Messages posted by Nugo Ka Chhota Bhai

I was bloody stupid moderators delete this thread please,....................................

How commit a painless suicide?I dont have lot of money and access to dangerous things.Please reply with details of method .No dnt do it sh**

Well da product given to dea first person isnt a prime no den he can gues da nos

now da sum given to da 2nd person cant be xpreesed as sum of prime no+1 since den da 2nd person mit not be able to noe for sure dat da first person is unaware of da nos

now each prime xpect 2&3 are xpressed as 6m-1 & 6m+1 ,so da sum doesnt equal 3,5or ne form 6m or 6m+2(ie prime+1)

letz da first person has been handed a certain no havin four factors ne1 pair of vich can be ritten as 3,4or 6m+1 6m+2 so da oder pair must be da nos........dat only one such pair shud xist amongst da all pair of factors...

if no given to p is 8....

den 8 either=4*2 or 8*1

now if nos are 4&2 den sum is 6 but six is ritten as 5+1 hence da nos must 8 &1.....

in dis way

Consider,

eliminating all numbers which have atleast two pairs of factors which cannot be written as 3,5,6m,6m+2

numbers between 1-40

so dere is more dan 1 pair of such nos..

lik 1,8 nd 1,15

Notice that you cant tear off any page without tearing off da page on da oder side as well so we tear off an odd numbered page and da page on da oder side gets torn(+1) even-numbered page(-1).....

let da page be odd---2n-1,2n

even-2n,2n-1

thus sum=4n-1

by hit nd trial 4n-1=75..thus nos are 37 nd 38

First to vote amongst many!!! i ll take ur advice bhaiya

using A.P. summation formula---first term of Ap's as a and a1 and common difference as d and d1

n/2(2a+(n-1)d)/n/2(2a1+(n-1)d1)=7n+1/4n+27

=>2a-d+dn/2a1-d1+d1n=ratio

d=7 d1=4

a=4

a1=15.5

ratio = a+10d/a1+10d1=4+70/15.5+40

74/55.5=148/111

x3-7x+6

rit as x3-8 -7x+14 (x-2)...common

x3+x2-24x+36

rit as x3-8+x2-24x+44 agn (x-2)...common

x3-23x2+142x-120

x3-1-23x2+142x-119

take (x-1) as common

remainder theorem f(x) is divisible by anoder exp (x-a) if f(a)=0

ATTENTION ALL GOIITIANS*SALUTE TO EVERY REPLYMOTIVATE ME TO STUDY REAL HARD TELL ME THINGS THAT WUD MAKE STUDY OUT OFMY SKIN

Designate the friends as a,b,c,d,e,f.

ATTENDEES(The man attends all lol....)

dinner no, -1)abcdef

2)abef

3)abcde

4)abcf

5) abcd

6)be

7)cdf

8)ce

9)adf

10)cfe

11)bd

12)af

13)de

14) I am falling of the sky and I am all alone the courage that's inside is gonna break my fall.

one of many possible combos notice that ab occurs 5 times(only "pair" to do so.......) and likewise for groups of three(abc) four(abcd) and five(abcde), each person occurs seven times.

i want to rate dis thing a 10^10000000 times

A wedge (right angled triangle) has a body of mass m on the midpoint of it's hypotenuse ,the hypotenuse makes an angle theta with da horizontal wat force shud be applied on da wedge so that the body just falls freely.Salute assured

Here goes no.3

x²y²+x²+y²=9(according to second eq)

(x+y)²-xy(2-xy)=9

again replace (x+y) by 3/(xy-1)

thus

9/(xy-1)² =9+xy(2-xy)

9(1/(xy-1)² -1)=xy(2-xy)

9(xy)(xy-2)/(xy-1)²=xy(2-xy).

from this we infer that either xy=2 or (xy-1)²=-9 which is impossible .

x^2+y^2+2xy+9-x^2y^2(=x^2+y^2)-x^2y^2+2xy=k(reqd value )+9-x^2y^2

we have 2(x+y)^2=K+9

=>2(3/xy-1)^2=K+9(xy=2)

=>k=9

feelin sleepy..mum mum

Ship A(Va)

---------------------------<--------------------|

|

v Ship B(-Vb)

` |

|

Consider two cars travelling along a straight line with velocities Va and Vb respectively thus relative velocity of A with respect to B is Va-Vb

the same concept applies here.Consider yourself as being on ship B then ship A would appear to head in the direction Va-Vb ie south-west direction obtained by adding vector A+ (-Vector B).Offcourse ship A appears to be stationary,now since there velocities are same and angle between them is 90 degree we have total resultant as sqr rt(20 sq+20 sq)=20 rt 2. The resultant vector heads in south west direction making 45 deg with both axes.

So our problem is simplified one ship (at the start 10 km due north of other is heading southwest at 20rt2 km/he whilst the other is stationary.

Now consider the isoceles right angled triangle formed by drawing a perpendicular from -Vb to Va-Vb we have

10 km as hypotenuse now solve the eqn for the distance and divide it by 20 rt 2 to obtain time.

a zero appears at the end everytime the pair 5 & 2 occurs now how many 5's are there(we can always find the 2 to pair with it)...the answer is not 70/5=14 since notice that 25 and 50 has more than one five ie two.....hence answer is 14 +2=16

what i am basically suggesting is write the entire factorial in this form 1*2*3*2*2........so the 0 comes from a 5*2 pair..

here basically we hav to find da normality of H+ ions.......so we gmeq of h+ ions in 200ml soln as 0.6+0.2=0.8gmeq in 200 ml ie 4gmeq in 1 litre hence normality is 4....

too easy bro......ac=(b+1)(b-1)=b^2-1

now....log(1+b^2-1)=log(b^2)=2logb...in general log(x^y)=y log x..proof .....we knw log(pq)=log p+log q...........nw think of log x^y as log x*x.....ytimes so......it is y(log x)....

the answers are correct .........the relative velocity is given by va-vb=20root2 and the angle betw da vectors is 45 deg..............

now draw a perp frm vb to da resultant vector........itz value is 10/root2 now =7.1(approx) and time reqd iz....10/20*2=1/4......if yu reqd detail xplanation den nudge

oh i am sorry i dint notice the fact tht 0 was to be overlooked.....bhatt' sir's method....iz best///

Just use the simple multiplication principle.....we have 10 shoes we can choose any one.....then out of the remaining any 8(the mathching one has to be avoided) can be chosen...so in this manner...10*8*6*4=1920 (i am a novice in these things so plz dnt make fun of me if it is wrong).................

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