CPhill

Let the  longer side of the  small rectangle  = x

Let the shorter side   =  y

For the width of the square we have  that

x + ( x - y)  + x  + (x - y) = 32

4x - 2y  = 32

2x - y  = 16  ⇒  -2x + y = -16    (1)

For the  height of the square we have that

y  + x  +  y  + x + y  = 32

2x + 3y  = 32   (2)

Add  (1) and (2)  and we have that

4y  = 16

y = 4

And x = 10

We can decompose the yellow area  into  2 rectangles

One has a width  of   (32 - x)  and a height of  (x+ y)

So...its area  =  (32 - 10) ( 10 + 4)  = (22) (14)  =  308

And the other rectangle has a width of (x - y)  and a height of (x + y)

So.....its area  =  (10 - 4) ( 10 + 4) =  (6)(14)  =  84

So....the yellow area =  308 + 84  =   392 units^2

a  = log₁₀ 3

b = log₁₀ 7

By the  change- of- base theorem we have that   

a = log 3 / log 10   =   log 3

b = log 7 / log 10 =   log 7                   [Note that log 10  = 1 ]

Also note that 

2a  =  2 log₁₀3   = log₁₀3^2   =  log₁₀ 9  = log 9 / log 10   =  log  9

So

log ₇ 9  =   log  9  / log 7   =  2a / b 

Find the centers of each of the 4  larger circles and connect them

This will form a square of 2 units on each side

And the distance from the center  of the circle at the top left to the  center of the circle on the bottom right  =

2√2  units

So......the diameter of the small circle =  2√2 - 2   = 2 [ √2 - 1]

And its radius is 1/2 of this  = [ √2 - 1 ]  units 

q(q + 1) (q + 2)(q + 3)  +  1   =

(q^2 + q)(q^2 + 5q + 6)  + 1   =

q^4  + 5q^3  + 6q^2

        +  q^3  + 5q^2 + 6q  + 1

_________________________

q^4  +  6q^3  + 11q^2 + 6q +  1

q^4  + 6q^3  + 9q^2 + 2q^2  + 6q +  1  =

(q^2 +3q)^2  + 2(q^2 + 3q)  +  1    =

([ q^2 + 3q ] + 1) ([ q^2  + 3q ]  + 1 )  =

(q^2  + 3q + 1)^2

Thanks, dude  !!!!

Since arc ST  = 170   and arc  QR  =  110

Then the sum  of  arcs  TR and SQ =  360 - 170 - 110  =  80

And  

Measure of P  =  (1/2)(arc TR - arc SQ)

30  = (1/2) (arc TR - arc SQ)    multiply both sides  by  2 

60  = arc TR  - arc SQ

So   we have these equations

arc TR  + arc SQ =   80

arc TR -  arc SQ  = 60         add these

2arc TR   =  140     divide both sides by  2

arc TR   = 70°

THX, whoisjoe.....!!!!

Correct, EP!!!

I just use this formula

d = 10^(8.6 - 7)  = 10^(1.6)  ≈  39.8  times as great

However.....your method actually conforms to what was asked   !!!!

∠A            ∠C      

___  =     ____       wihich implies  that  

   2             4

4∠A  = 2∠C       which implies that

2∠A    =  ∠C

But  ∠A  + ∠C   are supplementary......so....

∠A + 2∠A  = 180°

3∠A  = 180°

∠A  =60°

So  ∠C = 2∠A  =  2(60)  =120°

Further we have that

∠B           ∠C           120

___  =     ____  =   ____   =   30

  3              4             4

So

∠B

___  =   30

  3

∠B  = 90

And   ∠B  + ∠D   are supplementary

So

∠B   + ∠D   = 180

90  + ∠D  = 180

∠D  =90°

So  C is the largest angle  = 120°

See  here  :

https://www.desmos.com/calculator/umwyf6a0mf