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Dude, chill. Be patient. I won't answer your question because you need to learn that people won't answer you if you are so impolite.

Hmm, surprisingly, it does work for pi/45:

Perimeter of semi circle    1/2 pi * 8  cm

find hypotnuse of triangles   =   sqrt (10^2 + 4^2 ) = sqrt 116      there are two of them so   2 * sqrt 116    cm

Add the REDs .....      cm

Pretty straightforward....area:

area of SEMI-circle on the left =  1/2  pi (4²)    cm²

triangles (two of them)  area  =   2  *   1/2 *  10 * 4      cm²  

       add them together

If I understood your question, then:

a=1;b=2^a;c=2^(a+1)*b;printc,", ",;a++;if(a<24, goto1, 0)

1, 2, 8 , 32 , 128 , 512 , 2048 , 8192 , 32768 , 131072 , 524288 , 2097152 , 8388608 , 33554432 , 134217728 , 536870912 , 2147483648 , 8589934592 , 3 4359738368 , 13 7438953472 , 54 9755813888 , 219 9023255552 , 8,796,093,022,208, which is the 23rd term = 2^43

I believe this is a square with rounded corners resulting in the following (sketch it out)

(7 x 5)    +     (1 x 5)  *2      +    pi (1)^2

There are two different triangles: ΔABC  and  ΔADE   These triangles share the same angle,  ∠BAC  or  ∠DAE

1)       Use the Law of cosines to calculate an angle BAC of a triangle ABC.

2)       Use the Law of cosines to find the length of DE.    

Extand line AO to the other side of the circle   DAO then intercept 80 degrees of a half circle

    angle b intercepts anothe 60 degrees of this half cirlcle

        angle cod is a central angle and intercepts the rest of the half circle   180  - 80 - 60 = 40 degrees

3y + 9  =   2y+16    solve for 'y' 

and 

5x-4  =  10x - 24       solve for 'x'

'Guest'   question submissions are generally given lower priority....

Use distance formulat TWICE between    0, 0   and   -4,6      and then again on   -4,6   to   4,3   

d^2 =    (x1-x2)^2     +   (y1-y2)^2

then add the 'd' s together.....

Given line slope ,  m = 3/2       perpindicular slope  = - 1/m  =  -2/3

y = -2/3 x + b      sub in the given point to calculate 'b'

6 = -2/3 (-12) + b        shows b = -2

y = -2/3 x - 2

Still do not know what is being asked in this question AGAIN

(3 ,  5/3 x)

(-6, -1/2 x)      perhaps?

Perimeter     2l + 2w = 60

area               l * w   = 210       w = 210/l    sub this into the first equation

2l  + 2 * 210/l   = 60

2l^2  -60l + 420 = 0      Use quadratic fromula to find   l  &  w   =   15 + sqrt15      and     15 - sqrt 15   cm

The answer is 97, but I cheated and used a piece of software to find it!  (You could use Wolfram Alpha).

Try x = 2, y = 1, z = 1/2

This gives an even less reason for us to answer your questions.

If the lines AR and BS cross at the right angle, the result will be the same!

Let SR be 10 units long.

PS = 5 + sqrt(2)

So, we have the area of the rectangle    A = 10 * (5 + sqrt(2))  

I don't think that 16 is attainable.

My 'messy' calculations point to a minimum of 18.

Find the length of side AB.

Hello Guest!

\(\overline{AB}=12cm\cdot cos62^0\)

\(\overline{AB}=5.6cm\)

  !

Can I use your problems as a contest set? I'll cite you as the problem writer if you want,  just reply. 

Thanks! 

To find area, we need the radius of the circle. The radius of the red circle is 1/2 of the diameter 1, which is 1/2. The radius of the blue circle is 1/2 of the diameter 3, which is 3/2. The area of the red circle is (1/2)^2pi=1/4pi. The area of the blue circle is (3/2)^2pi=9/4pi. However, the blue area is 9/4pi-1/4pi=8/4pi=2pi. So the ratio fo the blue area to the red is 2pi:1/4pi=8.

There are 36 possible rolls

5 x      =   5 rolls

x 5     =    5 rolls

5 5      =   1 roll

11/36  

No, I ment \(\frac{\pi}{45}\)

That is what has been throwing me off.

Hello, Creampuff!    Do not use the method that I gave you. See my new post. 

Btw, this way is a whole lot easier.