ElectricPavlov

I'll TRY.....First here is a graph of the area of the base:

  area of equilateral triangle =  sqrt(3)/4 s^2

s in this case = 2x from 0 to 3

 from 0 to 1   this is    2y^2            from  1 to 3 this is     6-4y^2      now square s

                                  4 y⁴                                                     16y^4 -48y^2 + 36

sqrt( 3)/4  ( 4y^4 )                             +          sqrt (3)/4 ( 16 y^4 - 48y^2 + 36)   

      this part is y 0 to 1                                          this part is y is 1 to 0

Hmmmm...... I am a little stuck...a bit rusty......If I figure this out (this may not be the EXACT way to do it) ...I'll edit this post

    maybe asinus will answer correctly....

d = 5

a₁ = 16

a₉₀  = a₁    = 89 d

        = 16 + 89 (5) = .........

2/3 * batch = 4 cups

batch = 4 * 3/2   cups = ............   cups

2/5 * c0st = 240

cost =     240 * 5/2   = $ ..........

cross multiply the left side

gx^2 -g 4x + hx + 5 h  =  x^2 -2x+10         so    g = 1   h = 2      take it from here !

First 2x + 2y = 180     (angles next to each other in a parallelogram are supplementary)

x+y = 90

then   x+y + 180 -2z  = 180            (sum of angles in a triangle)

            90   = 2z

               z = 45^o

Welll.... we found that

x = 3/13 a

2 = 3/13 a

a = 26/3

40 days / 5 days/ half life = 8 half lives

13 mCi * (1/2 )⁸   = .05 mCi

1 hour is 3 half lives

25 x 1/2 x 1/2 x 1/2 = 3.125 mCi

dose remaining_8hrs = 15 (.84)⁸  = 3.72 mg