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The amount is

5000 ( 1 + .03)^16   ≈  $8023.53

If I didn't make an error...

Lateral surface area to be covered of  upright 11  posts =

Note that 1 m  = 100 cm

11  * [ 4* 100 cm * 2 cm]   = 

11 [800 cm^2 ] =  =   

8800 cm^2

Suface area of top rail to be coated  =

( Lateral surface area of rail ) -   area where the upright post meets the top rail

(Area at two ends of the top rail  + lateral surface area of rail )  -  11 (2cm *2cm)  =

2 ( 1cm* 4cm)  + 2*102 cm  ( 1cm +  4 cm  ) - 11 ( 4 cm^2)  =

2 (4cm^2) + 204cm (5cm)  - 44 cm^2  =

8cm^2  + 1020cm^2 - 44cm^2  = 984 cm^2

Total suface area to be covered  =   [ 8800+ 984] cm^2  = 9784 cm^2

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

Thank you!

Ok...Thanks!

THX, CentsLord....excellent    !!!

We can solve this problem by listing out the box's individual sides. There are 6 of them here.

Keep in mind that opposite sides have the same areas :D

There's one side, the side measuring 5 in. by 4 in. = 20 square inches. As there are two of those, they amount to a total of 40 square inches.

Then there's a lengthy rectangular side, measuring 12 in. by 4 in. = 48 square inches. Once again, there are two, so those two sides have 96 square inches total.

Then there's the bottom side, measuring 12 in. by 5 in. = 60 square inches. As the top and bottom are identical, those amount to 120 square inches total.

So, the total surface area of the shoebox will be 40 + 96 + 120 = 256 square inches :D

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   !!!!

cos (x/2)  =  -√ [ (1 + cosx) / 2 ]

cos (x/2)  = -√ [ (1 -1/5) / 2 ]   =  -√ [ 4/10]  =  -2 / √10  =    -2√10/10  =   -√10 / 5

I'm sure you can replace A by 8 and T by 4pi !!

Can you simplify that a bit please? I'm trying to write it in the form of the function/equation if possible.

The midline is at y = 2, so the amplitude, A = 10 - 2 = 8.  The period, T = 6pi - 2pi = 4pi.

The form of the equation will be y = 2 - A*cos(2*pi*x/T)

If you are graphing using rectangular coordinates then simply convert (r, theta) to (x, y) where x = r*cos(theta) and y = r*sin(theta).

"How many petals does the graph of the rose curve r=4sin(3θ) have?"

"Compute 1^2/2^1 + 2^2/2^2 + 3^2/2^3 + ... + n^2/2^n + ..."

As follows:

sumfor(n, 1, 1000, n^2 / 2^n) = it converges to 6

Substitute --> M=12*(4*10^-6)^2/6.67*10^-11*1

M =[12 * 1.6E-11] / 6.67E-11

M= 1.92E-10 / 6.67E-11

M = 2.87856 - Mass units

I suspect you dragged 180° into the 2nd box.  However, the second box requires the rotation under multiplication by i, not by i².  Multiplying by i is a 90° rotation.