CPhill

The semi-perimeter of this  triangle =  (8 + 12 + 17) /2  =  37/2  = 18.5

The area of this triangle   = sqrt  ( 18.5 * ( 18.5- 8) (18.5  -12) ( 18.5 - 17) )  =

sqrt [ 18.5 * 10.5 * 6.5 * 1.5 ]  = 

sqrt [  1893. 93.75 ]

The longest two altitudes will be drawn to the shortest two sides

So, using the area formula for a triangle,   we  have

sqrt (1893.9375 )  = (1/2)(8)  *altitude  1

sqrt(1893.9375) / 4    =    altitude 1      (1)

And

sqrt (1893.9375 ) = (1/2)(12)  * altitude 2

sqrt (1893.9375 ) / 6   = altitude 2          (2)

Adding  (1) and (2)  we get

sqrt (1893.9375) / 4  +  sqrt(1893.9375) / 6    ≈  18.133 units

Excellent  !!!!

Let's go through this

Slant height  =18

The side of the base =  sqrt (288)  =  12sqrt 2

The diagonal  length of the base  = 12sqrt (2)  * sqrt (2)  =  24

Half of this diagonal length  =  12

Using the Pythagorean Theorem, the  height of the pyramid  =  sqrt  [  slant height ^2  - (half of diagonal length)^2 ]   = 

sqrt [ 18^2 - 12^2 ] =  sqrt (180)  =6 sqrt (5)

The  volume of the pyramid  =   (1/3) ( base area) (height)  = 

(1/3) (288) (6 sqrt (5) )  =

2 * 288 * sqrt (5)  =

576sqrt (5) cm^3  ≈  1287.975 cm^3    (just as qjin27 found  !!!! )

Very nice, Vinculum  !!!!!

1.  The limit as we approach -4 from the left =   +inf  = undefined

2.  The limit as we approach -4 from the right  =  -inf  =  undefined

3. The limit as we approach -2 from both sides  = + inf  =undefined

4. The limit as x approaches + inf  =  -1      (this is the horizontal asymptote )

5. The limit as x approaches - inf =  undefined    (the function oscillates in this direction )

This might not be the easiest approach....but.....

By Leg - Leg ,  the two right triangles   are congruent.....so DC = DB   =11

Since angle ABC is bisected, then AB  =(25/11)BC 

And, since angles ADB and CDB are supplementary  the  cosine of  ADC  = - cosCDB

By the Law of Cosines

(  BC )^2  = 11^2  + 11^2  - 2 (11)(11) (cos CDB)

(BC)^2  = 242   - 242 ( cos CBD)

[(BC)^2 - 242 ] / [ -242]   = ( cos  CBD )         (1)

And

[ (25/11)BC]^2  = 11^2 + 25^2  - 2 (11)(25) (-cos CBD)

[ (625 /121)BC^2 ] = 746 + 550 (cos CBD)

[ (625/121)BC^2  - 746 ] / [550]  =  (cos CBD)   (2)

Equate  ( 1) and (2)

[ (BC)^2 - 242 ] / [ -242]  =  [ (625 /121) BC^2 - 746 ] / 550         cross-multiply

550 [ (BC)^2  - 242 ] =  (-242)[ (625/ 121) (BC^2) - 746]        simplify

550BC^2  - 133100 =  -1250BC^2 + 180532

1800BC^2  = 313632

BC^2  = 313632 / 1800

BC  =  sqrt [ 313632 / 1800 ]  =  13.2

So   AB  = (25/11) (13.2)   =  30

The semi-perimeter of   ABD  =  (25 + 11 + 30) / 2  =  33

So  ....using Heron's Formula, [ ABD ]  is

sqrt  [ 33 * ( 33-25) * (33 - 11) * ( 33-30) ] = 

sqrt [ 33 * 8 * 22 * 3]  =

sqrt [ 17424 ]   =

132  units ^2

Sorry, Viculum  !!!....but, at least our answers agree  !!!!

Circumradius =     [product of the sides ] /  [ 4 * area ]

Since we have a right triangle the area is easy  =  (1/2)(product of the legs)  =  (1/2)(60)  =  30

Product of the sides  = 5*12*13  =  780

Circumradius  =     780  / [ 4 * 30 ]   =  780 / 120  =    6.5

This brings up an important point :  In a right triangle the circumradius =  (1/2) ( hypotenuse)

So :   (1/2) (13)   = 6.5

See   EP's  excellent answer ,here  :

https://web2.0calc.com/questions/geometry_77281

Let's suppose  that we just begin to fill the tank

After one hour we have a cylinder of water that is 5 m  high  with a volume of 80pi m^3

So

Vcylinder  =   pi (r^2) * (h)     .....so

80 pi  = pi (r^2) 5               divide through by 5pi

16 = r^2

r = 4  (m)