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If I understand the question...

LCM [ 6, 8 ]  =  24

The multiples of 24 between 100 and 200 are  :

n = 120, 144, 168, 192    

So...n will have the same remainder  when divided by 6 or 8

(3a)

Note that angle CBD = angle CBA

And angle   CDB = angle BCA

So....by AA congruency, triangle BDC is similar to triangle BCA

(3b)

Note that angle CAB  =  angle CAD

And angle  ACB = angle  CDA

So....by AA congruency, triangle  ADC  is similar to triangle ACB

For the second one....it's easier to find y first

We have that   

12/5 = y/3      cross- multiply...

3*12 = 5*y 

36 = 5y      divide both sides by 5

36/5 = y = 7.2

And

[12 + 5 ] / x  = [y + 3 ] / 6

[ 17 ] / x = [ 7.2 + 3 ] / 6       cross-multiply

17*6  = 10.2 x  

102 = 10.2 x     divide both sides by 10.2

10 = x

You are saying that any link disengaged remains a singleton?

Or can it be left attached to some longer length or chain?

Challenge 4:

x:6 = 5:3 because of parallel lines.

we can write this in fraction form and solve:

x/6=5/3

6(x/6)=6(5/3)

x=30/3=10

5:3=12:y

5/3=12/y

y(5/3)=y(12/y)

5y/3=12

3(5y/3)=3(12)

5y=36

y=36/5

I revised it.

Sorry, can't read your "condition"!.

f(1)=2

f(2)=2^2=4

f(4)=2^4=16

f(16)=2^16

if you take the square root of 2^16, the answer is 2^(16/2)=2^8=16^2=256

I think the parallelogram has height 6 and the sides perpendicular to the height are 9, and the two remaining sides that are not parallel to the height are 2.5.

In that case, you use the formula a=bh, so the answer is 9*6=54.

Remember that, by the Triangle Inequality

11 + 17 > x

28 > x

So.....the longest possible integer side is 27

Really need to see a pic here, pigglepoox......hard for me to see what is being described

OK.....I'll see what I can do...!!!

We have two triangles....one with a height of 5 inches and the other with a height of 14 inches

And both have  the same base = 7.5 inches

So.....the total area is just the area of both triangles =

(1/2) base ( 14 + 5) =

(1/2) (7.5) ( 19)   =  71.25 square inches

The height of the parallelogram = 8

And the width = 5

So   [ ABCD ]  =    width * height =  5 * 8   =    40 ⇒  "D"

The bottom base = top base

So....the area =   (bottom base) (height) = (6) (7) =   42 square meters

f(g(x))  =  x^2 + 2

g(f(x) )   =  (x + 2)^2

So.....we want to know

x^2 + 2  =  (x + 2)^2       simplify

x^2 + 2  = x^2 + 4x + 4    subtract x^2 from both sides

2 = 4x + 4                      subtract  from both sides

-2 = 4x                       divide both sides by  4

-2 / 4 =   x =   - 1 /2

OK....glad to help

Yep.....40

Area of triangle   =  (1/2)(base)(height =  (1/2)(5) (16)  = 40 in ^2

Let the point on the graph be   [ a, (1/√2)(a^2   - 3) ]

So....if D is minimized....then so is D^2

So we have that

D^2  =   (a - 0)^2 + [ (1/√2)(a^2   - 3)   - 0)]^2

D^2 =   a^2    + (1/2)(a^2- 3)^2             take the derivative of this and set to 0

D^2 '   =     2a  +  (a^2 - 3)(2a)   = 0

D^2 ' =     2a [  1 + (a^2 - 3 ]  =  0

This implies that either

2a = 0

a = 0

Or that

1 + a^2 - 3 = 0

a^2  - 2 = 0

a = ±√2

When   a  = 0, D^2 =   0^2    + (1/2)(0^2- 3)^2  =  9/2

When a =  -√2   ,  D^2 =    (-√2)^2 +  (1/2) [ (-√2)^2 - 3 ]^2  = 5/2

The same result will resullt when a = √2

So D  will be  √[5/2]   =  √2 * √5  / 2   =  √10 / 2

And   a + b   =   12

-1/2