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Well, thanks for your help so far anyways. 

For part (b) someone told me to use parallel lines... using the fact that AB||CD could be helpful...

waits its supposed to be find \((r,\theta)\)  not whetever it is above

The perimeter  of the rectangle  = 

36  =  2 (W + L)

18  =  W + L      (1)

And  let the side of the square =  S   ....so.....

10W  = 2S      ⇒  5W  = S     ⇒  S/5  = W     (2)

2L  =  2S    ⇒    L  = S   ⇒          S = L           (3)

Sub (2) and (3)  into  (1)   and we have that

18 = S/5  + S             

18  =  (6/5)S

18 (5/6)  = S

15  = S

So....the perimeter of the square =  4S  = 4 (15)   =  60

Radius of Sphere  = R

Radius of cylinder  =  R

Height of cylinder  =  2R

Volume of Sphere                   (4/3) pi R^3             (4/3) R^3          (4/3)         2

_________________ =         ____________  =    _________  =   ____  =   ____

Volume of Cylinder                 pr * R^2 * 2R              2 R^3               2            3

We have a rectangular solid  surmounted by  the frustum of a pyramid

The volume of the  rectangular solid is easy   =  W * L * H =  8 * 8 * 2  = 128 m^3

The  volume of the  frustum is given by

(height / 3) [ Area  of lower base (AL)  + Area of upper base (AU) +  square  root  (  AL*AU )  ]

height = 4 cm

AL= 8 * 8  = 64  m^2

AU = 2*2 =  4 m^2

So....the volume is

(4/3)  ( 64  + 4  +   sqrt ( 64 * 4) )  =   (4/3) [ 68 +  16]  =  (4/3) ( 84)  =  112 m^3

So  the total volume =   [128 + 112] m^3  =   240 m^3

We can solve this through the method of  cross-section

See the following image  :

6sqrt (3)     = sqrt (108)      and 3sqrt (3)  = sqrt (27)

Note that triangle ABC is right......and we can find AB   as   sqrt  (  108 - 27)  =  sqrt (81)  =  9

Draw BE  perpendicular to AC

And triangles   CEB  and CBA  are similar....so

BC/ AC  = BE / AB

3sqrt (3) / 6sqrt (3) = BE / 9

3/ 6  =  BE / 9

1/2  = BE/ 9

BE  = 9/2  = 4.5

And triangle AFG  is similar to triangle  ABE

So

AF / FG =  AB / BE

9 - R                 9    

____     =     ______     cross-multiply

   R                 4.5

4.5 (9 - R)  =  9R

40.5  - 4.5R = 9R

40.5  =  13.5R

R  = 3     = the radius of the sphere

Note that    the line segment drawn through (3,2) and (6,0)   creates a triangle inside the square with a base of  3  and a height of 2

So.....the area of this triangle   = (1/2)(3)(2)  =  3 units^2

And the area of the square =  3^2  = 9 units ^2

So....the fraction of the area of the square above this line =  [ 9 - 3 ] / 9  =  6 / 9   = 2 / 3 

If there is a digit of five or more, then the string cannot be automatic.  With a digit that was five or higher appears any place in the string, say, 2352, then the machine prints out 0020.  We want the digits to be the same after put through the machine, so we cannot have a number that inserts a 0 when put through the machine.  We can use the same logic for a digit of 4.  With a digit of 4, say 4110, then the machine prints out 1200.  We want the digits to be the same after being put through the machine, so we cannot have a number that inserts a 0 when put through the machine, or else we would need to insert another number.  So the automatic string cannot have a 4.  The digit of 3 is a different case, because we can included 3 in an automatic string.  If the machine prints out a 3 after a string is input, then we need to insert two more digits, which then does not give us an automatic string.  For example, the machine turns 3303 into 1003, which is different from the number we put into the machine.  So an automatic string cannot have a 3 either.  We now know that we cannot have any digits 3 or higher, and we are now in a good place to find all automatic strings.  Since the automatic string cannot have a 3 or higher, the number at the end will be 0, so the number at the start must be 1 or 2.  If the number at the start is 1, then the automatic string cannot have any ones, so it would be a two.  Therefore, the automatic string must look like 1XX0.  The second digit is the number of ones.  If the second digit is 2, then we will not have any room for any more digits of 1, so the second digit cannot be a 2.  The second digit is 2 also is not possible.  If the second digit is 2, then the third digit must be a 1, but after the string is put through the machine, then we get two ones.  Now let 2 be the first digit.  If the second digit is 2, then we have two twos, which does not leave room for ones, so this does not work.  So let the third digit be 2.  The automatic string now looks like 2X20.  If X = 1, then we get 2120, which is not automatic, because there is one 1.  If X = 0, then we get 2020 which is an automatic string.  So the only automatic string is 2020.

Guest, are you and  madyl the same person.

It is confusing when you do not identify yourselves properly.

Nevermind, thanks for the input.

I also have only done part A

That solution in mathsisfun is very confusing, I am not surprised that people cannot follow it.

Hopefully I will put aside some serious time and effort for this question later.  I just have to find that time and drive which might be difficult.

Someone else might answer first. 

CPhill is usually pretty good with these and Omi67 does a lot of geometry questions too I think.

Five cannot be first ( to be less than 500)

2 or 4  must be last (to be even)

Position 1   4 possibles ( 1 2 3 4)

Position2    5 possibles (1 2 3 4 5)

sosition3     2 possibles (2 and 4)

4 x 5 x 2 = 40 possible numbers

I get 792

Let the number  be     a*10^2  + b*10  +  c

If the number is divisible by 22, it is also  divisible  by  11

This means (possibly)  that

a  -  b   +  c   =   0

And  b + c  =  11       subtract these

a - 2b  =  -11

2b - a   =  11

2b   =  11  +  a    

So....we want a  as large as possible    such that b  is a single digit....and  a  must be odd because 2b is even

The largest  value of a such that  b  is a single digit is when  a   = 7

So......this means  that  b  = 9

And

7 - 9 + 2  = 0

So...792 is the number

See the answer here:   https://web2.0calc.com/questions/help_27513#r1

The link doesn't work.

See the following image  :

Let the slope of AB  = -3/4

And the equation of the line  containing AB  is   y = (-3/4)x + 3

Put this line into standard form and we have that   

3x + 4y - 12   = 0 

And to find the distance  between  (0,0)  and this line we have that

l 3(0)  + 4(y)  - 12   l                12                     12

________________    =    _________   =       ___     =   2.4

   sqrt (3^2 + 4^2)                sqrt (25)                 5

Let  CE  be  perpendicular  to AB

And  let  the radius of  the  semicircle  = R

So....we  can construcst similar triangles   such that

  R                  2.4

_____   =     _____      cross-multiply

3 - R                 3

3R  = 2.4 ( 3 - R)

3R = 7.2 - 2.4R

5.4R  = 7.2

R =   7.2 / 5.4   =   4/3

We have the following list of triangles:

(2,2,sqrt(10))
(sqrt(10),sqrt(10),sqrt(2))
(2*sqrt(2),2*sqrt(2),3*sqrt(2))
(sqrt(2),sqrt(2),2)
(1,1,sqrt(2))
(2*sqrt(2),2*sqrt(2),2)
(2*sqrt(2),2*sqrt(2),2*sqrt(2))
(sqrt(5),sqrt(5),sqrt(10))
(sqrt(2),sqrt(2),sqrt(2))
(sqrt(5),sqrt(5),sqrt(2))
(sqrt(5),sqrt(5),2)
(sqrt(10),sqrt(10),sqrt(10))
(sqrt(5),sqrt(5),sqrt(5))

So there are 13 possible triangles.

We can break into cases:

111 - 1 number
11* - 10 numbers
1*1 - 10 numbers
*11 - 10 numbers
*1* - 100 numbers

This gives us a total of 131 memorable numbers.

Each day walking   1.5 hr

Each day drivin = 1.5 x 1/4 = .375 hr     time saving each day =  1.125 hr

1.125 hr/day x 5 days/wk x 37 weeks = ?

The area of a larger square is:   6 + 3*sqrt(3) = ≈ 11.196 u²    side = ≈3.346 

The area of a smaller square is:   sin15^° * 3.346 = 0.866

                                                      0.866 * 2 =   1.732

                                                       1.732² = 3.0 u²

The statements that are true and B, D.

xy  = 3x  + 2y  + z        rearrange as

xy - 3x  =   2y + z          factor out x on the left

x ( y - 3)  = 2y + z          divide both sides by  (y - 3)

          2y + z

x =    ______

          y  - 3  

(pq^3)^3 ( 4p^2q)^2

______________  =

   (2pq^2)^3

(p^3q^9)  ( 4^2  p^4  q^2)

___________________   =

    2^3 * p^3  * q^6

4^2  * p^7 * q^11

______________  =

    8 * p^3  * q^6

16 * p^4  q^5

___________  =

        8

2 p^4 q^5

ABCD is a square and K is the midpoint of BD.  E is on CK such that angle EDC = angle ECA.  What is the ratio of the area of trianlge AKE so the area of ABCD?

100 units - 90 units = 10 units

10/100=1/10

right, but what about the right side? is it just 2y+z / y-3?

Sorry...I messed up......see  my edited answer above.....