Recent content by grgrsanjay

$a^2 + b^2 = c^2$ if a and b are the lengths of the two short sides of a right triangle and c is its long side then this formula holds. Conversely, if the formula holds then a triangle whose sides have length a, b and c is a right triangle...

That was for fun i think though... I agree on achbach

I Wanted to know whether my logic holds good for every similar problem??

Thanks for your help! I Understood it :)

its a six digit number 1st digit can be 1,2,3,4,5 2nd digit can be 0-9 3rd digit can be 0-9 4th digit can be 0-9 5th digit can be 0-9 6th digit can be 1,3,5,7,9 total ways = 5(10)(10)(10)(10)(5)

First,i am ignoring the numbers on the seat, this is a round combination So, formula is (n-1)! no.of.ways is 5!(3!)(3!)= 4320 Now the seat are numbered, then i can more these combinations 1 seats,2seata,...9 seats apart from the original one so,number of ways is 43,200

Sorry...i did the same,typed wrongly i equated the denominator to zero.i got $ xe^{xy} = 1 $ then what? ---------------------------------------------- Yea,yea got it dy/dx=0, so the equation is of the form x = a y=0 substituting it at this $ xe^{xy} = 1 $,we getx=1 So, point is (1,0)...

The curve $\displaystyle y-e^{(xy)} + x=0 $ has a vertical tangent at which point?? I started to differentiate it, then equating dy/dx to 0, then how should i proceed??

I think none of my case were repeated...could you conform it whether any case is left?

Re: Calculus problem $\displaystyle f(x) = x + \int_0^1 (xy + x^2)f(y)dy$ Let me integrate it, then $\displaystyle \int_0^1 f(x)dx$ = $\displaystyle (1+\int_0^1 yf(y)dy)$. $\displaystyle \int_0^1 xdx$ + $\displaystyle { \int_0^1 f(y)dy}. \int_0^1 x^2 dx$ So , I get the equation 4A = 3 + 3B...

So,there must be less than 14C2 solutions??How do i eliminate them? Ok,could you check whether i can do it like this x + y + z = 12 Case 1:x=y=z No.of.ways = (4,4,4) = 1 Case 2:two of x,y,z are equal No.of ways = (1,1,10),(2,2,8),(3,3,6),(5,5,2),(6,6,0),(0,0,12) = 6 Case 3: x,y,z are all...

How many different garlands are possible with 3 identical beads of red color and 12 identical beads of black color?I was thinking to keep the no.of beads in between the 3 beads of red color as x,y,z So, x+y+z=12 no .of ways is 14C2...Was i Wrong somewhere??

Let $f:R \to R$ be a continuous and differential function given by $\displaystyle f(x) = x + \int_0^1 (xy + x^2)f(y)dy$ find $\displaystyle \int_0^1 f(x)dx$ and $\displaystyle \int_0^1 xf(x)dx$I wanted to know how i could start the problem.Please do not give full solution It would be good...