ElectricPavlov

35    and  12   work       by brute force

Last one:    the 19.3 and 14.2 dimensions will fit in the box without problems

   the WIDTH of the disc case is the questionable one   1.6 cm      sixteen of them would need 16 x 1.6 = 25.6 cm

     the width of the OUTSIDE of the box is 26 cm.... if the walls of the box are greater than .2 cm each, the interior dimension will not allow for 16 of the disc cases to fit inside.

First one:   (please don't post multiple questions in one post-----you may not get responses to all of them !!)

y = yellow cubes

b = blue = 3y

g= green = 5b = 5(3y) = 15 y

y / (y+3y+15y) =   y / 19y   =  1/19    chance of getting yellow

x = 20/(tan 38)   =  20/(.781286) = .....                      if 38 is in degrees

3000000 users/day    /    ( 3600 s/hr  *  24 hr/day )

3 000 000 / (3600*24) =.....

pi d   is circumference

11pi/18   /   2pi    is   fraction of circumference

(11pi /18)  /  2pi      *  pi *d    =   78.6       solve for 'd'     then radius is 1/2 d

x = score in 3 innings

x+5 = score in remaining innings    summed = 9

x      +     x+5             = 9       solve for x the 3 inning score, then the remaining innings score is  x+5

8x+3y=52

10x-3y=38     Add the two equations to get

18x     = 90    divide by 18

x = 5               Use this value of 'x' in one of the equations to compute 'y'

2m / 25 m/hr = .......    time to cross

(time to cross) * 17 m/hr   = distance downstream

Area of triangle = 1/2   base  *  height

Base and height are both   6 sqrt2

1/2 *(6sqrt2)(6sqrt2) =.......