ElectricPavlov

OK.....going to do some conversions into the metric system

11.58 ticks / 100 ticks/min x 60 sec/min =6.948 sec seconds

100 bar x 4/5 m/bar = 80 meters       His RATE is   80 meters/6.948sec = 11.514m/s

Usain rate = 100m/9.58sec = 10.43 m/s

So       10.43k = 11.514     k = 1.10

OK....the question was changed after I posted my answer   9.58 ticks was changed to 11.58 ticks

     changed my answer to reflect new numbers

OK.....what length of pipe is a cubic foot?

pi x r^2 x l = 1

3.14 x 2^2 x l = 1      l = .079577 ft of pipe is a cubic foot

1 foot per second is then    1/.079577 ft^3 of water per second = 12.566 ft^3/sec

12.566 ft^3/sec x 3600 sec/hr x 24 hr = 1,085,734 ft^3/day ~~~ 1,090,000 ft^3 /day

3L/8min    2L/5min     find(lowest) common denominator  (40)

15L/40min      16L/40min         At 40 minutes they will have combined 15+16 = 31 laps

(added 'lowest')

Does Sejes run 100 bar race in 9.58 seconds.....or 9.58 ticks?

ANother perspective;

the situation in the question is described by the following equation:

18/(x-9) = 18/x +4/60      where x = her usual speed in mph   (x-9) = new (slower speed) &  4 minutes = 4/60 hr

solve for x

18 = (18x-162)/x  +4/60x - 4/60(9)

18x = 18x-162  + 4/60 x^2 - 36/60 x

0 = .06667 x^2 -.6x - 162       Use Quadratic Formula to find x = 54 mph

Her new (slower) speed = 54-9 = 45 mph     so the trip will take her   18 m / 45m/hr = .6hr= 24 minutes

First, find the slope    = rise/run = (y1-y2)/(x1-x2) = (5-17)/ (4-8) = (-12)/-4 = 3    

   This is 'a'

  So the line , so far looks like

      y = 3 x + b      Now, let's put one of those x, y points in there to find 'b'

      5 = 3(4) + b

       5-12 = b    b = -7

So the equation for this line is   y = 3x -7      and    a-b = 3-(-7) = 3 + 7 = 10

Wooohoooo!  Nice, Chris !

Please post this 'it-shay' somewhere else....

    DId you get hacked?   Please delete your account...and start over with a new password and ID

I assume you mean the total area of the square + the triangle should be maximized

the square's area = l x l = l^2

the triangle's area = sqrt3/4 * a^2   

And    a = (5m - 4l)/3

Have to agree with above....if you want to maximize area, you would not put ANY wire in to a triangle...it would all be in the square......UNLESS you have a size for the triangle...or some sort of ratio of square to triangle....there really is no answer to this question. 

If you use the quadratic formula you will find   a= 52/8   or -20/8

    then you would have      (a-52/8)(a+20/8)        but you will see that   a x a = a^2   NOT 4 a^2.....

        so lets multiply both terms by 2

                  (2a-52/8 * 2)( 2a+20/8 *2)

                   (2a-13)(2a+5)