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The first card is a given

On the second draw, we have  a (39/52)  = (3/4)  probability of drawing a different suite from the first draw

On the third draw we have  a (26/52) = (1/2) probability of drawing a suite different from the first two draws

On the last draw we  have a (13/52) = (1/4) probability of drawing a different suite from the ones we  drew on the first three draws

So  the probability =    ( 3/4) ( 1/2) ( 1/4)  =   3 /32

The radius of the  circle can be found as

18pi  = 2pi * r

18pi / (2pi)  =  r   = 9

We can divide the  equilateral triangle into three congruent triangles.....the area of  of of these triangles is given by

(1/2) r^2  *sqrt (3) / 2  =    (1/2) 9^2  *  sqrt (3)  / 2  =   81 sqrt  (3) / 4  cm^2

So....the area of the equilateral triangle  =    3 (81)sqrt (3) / 4  cm^2  =  (243)sqrt (3) / 4  cm^2

Probably an easier way to do this....but.....

Let the bottom left vertex =  (0,0)  = A

Let the bottom right vertex =  (8,0)   = B

Let the  top left vertex  =  (0,10) = E

Let the top right vertex  ( 8,12) = C

Let the apex vertex =  D

CE =   sqrt ( 8^2 + 2^2)  = sqrt  ( 68)  = 2sqrt (17)

Semi-perimeter  of triangle   CED  =  ( 4 + 6 + 2sqrt (17) ) / 2 =    5 + sqrt (17)

Area of triangle   CED  =

sqrt  [ (5 + sqrt (17) )  ( 5 + sqrt (17) - 2sqrt (17)) (5 + sqrt (17) - 6)  ( 5 + sqrt (17) - 4 )  ]  =

sqrt [ (5 + sqrt (17) ) (  5 -sqrt (17)) ( sqrt (17) - 1) ( sqrt (17) + 1)  ]   = 

sqrt [ (5 + sqrt (17) ) (5-sqrt (17) ) (sqrt (17) -1) (sqrt (17 + 1)  ]   =

sqrt  [ (25 -17)  ( 17 - 1)  ]  =

sqrt [  8 * 16 ]  =  

sqrt ( 64 * 2)  =    8sqrt (2)

And  ABCE  is a trapezoid  with  bases of  10 and 12  and a height of 8

[ ABCE ] =    (1/2) [ 8] [ 10 + 12 ]=    4 * 22   =  88

So...area of the pentagon  = [  88  +  8 sqrt (2) ] in^2  =    8 (11 + sqrt (2))  in^2

2

 40

x 6

__

240

or 4*6*10

4*6=24

24*10=240

This should help:

I think it good 

                      -Thanks

Each side of a pentagon is sewn to a edge of a hexagon

5 x 12 = 60

Law of Sines

Angle B = 180 - 44 -53 = 83^o

sin B / 11  = sin 53 /BC

BC = (sin 53) / ( sin 83 / 11) 

If n = 3 then

mod(n^2,5)  =  4

mod(n^3,5)  =  2

mod(n , 5)  =   What do you think?

As follows:

But, more importantly, do you understand why these are the choices?

Can I ask you question? okay so the question is  this is a basic test question for you \(\mathbf{M} \begin{pmatrix} -3 & 4 & 0 \\ 5 & -7 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \mathbf{I}.\)

could it be basic math

or is it not correct?

is it correct?

0.000112832461083 is the answer

Sum of interior angles of pentagon = 540

angle 1 to angle 5 is four common differences and = 68    68/4 = d = 17

a + a + d  + a + 2d  + a + 3d  + a + 4d  = 540

5a   + 10d = 540     but d = 17

5a + 10(17) = 540

a = 74^o

1 - 3/8 = 5/8 - what remains of the pizza after John.

1/3 * 5/8  =5/24 - Tim's portion

1 - 3/8 - 5/24 =5 / 12 - Kevin's portion.

Check: 3/8 + 5/24 + 5/12 =1 Pizza.

The probability is 64/1225.

k = 15/2.

Archimedes should choose x = 1/4.

correct

But x=1/2 and y=1/2 leads to x-y = 0.

Try x = 1/2 and y = 0.   We still have 2(x^2 + y^2) = x + y, but now x - y = 1/2.

The quantities xxx and yyy are proportional.

xxxyyy

222222222

777777777

999999999

Find the constant of proportionality (r)(r)left parenthesis, r, right parenthesis in the equation y=rxy=rxy, equals, r, x.