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a

6!=$\boxed{720}$

b

6!/12=$\boxed{60}$

c

6!-60=$\boxed{660}$

Given this recursion equation, we can use the given $a_0=1$ and $a_1 = 5$ conditions to solve for $a_2,$ and then $a_3$. If we first let $n=1,$ we get $$(a_1)^2 - a_0 a_2 = (-1)^1,$$

$$25 - 1(a_2) = -1,$$

$$a_2 = 26.$$

Using this same logic, try to figure out $a_3$ by substituting into the relation $n=2.$

lets call FD x.

18^2+x^2=(24-x)^2

324+x^2=576+x^2-48x

48x=252

x=$\boxed{\frac{21}{4}}$

productfor(n, 1, 1000,  (2^n)^(3^-n))=1.6817928305 074290861==2^(3/4)==(2^3)^(1/4)

Draw a diagram, the base angles of an isosceles triangle are the same and all angles of a triangle add to 180 degrees.

All the angles in a triangle add to 180 and in an isosceles triangle 2 of the angles are the same. So JKI and JIK are the same number of degrees. So 180-J= JKI+JIK. 

Do not cheat on AoPS homework.

There is indeed a message board message 

The probability that A will win is 2/3 and that B will win is 1/3 so if A wins three times in a row, 2/3*2/3*2/3=8/27.

Pretty sure there is a messege board on aops you can use.

There is a message board... don't post the aops writing problems here please...

The ratio of $\triangle ABC$ to $\triangle ACD$ to $\triangle ADE, $ which is just the ratio of the bases $BC$ to $CD$ to $DE$ (because the height for each of the triangles is the same). Therefore, we can write equations corresponding to these ratios. $$\frac{BC}{[ABC]} = \frac{DE}{[ADE]}, $$  $$\frac{5-CD}{24} = \frac{4-CD}{18}.$$

Solving this set of equations, we find $$90-18CD = 96-24CD,$$ $$6CD=6,$$ $$CD = 1.$$

Therefore, we can use this value of $CD$ to substitute back into the original equation. $$\frac{CD}{[ACD]} = \frac{DE}{[ADE]}, $$ $$\frac{1}{[ACD]} = \frac{3}{18}, $$ $$[ACD] = 6.$$

Thus, our answer is $\boxed{[ACD] = 6}.$

(Note: The diagram is not drawn to scale).

Hi! I understand most of your answer, but I'm a bit confused as to how you calculated the cosine of 72?

Hello!

I love your pfp :))

So, lets turn everything where the base is 2. 

2^(1/3) is already in base 2.

4^(1/9) = (2^2)^(1/9) = 2^(2/9)

8^(1/27) = (2^3)^(1/27) = 2^(3/27) = 2^(1/9)

16^(1/81) = (2^4)^(1/81) = 2^(4/81)

Now that everything is in base 2, we can just add. :))

1/3+2/9+1/9+4/81 = 58/81

2^(58/81) = <81>sqrt(2^58)

I hope this helped. :))

=^._.^=

A single-elimination tournament, also known as a knockout tournament, is one where the loser of a match is automatically eliminated from the tournament. The winners of each round continue to play each other until a single winner is declared. Determine the total number of matches that are necessary for a chess tournament involving 1024 participants.     

After 1 round there are 512 participants.  

After 2 rounds there are 256 participants.  

After 3 rounds there are 128 participants.  

After 4 rounds there are 64 participants.  

After 5 rounds there are 32 participants.  

After 6 rounds there are 16 participants.  

After 7 rounds there are 8 participants.  

After 8 rounds there are 4 participants.  

After 9 rounds there are 2 participants.  

After the 10th round, there is one winner. 

_.

The tournament organizers deliberately arrange it so that there are 1024 deci-finalists.  Is that a word?  It is now. 

_.

Thanx , Melody.....I assumed it DIDN'T care about gender as  'without restrictions' was part of the Q ...

    Have a good one !

EP has assumed that you only care about gender.

I'd say it is 14C6     since I assume every member is an individual.

26*25*24*23=358800 without any uppercase letters and no repeated letters and 52*51*50*49= with uppercase letters

Not sure how to solve this but  

-9³  + -7³  +  -5³  + -3³  = -1224

See all of the red circles?   The answer is the x coordiante of all of those circles....the y coordinate = -1

As the payments are made at the BEGINNING of the period....this is an ANNUITY DUE calculation...

Here is one form of this calc:

FV = PMT * [  (1+i)ⁿ  - 1 ] / i       * (1+i)       PMT = 1000    i = .08     n =10       plug and play!    

Possible first flip outcomes  (2³)

ttt

tth

tht

thh

htt

hth

hht

hhh                  Flipping any of th h's will result in a t or h combo already listed     so 8 possible outcomes I believe

6 + 8 = 14

14 c 6 = 3003 ways       = 14!/ (6! 8!)

Is ΔJNM ~ ΔJLK? Support your answer with a postulate or theorem.

    Yes    S-A - S

Find the length of JN. Show all your work.

  kl is to mn      as     jl is to jn

kl / mn   =  jl/jn

8/10 =  3.2 / jn           use this to find   jn      then   ln =   jn - jl   =     jn - 3.2

Find the length of JK. Show all your work.

kl / mn   =   jk/ (jk+ mk)     

8/10      = jk/ (jk+1.5)                    solve for jk

Are these right triangles?      use what you have found and are given to see if pythagorean theorem is true

   if they ARE right triangles   then are of a right triangle  =  1/2  leg₁  * leg₂