CPhill

Actually......here's the reason for this.........

Let's suppose that  we have a rectangle divided into 4 smaller rectangles of equal  sides of  "m" length  and "n" height.....

And the exact center of the rectangle will be at  (n, m).....

Now...for instance...let us shift  this point down x units and to the right by  y  units

Drawing two lines parallel to the sides of our original rectangle through this point  will produce four new rectangles.......

So.......the perimeter of the original top left rectangle will increase by 2x + 2y  units

And the perimeter of the original bottom right rectangle will decrease by 2x + 2y units

And the height of the original rectangle on the top right will increase by y units but the width will decrease by x units.......so......the new perimeter is 2(m+n) - 2x + 2y units

Finally, the height of the original rectangle on the bottom left will decrease by y units but the width will increase by x units..........so......the new perimeter is 2(m+n) + 2x - 2y units

So.....the sum  of the perimeters of the top left and bottom right rectangles will be :

2(m +n) + 2x + 2y + 2(m + n) -2x - 2y  =   4 (m + n)

And.....the sum  of the perimeters of the top right and bottom left rectangles will be :

2(m +n) - 2x + 2y + 2(m + n) +2x - 2y  =   4 (m + n)

So........the sum of the perimeters of the diagonally situated rectangles will be exactly the same !!!

I happened to notice something else interesting about this last problem.....the sums of the two intermediate perimeters = the sum of the two "extreme" perimeters......

Probably  just a coincidence  ???

But maybe not  ???

Note .....perimeter of  AEJF = 18......perimeter of JHDG  = 8

Perimeter of ECHJ  = 10  ......perimeter of FJGB = 16

So.....the sum of the extreme perimeters  = 26

And the sum of the intermediate perimeters  = 26

Does anyone know if this is a "theorem"  ????

If not.....I posit it as "The CPHill  Conjecture of Mean and Extreme Perimeters".....LOL!!!!!!

Call the amount at 60%  =  x

Call the amount at 15%  =  450 - x

So

.60x + .15(450 - x)  =  .25 (450)   simplify

.60x + 67.5  -  .15x =   112.5

.45x  =  45    divide both sides by .45

x = 100 gallons ......this is the amount at 60%

450 - x  =  450 - 100  =  350 gallons  = amount at 15%

You have 4 variables but only 2 equations....we can't derive any solutions from this

{ }   means no solutions exist

Are you sure this isn't meant to be

xy = 76  and    x  +  y  =  18    ???

If so.......   xy = 76   ⇒   y = 76/x

Putting this into the second equation, we have

x +  76/x  =  18       multiply through by x

x^2  + 76  =18x      rearrange as

x^2 - 18x + 76  = 0

The two solutions are     x =  9 - √5     and x = 9 + √5

Whichever you choose for x, y will be the other one

1.   We can't take the log of a negative or 0...... so.....domain ⇒ x > -5

2.   all real numbers

3.   all real numbers

4.   all real numbers 

5.   all real numbers  

6.   6 - 3x  must be ≥ 0  .....so...... x ≤ 2

7.   all reals except  x = 3    [ 3 makes the denominator  = 0 ]

TR has recently re-surfaced........AWESOMEE is probably gone forever......

Very nice, hectictar.....this one stumped me, too!!!!

All triangles are under the same height

If AWX  = 30......and AYZ  = 20....we have that

(1/2)WX * height   =  30

(1/2)YZ * height  = 20

Which implies that    WX =  (3/2)YZ

Which implies that

WX + XY = 9       and

YZ +  XY = 7        so

(3/2)YZ + XY  = 9

YZ        + XY = 7     subtract these two equations

(1/2)YZ  =  2      ⇒  YZ  =  4       ⇒ XY = 3

And the height is   (1/2)YZ  * height = 20    ⇒  (1/2)4 * height = 20  ⇒ height = 10

So.....the area of AXY  =  (1/2) XY * height  = (1/2) * 3 * 10   =  15 units^2

Hectitar's answer was more concise.....I told my ghost not to be so long-winded!!!!!

Thanks, hectitctar!!!......there is a sligtly faster way of finding the missing area

Using hectictar's diagram [ along with a theorem from Euclid ]

        area of ab *  area of dc =   area of bd * area of ac

               4        *         5            =         3        *    area of ac

                      20                      =          3        *    area of ac      

                                Divide both sides by 3

                                20 / 3   =  area of ac