) span R^3.
Hmmm. All is well.

let u, v and w be elements of S, the span of R^3, then there exist x, y and z € R such that
xU+yV+zW=Q, for some Q€S, so
Q=x(1,2,3)+y(0,1,2)+z(0,0,1)=
(x, 2x+y, 3x+2y+x) hence S is contained in R^3. Now suppose
u, v and w are in R^3, then we show that there exist x, y and z for every vector
(p, q, r) for which u, v and w combine linearly...let
xU+yV+zW=(p,q,r) then
x(1,2,3)+y(0,1,2)+z(0,0,1)=(p,q,r) or
(x, 2x+y, 3x+2y+z)=(p,q,r) so that
x=p
2x+y=q
3x+2y+z=r, hence for any choosen vector (p,q,r) we can find x=p,
y=q-2p and z=r+p-2q for which the vectors u, v and w are combined linearly, so R^3 is contained in S...hence
SpanR^3=(u,v,w)....all is well

...pls Maths Generals i'm so curious 2 knw d relation connectn d derivative Ux and inverse derivative Xu of a PDE...help pls!!!!!!

jackpot: lol, I bet you that solving for Ux, Vx and Qx simultaneously from what you posted earlier will take more than a page.

Next, you will still apply your same approach to obtain Uy, Vy and Qy and it will take up to the same number of pages.

Next, you still do same for Uz, Vz and Qz. This will equally take its own page.

At least, 3 pages don go oooooo

And that is just for first partial derivatives ooo.

We neva consider how much pages the second partial derivatives (Uxx, Vxx, Qxx, Uyy, Vyy, Qyy, Uzz, Vzz, Qzz) will take. For each, you will apply quotient rule which takes, at least, 4 lines for each.

You will still obtain chain rules formula for Fxx, Fyy and Fzz.

Then, add and come simplify.(at this point, you will basically do the same thing I did, same fixed step. My own addition and simplification is 3 pages.)

All that in 4 pages! Bro, come on! Small salt dey am

hahahahhaahhahah!!!! abeg no kill me oooo...na 4pgs d bring plenty analysis like dis?...na 12pgs...ahhahhahahahaha

jackpot: hehehe. Score kwa? No be competition oooo
you be my oga na.

Oya, feel free. Temper justice with mercy though

na learn i want learn ooo..no vex..since
Ux#1/Xu, which relation d connect Ux and Xu

jackpot: eeya, your phone doesnt have PDF reader? You can still view it via a PC.

Description?
I obtained Ux, Uxx, Vx, Vxx, Qx, Qxx, Uy, Uyy, Vy, Vyy, Qy, Qyy, Uz, Uzz, Vz, Vzz, Qz, Qzz.

Next, I obtained the expressions for Fxx, Fyy and Fzz which I added and simplified. Later, I equated everything to zero.

there's one more score 2 settle

jackpot: Here comes the full solution in PDF format.


@Readers, pls download it. I need your reviews

pls can u give a description?...my mobil phone cant read dat

jackpot: ^



No, am just curious to know how many sheets/pages it took you, and not really you posting it.

Is it up to 10?

wit my 40leaves notebook, it barely took 4pgs

jackpot: well, the only lil' harmless mistake is the bolded, cos you swapped the definitions of x and y. apart from that, your method is correct so far.

Why not, for fun-sake, complete the solution? I really want to know whether the solution will be shorter

u certainly dont xpect me 2 do all dat computatin here...like i told u b4, d solutn is right here wit me...just follow d recipe...u can as well describ ur own method if u dont mind...

Cashio: ..bro,thumbs up....pls once again let us all stop this unecessary enemity here..i learn enough from this thread at least if no one else does..............can someone help me find the 10th term of the expression..(1+x)^15

15Combination9

jackpot: @Laplacian

am happy. Post the solution, please.

first i need to point out 2 u dat in PDE one can take d differental of an eqn b4 proceedin 2 d limt witout fear of error...it is rigorous & not an assumption...wit our former notation; now x=uvsinq, y=uvcosq, z=0.5(u^2-v^2), like b4,
Fx=Fu*Ux+Fv*Vx+Fq*Qx...to get Ux, Vx, and Qx we take d differential of d transformation equations; first we take Log of x=uvsinq,
Logx=logu+logv+logsinq, let d denot delta, so dat
1/x*dx=1/u*du+1/v*dv+1/sinq*cosq*dq, divide thru by dx and proceed 2 d limt;
1/x=1/u*Ux+1/v*Vx+cotq*Qx, similarly for y=uvcosq,
logy=logu+logv+logcosq,
1/y*dy=1/u*du+1/v*dv-1/cosq*sinq*dq, divide thru by dx but this time partial derivativ of y w.r.t.x is zero since x and y are independent variables;
0=1/u*Ux+1/v*Vx-tanq*Qx...again for z=0.5(u^2-v^2),
dz=udu-vdv, dividin by dx;
0=uUx-vVx...from d 3 eqns above solve 4 Ux, Vx and Qx...in d eqns above, divide, respectively, by dy and dz to obtain Uy, Vy, Qy and Uz, Vz, Qz...substitute and differentiat Fx,Fy,Fz respectively...u 'll obtain ur answer as i 've done here after simplification...

jackpot: @d citizen

Thank God that you fully recovered. While you were away, Sire, we your boys and girl couldn't resolve this. So we said make we wait for our only 5 STAR Math General to helep us scatter am. Expecting your answer, Sire!

...i hav d solution 2 ur problm...

...to integrat sec@=1/cos@, multiply numerato & denominato by cos@ to get cos@/cos^2@ or
cos@/(1-sin^2@), let p=sin@,
dp/d@=cos@, so that §sec@d@=§dp/(1-p^2)=1/2*Log[(1+p)/(1-p)] ...after resolvin 2 partial fractio n integratn...
§sec@d@=1/2*Log[(1+sin@)/(1-sin@)]

Osoayobami: Pls help me on this integral question..square root of (x^2-36).using trig substitution pls.i can do it using hyperbolic substitution of cosh@.i really need your help pls.with including my maths recruit o.you know at times recruits are talented with skills.lol.i love that ben..

y=§sqr(x^2-36)dx, let x=6sec@, dx/d@=6sec@tan@, makin substitutn, y=§sqr(36sec^2@-36)*6sec@tan@d@, it simplifies to, y=§sqr(36tan^2@)*6sec@tan@d@ or y=§6tan@*6sec@tan@d@ or
y=§36sec@tan^2@d@ or
y=§36sec@*(sec^2@-1)d@ or
y=§36sec^3@d@-§36sec@d@ we now integrat d first integral by part...
Y=36sec@*tan@-§36tan^2@sec@d@-§36sec@d@
...or
y=36sec@tan@-y-§36sec@d@
2y=36sec@tan@-§36sec@d@...u only have to integrat sec@ and obtain ur answer...

Alpha Maximus: is it the product of any two plus one or each of the two numbers plus one and their product?

the product of any two plus one

...three numbers are in arithemetic progression, the product of any two when increased by one is a perfect square...find d numbers...

...we need understandin 2 make progress...i suppose we 're all adults here...we should b a family...what i xpect 4rm maturd minds is cooperation...let's all learn 2 accept our mistaks, cos dats d surest way 2 upliftment....

...Hello Jackpot, i am more than glad 2 recieve dis correction from u, sometin i'd neva thought of...i should infer dat u're by no means an amature. 4 if u were, u wouldn't 've debugd dat intricate error...now, usin d same notation,
Fu=Fx*Xu+Fy*Yu+Fz*Zu...u can make direct differentiation now...Fuu, Fv, Fvv, Fq and Fqq may then b obtaind similarly...u'll have three equations, i.e Fuu, Fvv and Fqq, in three variables; Fxx, Fyy and Fzz which u can solve (preferably by Matrix) and make substitution into Laplace equation and simplify...i know u posses a shorter method...if u do, relate it to us....tanx

jackpot: dear Maths generals, pls help me with this little problem.


[size=14pt]By using the Cartesian-paraboloidal coordinate transformation equations:
(x=uv sin φ, y=uv cos φ, z=¹/₂(u²-v²),
show that the 3-dimensional Laplace's equation
F_xx+F_yy+F_zz=0
reduces to
¹/_(u²+v²) (F_uu+F_vv+ ¹/_u F_u+¹/_v F_v)+¹/_(u²v²)F_φφ = 0.[/size]

i use Fx to denot derivatv of F w.r.t.x....let q denot phi now
Fx=Fu*Ux+Fv*Vx+Fq*Qx, now
Ux=1/Xu=1/(v*sinq), similarly, Vx=1/(u*sinq), Qx=1/(uv*cosq)....substitut into the eqn for Fx and differentiat partially w.r.t.x again to get Fxx....do the same for Fyy and Fqq, then put the result in the Laplace eqn, simplify and get the result...the result is too weird to b includd here...

jackpot: dear Maths generals, pls help me with this little problem.


[size=14pt]By using the Cartesian-paraboloidal coordinate transformation equations:
(x=uv sin φ, y=uv cos φ, z=¹/₂(u²-v²),
show that the 3-dimensional Laplace's equation
F_xx+F_yy+F_zz=0
reduces to
¹/_(u²+v²) (F_uu+F_vv+ ¹/_u F_u+¹/_v F_v)+¹/_(u²v²)F_φφ = 0.[/size]

i use Fx to denot derivatv of F w.r.t.x....let q denot phi now
Fx=Fu*Ux+Fv*Vx+Fq*Qx, now
Ux=1/Xu=1/(v*sinq), similarly, Vx=1/(u*sinq), Qx=1/(uv*cosq)....substitut into the eqn for Fx and differentiat partially w.r.t.x again to get Fxx....do the same for Fyy and Fqq, then put the result in the Laplace eqn, simplify and get the result...the result is too weird to b includd here...

benbuks: If arccos@ +arccos & + arccos¥ =pi, show that @^2 +&^2 +¥^2 +2@&¥=1...

arccos@+arccos&=pi-arccos¥...now take d cos of both sides....
Cos(arccos@+arccos&=
cos(pi-arccos¥)...or
@&-sin(arccos@)*sin(arccos&=
-¥...or sin(arccos@)*sin(arccos&=@&+¥...square bothsides;
sin^2(arccos@)*sin^2(arccos&=
(@&+¥)^2...or [1-cos^2(arccos@)]*[1-cos^2(arccos&]=(@&+¥)^2...or
(1-@^2)*(1-&^2)=(@&+¥)^2....or
1-&^2-@^2+(@&^2=(@&^2+2@&¥+¥^2...rearrangmnt gives...@^2+&^2+¥^2+2@&¥=1....proved

benbuks: 1) arcsin(a/x) + arcsin(b/x)=pi/2
2) arctan(x+1)=3artan(x-1)...solve...

1.) arcsin(a/x)=pi/2-arcsin(b/x)...take sine of both sides; a/x=cos[arcsin(b/x)] or arccos(a/x)=arcsin(b/x)=p, say, then a/x=cosp and b/x=sinp, sqr both eqn and add; (b/x)^2+(a/x)^2=1 or
x^2=a^2+b^2
2.) arctan(x+1)=3arctan(x-1)=3p, say...then x+1=tan3p and x-1=tanp....now
(1-tanp*tan2p)*tan3p=tanp+tan2p....but (1-tan^2(p))*tan2p=2tanp...so that (1-(x-1)^2)*tan2p=2(x-1) or
x(x-2)tan2p=2(x-1), substitutin for tan2p in d first eqn, we get..
(1-(x-1)*2(x-1)/x(x-2))(x+1)=(x-1)+2(x-1)/x(x-2)...multiply thru by x(x-2) to get....
x(x-2)(x+1)-2(x-1)^2*(x+1)=
x(x-2)(x-1)+2(x-1)...or...(x^3-x^2-2x)-2(x^3-x^2-x+1)=x^3-3x^2+4x-2...or...
2x^3-4x^2+4x=0...or...x=0 or
x^2-2x+2=0

muyhee: @laplacian wel dats u.odas wil undastnd mind u am nt arguing.wat is cos90,sin90 nd d rest.they are d values of d letas.

pls i dnt mean 2 b rude...hop no offenc...just wish dat anyone who understnds shud throw more light...

muyhee: d provin.theorem:a numba raisd to an exponent(square)is assumd to b at an acute angle 0<theta<=90.4rm binomial expansion of non-+ integer.we av dat.(d alphabet ar insertd to fit)
(1+x)^n=1.a+nx.b+(n(n-1)x^2.c)/2!+(n(n-1)(n-2)x^3.d)/3!+...
Let x=90(since n is at an angle to x).5 cn stil b like 5+0.0001.(0.0001 is error assumption)
Usin d maclaurin series diffrntn.
Sin90=a. There4 a=1(al odas wil b canceld out if u carfuly diffrnt)
Cos90=bn( x is gone).there4 b=0
-sin90=cn(n-1).ther4 c=-1
Note dat.a=knx^n-1 as a=x^n
Lets derive.diffrnt x^n
X^n=knx^n-1.nw lets lets input d letas values.
X^n=a+bnx+(cn(n-1)x^2)2!+...
X^n=a-(n(n-1)x^2)2!.recal dat a=knx^n-1.so,cumulatin altogeda
X^n=(knx^n-1)-(kn(n-1)x^2)/2!+...this is it guys.chek out nd cment.

i wish i unaderstnd a line of ur argument

@muyhee...show us d proof of ur formula

[quote author=muyhee][/quote]in my opinion, d formula is completely unuseful for evaluatin squares since d R.H.S contains d same square u 're evalutin

rhydex 247: Breakfast questn
1. Solve. (x^2-16)(x-3)^2+9x^2=0.

@calculus f(x)...u have very concise methods dat i admire, i'll need u 2 attempt d questn...

Mbahchiboy: i will b glad if u do d no 1 hear caus am not so gud in maths and ur soln to no 2 is somehow

even me am not very gud in maths, here's my attempt
From Maclaurin's series;
f(x)=f+xf'+x^2f"/2!+x^3f"'/3!+...
Where all derivatives are @ x=0,
now f(x)=tanhx=sinhx/coshx so f(0)=0, f'(x)=(cosechx)^-2 so that f'(0)=1,
f"(x)=-2tanhx*(cosechx)^-2 so that f"(0)=0
f"'(x)=-2tanhx*f"(x)-2[f'(x)]^2...so that f"'(0)=-2 and so on, substitutn now gives;
tanhx=x-2x^3/3!+...
Pls ignore my previous solution for ur first questn....the second solution remains d same and i left d answer in polar form...

d citizen:
The five star general,

Here is the clue to the analytical solution to the problem: 3^x + 4^x = 5^x

Divide both side of the equation by 4^x

3^x/4^x + 4^x/4^x = 5^x/4^x

0.75^x + 1 = 1.25^x

(1-0.25)^x + 1 = (1 + 0.25)^x

(1 + (-0.25))^x + 1 = (1 +0.25)^x
U will use binomial theorem to do the expansion and limit the expansion to x^2

(1 +x)^n = 1+ nx + n(n-1)x^2/2!. Use the expansion to get the answer to x by quadratic equation.

why limit it to second order?...why not third or fourth order because in those cases u have differnt results