CPhill

cos^2x-sin^2x/ctg^2x-tg^2x=sin^2x*cos^2x

First, notice that :

ctg^2x - tan^2x =    cos^2x/sin^2x - sin^2x/cos^2x  = [cos^4x - sin^4x]/[sin^2x cos^2x)..so we have..

[cos^2x - sin^2x] [sin^2xcos^2x]  / [cos^4x - sin^4x]  = 

[cos^2x - sin^2x] [sin^2xcos^2x] / [ (cos^2x - sin^2x) (cos^2x + sin^2x) ] =

[sin^2xcos^2x] / [sin^2x + cos^2x]              and [sin^2x + cos^2x ]  = 1  .....so....

sin^2xcos^2x  =  sin^2xcos^2x

Nah......that seems pretty logical to me, Melody.

7 different groups each seated 5! ways......

I agree with your answer......and we're NEVER incorrect on these counting probs, are we ???  LOL!!!

Hey, Melody....a brief question......what was the motivation for multiplying by "2" on the left hand side in the top equation ???

Go slow with that explanation......you know me and "counting" problems....LOL!!!!

Notice that, if A comes first, there are 24 ways to form a two-letter combo. {We can't use O}

And if B comes first we have 23 possibilities.

Up to the letter O, we have :

24 + 23 + 22+ .....+ 12 + 11 possibiilities

And after O, we have  10 + 9 + 8 +....+ 2 + 1 possibilities

So, the total possibilities is just (24)(25)/ 2 = 300 possibilities

And for the numbers, since 0 can't be used, we have :

8 + 7 + 6 + ......2 + 1  =  (8)(9)/ 2 = 36 possibilities

So, the total ways of forming a license plate is given by ;

300 x 36 = 10,800 combos

csc^2 (x)- csc(x) - 2=0    factor

(cscx + 1) (cscx -2) = 0      set each factor to 0

So either

cscx  = -1      and this happens at  -3pi/2 ± n 2pi     where n is an integer

OR

cscx = 2        and this happens at pi/6 ± n 2pi     and  at  5pi/6  ± n2pi     agian, where n is an integer

I actually can't take much credit for this one.......the answer comes straight from a "formula" supplied by  Nauseated....

For any R x C grid arrangement where R represents the rows and C represents the columns.....the number of distinct paths from one corner of the grid to a diagonally opposite corner is given by :

(R + C ) !   / [R! * C! ]

Yeah.....and a lot more, too.....it's a non-repeating, non-terminating decimal......

2\5\8 * 5\1\3  =

(2/5) / (8/1) *  (5/1) /(3/1)  =

(2/5) * (1/8)  *   (5/1) * (1/3)  =

10/120 =

1/12

4-5[8-2((-4)^2+5)]    simplify

4 -5[8 -2( 16 + 5) ] =

4 -5[8 - 2(21)]  =

4 -5 [8 - 42] =

4 - 5[ -34]  =

4 + 170  =

174

They have no common factors.....{they are relatively prime }

If you mean "least common multiple,"  it's 36