CPhill

If  (a + b)^2  = 49......then either

a + b  = 7               or          a +  b  =  -7

b = 7 - a                or           b = -7 - a

And if ab  = 10, then either

a(7 - a)  = 10        or        a (-7 - a)  = 10

7a - a^2  = 10      or        -7a - a^2  = 10    

Rearrange both as

a^2 - 7a + 10  = 0            or            a^2 + 7a + 10 = 0

Factor

(a - 5) (a -2)  = 0            or            (a + 5) (a + 2)  = 0

Setting each factor to 0, a takes on 4 values

a = -5, -2, 2 or 5

And using ab = 10, b = 10/a....so.......b = -10/5, 10/-2, 10/2 or 10/5   .......   and these values are b in terms of the value for a

Thanks, Guest.......here's a slightly quicker method....

x + 2y = 20    →   y  = [20 - x ] /2     (1)

2x + z = 22   →   z  =  22 - 2x            (2)

y + 2z = 9    (3)

Substituting (1) and (2)  into (3), we have

[20 - x] / 2 + 2 [22 - 2x] = 9   multiply through by 2

20 - x  + 88 - 8x  = 18        simplify

-9x + 108  = 18      subtract 108 from each side

-9x = -90           divide both sides by -9

x = 10

And using (1) and (2)        y  = [20 - 10] / 2   = 5        and     z = 22 - 2(10)  =  2

Like this :  https://www.desmos.com/calculator/lcofxardxh

The temperature at 11000 ft is given by :

59.7 - 1.1 (11)  =

59.7  - 12.1  =

47.6° F

The same problem at my local library.......I used to be able to log on there, before the new format......now........when I log on, it throws me out when I try to do anything....

My laptop at home works fine........

Odd.......

Thanks for that video, Melody........

Unfortunately.........Sir La-Tex    doesn't fully grasp "imperial" measurements......

He deals easily with "imperial minims,"  however........

Also.....the problem with trying to ascertain  a good estimate for "thumb and finger hole volume" is that this is equivalent to trying to determine how much of a pie is displaced by having one's thumb and fingers in it..........

Thus, there is probably some other  "pie" component to our spherical volume estimation.......

Where is the decimal supposed to be placed in your %.......is it  1.515% or 15.15%.......or something else ???

Anyway........the general answer is :

C(2020, 22) * (percent of defect expressed as a decimal)^22 * ( 1 - percent of defect expressed as a decimal)^1998

85.4 kg to 78.7 kg.....the percent is given by :

  ( [ Old Weight - New Weight ] / Old Weight  x 100 ) %    .....so we have

 ( [ 85.4 - 78.7] / 85.4 x 100 ) %  =   about 7.85%

LOLOLOL!!!!!

Good point......we'll consider it "pre-drilled".........only because I couldn't find any "finger hole" formulas to reduce the volume.......

I may see if Sir LA-Tex knows  of any such formulae........he's pretty good at researching the obscure........!!!!

Stay tuned.......

The side  = 56/4   = 14

And the diagonal of a square is sqrt(2) times as long as the side....so....

The diagonal =  14*sqrt(2)   = about 19.8