CPhill

rotation of 90° counterclockwise about the origin   ( a, b)  →  (-b. a)

translation of 1 unit up   (-b, a) → (-b, a + 1)

Example

R = (-3,-1)

First transformation yields ( 1, -3)

Second  transformation  = (1, -2)  = R'

And so  forth  ( check the other points for yourself)

3xz + 6xyz + 15x + 30xy + 8yz + 40y + 4z    =  127

3xz ( 1 + 2y)  + 15x ( 1 + 2y)  + 4z (1 + 2y) + 40y =  127

(1 + 2y) ( 3xz + 15x + 4z) + 40y = 127

Let y  = 1

(3) (3xz + 15x + 4z)  + 40   = 127

(3) (3xz +15x + 4z)  =  87

3xz + 15x + 4z  =  29

Let   x  = 1    and  z  = 2

3(1)(2) + 15(1)  + 4(2)  =  29

6 +  15  + 8   = 29

29  = 29

x + y + z =  4

We have  these intervals to consider

(-inf, -5)     no good

(-5, 3)        good

(3, 5)         no good

(5, inf)     good

The smallest integer  that satifies this  is   x =    -4

The diagonal of the rectangle  = sqrt [ 5^2 + 4^2] = sqrt [ 41] in

Call the radius of the table  = r

By the Pythagorean Theorem

r^2 + r^2  =  [ r + sqrt 41]^2

2r^2  = r^2 + 2rsqrt 41 + 41

r^2 - 2sqrt 41 r  - 41  =  0

Solving for r gives that  r =  sqrt (41) + sqrt (82) =   sqrt (41) ( 1 + sqrt 2)  in  ≈ 15.46 in

Draw AC

Triangle BAC is right with angle ABC  = 90

AC  = sqrt [AB^2 + BC^2] =  sqrt [ 6^2 + 4^2 ]  = sqrt [ 52 ] 

sin [angle BCA ]  =  6/sqrt [52]

Angle BCD =  90

Angle BCD = angle BCA + angle ACD

sin [ angle BCA ]  = cos angle [ACD ]

cos [angle ACD ] =  6/sqrt 52

sin [angle ACD ] =  sqrt [  1  - (6/sqrt 52)^2 ] =  sqrt [ 1 - 36/52 ] = sqrt [ 16/52]  = 4 / sqrt 52

[ ABCD]  = 

[ABC] + [ACD]  =

(1/2)(BA)(BC)  + (1/2) (AC) (CD) ( 4 /sqrt 52)  = 

(1/2)(6) (4)  + (1/2) (sqrt 52) ( 10) (4 /sqrt 52)  =

12  +   20  = 

32

The major axis length =  46 +  70  = 116 =  2a

So a = the semi-major axis length =   116 / 2 =  58  =  a

The focal length =   58  - 46  =  12  =    c

We can find the  length of the semi-minor axis as

sqrt [ a^2 -c^2 ] = sqrt [ 58^2 - 12^2]  = sqrt [3220] = b   →   b^2  = 3220

The eccentricity is  given by

sqrt  [  1 -  b^2 / a^2 ] =

sqrt [ 1  -   3220 / 58^2 ] = 

sqrt [ 1 - 805/841  ] =

sqrt [ 36 / 841 ]  = 

6 / 29    ≈     .21

 4/(5/3 - 3/5)  = 

4 / [ ( 25 - 9) / 15 ]  =

4 / [ 16 / 15 ]  =

4 * 15  /  16  =  

60  / 16  =

15 / 4

Here :

https://web2.0calc.com/questions/help-with-range-plz_1

Through the use of similar triangles, we can find the radius, r,  of the  sphere thusly :

Let one right triangle CDB be composed  of  two  legs.....DB =  the base radius  of the  cone = 3 and  the other DC = the height of the cone = 3

A simiar  second  right riangle CFE  can be  formed   (see the illustration) with legs EF =  r and FC =  r

The hypotenuse of this triangle = CE =   sqrt [ r^2 + (r^2) ] = r sqrt 2

And we  can  find the  radius of the  sphere thusly :

CD = 3

ED + CE  =  3

r + r sqrt 2  = 3

r ( 1 + sqrt 2)  = 3

r = 3 / ( 1 + sqrt 2)

Volume of sphere to volume of the cone  =

(4/3) pi ( 3 / (1+sqrt 2) )^3                        4

_____________________  =      ____________ ≈   .284

(1/3) pi ( 3)^2 * 3                           (1 + sqrt 2)^3

A

         12

                   X

                          4

B                               C

BC^2 - XC^2 = BX^2

AB^2 - AX^2 =  BX^2

BC^2  - 4^2  = AB^2 - 12^2

BC^2 = AB^2 - 144 + 16

BC^2  = AB^2 - 128

BC^2 + AB^2  = AC^2

AB^2 -128 + AB^2  = 16^2

2AB^2 - 128 = 256

2AB^2  = 384

AB^2 = 192

AB^2 = AX^2 + BX^2

192 = 144 + BX^2

48 = BX^2

sqrt (48) = BX  =  4 sqrt (3)