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Let the circle have a radius of 4

Let A = (0,4)

Let B =  ( 2sqrt 3 , -2 )

AB = sqrt [ (2sqrt 3)^2 + (4--2)^2] = sqrt 48 = 4sqrt (3)  side length of triangle

MC = = (1/2) side length = 2sqrt 3

MB = side length * sqrt 3 =  2 sqrt 3 * sqrt 3 =  6

MB /MC  =  6 / [ 2sqrt 3]  =  3 /sqrt 3 =  sqrt 3

Each term   will  be -1 < term value  < 0

So the floor for each will be -1

Sum = 1000 (-1)  = -1000

r =  .016 / .4  =  .04

Sum  =     .016 / [ 1 -.04] =  .016 / .96  =  1/ 60

a / ( 1-r)  = 3

a^2 / (1 - r^2) = 10  

 [ a / (1-r) ] * [ a/ (1+ r)  ]  =10

[ 3] * [ a / ( 1 + r) ] =10

a/ [ (1+ r)] =10/3

a = (10/3)(1 + r)

So

(10/3)(1 + r) / (1-r)  = 3

(10/3) (1+r) = 4 (1 - r)

10/3 + (10/3)r = 4 -4r

(10/3 + 4)r = 4 -10/3

r =  [ 4 - 10/3 ]/ [ 4 + 10/3] =  [ 2] / [ 22]  =  1 / 11

(n+ 2)/(n ) * (n + 3) / (n + 1) * (n + 4) / ( n + 2)  = 4          simplify as

 (n + 3) *(n + 4)  =  4 (n)(n + 1)

n^2 +7n +12  =  4n^2 + 4n

3n^2 - 3n - 12 = 0

n^2 - n - 4  = 0

n^2 - n +1/4  =   4

(n - 1/2)^2  =  +/- 17/4

n =  [1 + sqrt (17)] /2                     or     n = [ 1 -sqrt (17) ] /2

No integer values for numerator/denominator  are possible

a^3 + 3ab^2  = 675

(675 - a^3) / 3 = ab^2

Subbing this into the second equation we have 

3a^3 - (675 - a^3)/3 = 615      multiply through by 3

9a^3 -675 -+ a^3 = 1845

10a^3  = 2520

a^3 = 2520/10

a = (252)^(1/3) ≈ 6.316

Using the first equation

252 + 3(6.316)*b^2  = 675

18.948b^2  = 423

b = sqrt [ 423 / 18.948 ] ≈ 4.725  or -4.725

a - b =  6.316 - 4.725  ≈ 1.591

or 

a - b = 6.316 - (-4.725) ≈ 11.041

1/(1*4)+1/(4*7)+1/(7*10)

= 1/4+1/28+1/70

Using the common denominator of 140:

35/140+5/140+2/140

= 42/140

= 3/10

3/10 or 0.3

The answer to both questions is yes.  You can use geometric series.

1. Find the formula for the sum of an arithmetic series:

The sum of an arithmetic series can be found using the formula:

Sum = (n/2) * (first term + last term)

where:

n = number of terms

first term = 2

last term = 2 + 5(n-1) = 5n - 3

2. Set up the inequality:

We want to find the 'n' where the sum exceeds 1000:

(n/2) * (2 + 5n - 3) > 1000

(n/2) * (5n - 1) > 1000

5n^2 - n > 1000

5n^2 - n - 1000 > 0

3. Solve the quadratic inequality:

Use the quadratic formula to find the roots of 5n^2 - n - 1000 = 0

n = [1 ± √(1 + 4 * 5 * 1000)] / (2 * 5)

n ≈ 28.3 or n ≈ -28.2

Since 'n' represents the number of terms, it must be positive.

Therefore, n > 28.3

4. Find the last number Laverne says:

Last term = 5n - 3

Since n > 28.3, the smallest integer value for 'n' is 29.

Last term = 5 * 29 - 3 = 142

Answer:

Laverne says the number 142 that sends Shirley screaming and running.

The answer is MB/MC = 6/5.

The smallest possible value of RX is 15.

The area of triangle ABC is 10*sqrt(3).

The value of TU is 14*sqrt(2).

The radius of the circle is 3/5.

The answer is 22.

The answer for AC/BC is 30/7.

Ok I did the problem again and I disagree. Here's my approach. 

For the first transformation, any of the 11 options will achieve a satisfactory result (none of the 6 points has yet been transformed to). After that, the remaining 5 points can be transformed to in any order, as long as there is no repetition, so 5! = 120 orders for visiting points. At each step after the first, there are two transformations (one rotation and one reflection) that will get from one point to the next [distinct] point. Thus, the overall probability of success is

\(120\times\frac{11}{11}\times\frac{2}{11}\times\frac{2}{11}\times\frac{2}{11}\times\frac{2}{11}\times\frac{2}{11} = \frac{42,240}{11^6}\)

so k = 42240

EDIT -------

I think you did 6! instead of 5!

Since you counted that the first transformation can be any of the 11 moves, then there are only 5! ways to arrange the other 5

To solve this problem, we need to consider the transformations that can move point P to each vertex exactly once.

The value of k in the probability $\frac{k}{11^6}$ can be calculated as follows:

There are 6! = 720 ways to arrange the six vertices in a sequence.

For each sequence, we need to count the number of ways to choose transformations that achieve that sequence:

The first transformation can be any of the 11 that moves P to the first vertex in the sequence.

The second transformation must be one of the 2 that moves the current position to the second vertex.

Each subsequent transformation also has 2 choices to move to the next vertex.

Therefore, for each sequence, there are 11 * 2^5 = 352 ways to choose the transformations.

The total number of favorable outcomes is thus 720 * 352 = 253,440.

Therefore, k = 253,440.