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Thank you CPhill,

here is a nice youtube video

How To prove that 3=sqrt(1+2sqrt(1+3sqrt(1+4sqrt(1+...))))

As clarification, I would say a rhombus satisfies the conditions since they have equal parallel sides (so when one vertex is mapped to its corresponding one, the other will also. For the second condition, rhombus have reflection all symmetry along their diagonals, directly implying the result.

same guest here btw :)

The  centers  of the circles will be  (r, 0)   and (-r, 0)

The  equations of the  circles are

(x -r)^2  + y^2  = r^2   ⇒  x^2  - 2xr  + r^2  +  y^2  =  r^2   ⇒  y^2  = 2xr - x^2

(x + r)^2 + y^2  = r^2  ⇒  x^2 + 2xr + r^2  +   y^2  = r^2     ⇒ y^2  = -2xr - x^2 =  -(2xr + x^2)

Subbing  in  y^2   in the equation of  the ellipse  we  have this system 

x^2  + 2(2xr  - x^2)  =  6

x^2  -  2 (2xr + x^2) =  6

x^2  + 4xr   -  2x^2  = 6

x^2  -  4xr   + 2x^2  = 6        add  these

2x^2  = 12

x^2  =  6

x = sqrt (6)

Using   x^2  + 4xr  - 2x^2 = 6  to solve for r

(sqrt (6))^2  + 4 (sqrt 6) r  - 2(sqrt 6)^2  =  6

6  +  4sqrt(6) r  -  12 =  6

4sqrt (6) r  = 12

sqrt (6)r   = 3

r =  3 / sqrt (6)   =  3/ ( sqrt 3  * sqrt 2)   =   sqrt  (3/2)

The  graph below shows  what we have.....the circles are   tangent to each other and  internally tangent to the ellipse   at the endpoints  of the major axis of the ellipse......

14p^2   +  21pg  =

7p  ( 2p  + 3g)

The  bottom part of  the shaded area   =  8

The length of  one  of the  curved  sides =   1/4  circumference of a  circle with a radius of 4  =

(2 pi * 4)  / 4   =  2pi

So.....the perimeter of  the  shaded  area  =   8  + 2 (2pi)   =  8 + 4pi    units

See  here  :   https://web2.0calc.com/questions/counting-question_35

Correct  answer    =  14

number of 1x1 squares that contain the black square  =  1

number of 2x2 squares that contain the black square  =  4

number of 3x3 squares that contain the black square  =  4

number of 4x4 squares that contain the black square  =  4

number of 5x5 squares that contain the black square  =  1

I counted a total of  14  squares that contain the black square.

Nice, heureka   !!!!!

Thx, heureka.......this  one  is  very difficult    !!!!

5            14              29              59             113              200            329

        9            15               30               54             87              129

                 6              15             24              33              42

                          9              9                9               9

We  have   three rows of  non-zero   differences

So.....we  can  create a  3rd degree polynomial   ⇒  an^3 +  bn^2 + cn  + d

(Where  n is the nth term )

We  have  this  system

a  + b  +  c   +  d     =  5

8a + 4b + 2c  + d  =  14

27a + 9b + 3c + d  =   29

64a + 16b + 4c + d  =   59

This is a little tedious to solve ( but not impossible), so I'll enlist a little  help  with technology here  :

https://matrix.reshish.com/gauss-jordanElimination.php

a = 3/2     b   = -6     c   =  33/2    d  =  -7

So.....the  polynomial  that will generate  this series is

P(n)   = (3/2)n^3  -6n^2  + (33/2)n  -  7

The 8th term  =  509

Prove that \(3=\sqrt{1+2\sqrt{1+3 \sqrt{1+4 \sqrt{1+ \cdots}}}}\).

There exist several positive integers x such that
\(\dfrac{1}{x^2 + 2x} \) is a terminating decimal.
What is the second smallest such integer?

Terminating decimal, if \(2,5 | x^2 + 2x\)

\(\begin{array}{|rcll|} \hline x^2 + 2x &=& 0 \pmod{2} \quad | \quad +1 \\ x^2+2x+1 &=& 1 \pmod{2} \\ (x+1)^2 &=& 1 \pmod{2} \\ \sqrt{ (x+1)^2 } &=& \sqrt{ 1 } \pmod{2} \\ x+1 &=& \pm 1 \pmod{2} \\ \hline x+1 &=& +1 \pmod{2} \quad | \quad -1 \\ x &=& 0 \pmod{2} \qquad \text{All even numbers: } x=0,~2,~4,~6,~\dots \\ \hline x+1 &=& -1 \pmod{2} \quad | \quad -1 \\ x &=& -2 \pmod{2} \\ x &=& -2+2 \pmod{2} \\ x &=& 0 \pmod{2} \qquad \text{All even numbers: } x=0,~2,~4,~6,~\dots \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x^2 + 2x &=& 0 \pmod{5} \quad | \quad +1 \\ x^2+2x+1 &=& 1 \pmod{5} \\ (x+1)^2 &=& 1 \pmod{5} \\ \sqrt{ (x+1)^2 } &=& \sqrt{ 1 } \pmod{5} \\ x+1 &=& \pm 1 \pmod{5} \\ \hline x+1 &=& +1 \pmod{5} \quad | \quad -1 \\ x &=& 0 \pmod{5} \qquad x= 0,~5,~10,~15,~\dots \\ \hline x+1 &=& -1 \pmod{5} \quad | \quad -1 \\ x &=& -2 \pmod{5} \\ x &=& -2+5 \pmod{5} \\ x &=& 3 \pmod{5} \qquad x=3+5n,~ n \in \mathbb{Z} \\ \hline \end{array}\)

positive integers \(x= 2,~ 3,~ 4,~ 5,~ 6,~ 8,~ \dots\)

The second smallest such integer is 3

\(\frac{2x^{2}-3x}{x^{2}-x}+5x-11=\frac{3x^{2}+4}{x^{2}-1}\) (Problem)

\(\frac{x(2x-3)}{x(x-1)}+5x-11=\frac{3x^{2}+4}{(x+1)(x-1)}\)  (Factor out the x and expand \(x^{2}-1\))

\(\frac{2x-3}{x-1}+5x-11=\frac{3x^{2}+4}{(x+1)(x-1)}\)   (Cancel the x)

\((2x-3)(x+1)+5x(x-1)(x+1)-11(x-1)(x+1)=3x^{2}+4\)  (Multiply everything by (x-1)(x+1) to get rid of the denominator on the right hand side)

\((2x^{2}-x-3)+(5x^{3}-5x)-(11x^{2}-11)=3x^{2}+4\)  (Expand)

\(5x^{3}-9x^{2}-6x+8=3x^{2}+4\)   (Simplify)

\(5x^{3}-12x^{2}-6x-4=0\)   (Simplify)

I hope you can take it from here!

Wow! That sure is an interesting question. I plugged the first numbers into my calculator and it came close to 3, so I think it is correct, but "It works because my calculator said so" isn't a good proof (unfortunately).

I do know that we can solve for x when we have: \(x = \sqrt{y+\sqrt{y+\sqrt{y+...}}}\) because we can square both sides, subtract by y, and then the RHS of the equation is x, so we are left with \(x^{2}-y=x\).

However, there is no reoccurring pattern in your problem, it is just an ongoing arithmetic sequence. So I'm not really not sure how to find it. However, if you set the RHS equal to x, square both sides of your equation, and then play with it to get the RHS of your equation to come up again, then you can solve it.

It would be nice to see the table.

\(A\cap B\) includes the points that are in set A and set B. The only numbers in both sets are 1 and 2. 
 

Edit: I got sniped!

A={1, 2, 5, 7}​, and B={1, 2, 8}.                    1,2    is the intersection of A and B

Thanks guys  

I do not think we can use similar triangles here, we aren't given enough information to make an assumption.

Okay, so let's break this down.

Note: != means not equal to. This would be an equal sign with a slash between it in written form. 

1) We have the equation (x-1)/(x-2) = (x-k)/(x+6) 

2) We can identitify the non permissible values, aka the values that make the denominator equal to 0 (we cannot divide by 0, gives us an error). Now, if we set both denominators to equal 0 and solve, we will get the following: x-2 = 0; x != 2 and x+6 = 0; x != -6. This means that x cannot equal -6 or 2

3) Okay, now we know what x cannot equal to. But the question asks what are the non permissible values of k. We can solve for x in this case to create an equation that will allow us to find the non-permissible values of k. 

--

1) We start with the equation (x-1)/(x-2) = (x-k)/(x+6). 

2) We can cross multiply which will give us (x+6)(x-1) = (x-2)(x-k) 

3) Foil and combine like terms for both sides. x^2 + 5x - 6 = x^2 - xk - 2x + 2k 

4) Subtract x^2 from both sides. x^2 + 5x - 6 - x^2 = -xk - 2x + 2k

5) We can then isolate the other x^2 on the left side (+ x^2 - x^2). Then we get 5x - 6 = -xk - 2x + 2k 

6) We can add xk to both sides. Then we get 5x - 6 + xk = -2x + 2k 

7) Then we can add 2x to both sides. We then get 5x + 2x - 6 + xk  = 2k

8) Combine 5x and 2x. We get 7x - 6 + xk = 2k

9) Add 6 to both sides. We get 7x + xk = 2k + 6

10) Factor 7x + xk. We get x(k+7) = 2k + 6. Or you could arrange it like x(7+k) = 2k + 6. Whatever works for you. 

11) We can now comfortably solve for x. Divide both sides by (k+7). We get x = 2k+6/k+7.

12) Now we can find the non permissible values of k. Set the denominator to equal to 0. k + 7  = 0 and solve. Subtract 7 from both sides and we get k = -7. 

Solution (word it however you want but these would work): 

1) k != -7 

2) The non permissible values of k for the equation (x-1)/(x-2) = (x-k)(x+6) is k != -7

3) The equation (x-1)/(x-2) = (x-k)(x+6) has no solution for k when k = -7

By the Fundamental Theorem of Lack of Information, this problem becomes trival: no solutions.

Create similar triangles

theres no b