heat is a random form of energy ......................................

itz

so nw put tan x=t

u get sec^2x dx=dt

thus ur question reduces to

int of (t)^n/3  *  dt

so nw u can integrate it easily

hope it helps............

int areac(int );.................................fucntion prototype

int areas(int);...............................fucntion prototype

#include<iostream.h>

#include<conio.h>

void main()

{

clrsrc ();

int  r,l,area,areaa;

cout<<"entre d radius and d side of d square";

cin>>r>>l;

area=areac(r);.......................function call

areaa=areas(l);........................function call

cout<<" d area of d square is "<<areaa;

cout<<" d area of d circle is<<area;

getch ();

}

int areac (int r)...........................function defination

{ int x,a;

a=pi*r*r;

return a;

}

int areas(intl)...................function defination

{

int y,b;

b=y*y;

return b;

}

hope u get it ..........

i guess itz non integrable

substitutin pi/2e^x = t

thus pi/2e^x dx=dt

in d question multiply and divide by e to d power x

thus u get d integral as ..2/pi(sint/t)dt

sint/t is nt cont at t =0 .....................

thus itz nt integrable

correct me if i m wrng ....

http://www.goiit.com/chapters/tutorial/chemistry/nucleophilic_substitution.htm

hav a luk at this!!

hope it helps .........

koi toh bata do.......kissi ko ni aata kya.........??

consider d resistance betwn a nd b as r

as itz a infinte loop d resistance for all d others after first loop is considered same as r

thus c 2 R's and one r .. are in series thus their equivalent resistance is (2R+r)

this equivalent resistance is in paralell wid R thus u get

1/r=1/R+1/(2R+r)

solvin this u get a quadratic equation and then solve it for d answer ..........

for  d 5th q

d ansewr is 7x^2+4xy-2y^2=0

c if ther's ne combined equation of straigth lines ax^2-2hxy+by^2=0.....which passes thru d origin then d equation of line perpendicular 2 it is given by bx^2+2hxy+ay^2=0

yeh raata mar lo.......wat u hav 2 do is just change d sign of d middle term and change d coefficients of d other two...

note .this is only and only if they pass thru origin

nope.

for an xperssion D is always less then zero

taht's d condition

if there's an xpression (quadratic) then we always consider that itz discriminant value is always less then 0 .........

yeah..d second is rite ,........... i guess that's rite coz c in d first one itz a tertiary carbocation thus d xtend or d positive charge is less coz of inductiev effct weras on d alkyne wid a sp hybridised anion d negatiev charge density is high thus hw can they react ..........

this also happns c wen we use a LiAlH4 wer d charge density on H is high(negative) and we use a tertiary carbocation .in this case instead of formin an alkane it forms an alkene ..remember???

thus then we use TPH ......

correct em if i m wrng

d ansewr is B .....

this is an xpression thus b^2-4ac <0

thus u get ...........

(2*(4k-1))^2 - 4(15k^2-2k-7)<0

thus u get .........k^2-6k+8<0

solvin u get (k-2)*(k-4)<0

thus 2<k<4

thus k=3.............

hope it helps..................

is d answer A................... ??

Tnth term of an AP=A+(n-1)D

thus if d nos r in AP their reciprocals shud b in HP

thus ...................1/a=A+(p-1)D.......................................1

                             1/b=A+(q-1)D..........................................2

                             1/c=A+(r-1)D.....................................3

nw (2-3)equations ,(3-1)equations,(1-2)equationss

thus u get (2-3)equations as

(q-r)*D*bc=c-b

similarly u'll egt 4 d other equations ......then add all of them and a,b,c get cancelled thsu u get d final answer as zeroooooo