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They can start with a  1 3 5 7 or 9     5 numbers odd    or   2 4 6 8      4 numbers even  

   the last number can be even or odd with same frequency       

    4 out of 9    x    1 out of 2

4/9 x 1/2 = 4/18 = 2 / 9

x = square side length

Triangle in bottom right has base length of     (8 - 2x)/2   

Tan 60 = opp/adj  = x / (8-2x)/2   = 2x/(8-2x)     tan 60 = sqrt 3

sqrt 3 = 2x/(8-2x)

8 sqrt 3 - 2x sqrt 3 = 2x

8 sqrt 3 = 2x + 2x sqrt 3

8 sqrt 3 = x (2+ 2 sqrt 3)

( 8 sqrt 3) / (2+2 sqrt 3) = x

 rationalize the denominator to get

x = 6- 2 sqrt 3

With four tangent unit circles, we can form two squares. The large gray square is formed by the tangency points between the circles. The small blue one is formed by joining the midpoints of each arc that are formed by tangency points.

Find the difference of between the area of the gray square and that of the blue square.

"Two red triangles the sum of them is 12, let the red triangle be x, x+x=12, 2x=12, x=6, one red triangle=6) "

"Yellow rectangle and half-shaded triangle, logically I'd give that triangle value of 3 rather than 6. , so let yellow rectangle=f, f-3=1, f=4, yellow rectangle=4

"Green circle added to blue hexagon=17", "Blue hexagon - Green circle =7" hm let's use variables, let the green circle be q and blue hexagon be p" 

q+p=17

p-q=7   system of equations, add both equations,

2p=24

p=12, remember that hexagon is p so hexagon value is 12, substitute, green circle +12=17, green circle=5

(Other solution than the system of equations (A logical one rather than mathematical), Notice the pattern that each figure's value number is represented by the number of triangles inside except the circle. I.e. look at the yellow rectangle, draw to diagonals we have 4 triangles, thus it values is 4. half-shaded triangle, only the shaded area so it is half of the full triangle which is 6 if you draw diagonals. 

E.g. the blue hexagon, by drawing it's diagonal you will have 12 triangles and its value is 12.)

anyways let's continue, 

So far we have red triangle= 6

Yellow rectangle=4

green circle(q)=5

blue hexagon(p)=12

Thus last line,

6+4+5+uh?=?

well, it is hexagon leaving 2 triangles outside thus its value is 10?(Logical approach)

The mathematical approach would be 4.. as it is a rectangle. 

so

6+4+5+10=25

6+4+5+4=19 

Which one is correct? Check answer..

how do I add the degree sign after a number    

You mean here in the forum?  What I do is type a lower shift letter "o" , put a couple of spaces after it, then go back and select it.  By select it, I mean hold down the Shift key and swipe across that one letter.  Then I go up to the top of this field and click the x² button.  

_.

Find the units digit of 1^5 + 2^5 + 3^5 + ... + 123^5.

looking at the last digit for each one

1+2+3+4+5+6+7+8+9+0 = 45

45*12 = 540

540+1+2+3=546

So I think the last digit is 6

Where did that squareroot come from CalculatorUser?

It is interesting that you can answer a question when half of it is missing.

Half your question is missing.

Try proofing it before you post!

The hexagon below has interior angles that measure all the same. As shown, only five of the six side lengths are known: 10, 15, 4, 18, 7. What is the unknown side length?

Thanks Hellospeedmind  

In the figure below, there are 6 squares: 4 small, 1 medium (in blue), and 1 large. There are 9 circles, which will be filled with all the integers 1 through 9 such that the sum of the vertices of each square is 20.

           What is the product of the 4 numbers along the blue square?          

                                     The product   P = [sqrt(2025 * 7⁰) /2] + sin(30°)     

Btw, on this concept, smart toys could be made.

2 , 2 , 2 , 5 , 11 , 73 , 137 , Total Prime Factors = 7
Distinct Number of Factors = 5
Euler's Totient = 1566720
(1, 2, 4, 5, 8, 10, 11, 20, 22, 40, 44, 55, 73, 88, 110, 137, 146, 220, 274, 292, 365, 440, 548, 584, 685, 730, 803, 1096, 1370, 1460, 1507, 1606, 2740, 2920, 3014, 3212, 4015, 5480, 6028, 6424, 7535, 8030, 10001, 12056, 15070, 16060, 20002, 30140, 32120, 40004, 50005, 60280, 80008, 100010, 110011, 200020, 220022, 400040, 440044, 550055, 880088, 1100110, 2200220, 4400440) >>Total = 64 Divisors

There are 7! = 5040 ways to label the heptagon.  There are 4! = 24 ways where the O's are next to an N, so the number of ways where N is not next to an O is 5040 - 24 = 5016.

There are 13*3 possibilities for the first pair, leaving 12*3 possibilities for the 2nd pair, leaving 11*4 possibilities for the singleton.

So:  13*3*12*3*11*4 = 61776 in total.

First, we can choose the seats for the four mathematicians.  Then we seat the physicits, then the engineer.  There are 4! ways to seat the mathematicians, then 3! ways to seat the physicists, then 1! ways to seat the engineer.  But then we must divide by 8, because we can rotate the table, so there are 8!*4!*3!/8 = 725760 ways to seat everyone.

Well Im going to give ya my method:

The diagonal of the square is equal to the diameter of the circle around it.

The Diameter of that circle is equal to the side length of the square around it.

So you just keep calculating. Quite a tedious problem, once you know the method.

Find units digit of each term

1^5 is 1

2^5 is 2

3^5 is 3

See the pattern here?

Its 1 + 2 + 3... 121 + 122 + 123

So (123)*(124)/2 = 7626.

The units digit of that number is \(\boxed{\sqrt{6}}\)

a^b * b^a  mod 2000 = 0 

a = 5   and   b=4

a  *  b = 20

sumfor(n, 1, 123, n^5) = 591,309,561,876

Circle B radius     >   R_b = 1

Circle A radius     >   R_a = ?                  R_a = sqrt [ (R_b)² + (C/2)² ]       R_a = 2.236 u² 

Chord length        >   C = 2

Area of  A             >   A₁ = ?                   A₁ = (R_a)² * pai                 A₁ = 15.708 u² 

Area of  B             >   A₂ = ?                   A₂ = (R_b)²* pai                 A₂ = 3.14159 u² 

Ratio                     >  A : B = ?               A  :  B =  A₁ / A₂               A : B = 5 : 1     

The original  total of the numbers will be    80 n

then remove one number, 100 and the average becomes 78

  the new total will be   78 (n-1)        which is 100 less than the old total  80 n

    soo   78 (n-1) = 80n - 100

             78n-78 = 80n - 100

                   22 = 2n

                     n=11

The table forms an AP
55, 110, 165, 220..........etc.
Sum = 10/2 * [2*55 + 55(10 -1)]
Sum =   5 *    [110 + (55*9) ]
Sum =   5 *    [ 605 ]
Sum = 3,025