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Nice, hectictar   !!!!

1234  +  4321   =   5555

2345  +  5432   =   7777

3456  +  6543   =   9999

4567  +  7654   =   12221

5678  +  8765   =   14443

6789  +  9876   =   16665

And....

gcd(5555, 7777, 9999, 12221, 14443, 16665)  =  101 * 11  =  1111

Look at the following diagram :

Let CD  = the bisector of angle BCA

We want to prove that  XLB = 90 - BCA/2 

And BCA/2  = (1/2)C  =  XCL   

So  

XLB = 90 - XCL

XLB + XCL = 90

Prove that  XLB + XCL  = 90

A+ B + C = 180 

(1/2)A + (1/2)B + (1/2)C = 90

XLB  = YLA     (vertical angles)

YLA + ALB= 180

BAL + ABL + ALB = 180

Therefore

YLA + ALB  = BAL + ABL + ALB      (subtract ALB from both sides)

YLA =  BAL + ABL  

YLA = (1/2)A + (1/2)B

So....by substitution

[(1/2)A + (1/2)B] + (1/2)C  = 90

[ YLA ] + (1/2)C = 90

[ XLB ] + (1/2)C = 90

[ XLB ] + [ XCL]  = 90

For a number end in \(x\) zeros, it must be divisible by \(10^x\)

So for N! to have 3 zeros on the end of it, it must be divisible by 10*10*10 or \(5^3\cdot2^3\)

To get 3 fives when multiplying out N! there must be 3 multiples of 5 less that or equal to N.

15!=15*14*13*12*11*10*9*8*7*6*5*4*3*2*1

15, 10 and 5 are multiples of 5. There are also three 2s in 15!.

\(\boxed{15}\) is the smallest value N can be because 15! will have 3 sets of 2s and 5s to make zeros.

Here is another diagram:

The person is sitting at point  B .

height of person  =  CD + DE

                                                And  DE = 1  because the bottom of the wheel is 1 m off the ground.

height of person  =  CD + 1

Now we just have to find the length of  CD .

CD  =  AD - AC

                               And we know  AD = 8  because the radius of the Ferris wheel is  8 m.

CD  =  8 - AC

And we can find the length of AC by looking at △ABC.

cos θ  =  adjacent / hypotenuse

cos θ  =  AC / AB

                               And we know  AB  =  8  beause the radius of the Ferris wheel is  8 m.

cos θ  =  AC / 8

                               Multiply both sides of the equation by  8 .

8 cos θ  =  AC

So........

height of person   =   CD + DE   =   (AD - AC) + DE   =   ( 8 - 8 cos θ ) + 1  =  -8 cos θ + 9

And....if you don't understand how to relate  θ  and  t .....this might help: try to imagine what would that angle be after 10 seconds? (10 seconds is half of 20 seconds!) What would that angle be after 5 seconds? (5 seconds is a fourth of 20 seconds.)

Here are all the triangular numbers that are squares up to 1,000,000:
1                   = 1
6                   = 36
35                 = 1,225
204               = 41,616
1189             = 1,413,721
6930             = 48,024,900
40391           = 1,631,432,881
235416         = 55,420,693,056

Here is a short computer code to check the first 165 prime numbers between 7 and 1,000:
n=7; c=0;m=1000; a=if(isprime(n), goto4, goto6);print(n^2)%30,", ",;c=c+1; n++; if(n >Total = ",c
[19 , 1 , 19 , 19 , 1 , 19 , 1 , 1 , 19 , 1 , 19 , 19 , 19 , 1 , 1 , 19 , 1 , 19 , 1 , 19 , 1 , 19 , 1 , 19 , 19 , 1 , 19 , 19 , 1 , 19 , 1 , 1 , 1 , 19 , 19 , 19 , 19 , 1 , 1 , 1 , 19 , 19 , 1 , 1 , 19 , 19 , 1 , 19 , 1 , 1 , 1 , 19 , 19 , 1 , 1 , 19 , 1 , 19 , 19 , 19 , 1 , 19 , 19 , 1 , 19 , 19 , 1 , 19 , 1 , 19 , 19 , 1 , 19 , 1 , 19 , 1 , 1 , 1 , 1 , 1 , 19 , 1 , 19 , 1 , 19 , 1 , 19 , 19 , 1 , 19 , 1 , 1 , 19 , 1 , 1 , 19 , 1 , 19 , 19 , 19 , 1 , 1 , 19 , 19 , 19 , 1 , 1 , 19 , 19 , 19 , 1 , 1 , 1 , 19 , 19 , 19 , 1 , 1 , 19 , 19 , 19 , 1 , 1 , 1 , 1 , 19 , 19 , 1 , 19 , 1 , 19 , 1 , 1 , 19 , 19 , 19 , 1 , 1 , 1 , 19 , 19 , 1 , 1 , 19 , 19 , 1 , 19 , 19 , 1 , 19 , 19 , 19 , 1 , 1 , 1 , 19 , 1 , 19 , 19 , 19 , 1 , 19 , 19 , 1 , 19]>> Total =  165

See here, Manuel :

https://www.chegg.com/homework-help/prove-locus-points-centers-circles-tangent-two-parallel-line-chapter-6.6-problem-13e-solution-9780321700209-exc

But   4 pi^3 / 2    is just 1/2 of the area

B

12                        15

A                    D                        C

We have a 9 - 12 - 15   right triangle

So   AC   = 9

Let  AD = x    and DC = 9 -x

And since BD bisects angle ABC,we have the following relationship

AB/ BC  =  AD / DC

12/ 15 =   x / (9 - x)    cross-multiply

12 ( 9 - x)  =  15x

108 - 12x  = 15x

108  = 27x

x = 4  =   AD

And  BD  = √[ AD^2  +   BA^2]  =  √[ 4^2 + 12^2 ]  = √160  = 4√10

And since  angles  BDC  and ADB  are supplementary

cos (BDA)  = -cos (BDC)

- cos(BDA) = cos(BDC)

And cos  (BDA)  = AD/ BD   =    4 / [ 4√10]  =  1 / √10  =  √ 10  /  10

So   cos (BDC) = - cos (BDA) =  - √10/ 10

There are a total of four:

1 - 35, 21, 28
2 - 53, 45, 28
3 - 100, 96, 28
4 - 197, 195, 28

But, numbers 1 and 3 are NOT relatively prime.

n=1; x=1;d=0;cycle:a = (9*n +11*x); if(a==2007, goto5, goto6);printa,"=", x," ", n;x++;if(x<200, goto cycle,discard=0;x=1;n++;if(n<300, goto cycle,0)

  a    x      n

1 = 180   3
2 = 171   14
3 = 162   25
4 = 153   36
5 = 144   47
6 = 135   58
7 = 126   69
8 = 117   80
9 = 108   91
10 = 99   102
11 = 90   113
12 = 81   124
13 = 72   135
14 = 63   146
15 = 54   157
16 = 45   168
17 = 36   179
18 = 27   190
19 = 18   201
20 = 9   212
 

Could it be formatted as 4pi^3/2 my math review has specific rules for how to write it

Circumference  =   2pi * r  ....so....

4pi^2  = 2 pi * r        divide both sides by  2 pi

2 pi  =  r

So...the area  =

pi * r^2  =

pi * (2 pi)^2

4 pi ^3    units^2

  The square  triangular  numbers  , after 1, are

So   [ C(9, 7)]  =    9! / [ 2! * 7! ] =  (9 * 8) / 2  =   9* 4  = 36   

The next number occurs at   C(50,48)  =  (50!) /[ 48! * 2! ]  =  [50* 49] / 2 = 25*49  = 1225

CALCULATE THE VELOCITY AT A AND THE DISTANCE AB AND BC.

Here are all the relatively prime sides of a right triangle up to a 1,000,000:
53, 28, 45
197, 28, 195

Sums of Hypotenuse: 53 + 197 = 250

All the primes > 5 when squared mod 30 give a remainder of either 1 or 19.

Here are all the Pythagorean triples up to a 1,000,000:
5, 3, 4
29, 20, 21
169, 119, 120
985, 696, 697
5741, 4059, 4060
33461, 23660, 23661
195025, 137903, 137904
1136689, 803760, 803761

\(\displaystyle \omega =-\frac{1}{2} \pm i\frac{\sqrt{3}}{2}.\)

Consider a general term on the lhs,

 \(\displaystyle \frac{1}{a_{k}+\omega}=\frac{1}{(a_{k}-1/2) \pm i \sqrt{3}/2}=\frac{1}{(a_{k}-1/2) \pm i \sqrt{3}/2}.\frac{(a_{k}-1/2) \mp i \sqrt{3}/2}{(a_{k}-1/2) \pm i \sqrt{3}/2}\)

\(\displaystyle = \frac{(a_{k}-1/2) \mp i \sqrt{3}/2}{(a_{k}-1/2)^{2}+3/4}=\frac{(a_{k}-1/2) \mp i \sqrt{3}/2}{a_{k}^{2}-a_{k}+1}.\)

Substitute and equate reals,

\(\displaystyle \frac{a_{1}-1/2}{a_{1}^{2}-a_{1}+1}+\frac{a_{2}-1/2}{a_{2}^{2}-a_{2}+1}+ \dots + \frac{a_{n}-1/2}{a_{n}^{2}-a_{n}+1}=2,\)

and multiplying that by 2,

\(\displaystyle \frac{2a_{1}-1}{a_{1}^{2}-a_{1}+1}+\frac{2a_{2}-1}{a_{2}^{2}-a_{2}+1}+ \dots + \frac{2a_{n}-1}{a_{n}^{2}-a_{n}+1}=4.\)

Good,Good, Yes school is not the greatest! 

3.Find the value of n that satisfies   \(\frac{1}{n+1} + \frac{2}{n+1} + \frac{n}{n+1} = 3\)

[ 1 + 2 + n ]

__________   =    3                   multiply through by n + 1

     n + 1 

3 + n  =  3n + 3      subtract n, 3 from both sides

0 = 2n   

n = 0

2. What is the y-intercept of the line x-2y=5?

The y intercept occurs where x = 0......so....

(0) - 2y = 5

-2y = 5        divide both sides by  -2

y = -5/2   = the y intercept

1.   4x + ay  = -8  (1)

      2x + y  = b           multiply through by 2  ⇒  4x + 2y = 2b     (2)

This will have infinite solutions  when (1) and (2) are the same

So    a  = 2

And  -8 = 2b  ⇒  b = -4

So  (a, b)  =  ( 2, - 4)

Day is going great! Happy to be alive.