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thnx dear

#2 Help Me ! » vectors » 2014-02-27 17:05:19

rzaidan
Replies: 2

find a unit vector parallel to a plane determined by the two vectors     3i -2j +k   ,     i + j - 2 k   and perpendicular to the vector   2i+2j-k
plz  help as soon as possible

#3 Re: Help Me ! » Surface integral over a plane » 2010-07-24 08:49:34

1)find two vectors(a,0,-a) , (-a,2a,0)
2)find their cross product which is the normal n=(2a² , a² ,2a²)  to the plane.
3) compute the surface integral=∫∫v.n ds=∫∫(yk).n (dxdy/|n.k|)
=∫∫v.n ds=∫∫2a²(dxdy/2a²=∫∫dxdy=area of triangle in the xy-plane with vertices(0,0) , ( a, 0), (0,2a)=(½) (a)(2a)=a²
I hope I am right
Best Wishes

#4 Re: Help Me ! » Integration- Trigonometric functions » 2010-07-08 23:54:56

hi1  a2b3c2212
bobbym Moderator is right

#5 Re: Help Me ! » integral problem » 2010-05-18 17:41:32

Dear soroban
THanks a lot, really I was having the idea, but I forgot it because of my getting old. Anyhow thanks again.
Best Regards

mit open courses

#7 Help Me ! » integral problem » 2010-05-18 02:46:13

rzaidan
Replies: 2

ps if we can colaborate to find the integral
how to integrate dx /(sin x +cos x + 1 )

#8 Re: Maths Teaching Resources » probability question » 2010-05-17 19:52:16

Dear rosalie
1) P(Arrive only)= P(A-D)=P(A)_P(A and D)
=.82-.78=.04
2)P(D/A)=P(D and A)/P(A)=.78/.82=0.95
I hope I am right
Best Wishes

#9 Re: Ganesh's Puzzles » Logarithms » 2010-04-18 21:14:13

Hi ganesh
for  L # 2
I think that  the following proof  is easier:
Assume Log(x)/(b-c)=Log(y)/(c-a)=Log(z)/(a-b)=t
So Log(x)=t(b-c),Log(y)=t(c-a)  ,  Log(z)=t(a-b) So Log(x)+Log(y)+Log(z)=tb-tc+tc-ta+ta-tb=0
So Log(xyz)=0 so   xyz=1   Q.E.D
Best Regards

#10 Re: Ganesh's Puzzles » Logarithms » 2010-04-18 21:04:41

Hi ganesh
for  L # 1
since log(a)= 1 / log(b),    log(a)=1
b               a            a
we have
1/log(abc)+1/log(abc)+1/log(abc)=
a                 b                c
log(a)+log(b)+log(c)= log(abc)=1
abc      abc          abc    abc
Best Regards

#11 Re: Help Me ! » partial fraction » 2010-04-13 07:56:26

Dear Onyx
As you wrote you will have as  A =0  and  B=1
salam

#12 Re: Help Me ! » partial fraction » 2010-04-13 07:54:43

Dear ZHero
I think that it is not easy (or say impossible) to write this fraction into partial fractions as you did,since the fraction which contains in its numerator Cx is not allowed because the denominator contains a linear factor repeated twice , and the appropriate fraction must contain a constant in its numerator, and the purpose of this stratigy is to find the integral of this fraction, and it is easy to find the integral of this fraction by means of substitution u = (aX+b) and continue
Best Wishes

#13 Re: Help Me ! » Vec Vec Vector!! » 2010-03-27 17:38:19

Hi GOKILL ;
you can take the dot product
(c1X1+c2X2+...c1X1).(c1X1+c2X2+...c1X1)=0  by using the associative and commutative properties,  and since each
Xi.Xj =0 for different i and j the product contains
the square of the constant times the square of the length of the corresponding vector , but this sum is zero , therefore each term is zero so each coefficient is zero so each ci is zero
Best Wishes

#14 Re: Help Me ! » integral of e^x » 2010-01-25 02:10:50

Hi henrybrice
Novice
You can use integration by substitution u=3x   and continue

#15 Re: Help Me ! » surface integral with stokes » 2010-01-16 08:15:26

Hi Identity :
you are to find curl F which is the crosss product of del × F to get
del × F = 3x i -(3y-x) j +2y k
then find the normal to the surface to get n = (1/√11)(3i+j+k)
then find ds = dydx/(n.k)=√11 dx dy
afterthat apply stokes  theorem:
∫F.dr = ∫ ∫del  × F .n ds =1/√11 ∫ ∫ 10 x -y)dydx   on R where R is the tringle x + 2y = 3 that is the integral limits are : x from 0 to 3-2y and y from 0 to 3/2
and complete
I think it will be easy
Best Wishes

#16 Re: Help Me ! » conditional statment » 2009-12-13 06:16:59

Hi dheeraj;
this is a conditional statment which is true except when the condition is true and the result is false(i.e the statment p⇒q is always true except when p is true and q is false ) , but this case does not hold since when Эx(P(x)^Q(x)) is true means that  Э(x)P(x) ^Э(x)Q(x)  is true too . Therefore Эx(P(x)^Q(x)) →Э(x)P(x) ^Э(x)Q(x) is a tautology .is a tautology .
Best Regards

#17 Re: Help Me ! » Ring Integrals » 2009-12-06 20:55:56

Hi JamesStirlingJr.:
This integral  is not called a ring integral , it is a line integral
(in vector analysis) when taken over a closed region ((as a circle, an ellipse,...) and it has some determined value.
With Best Wishes

#18 Re: Help Me ! » Fibonacci's Sequence » 2009-11-16 18:41:16

Hi simplyjasper,
You can prove that the sequence is less than 2 by mathematical induction together with finding the limit.

#19 Re: Help Me ! » Integral » 2009-11-06 05:08:40

Hi Identity
In order to find this integral we do as the following:
∫1/(x ²+a ²) ^(3/2)   dx =  ∫(x ²+a ²)^-(3/2) dx
= ∫((x²(1+a ²* x^(-2))^(-(3/2)) dx by taking x ² az a common factor inside
=∫(x²)^-(3/2)(1+a ²* x^(-2))^-(3/2)  dx
=∫x^(-3)(1+a ²* x^(-2))^-(3/2)  dx then use substitution   u = 1+a ²* x^(-2) to complete
Best Wishes

#20 Re: Guestbook » 196 » 2009-10-12 07:44:39

Hi hy
on this site we discuss math problems,exams, and any issue concerning mathematics. We are very pleased to join you and other freinds here, so you are welcomed here.
Best Regards

Thanks for all

#22 Re: Help Me ! » Modulus inequalities » 2009-09-24 08:20:04

Hi bobbym
When taking the square root for x² you must take the absolute value of x and so you will have two inequalities, and youmust solve these two inequalities
-x+ 1+ 2x  1 > 0 when x<0 ⇒ x>0 which is empty sunce x<0 and
x+ 1+ 2x  1 > 0 when x>0⇒3x>0 ⇒x>0 therefore the solution of this inequality is {x∈R: x>0} mainly the positive reals.
Best Regards

#23 Re: Help Me ! » Modulus inequalities » 2009-09-23 07:46:34

I think that you must redefine these inequalities without using the absolute value notation , after that you deal with  usual inequaliteis  containing functions of more than one rule.
Best Wishes

#24 Re: Help Me ! » solve me this graph of Function plzzz » 2009-09-14 07:33:34

Hi dear Avva
اريد التعرف على حضرتكم ان امكن؟!!!!!!!!!!
مع تمناتي الحارة بالتوفيق
رياض زيدان

Dear Identity
Thanks