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That's exactly what jimkey17 said 

it is 0

Thanks jimkey17

we simplify it to -1*11*0=0 so our final answer is 0

Thanks

we find that sin(270)=-1

1L=1000cm³ so  it is 8*1000=8000cm³ so we have 8000/10/25=800/25=32 so the length of the rectangular container is 32

Remember   tan = sin / cos      so for   60^o        tan 60 =   sqrt3/2  /  1/2  =  sqrt 3

                                                                           tan 300 =  -sqr3/2  / 1/2   = - sqrt 3

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We solve that cos(180)=-1

This is correct thanks jimkey17

Thanks jimkey17

This question was answered here 

Its always a good idea to search your problem before you post it. 

A lot of problems have already been answered so there is no point in posting it here.

https://web2.0calc.com/questions/help-please-thx_4#r4

it got reported again

is this correct

tan300=-sqrt3 and tan60=sqrt3 so we have \(\tan 300=-\sqrt3~~and~~\tan 60=\sqrt3\)

By similar triangles, YZ = 23/4.

since 0!=1

k can also equal 0

tan90^o is undefined

mmm

If the maths I have presented above is correct I do not see how this could work because you would be left with some primes example 37 that do not have a pair.

Maybe there is something wrong with my logic that  S = 100!*100!

I am not doubting your output but it does not make sense to me.

Maybe another mathematician will comment.

Hi Melody:  I wrote a very short computer code and it indicates that  S/50!  gives a perfect square, which is 3,438 digits long! It begins: 9 427 570 616 109 075 150 705 396 694 534 826 731 504.........etc. Of course 50! sits smack in the middle of [1! 2! 3!.....100!],  and it probably is no surprise! But how do you prove it mathematically? I don't know! Maybe Alan, heureka or yourself can do so.