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Knockout is correct....!!!!

Note  that  all the possible   positive differences are generated by

3 -1      =  2

5 -1     =  4

7-1      =  6

9-1       = 8

11 - 1  = 10

13 - 1  = 12

All the other positive differences  are just subsets of these 6

The explanation is sort of complicated, and I'll try to figure out a way to simplify what I want to say into a few short words, but in the meantime, I am going to tell you that the answer I got for this problem was \(\boxed 6\).

222

225

252

255

522

525

552

555

A B C   =  3!   = 6

A B D  =   3!  = 6

A C D  =   3!  = 6

A B B  =  3! / 2!  = 3

A C C  =  3! / 2!  = 3

A D D  =  3! / 2!  = 3

A A B  = 3

A A C  =  3

A A D  = 3

A A A  = 1

So....I get   37

Anyone get something different  ???

Triangles AZY and  ACY have the same height

But base AC is 5/2  of base AZ..so area of ACY  = (5/2) AZY

Similary....triangles ZYC and  AZY  have the same height

But base ZC is  (3/2) base AZ...so  area of ZYC  = (3/2) AZY

And triangles ACY  and  YBC  have the same  height

But base AY is 3 times base  BY....so area of ACY = (3)YBC

So.....

ACY  = (3)YBC    (1)

ACY = (5/2)AZY  (2)

ZYC = (3/2)AZY    (3)

From (1)  and (2)

(3)YBC = (5/2)AZY

(6/5)YBC  = AZY    (4)

And from (3)

(2/3)ZYC  = AZY  (5)

So...from  (4)  and (5)

(6/5)YBC =  (2 / 3)  ZYC

So

[ YBC ] / [ ZYC ]  =    [  2/3] / [ 6/5]

[ YBC] / [ZYC ] =  (2/3) * (5/6)   =  10/ 18  =  5 / 9

Thank you so much, CPhill!

f^-1(4), f(x)=4x-2, 4=4x-2, 4x=6, x=6/4, x=3/2. So, \(f^-1(4)=\frac{3}{2}\).

Expanding, gives us: 4z+20=32, 4z=12, z=3.

Excellent,  tertre  !!!

Draw QT  and QS

Angles TQP, SQT  and  SQR  will trisect the interior angle PQR

And PQR  = 108

So  ...angle  SQT  = 108 / 3   = 36

But...by symmetry....angle XQS  =  (1/2)angle SQT  = (1/2)36  = 18°

AWX  and AYZ  are under the same height

So....they are to each other as teir bases

So  WX / YZ  =  30/20   = 3/2

WX =  (3/2)YZ

So

WX + XY   = 9    ⇒  (3/2)YZ + XY  = 9       (1)

XY + YZ  =  7     ⇒  YZ + XY  =  7         (2)

Subtract  (2)  from (1)  and we have that

(1/2)YZ  = 2

YZ  = 4

So XY = 3

And WX = (3/2)YZ  = (3/2)4  = 6

So......the height of   AWX  can be found as

30  = (1/2) WX * height

30  = 3 * height

10 =  height  of  AWX  = height of  AXY

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

We can solve this with a proportion

32/ 60  =  L / 15         divide top bottom of the fraction on the left side by 4

8 / 15  = L  / 15

L = 8 ft = shadow cast by radio tower

See my corrected answer, shreyas1

The rectangle at the top left has the perimeter of 24

The rectangle at the top  right has the perimeter of 42

The rectangle at the  bottom right has the perimeter of 32

So...we are trying to find the perimeter of the rectangle at the bottom left

Since the top rectangles are under the same height...call this height x

Call the width of the top left rectangle, y   and the width of the top/bottom right rectangles, z

Call the  height of the bottom left/right rectangles, w

So  

2x + 2y  = 24           2x + 2z =  42

x + y  = 12     (1)              x + z  = 21    (2)

Subtract (1) from (2)  and we have

 So    z - y = 9  ⇒   z  = y + 9    (3)

So...for the bottom right rectangle

2 ( w + z )  = 32

w + z  = 16

z = 16 - w      (4)

So...equate  (3)  and (4)

y + 9 =  16 - w

y + w  = 7

So  the perimeter of the bottom left rectangle  =

2 ( y + w)  =

2 (7)  =  14

Done, sorry about the Time limit as well

LOL    

Thank you for that answer Omi67

Thank you guest and CPhill

Thank you CPhill

HAHAHA!!!...on these geometry problems....I'm never too sure, either...!!!

Thank you CPhill

Thank you so much Melody.

Thank you so much guys.

I have already started doing that. Thank you and sorry again.