CPhill

We have this form :

y  =  Acos [ B (x + C) ] + D

A = the amplitude  =  [ 10 - -6] / 2  = 8

B  = the  number of periods in 2pi   =  2pi / the period

The period  = 4 pi   ....so...

B =  2pi / [ 4pi]   =  1/2

C  = the phase shift

The normal cosine curve  has its first max at x = 0

This one has its first max at 2pi

So.....this is a shift to the right of 2pi

So...   C   =  -2pi

D = the midline  =   [ 10 - 6] / 2   =  4/2  = 2

So....the function is

y = 8cos [ (1/2) (x - 2pi) ] + 2

See the graph here : https://www.desmos.com/calculator/bztp33f42u

THX, CentsLord....excellent    !!!

Since x lies in Q3......then x/2 will fall into Q2...so....the cosine is negative there

cos (x/2)  =  -√ [  (1 + cos x)  / 2 ]

cos (x/2)  =  - √ [ ( 1 - 2/3 ) / 2 ]   =   - √ [ 1/6]  = - √6 / 6

Mmmm....this is an interesting question  !!!!

Note that 20 ft^2  =  20 * 12^2  =  2880 in^2

So 5 gallons of paint must cover 5 * 2880  =  14400 in^2

Each revolution of the roller represents the lateral  surface area of a cylinder with a  radius of 1.5 in  and and a height of 8 in

So...the  surface area covered in one revolution  =  2pi * radius * height   =  2 pi (1.5) (8)  = 24 pi in^2

So.....the  number of approximate revolutions  is

Total area covered by 5 gallons of paint          14400

________________________________   = ________  ≈  191 revolutions

  Area covered by  one revolution                     24 pi 

4  was answered  by geno, here : https://web2.0calc.com/questions/divide_11

cosx +cosx * cot^2 x  =      cot x csc x

cos x (  1 + cot^2 x)  =      cotx csc x

cos x  ( csc^2 x)    =     cotx csc x

cosx   ( 1 / sin^2 x)  =   cotx csc x

cosx         1

____ *  ______   =     cot x csc x

sinx        sin x

cot x csc x     =  cot x csc x

3.    -3 + 3i

z = √ [ (-3)^2 + 3^2]  =  √ 18   =  3√2

arctan (3/-3)  =  arctan (-1)  = -  pi/4   = θ

However....this is located in Q2  so  θ =  3pi/4

So

z = 3√2 cis( 3pi/4)

The second is easy....

If she has a 20% chance of winning she bought   60 (.20)  =  12 tickets

For the first one we need to find the  probability of a "3"

This is    1 -  ( .13 +.05 + .21 + .19 + .3)    = .12

P (3 or 4)   =  [ .12 + .21]  =   .33

If rolled 300 times  we would expect a 3 or 4    =   300 (.33)  =   99 times

Thanks, Alan   !!!!