$a^2 + b^2 = c^2$
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...
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?
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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??
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...