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If you know how to code, you can write a computer code that will solve them all for you. On the other hand, there are online inequality calculators that will solve them for you one at a time.

it's the answer to the life, the universe, and everything.

and yes, I know you do aops.

N = 27 mod[LCM(4, 5, 7)]

N = 27 mod 140

N =140n +27, where n=0, 1, 2, 3.......etc.

So, the smallest N = 27.

n=1;m=5000; c=0;a=if(n % 13==11, goto4,goto5);c=c+1;n++;if(n<5000, goto3, discard=0;print"Total = ",c

Total =  384

1560  factors as  2^3 * 3 * 5 * 13

So....the smallest n   =   13

Perhaps search your e- mail....when yo signe up this was in effect:

* You will receive an activation email to this address

Here's my best attempt : https://web2.0calc.com/questions/did-i-feel-out-my-information-correctly

is the period 2 

and the amplitude 

7000-7000/2 ?

\(f(x,y) = \arctan\left(\dfrac x y\right)\\ f_x = \dfrac{1}{1+\left(\dfrac x y\right)^2}\cdot \dfrac 1 y = \dfrac{y}{x^2+y^2}\)

\(f_y = \dfrac{1}{1+\left(\dfrac x y \right)^2}\cdot \dfrac{-x}{y^2} = -\dfrac{x}{x^2+y^2}\)

\(f_{xx} = -\dfrac{y}{(x^2+y^2)^2}\cdot 2x = -\dfrac{2xy}{(x^2+y^2)^2}\)

\(f_{yy} = \dfrac{2xy}{(x^2+y^2)^2}\)

\(f_{xy}=f_{yx} =\dfrac{ 1(x^2+y^2)-2y(y)}{(x^2+y^2)^2} = \dfrac{x^2-y^2}{(x^2+y^2)^2}\)

52 x 51 = 2652

You multiplied an even number by an integer and obtained an odd number... Impressive!

I think the reason why it works for 10,000, but not for 5,000 is as follows:
1 - Each of the 10 digits[0 to 9] must be EQUALLY represented up to a 1000, since there are: 10 digits x 10 blocks of 100 numbers each =1000. As you can see in this breakdown of all the 10 digits between 1 and 1000:
[189, 300, 300, 300, 300, 300, 300, 300, 300, 300]. Zero(189) is an exception due to the peculiarity of the decimal system. 

2- In order for all 10 digits to be equally distributed, you have to go up the numberline in powers of 10. Therefore, the next block will be:1000 x 10^1 =10,000, when all the 10 digits will be EQUALLY represented again. The following breakdown illustrates the case.
[2889, 4000, 4000, 4000, 4000, 4000, 4000, 4000, 4000, 4000]. Again, zero is the exception.

3- To show the distribtion of all 10 digits up to limit that is NOT a power of 10, the following couple of example illustrates that very clearly: From 1 to 5000 you have the following: [1389, 2500, 2500, 2500, 2500, 1500, 1500, 1500, 1500, 1500]. As can be clearly seen, the distribution of the 10 digits is anything but equal, the first 5 being quite different from the last 5.
From 1 to 15,000 the distribution is skewed like this:
[5389, 11500, 6500, 6500, 6500, 5500, 5500, 5500, 5500, 5500]. Again, they are anything but equal.

4 - The next EQUAL distributions would be powers of 10, namely:10,000 x 10^1 =100,000 which has the following EQUAL distribution:[38889, 50000, 50000, 50000, 50000, 50000, 50000, 50000, 50000, 50000]. And the last, for this pupose, would be:100,000 x 10^1 =1,000,000, which has the following EQUAL distribution: [488889, 600000, 600000, 600000, 600000, 600000, 600000, 600000, 600000, 600000].

5- I hope that this is a satisfactory explanation to your question.

Actually, they are only counted once. If you saw a door connecting two rooms in real life, you wouldn't say "there are two doors".

Sorry if this was unclear.

i am not rogr

y  =  (x - 3)  / ( x + 7)                               (x + a)  ( y + b)  =  c

y(x + 7)  = x - 3                                         xy + ay + bx + ab =  c  

xy +7y  = x - 3                                         xy  + ay  + bx   =  c  - ab  (2)

xy + 7y - x  = - 3     (1)

Equateing terms in  (1)  and (2)   and we have that

a = 7,  b = -1

And

-3  =  c - ab

-3  = c - (7)(-1)

-3  = c + 7

-10 = c

So  { a, b , c }  = { 7, - 1,  -10 }

Can you answer number 2 please?

I have decided to leave my answer visable because you have made an attempt at the question and I have not found your error that you are supposed to find for yourself.

And

I know answers similar to mine would be all over the internet. It is a common question.

"Thanks for any answer I may receive, as I understand this post is long.

For anonimity reasons, I might delete this later. I hope you're ok with that."

No it is absolutely not ok to delete your question.

That just means that you are worried that your teacher will see that you are cheating.

f(h)=45-7t-6t^2
Sub 25 for h and rearrange:
25 =45 - 7t - 6t^2
6t^2 + 7t - 45 + 25
6t^2 + 7t - 20 = 0
Solve for t using the Quadratic Formula
t = 4 /3 = 1 1/3 seconds.  t =- 5/2 (discard)

Problem: Suppose we have a house (with finitely many rooms) in which every room has an even number of doors. Prove that the number of doors from the house to the outside world is also even.

I will assume that there are no doors to cupboards.  All doors go either to another room or to outside.

My logic on this is quite simple.  

I expect you want me to look at your logic.  Maybe I will do that afterwards.  

Let 2n be the number of internal door SIDES.

Let k be the number of external door sides.

The number of door sides altogether is  2n+k

Every door has 2 sides so the number of doors is   (2n+k)/2 = n+k/2

There must be a whole number of doors so k must be divisable by 2

therefore

If there is an even number of doors in each room then there must be an even number of doors going outside as well.

Thanks Alan 

I find these difficult to get my head around too.

I find it easier if I draw a graph and I can see what I am doing visually.

g(10.5)=-20

g(-20)=20

g(20)=-1

          Now going backwards.

         g(what)=20    there are two answers, -20 and 30.5   (this is easy to see from the graph)

          so

          g(-20)=20      and     g(30.5)=20

                   now go backwards again

                   g(what)=-20      and     g(what)=30.5  looking at the graph I can see that

                   g(10.5)=-20      and      g(-30.5)=30.5    and   g(35.75)=30.5

so

g(10.5) is what we started with

You are told that a is negative so  35.75 is no good

a=-30.5

check

g(-30.5)=30.5

g(30.5)=20

g(20)=-1  = g(g(g(10.5)))

ie

g(g(g(-30.5)=g(g(g(10.5)))

a=-30.5

yes that is correct.  internal doors are counted twice.

Very clever, Tiggsy!.

Let k = -p/q, (where p and q are integers), then (x + p/q) will be a factor of both polynomials, so

\(\displaystyle (x+p/q)(sx^{3}+\dots+ t)=0,\text{ and }\\(x+p/q)(ux^{4} +\dots +v)=0. \)

Multiplying both by q,

\(\displaystyle (qx+p)(sx^{3}+\dots+t)=0,\text{ and}\\(qx+p)(ux^{4}+\dots+v)=0.\)

Equating leading coefficients, qs = 75 so q = 1,3,5,25 or 75  and qu = 12 so q = 1,2,3,4,6 or 12.

q can't equal 1 since that would make k an integer, so q = 3.

Equating constants, pt = 12 so p = 1,2,3,4,6 or 12 and pv = 75 so p = 1,3,5,25 or 75.

p can't equal 3 since that would make k an integer, so p = 1, implying that k = -1/3.

Assuming that ABC is a three digit number (rather than the product A*B*C), 

145 = 1! + 4! + 5! = 1 + 24 + 120 = 145.

 a = -30.5:

g(10.5) = -20; g(-20) = 20; g(20) = -1 

a = -30.5:  g(-30.5) = 30.5;  g(30.5) = 20;  g(20) = -1

Edited to correct silly mistake!  Thanks for pointing this out Melody.