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Hey max 

\(3\sqrt{x - 2y}+\dfrac{3}{\sqrt{x-2y}}=10\\ 3\left(\sqrt{x - 2y}\right)^2-10\sqrt{x-2y}+3=0, x \neq 2y\\ \sqrt{x - 2y} = \dfrac{1}{3} \text{ or }\sqrt{x - 2y} = 3\\ x - 2y = \dfrac{1}{9}\text{ or }x - 2y = 9\\ (a - 2)y + b = \dfrac{1}{9}\text{ or }(a - 2)y + b = 9\\ \text{Assuming the above equations are true for all values of }y,\\ (a, b)=\left(2,\dfrac{1}{9}\right)\text{ or }(2,9)\)

\(\begin{array}{cll} &\text{Sum}\\ =&\displaystyle \sum^{\infty}_{n=1} \dfrac{7^{n - 1}}{10^n}\\ =&\dfrac{1}{7}\displaystyle\sum^{\infty}_{n = 1} \left(\dfrac{7}{10}\right)^n\\ \stackrel{\text{sum of GS}}{=} &\dfrac{1}{7}\left(\dfrac{\dfrac{7}{10}}{1-\dfrac{7}{10}}\right)\\ =&\dfrac{1}{3} \end{array}\)

\(a^2 + b^2 = c^2\\ a^2 + b^2 = (b + 1)^2\\ 2b + 1 = a^2\\ \text{Let }a = k^3,\\ k^6 - 1 = 2b\\ k^6 - 1 \equiv 0 \pmod 2\implies k\equiv 1 \pmod 2\\ \text{Exhaust }k.\\ \text{When }k = 1, b = 0\text{(rejected)}\\ \text{When }k = 3, b = 364\\ c = 364 + 1 = \boxed{365}\\ \text{Check:}\\ a^2 + 364^2 = 365^2\\ a = 729 = 9^3\)

From x + y + z = 8 and x - y + z = 4,

\(\color{red}(x+y+z)\color{black}-\color{blue}(x-y+z)\color{black} = \color{red}8\color{black} -\color{blue}4\color{black} = 4\\ 2y = 4\\ y = 2\)
\(x + 2 + z = 8 \\ x + z = 6\)

\(\sqrt{x+z-2}+\sqrt2 = \sqrt x + \sqrt z\\ 2 + \sqrt{2} = \sqrt{x} + \sqrt{z}\\ \text{Let }z = 6-x\\ 2+\sqrt{2} = \sqrt{x} + \sqrt{6 - x}\\ \text{Square both sides,}\\ 6 + 4\sqrt 2 = 6 +2\sqrt{6x-x^2}\\ 2\sqrt 2 = \sqrt{6x - x^2}\\ \text{Square both sides again,}\\ 8 = 6x - x^2\\ x^2 - 6x + 8 = 0\\ (x - 2)(x - 4) = 0\\ x = 2 \text{ or } x = 4\\ \text{When }x = 2,z=4\\ \text{When }x=4,z=2\\ \text{The required ordered triples are }(2,2,4)\text{ and }(4,2,2)\)

\(g(x)=ax^5+bx^4+cx^3+dx^2+ex+f \\~\\ Find \;\;f\\~\\ g(1)=a+b+c+d+e+f =2\\ g(2)=32a+16b+8c+4d+2e+f =3\\ g(3)=243a+81b+27c+9d+3e+f =4\\ etc\\ \)

This appears to me to be a very long matrix problem (if done by hand)

I am wondering if there is a shortcut that I have not seen.....

CODING

g(x)=ax^5+bx^4+cx^3+dx^2+ex+f \\~\\
Find \;\;f\\~\\
g(1)=a+b+c+d+e+f  =2\\
g(2)=32a+16b+8c+4d+2e+f =3\\
g(3)=243a+81b+27c+9d+3e+f =4\\
etc\\

It is a standard combination question.

At least 2 means  P(2) + P(3) + P(4)+P(5)+P(6)

Thie is exactly the same as   1 - [P(0)+P(1)]

P(n) = 6Cr  *  (211/243)^r  * (1- 211/243)^(6-r)

Now you have all the info you need to answer the question.

Explanation of late edit:

I have changed the letter to r becasue r is the one normally used and it was confusing for you before.

I appologize for that.

Hint:  Let the number be 10a+b.     When writen it will look like ab

For example, if the number was 27 then a=2 b=7   and the number is 10a+b

Use this fact to set up an equation which you must then solve. 

Find a polynomial q(x) such that (x+1)^3+x^2*q(x) has degree less than 2.

You just have to cancel out the x^3 term.

q(x)= -x+k    where k is a constant.  Any real value of k will work fine.

In the five-digit base ten numeral ABCDE, different letters do not necessasrily represent different digits.  If this numeral is the fourth power of an integer and if  A + C + E = B + D, find the digit C.

10^4=10000    

17^4=83521

19^4 is too big.  so there are only  17-10+1= 8 possible numbers

Since there is only 8 possibilities I would just list them and see which one works for the given equation.

This is what you should be doing :

\(|x^2-20x-1|=20\\~\\ \color{blue}x^2-20x-1=20 \qquad or \qquad x^2-20x-1=-20 \\\)

       Think about this simple example

       |x|=5

       What this says it that the distance x is from 0 is 5.  So x could be 5 or maybe -5.

If you let    \(y=x^2-20x-1\)

then your question is asking, for what values of x is   y  20 units away from 0.

So y can equal 20  or   y can equal -20.

You will get 4 answers.  They will not necessarily all be real.

I expect you are only interested in real answers so if some are not real you can discard them.

Now you must substitute the real ones into the original equation to check that they work properly.

Discard any that do not work.

Is that enough help ?

You are very welcome EpicWaters.

Let AB be a leg of the right triangle of least perimeter whose sides have integral lengths, whose hypotenuse is one unit longer than AB, and in which AB > 100.  Find AB.  

Let's call the sides a, b, and c, for my convenience.  Let AB be a, the other leg b, and the hypotenuse c.

By Pythagoras' Theorem                                 a² + b² = c²               

                                                                        a must be larger than 100    

                                                                        c must be one unit larger than a

                                                                        all values must be integers

I don't have a general proof for this;

I had to do it by trial and error.                          AB (that is, a) equals 112    

The three sides of the triangle are                    112, 15, and 113     

                                                                          112² + 15² = 113²      

Since AB had to be larger than 100, I started with 101, squared it, subtracted that from 102 squared, and checked if the remainder were a perfect square.  It wasn't.  I moved on to 103² minus 102² and the remainder was not a perfect square.  When I tried 104² minus 103² I noticed a pattern to the remainders so I skipped ahead to 111² minus 110² and that wasn't it either but I was approaching 225.  I finally found it at 113² minus 112² .... whew.

_.

 12 , 24 , 36 , 48 , Total =  4

Oof I was just going to change it to AC equals 4 sqrt2 and BM equals 2 sqrt 5, then I saw your hint.

I want to know like where the point o is...

From there, Maybe I could work it out?

yay geometry!!

median 1=9

median 2=12

median 3=15

you know the three medians, and the medians meet at a point called the centroid. the centroid cuts up the median into a 2:1 ratio, so median 1's two lengths are 3 and 6, median 2's lengths are 4 and 8, and median 3's length's are 5 and 10.

it would also be helpful if you made a drawing! it would also also be helpful if you included your work or where your head's at with this problem so that i can figure out what to do :)

i made a square too. 

for me, AC split through the center of the square, so AC=4sqrt2 by a^2+b^2=c^2. now that you know the length, can you take it from here? because i got stuck after that :( 

well actually we also know that BM=2sqrt5 by the same theorem above, but that's all i've got :'(

lol i'm actually trying to answer your question too :D

i thought what you said too, i just don't know if absolute value is the same as plus or minus

Split it.

So just go like:

\(x^2-20-1=20\)

Solve that AND

\(-x^2+20+1=-20\)

I think.

Can somebody verify me?

Using the multi-Factor Theorem for repeated roots, the remainder when x^{1000} is divided by (x^2 + 1)(x + 1) is 2x - 3.

The number is =14,641. It is = 11^4

A + C + E = B + D

1 + 6 + 1 = 4 + 4

8  =  8

Thank you so much, Melody. I understand this problem.

WHoops!

Thank you!!!!!!!