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I think it's a pencil...

BTW this is Tommarvoloriddle

I'm just at school ig.

cos x = 1/2       60 degrees    -60 degrees (or 300 degrees)   pi/3 R     5 pi/3   or   - pi/3 R

it is!

The top left graph  is correct

cos  (41 pi / 12)

Convert radians to degrees  =     ( 41pi / 12 ) * 180  /pi   -  (41/12)*180  = 615°

615  -  360  =   255°  

cos (255°)   =  cos (210 + 45)   =

cos 210 cos 45  - sin 210 sin 45  =

(- √3/2 )  (  √2/ 2)  - (- 1/2)( √2/2)  =

- √6/4 +  √2/4  =

 √2 -  √6

________

      4

cos  ( x - pi/2)   =

cos x  * cos pi /2  +  sin x * sin pi/2    =

cos x * (0)  +  sin x * (1)  =

sin x

First one  :

88  is  22%   of    the total individuals  surveyed,  (T)  ....so.....

88  =  .22 T        divide both sides by  .22

88/ .22  = T   =   400

Second one.....multiply   4 x 3 x 2

sin (43 pi / 12)

Convert this to degrees

43 pi/ 12  * 180  /pi  =

(43/12) (180)  =  645°  =   645  - 360   =  285°

sin (285)   =   sin (150 + 135)  =

sin150*cos 135  +  sin 135 * cos 150   =

sin 30 cos 135  + sin 45 cos 150   =

(1/2) * (- √2/2)  + ( √2/2) (- √3/2)  =

- √2 -  √6               - [  √2 +   √6 ]

________   =      _____________

      4                              4

cos ( 5 pi  /12)   =

It is easiest to convert   the radians to degrees....so

5pi/ 12  * 180  /pi   =

(5/12) (180)  =   75°

So

cos  (75°)   =   

cos (30  + 45)  =

cos30*cos45  - sin 30 *sin 45  =

( √3/2 ) (  √2/ 2)   -  (1/2) ( √2/2)  =

 √6/4  -  √2/4  =

 √6 -  √2

_______

    4

P(Brown )   =     Number of Brown Candies  /  Total Number of Candies

Second one.....see if you can figure this one out  .....

For the first we need something known as Stirling Numbers of the second kind

S (8, 3)   =  966  ways

For the second we want to  partition  the integer  8   into 3 parts

This is

P(8,3)  = 

8   0    0

7   1    0

6   2    0

5   3    0

4   4    0

6   1    1

5   2    1

4   3    1

4   2    2

3   3    2

10  ways

The largest possible value of n is 13.

(a) For a nonagon, the probability that the triangle contains the center of the circle is 3/16.

(b) For a (2n + 1)-gon, the probability that the triangle contains the center of the circle is (n - 1)/(4n).

I think this is supposed to be

f(x)  =  log   (  1  / ( x + 12) )

We cannot take the log  of  0  or a negative.....thus......the domain  is  all real nubers  >  -12   ⇒  "C"

Second one

 (-1  +  2i ) * 5i    =     -5i + 10i^2  =  -5i - 10  =   -10  - 5i

Angle APB is 103 degrees.

Here's the graph  :

https://www.desmos.com/calculator/rpczckxvae

First part    1/ csc² x   = 1 /  1/sin²x  = sin²x

    so     1 - sin²x       (remember: this = cos²x )

        can you try it from there?

cot y

_____   =

sec y

cos y  / sin y

___________  =

    1 / cos y

cos y          cos y

_____  *   ______  =

sin y             1

cos^2  y

________  =

  sin y

1 -  sin^2 y

_________   =

  sin y

 1

____   -  sin y   =

sin y

csc y  -  sin y

Correct   !!!!

∑[ (3 n^2 + 3 n + 1), n, 1, 99] = 999,999

The problem might be that there are many ways to express  cos2θ  +  sin2θ 

You might also try these

1 - 2sin^2 θ  + 2sinθcosθ

2 cos^2θ  - 1 + 2sinθcosθ

sin x             cos x

____   +      ______    =

csc x            sec x

sin x                cos x

_______ +   _________    =

1/sin x           1 / cos x

sin^2 x   +   cos^2x    = 

1

IT IS   !!!!!!!!

Good job  !!!!!