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See the following image :

If we draw  a  line segment  between the two intersection points  of the semi-circles....the length of  this segment - AC -  is given by

2√  [ 7^2  - 3.5^2]  =  (2 * 3.5)√ [ 4 - 1] =  7√3

Using the Law of Cosines

(7√3)^2   = 7^2  + 7^2  -  2 (49) cos (ABC)

147  =  98  - 2(49)cos (ABC)

49  =  -2(49) cos(ABC)

-1/2  =  cos (ABC)

arccos (-1/2)  =  ABC  = 

ABC  =  120°  =  (2/3)pi

So  (1/2)  of the blue shaded area  =

Area of sector ABC - area of triangle ABC

(1/2) (7^2)(2/3)pi - (1/2)(7^2) sin (120°)

(1/2)(7^2)(2/3)pi  -  (1/2)(7^2)*√3/2

(1/2) (7^2) [ (2/3) pi  - √3/2 ]

So....twice this area =  the blue shaded area  =

49  [ (2/3)pi - √3/2 ]  units^2 

I finished part of my essay so I decided to take a little break with math.

So the center of the circle is E.

1. We draw EA, which is the radius of the circle.

2. Now we connect the center to the midpoint of AD.

3. We set the hypotenuse of the triangle equal to X.

4. We know that the longer leg of the triangle is 5

5. We draw a line from center E to the midpoint of BC.

6. We know that is a radius. So the length of that is also X.

7. Since the side of the square is 10, we know that the line from center E to the midpoint of AD is 10-x

8. Using the pythagorean theorem, you set up the equation

\(5^2+(10-x)^2=x^2\)

9. Solve for X.

10. Use the circle formula to find the area of the circle using your value of X. Then  you have the answer.

Not giving my answer, but when I solved it, I got a decimal (or fraction).

26^3 *  1000   =  number of  plates  in old format

26^4 * 100  = number  of plates in new format

So

26^4 * 100  -  26^3 * 1000  =

28,121,600   =  difference 

f(1)  =  -1

f (-1)  =  7

f(7)  = -49

f(-49)  =  - 41

f (-41)  =   - 33

Very nice, Atroshus   !!!!

Multiply both sides by (x-4)(2x+1)

then we get (2x + 3)(2x + 1) - (2x - 8)(x - 4) = (x - 4)(2x + 1)

Simplifies to 31x - 25 = 0

so x=25/31

We try to rewrite the given equation in the standard form for an ellipse. Completing the square in both variables, we have

2(x^2+4x) + (y^2-10y) + c = 0

2(x^2+4x+4) + (y^2-10y+25) + c = 33

2(x+2)^2 + (y-5)^2 = 33-c

we need 2(x+2)^2 + (y-5)^2 to be 0 (degenerate ellipse)

so c=33

The identity  is    

C(n, k)  =   C(n - 1, k)  + C(n - 1, k -1)

So we have

[ C(5,0)  + C(5,1) ] + C(6,2) + C(7,1) + C(8,3) + C(9,2) + C(10,4) + C(11,3)

[C(6,1)] + C(6,2)  + C(7,1)  + C(8,3)  + C(9,2)  + C(10,4) + C(11,3)

[C(6,1) + C(6,2) ]  + C(7,1)  + C(8,3)  + C(9,2)  + C(10,4) + C(11,3)

[C(7,2)] + C(7,1)  + C(8,3)  + C(9,2)  + C(10,4) + C(11,3)

[C(7,2)+ C(7,1)]  + C(8,3)  + C(9,2)  + C(10,4) + C(11,3)

C(8,2)  + C(8,3)  + C(9,2)  + C(10,4) + C(11,3)

[C(8,2)  + C(8,3)]  + C(9,2)  + C(10,4) + C(11,3)

[C(9,3)] + C(9,2)  + C(10,4) + C(11,3)

[C(9,3)+ C(9,2) ]  + C(10,4) + C(11,3)

[C(10,3 ) ] + C(10,4) + C(11,3)

[C(10,3 )  + C(10,4)] + C(11,3)

[C(11,4) ] + C(11,3)  =

C(12.4) =  

495

*****

Find the positive integer n for which 3^9+3^12+3^15+3^n is a perfect cube of an integer.

n = 14 which gives =  19,683,000^(1/3) = 270

tysm, I understand how to do this problem now after reading the solution!!!

3)

We  actually have the  series

(1 + 1/9)    +  (1/9)  +   (1/90)   + 1/900  + ......+     =

(10/9)   +   (1/9)    +  (1/90) +  1/900 

The common ratio, r,  =    (1/10)    and the first term  is  (10/9)

So....the sum of this infinite series   =

(10/9)              (10/9)               (10/9)            10       10           100

______   =     ________   =    ______  =      ___ *  ___   =   _____  

 1   -  r              1  -1/10            (9/10)             9         9             81

1)

     1                     A               B

_______  =      _____  +   _____                       multiply through by  k (k + 3)

   k(k+3)               k             k +  3

1  =   A(k + 3)   + Bk        simplify

1 +  0k  =   ( A + B)k  +  3A

Equating like terms

3A  =   1                        

A = (1/3)

And

A + B  =  0

B = -1/3

So

(A, B)    =  (1/3, - 1/3)

Here's another way.  This way works not only here, but also in any other Windows application.  

Turn on the numeric keypad on your keyboard.   You're going to use numbers, but the keys across the top don't work. 

Locate your cursor at the spot where you want to insert the degrees symbol.

While holding down the ALT key, on the keypad type 0186 then release the ALT key.

Comparison of appearance:  

Using the ALT-0186 method:  98.6º

Using the x² button above:     98.6^o   

The x² button produces an easier-to-see symbol, but remember, you can use ALT-0186 in any Windows program.

_.

Ok. Thank you

From AMC 10 Website:

Permitted Materials

AMC 8 - AMC 10/12 - AIME
The only materials that students are allowed to have on themselves during the competition are blank scratch paper, blank graph paper, rulers, protractors, compasses, and erasers. NO CALCULATORS OR ELECTRONIC DEVICES OF ANY KIND ARE ALLOWED. No questions require the use of a calculator.

There are MANY solutions to your question, most of them "complex". There are also some real solutions, one of which is:  u = 7  and v = - 2

u^2  +  v^2 = 53

https://web2.0calc.com/questions/help-please_58204

I get x = 11

sqrt(x-2) = 14 - sqrt(11x)   square both sides

x-2 = 196 - 28 sqrt(11x) + 11x

-10x -198 = -28 sqrt (11x)          square both sides AGAIN

100x^2 + 3960x + 39204 = 8624x     rearrange

100x^2 -4664x + 39204 = 0     Use Quadratic Formula to find x = 11

The Big and Bad wolf is moving (10-6 m/s) faster.     4 m/s faster

to cover the 10 meters between them it will take      10 m / 4 m/s = 2.5 seconds to a steak dinner ! 

....you can just click the little  x² before you type the letter 'o' so you do not have to go back and highlight....but do no  forget to 'un-click' the x²

        ^{or you will get this !}