CPhill

I was away about nine months, too....it was a little slow when I first came  back, but it appears to have "revived" some !!!!

Whoopee !!!

Also.....good to see you on here, Max  !!!!!

Let the original volume =  pi * r^2 * original height

Note that if the diameter is increased by 36%, then so is the radius

So

pi * r^2 *  original height      =   pi *  (1.36r) ^2  * new height

original height =   (1.36)^2  * new height

original height / (1.36)^2   =new height

original height  * ( 1 /1.36^2)  =  new height

.54 original height = new height

The original height  should be reduced by  ≈   (1 - .54)  =  .46 ≈  46%

Two cakes are better than just one   !!!!!

13  is the circumradius  = C

C  = 13  =    product  of the sides / [  4 Area ]

Let x =  BC

13  =           (22 * 22 * x)

             _________________________________________________

            sqrt  [  (22 + 22 + x ) ( 22 + 22 - x) ( x + 22 - 22) (x + 22 - 22) ]

13  =         [ 22 * 22 * x  ]

           ________________________

             sqrt  [ ( 44 + x)  ( 44 - x)  (x^2) ]  

13 =     [ 22 * 22 * x ] 

          ________________________

            x * sqrt [ (44 + x) * (44 - x) ]

sqrt [ (44 + x) (44 - x) ] =    484   /  13

sqrt [ 1936 - x^2 ]  =  484/13       square both sides

1936  -  x^2   =    (484)^2 / 169

x^2  =   1936   -  (484)^2 /  169

x^2  = 92928  / 169

x = sqrt [ 92928 / 169 ]  ≈  23.449  = BC

\($\sqrt{21}$, $\sqrt[3]{27}$, $-\sqrt{64}$, $\sqrt[4]{25}$, $\sqrt{40}$, $-\sqrt[3]{81}$ \)

sqrt (21)  =  (4. and some  decimal)

cube root  27  = 3

-sqrt (64)     = -8

4th root of 25  =  (5^2)^(1/4)  = 5^(1/2)  = sqrt 5  =  ( 2. and  some  decimal)  

sqrt 40  = 6.and some  decimal

-cube root of 81  =    ( -4. and some decimal )

You should be able to  do this, now

x (3x - 7 )   =  -1 + 4x

3x^2 - 7x  = -1  + 4x           simplify as

3x^2 - 11x + 1    = 0

By  the Q formula  we  haxe

solutions =       11 ± sqrt [ 11^2  - ( 4 * 3 * 1 ) ]                    11 ± sqrt [ 109 ]

                       _______________________  =                ______________

                                         2 * 3                                                     6

m = 11

n = 109

p = 6

m + n + p  =   126

Your base is  correct   !!!!

Draw a an altitude from the top right vertex to the base

This forms a 30-60-90  right triangle......which has sides of 1  , sqrt 3   and 2

sqrt 3   is the altitude  (height)

Can you take it  from  there ???

Note

1/a^2   +   1/b^2  =     [ a^2 + b^2 ] /  [ ab]^2

5x^2   -11x + 4   =  0

By Vieta

Sum of the roots =  11/5

a + b = 11/5       square both sides

a^2 + 2ab  + b^2   =  121/ 25 

a^2 + b^2  =  121/25 - 2ab      (1)

And product of the  roots  =  4/5

So

ab = 4/5    ⇒  [ab]^2  = 16/25

2ab = 8/5

So (1)  becomes a^2 + b^2 =  121/25 - 8/5  =   121/25 - 40 /25  =   81/25

So

1/a^2  + 1/b^2   =     [ a^2 + b^2]  / [  ab]^2   =   [ 81/25] / [ 16/25]  =   81/16

Whatever x  makes the denominator  = 0  is  not in the  domain

x - 64  =0 

64 is  not in the  domain

x^2 - 64  =  0     so.... ( x + 8) ( x -8)    = 0     ....so    x = -8  and  x = 8  are not in the domain

x^3 - 64  =  0

x^3 = 64

x = 4    is not in the  domain

x^4 - 64   =  0 

(x^2 - 8) (x^2 + 8)   = 0

x^2  - 8  =0  ⇒   x =  ±sqrt (8)

The second factor set to 0  produces a non-real solution 

So....not in the  domain  =  x =   -8,  -sqrt (8) , 4 , sqrt (8), 8 , 64