Adegite5: Pls friends, i need quick answer to this question,am abit confused. The nitrogen gas obtained by burning an excess phosphorus in air,has a density higher dan d nitrogen frm nitrogen compound bcus d nitrogen frm air is contaminated with ( A) unburnt phosphorus (B) P2O5 vapour (C) some oxygen (D) rare gases (E) carbon dioxide

Richiez: Solve Lim. (SinX)/X
(X->0)

emmyeuler1: dehydrohalogenation of alkyl-halides yield what product?

2nioshine: Most of us r always bn encouraged nd most tyms luv 2 post ..but our engagement often times prevents us...2 al d general nd guruz n d houx..."hat's off" 4 al....sti following

Calculusf(x):
What's the centre of gravity of an equilateral triangle of side 24?...

Richiez: Seems we have more gurus popping into the house...
Sorry guys for being away for a while, let's get back to business. But firstly, I must appreciate the efforts of my generals and other gurus in the house...I'm marvelled at the way you all keep the thread lively. Kudos!

johnpaul1101: gudday everyone, plz i have this question:
(1)Does the sun moves? If yes, How?
(2)can shadow be called an object or the image formed? This is my number08137456323

Mr Calculus: I really need ur help hear guys.
If a+b+c=0,simplify::
[{x^(a^2 b^-1 c^-1)}]*[{x^(a^-1 b^2 c-1)}]*[{x^(a^-1 b^-1 c^2)}]
Given that::a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
NOTE: * means multiply

Mr Calculus: I really need ur help hear guys.
If a+b+c=0,simplify::
[{x^(a^2 b^-1 c^-1)}]*[{x^(a^-1 b^2 c-1)}]*[{x^(a^-1 b^-1 c^2)}]
Given that::a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
NOTE: * means multiply

Calculusf(x):
...i understand this bro but it's not the answer coz you know u=sinx therefore du/dx=cosx...and dx=du/cosx...so the question becomes integral 7^u.du/cosx...you just took the cosx out of the integral function as if it's a constant...so,it can't be that bro...how can one integrate it from there...

Edis4christ:
If its 81^1/(log3)/ 36^1(log6) all in base 2 den it will b
From laws of logarithn, 1/logb base a=loga base b
So we ave 81^log2 base 3/ 36^log2 base 6
3^4log2 base3/ 6^2log2 base 6
But b^ylogx bseb=ylogx baseb=logx^y baseb
3^log2^4 bsae3/6^log2^2 base6
4rm laws of log a^logb base a=b
2^4/2^2=16/4
=4

echibuzor:

I am assuming that the question says 81^(1/log3) / 36^(1/log6)
.
next step is divide the 81 by 36 then subtract the powers...
.
(81/36)^{1/log3 - 1/log6}

(9/4)^{1/log3 - 1/log6}
.
using LCM for the Powers,,, (log6 - log3) /(log3.log6).........(log6 = log(2*3) = log2 + log3)
.
(log2 + log3 - log3) / log3.log6
(Log3 cancel Log3 {LOL}) and the question says the logs are of base 2 therefore log2 = 1
.
That gives the power as 1/log3 (log3 + log2)

= 1/log3 (log3 + 1)

The whole solution now becomes...
(9/4)^{1/log3 (log3 + 1)}

Thats the point I could go no further.. Additions are welcome.

Boladearo: (81^1/log3) / (36^1/log6). The logs are in base 2.

Calculusf(x):
Pls.help me on this integration...integral 7^sinx dx

oladoya: oga talk say make we no use dat word again, u no dey hear

ladokuntlad: pls help me find d value of x if x^2=4x. show workings oooo. hint x=4

Ortarico:

no bro I don't think it's +105, it's -105.
Coined from unilag's post-utme past question.

labodinho: yea,i jst got dah,kan u n @ all check on these....(1)solve the equation _/x+7=x-5 (2).....Let Y=4x+3/2x-5,write x as a function of y. (3).....If dy/dx=6xrtp2 + 15xrtp4 and y=7 when x=2.Find y N.B:rtp=rays to power....tanx

doubleDx: ^
If the question is =>>

∫7^{sin x}dx

Then, I don't think there is any solution to it in terms of standard mathematical functions!

Mr Calculus:
Let u=sinx
-:du/dx=-cosx
-:dx=du/-cosx
integra7^sinx=integra7^u*du/-cosx,,
1/-cosxintegra7^u du recall::integra a^x=a^x/Ina,,,
:1/-cosx*7^sinx/In7+C =-7^sinx/cosxIn7

2nioshine:
few tips*
1 draw a 4 sided figure

2 divide into 4 halves taking its diagonals as PQ=13 and SR=?..respectively(this form set of triangles)

3.Apply pytagoras theorem 2 get pR=5

4.APLY same principle to get sQ=12.76
then solve 4 SR using 12 and12.76 as d syd of d triangle using pytagoras thrm....

Sti confuse?
I just hope dis helps

doubleDx: ^
Yes, whatever works Chief!

Thanks.

y = cos^2t .....(1)
then dy/dt = -2costsint
x = sin^2t ........(2)
then, dx/dt = 2costsint
dy/dx = dy/dt/dx/dt
= -2costsint/2costsint
dy/dx = -1
:. d^2y/dx^2 = 0