CPhill

5)  All of these work the same, pretty much....

% of 25% protein milk * unkown amt  + % of 40% protein milk * [ 12 - unknown amt ]   =  40% protein milk * 12 L

.25x  + .40(12 - x) = .35(12)    simplify

.25x + 4.8 - .40x =  4.2         

-.15x + 4.8 = 4.2      subtract  4.8 from both sides

-.15x  = -0.6      divide both sides by  -.15

x = 4  L of 25%    and    12 - 4   = 8L  of 40%

4).....How much water = x  .....it's easier if we express everything in terms of % water....... note that 32% salt solution = 68% water solution    and pure water = 100%  = 1     and a 50% solution = .50 water

.68(20)  -  1x  =  .50 [ 20 - x]    simplify

13.6 - 1x  = 10 - .50x      subtract 10 from both sides.....add 1x  to both sides

3.6 =  .50x       divide both sides by .50

x  = 3.6 / .50   = 7.2 L

3) Note.... 1/2 = .50       3/4 = .75     3/5 = .60   ....so we have

%purity * known amount of XRD  + %purity * unknown amount of QXJ  =  final %purity [ known + unknown amts. ]

.50(1)  + .75x  = .60 (1 + x)      simplify

.50  + .75x  = .60 + .60x        subtract .60x, 50   from both sides

.15x  =   .10      divide both sides by .15

x = [.10 / .15 ] L  =  [10/15 ] L =  [2/3]L

2)   Note that pure octane = 100% = 1.00   

.95(1)   + 1.00 (x)   = .96(1 + x)   simplify

.95  + x   =  .96 +  .96x      subtract  .96x,  .95 from both sides

.04x  = .01   divide both sides by  .04

x = 1/4 L

I think we need a logisitc growth model here......this is given by:

P = [ P₀K ] /  [ P₀ + (K  - P₀)e^(-rm) ]    

Where  P₀   = 29  = the initial no. of people       K  = 87  = the max no. of people     r = 10%  = .1     and m is the number of months   .....so we have

P  = [29 * 87] / [ 29 + ( 87  - 29)e^(-.1m) ]   = [2523] / [58e^(-.1m) + 29 ]

Here's the graph......notice that 87 is the "carrying capacity"   ........https://www.desmos.com/calculator/bs1mlbrbrc

[ Also....the graph is only valid for m >= 0 ]

For the second part, I think we have this

1800  = 5040 / [ 71 e^(-.182t)+1]   rearrange as

71 e^(-.182t)+1  =  5040/ 1800        subtract 1 from both sides

71e^(-.182t)  = [5040 - 1800] / 1800   simplify

71e^(-.182t)   = 1.8       divide both sides by 71

e^(-.182t)   = 1.8/ 71   take the ln of both sides

ln e^(-.182t)  = ln[ 1.8 / 71  ]      and we can write

-.182t ln e   =  ln [ 1.8/ 71]      and ln e = 1    so we can ignore this....and we have

-.182t   = ln [ 1.8 / 71]          divide both sides by -.182

t=  ln [ 1.8 / 71] /   [-.182 ]   = about   20.19 years

Here's the proof........good luck!!!!  [ It's way above my pay grade......!!!! ]

http://www.artofproblemsolving.com/wiki/index.php/Fermat's_Two_Squares_Theorem

Very nice, Melody.....but....you stole my thunder on this one.....curses on you.....!!!!    [LOL!!!]

I had forgotten  - until  I thought about it awhile ago -  that  we could have had   7p - 24 =  5p  or  that 7p - 24  = -5p    and  it's obviously the second one that's correct

I might give myself points for just the thought......plus....you didn't even share any of your BBQ with me.....!!!!!

I still gave you 5 pts.....

Here's 6

1.   angle  XUW  = angle 1  = 90                                      given

2.   angle XUY   = 180                                                      XUY  is a straight angle = 180 

3.  angle XUY =  angle YUW + angle XUW                      angle addition theorem

4.  angle XUY - angle XUW  = angle YUW                      subtract angle XUW from both sides of 3

5.  180  -  90     = angle YUW                                           substituting 1 and 2 into 3

6.  180  =   angle YUW + 90                                              add 90 to both sides of 4

7.  180  =  angle 2  +  90                                                    angle YUW =  angle 2

Here's 5

1.     angle 1  = angle 3                                                                               given

2.   angle JOK  = angle LOM                                                                      given

3.  angle JOK + angle KOL  = angle KOL +  angle LOM                        add angle KOL to  each side of 2

4.  angle JOL       =   angle  KOM                                                          angle addition theorem