Messages posted by akhil

pls send me papers of aieee

I've played the game so many times but it's the first time that girls seem to be winning.

anyway, 70-2=68.
Come on Boyz!!

tan(a+b+c)=tan a +tan b +tanc-tanatanbtanc/1-tanatanb - tanbtanc-tanctana.
tan(a+b+c)=0/1.
tana + tan b + tanc comes out as 4.
tanatanb+tanbtanc+tanctana comes 0.
now consider (tana+tanb+tanc)^2.
We get the answer as 16.

I think the question is quite absurd.
Wat do we have to do?

I ask u find x. Is it a proper ques?

u need to tell in what terms u want the answer.
Praveen if u answer my ques as x belongs to R it's not a valid answer.

integrate it using partial fractions.

writting 5^1/2 as c.

The ans is (ln2)/3 +c*ln((7-3c)/2)/6

=(1/(r/n)^3 +1) / n

let r/n=x
f(x)=1/(x^3+1)

So the given ques becomes:

integration from 0 to 1 f(x).

s/t is not v but is v/2.
This is how it is:

s=1/2at^2
at=v;
s=1/2vt.
s/t=1/2v.

now why is it so:

This is because it is uniform acceleration and not uniform velocity.
if uniform velocity then only s/t=v else if uniform acc. then s/t=(v+u)/2. now u=0 so s/t=v/2.

Got it!!

By fundoonick's answer it seems the question was:

(1+x)log(1+x)/x^2 - 1/x

write it as:
[(1+x)log(1+x)-x]/x^2.
Now apply L'hopitale rule twice to get the answer 1/2.

an observer revolving in a horizontal circle of radius R with uniform angular velocity w throws a ball with velocity v at an angle

with the horizontal. assuming ball remains in air, magnitude of avg acc of ball over duration of one complete revolution is ?

FOR ALL SUCH PROBLEMS ANSWER IS ^n-1C_r-1

THUS ANS IS ²⁹C₂^.

^{_{BUT BOXES R IDENTICAL SO ANS IS (29C2)/3}}

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