So, the number of ways of distributing the prize
1) If the prizes are distinguishable:
no. of prizes, ways of choosing them WAYS OF ARRANGING THEM TOTAL
I II III IV...
Then if I take all the prizes not distinct then my answer would be correct? What changes then I would have to apply in the answer 4^5 to get the same for non-distinct prizes. Is it 4^5/5!
I fif it that way and since each price have prize will have 4 ways so 5 prizes will have 4^5=1024 ways.
But can you please tell the faults or mistakes I made in the previous approach
the first boy has 6 ways,
either he wins 0 or 1 or 2 or 3 or 4 or 5 ways
According to the first boy, the second boy also has 0 or 1 or 2 or 3 or 4 or 5 chances
According to the first and second boy, the other boy also has 0 or 1 or 2 or 3 or 4 or 5 chances.
So, the answer should be 4(either of...
How come the L.H.S of an equation is not defined for a point but the RHS value is. Then both the functions are not equal and our equality shouldn't hold
The answer 2sin^2(A) was given as the answer in the book. But when I put any value of A, like 90 I don't get the LHS and RHS EQUAL. Also when I put A=90 in the answer I got I didn't get LHS = RHS
[cos^2(2A)-1 +2sin^2(A)cos^2(A)]/sinA(cos2A+1)
=[cos^2(2A)-1 + sin^2(2A)/2]/sinA(cos2A+1)
=[cos^2(2A)-1]/2sinA(cos2A+1)
=[cos(2A)-1]/2sinA
=-2sin^2A/2sin(A)
=-sinA
But this isn't correct when I put some values of A for eg. 90 degrees and 45 degrees
cos2A+1=2cos^2A
cos2A-1=2sin^2A
Yeah I typed it wrong.
But I am still not getting the answer.
Give me a hint on how to proceed.
(cos2A -1)/sinA +sin2A/(cos2A+1)*cosA
[cos^2(2A)-1 +2sin^2(A)cos^2(A)]/sinA(cos2A+1)
[-sin^2(2A) + 2sin^2(A)cos^2(A)]/sinA(cos2A+1)
b^2-1= tan^2(x) + cot^2(x) + 2 -1
b^2-1= sin^2(x)/cos^2(x) +cos^2(x)/sin^2(x) -1
b^2-1=[sin^4(x) +cos^4(x)]/sin^2(x)cos^2(x) -1
b^2-1=[1-sin^2(x)cos^2(x)]/sin^2(x)cos^2(x) -1
a(b^2-1)=sinx+cosx {[1-sin^2(x)cos^2(x)]/sin^2(x)cos^2(x) -1 }
I am not able to go any further than this step to reach...
why the overlap will be smaller. I am just saying than instead of overlapping of the carbon's unhybridized orbital pz with that of another carbon. Can't the two carbon have their orbitals overlapped in the hybridized state? the energy of hybridized orbital will definitely be less than that of...
I know the geometry of sp3 orbitals is tetrahedral and sp2 is trigonal planar which is the geometry of ethene.
So is it purely based on experimental value than on p orbital shouldn't be hybridized to form a trigonal planar shape
why is sp2 hybridized carbon is need with only 3 hybridized orbitals and one unhybridized orbital? The unhybridized orbital forms a bond anyway. Then why can't they be hybridized and form the same bond
Work done in moving a charge around the circular path is 0 because the distance between the charges remain the same so there is no potential difference and delta(V)=0 but I thinkk some torque or external force must be required to move the charge around the circula path
shouldn't lower bound of summation i=0 rather than i=1 since you have kept the function f(xi)=(a+i(Δx))^2
as for the area under the function we had a terms (a+a+Δx+a+2Δx+.....+a+9(N-1)Δx)Δx