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The answer is 4.

sumfor(n, 1,  46, n*6 - 5) mod 6 ==4 - the remainder

CM =  OM - OC =  36  - 32 =  4

So MB =  4

So OB  =   OM + MB  =  36 + 4 =  40   = the radius of the largest circle

So AO = 40

And AO +  OC =     40 + 32  =  72 =   AC 

Arc length  =      (55/360) * 28   ≈   4.28 cm

22^22 * 11^11  =

[ 2 * 11 ] ^22 * 11^11  =

2^22 *  11^22 * 11^33  =

2^22 *  11^33

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 16    and repeats in a cycle of four

2^22 will end  in  a 4

11^any positive integer power ends in a 1

So

22^22 * 11^11   ends in  a  4

M = 90  - 10  =  80°

Supplement of M  =   180  - 80  =   100°

x + x = 180°

2x = 180°

x =  180 / 2 =    90°

I only see one shaded region........half of the larger circle

Area =  32 pi

Area of smaller square =  x^2

Area of larger square=  (2x)^2 = 4x^2

Ratio of areas =   x^2 /  4 x^2  =     1  / 4

4 feet  =  48 inches

12 feet = 144 inches

On two oppositie sides of the framing we will have   48 / 2 =  24 tiles on each side

On the other two sides of the framing we will have  144 / 2 =  72 tiles on  each side

Plus.....we have 4 tiles on the  corners

Total cost =  [ 2 * 24 + 2 * 72  +  4 ]   *  .10  =    $19.60

That problem is too easy.  I think the OP made a typo and it should have asked "..a heart or a jack?"

Also, "a heart of a jack" is an unusual way to say it.  The expression would have been "the jack of hearts." 

_.

you can also think about it this way

98 = 2*7*7
k=(2^4 x 7^4)/2x7x7

k=2^3 x 7^2 = 392

The smallest n ==392, because:

98 * 392 ==38,416^(1/4) ==14

My co-worker Larry only likes numbers that are divisible by 4, such as 20, or 4,004. How many different tens digits are possible in numbers that Larry likes? 

Just intuitively, I'd say all 10 of them.  

Here are examples

  12

  20

  32

  40

  52

  60

  72

  80

  92

100

_.

here is a website:

z^2 - z = 5 + 5i
z^2 - z + 1/4 = 5 + 5i + 1/4
(z - 1/2)^2 = 21/4 + 5i
z - 1/2 = ±√(21/4 + 5i)
z = 1/2 ± (5/2 + i)

z1 = 1/2 + 5/2 + i = 3 + i
z2 = 1/2 - 5/2 - i = - 2 - i

Product of roots :
(3 + i)(- 2 - i) = - 5 - 5i

The real part of the product of the roots is - 5

favorite?

Hi Melody,

 But I have to subtract all the ones that are in both every row and every column (becasue they have been added twice.)This number is 8!

You’ve subtracted (8!) from the total successful arrangements. That is a lot duplication that is NOT indicated in the positional analysis of the rooks. The only time there is a rook in every row and every column is when they are positioned diagonally across the board.  Other than this, one or more rows and/or one more columns are unoccupied, but not unattacked.   

As I described above, this diagonal arraignment occurs twice: once, descending from the top right to the bottom left, and again from the bottom left to the top right. These diagonal arraignments are the ONLY two arrangements that occur in both the vertical movement progression and the horizontal movement progression.  There are no other duplicates, so where does the (8!) come from?

Appended to this post are images depicting the chessboard with eight rooks in horizontal and vertical formats, demonstrating their respective movements on the chessboard and points of interest.

If I am missing (8!-2) arrangements where these two formats intersect, then I have royally rooked myself. I would really like to see them. Can you graphically demonstrate where some of these duplications occur?

GA

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Let the roots be x and y. Note that \(x^2 + y^2 = ( x+ y)^2 - 2xy\)

By Vieta's, \(x + y = -{b \over a} = {5 \over 3}\) and \(xy = {c \over a} = {8 \over 3}\)

So, \(x^2 + y^2 = {25 \over 9} - {16 \over 3} = {25 \over 9} - {48 \over 9} = \color{brown}\boxed{-{ 23 \over 9}}\)

There is exactly 1 heart of jack, and 52 cards, so the probability is \(\color{brown}\boxed{1 \over 52}\)

Let r be the radius of the larger circle. Because the area is \(64 \pi\), the radius is 8. 

So, the area of the larger shaded region is half the circumference plus the diameter, which is \(\pi r + 2r = 16 + 8 \pi\).

Now, note that the diameter of the smaller circle is the radius of the larger circle, so the radius of the smaller circle is \(4\). 

This means that the perimeter of the shaded region is \(8 + 4 \pi\), by the formula above. 

So, the total perimeter is \(8 + 4 \pi + 16 + 8 \pi = \color{brown}\boxed{24 + 12 \pi}\)

There are \({8 \choose 2 } = 28\) combinations. 

Now, just count the number of pairs that work and find the probability 

https://web2.0calc.com/questions/probability_87332

The first 10 digits are 2, 3, 5, 6, 7, 11, 12, 13, 14, 15.

The 10th digit is 15. Can you take it from here?