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sin (x)  =   1  - 2cos^2 (x)

Note  that   cos^2(x)   =  1 - sin^2(x)

So  we have

sin x  =   1 - 2  [  1  - sin^2 (x) ]

sin x  =  1  - 2  +  2sin^2(x)     

sin (x)  =  - 1  +  2sin^2(x)         rearrange as

2sin^2(x)  -  sin (x)  - 1  =  0        factor as

(2sin (x)  + 1) (sin (x) - 1)  =  0

Se each factor to 0 and solve

2sin x  + 1  =  0      subtract  1 from  both sides

2sin x  = -1         divide both sides by 2

sinx  =  -1/2    and this occurs  at    7 pi / 6    + n*2pi   and  11pi / 6  + n*2pi     

[ where n is an integer]

And..for the other solution we have that

sin x  - 1   = 0   add 1 to both sides

sin x  = 1     and this occurs at   pi/2 + n*2pi 

Possible outcomes

P          N          Q

H        H           H

H        H           T

H        T           T 

T         T          T

T         T          H  

H        T           H

T        H           T

T        H           H

Note there is only one outcme with no heads....so...the probability of at least one head is :

7/8

The probability of getting NO heads =1/2^3 =0.125. Then the probability of getting at least 1 head is:

1 - 0.125 =0.875

In triangle MNP, Sine of angle MPN =46.2/72.6 = Sine of angle ABM =0.636363.......etc.

Side MA =46.2 - 14 =32.2 cm, So you have:

0.636363..... =32.2 /x and x =32.2 /0.636363.... =50.6 cm

Here's the "formula"  to compute each new point

New Point  =  [ Given Point  - Dilation Center ] * Scale Factor  + Dilation Center

Note that   

(a, b )  - (c, d)  =  ( a - c, b - d)

(a, b)   + ( c, d)  =  ( a + c, b + d)

(a, b) * N  =   ( N*a, N*b )

So....the new point associated with ( 3, - 3)  is

[  (3, - 3)  - (2, -7) ] * 3  +  (2, -7)  =

[  ( 1, 4 ) ]  *3   +  (2, - 7)  =

 (3, 12 )  +  (2, - 7)  =

(5, 5 )  

You should be able to compute the other three

That would be great.  Thanks for helping out!

Total possibilities for 1st row

We have  10  choices for each   square  in the row

So...   10⁵  =   100,000 possibilities

%  of  Ni  =  33.1 / 41.9   ≈   79%

%  of O   =   100 -  79  ≈  21%

Well..we have  7 shakes where the same flavor is used for both scoops

Then....we can choose  any  2 of 7 for the rest  = C(7,2)  = 21

So....the total  possible number of shakes is  7 +  21  =   28

∑2/[n(n+1)(n+2)] from n=1 to infinity, then it converges to 1/2.

Sorry, same here, except that the Partial Sum Formula would be:  (n^2 + 3n)/(2 (n + 1) (n + 2)). So, if n=1,000,000, then:[1,003,000/2,006,004] =0.49999900.............etc.

Sorry, ACG...I know that the infinite sum of the second series is  1/2, but I don't know how to prove it......heureka  may know how to prove this.....I'll  send him a message....

−4y^2+3xy−12y+9x

y (-4y + 3x)  + 3 ( -4y + 3x)       the common factor  is  (-4y + 3x)

(-4y  + 3x)  (y + 3)  =

(3x - 4y)  (y + 3 )

MN  - AN  =  MA   ...so...

46.2  - 14  =  32.2  =  MA

And we have that

MA / MN  =  32.2 / 46.2    =  23/33

And the other side has a similar proportion such that

x / 72.6  =   23 / 33       cross-multiply

33x  =  23 * 72.6        divide both side by 33

x  =  [ 23 * 72.6]  / 33  =  50.6

Thanks!  I now realize I forgot to add this to the second problem.  Suppose that the second sequence was infinitely repeating (+ … ), what would be the sum then?

thank you Cphill

2.26, 12/5, 26.8%, 2.26, 271  =

[ 12/5  =  2 + 2/5  = 2.4 ]    [  26.8%  =  .268  ]

2.26, 2.4, .268, 2.26 , 271  

Ordered from  least to greatest

.268, 2.26, 2.4, 271

Here's the number line : https://www.desmos.com/calculator/b47rzyrxmn

[ The last value will be far to the right  ]

Sorry CPhill, I get a slightly different answer as follows:

It is true that there are 128 matches, but that icludes both men and women.

There are seven rounds in every Slam tournament, including the Australian Open:

First Round, with 128 players (sixty-four matches)

Second Round, with 64 players (thirty-two matches)

Third Round, with 32 players (sixteen matches)

Fourth Round, with 16 players (eight matches)

Quarterfinals, with 8 players (four matches)

Semifinals, with 4 players (two matches)

Final, with the last two players playing for the title

If you add the above number of "matches" NOT "players" you get:

128 - 1 = 127 matches to crown men's champion, and:

128 - 1 = 127 matches to crown women's champion.

In general, if the number of matches is n, then the number of matches to crown the winner

is: n - 1, or 2^R - 1. R for rounds.

Height  =  2.9 cm

Base  = 2 cm

Area  of parallelogram  =  Base * Height  =  2 * 2.9   =  5.8 cm^2

First one

[1 - 2] + [ 3 - 4] + ... + [2005 - 2006] + 2007

Note  that  each  pair of terms in the brackets  sum to  -1

And we have   2006/2  =  1003 of these pairs....so the series sums to  :

1003 (-1)  +  2007    =  1004  

Second one.....

2/(1*2*3) + 2/(2*3*4) + 2/(3*4*5) =

2/6   +  2/24   +  2/60  =

1/3   +  1/12  +  1/30  =        

20/60   +  5/60  +  2/60   =

27  / 60 

Corrected now, SMM....

Thanks so much!

good thing i checked to make sure evrything was correct

your answer was incorrect because you accidentally wrote 3 lips when it was two lips

We have

4 hairstyles  [ including bald ]

2 eyebrows

1 eyes

2 ears

2 lips

2 shoes

So....the total  "personalities"  =  4 * 2 * 1 * 2 *  2 * 2  =   64

EDIT TO CORRECT A PREVIOUS ERROR

Assuming  that letters can be repeated......we have 7 letters to hose from in each case....so.....

7  * 7  * 7   =   343  possibilities