ElectricPavlov

9    and  3   o'clock

When the dolphin re-enters the pool   height (y) = 0

  solve   0 = -16t^2 + 24t    for t     

     another way......note that the dolphin reaches a peak at .75 sec.....  another .75 sec brings the dolphin back down .75 + .75 = 1.5 sec

   Max height occurs at .75 sec       max height = -16(.75)^2 + 24 (.75)   feet

Center at   -1,1

(x+1)^2   + (y-1)^2   = r^2         r^2 = distance between center and the point on the circumf     r^2 = ( 3^2 + 2^2) = 13

(x+1)^2 + ( y-1)^2 = 13      expand to get to the form requested ....

x^2 + 2x + 1    +  y^2  -2y + 1   - 13 = 0

x^2   + 2x        +  y^2  - 2y     -11   = 0                 <========= take if from here !

=   4*8  * n    =   32 n

Using sin law

(first note that angle C = 180 - 45-30 = 105 degrees)

sin A / sqrt 6 =  sin 105 / AB

AB = sqrt 6   sin 105 / sin 45

10 #/ft   *    1 ft / 12 inches    *    4 inches    *    16 oz / #    =    53.333 oz              (as proyaop found !)

See this:

h = 8

8 = -4.9t^2 + 14 t - .4

0 = -4.9t^2 + 14t - 8.4     Use quadratic formula to solve for t     the difference in the two t's is the time above 8 m

                                            a = - 4.9       b = 14       c = -8.4

\(t = {-14 \pm \sqrt{14^2-4(-4.9)(-8.4)} \over 2(-4.9)}\)                reults in t = 2   and  .8571      difference = 1.143 sec        <==== time = or above 8 m

30 km / 9 km / hr =   10 / 3  hr

(30 km + 20 km )   = 50 km       50 km / 10km/hr = 5 hours

5 - 10/3 =   5/ 3 hr   to cover 20 km

20 km / 5/3 hr =   12 km / hr

+x³ - x³  =  0 x³