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Did you look under your first post of yesterday, just about a dozen questions below this one? Look under it and see the answers!!!!. We figured the monthly payment for 30 years or 360 months. Just follow the exact same procedure and change t from 360 to 180 months or 15 years. You should get the monthly payment for 15 years to be =$1,170.44.

Umm... this isn't a complete question...?

You have picked up a 

BR, or RR or RR card

You can see one red side the other side could be

B,R,R,R,R

So I think the probability that the other side is red is     4/5

Thanks Chris :)

I think EP and Melody  have secret powers.....!!!!

That is incorrect.

The greatest GCD is when:

(n + 7) =(2n + 1), solve for n

2n -n = 7 - 1

n = 6

(6 + 7) =(2*6 + 1). So GCD(13, 13) = 13 - greatest possible value of the GCD.

Here is a "proof" or verification for 3 terms only but it holds true for any n terms:

Expand the following:
binomial(n, 1) + 2 binomial(n, 2) + 3 binomial(n, 3) = n×2^(n - 1)

n (n - 1) = n n - n:
binomial(n, 1) + n n - n + 3 binomial(n, 3) = n×2^(n - 1)

n n = n^2:
binomial(n, 1) - n + n^2 + 3 binomial(n, 3) = n×2^(n - 1)

(n - 2) (n - 1) = (n) (n) + (n) (-1) + (-2) (n) + (-2) (-1) = n^2 - n - 2 n + 2 = n^2 - 3 n + 2:

binomial(n, 1) - n + n^2 + (n^2 - 3 n + 2 n)/2 = n×2^(n - 1)

n/2 (n^2 - 3 n + 2) = (n n^2)/2 + 1/2 n (-3 n) + (n 2)/2:

(n n^2)/2 - (3 n n)/2 + (2 n)/2 + n^2 + binomial(n, 1) - n = n×2^(n - 1)

(n (-3) n)/2 = (-3 n^2)/2:

(n n^2)/2 + n^2 + (-3 n^2)/2 + (2 n)/2 + binomial(n, 1) - n = n×2^(n - 1)

(n n^2)/2 = n^(1 + 2)/2:

n^(1 + 2)/2 + n^2 - (3 n^2)/2 + (2 n)/2 + binomial(n, 1) - n = n×2^(n - 1)

1 + 2 = 3:

n^3/2 + n^2 - (3 n^2)/2 + (2 n)/2 + binomial(n, 1) - n = n×2^(n - 1)

Grouping like terms, n^3/2 + n^2 + 1/2 n^2 (-3) + (n 2)/2 + binomial(n, 1) + n (-1) = n^3/2 + (n^2 - (3 n^2)/2) + (n - n + n):

n^3/2 + (n^2 - (3 n^2)/2) + (n - n + n) = n×2^(n - 1)

n^2 - (3 n^2)/2 = -n^2/2:

n^3/2 + -(n^2)/2 + (n - n + n) = n×2^(n - 1)

n - n + n = n:

n^3/2 - n^2/2 + n = n×2^(n - 1)       Courtesy of Mathematica 11 Home Edition.

I can do it too LOL

Yours is a little more impressive than mine though since you did not do it my way.

At least I do not think that you did. 

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q2wer345

Thanks guest, my method is the same as yours but i have been asked to recreate in the way i have done in the past.

\(164x+37y=1\)

The method that guest and I both use was devised by the ancient Greek mathematician Euclid.

\(\text{Euclidean algorithm}\\ 164=4*(37)+16\\ 37=2*(16)+5\\ 16=3*(5)+1\\~\\ \text{Extended Euclidean algorith}\\ 1=16-3*(5)\\ 1=16-3*(37-2*(16))\\ 1=16-3*37+6*(16)\\ 1=-3*37+7*(16)\\ 1=-3*37+7*(164-4*(37))\\ 1=-3*37+7*164-28*(37)\\ 1=7*164-31*(37)\\~\\ 1=164(7)+37(-31)\\\)

So    x=7  and  y=-31 is one integer solution.

The question is now answered but I will continue in order to find all the possible solutions

\(1=164(7)+37(-31)\\ 1=164(7+37n)+37(-31-164n)\\ \text{I have added and then subtracted 164*37n}\\~\\ x=7+37n\qquad and \qquad y=-31-164n \qquad For\;\; all \;\;n\in Z\)

I already tried it and it didn't work... I think EP is practicing dark magic

This is 5 years old, but yeah, I understand.

Mmmmm... a picture of nothing....that's an interesting concept  !!!!

Let AX  =  x    and  BX  = 50 - x

Angle C is bisected...so....we have this relationship

CA / CB   =  AX / BX

28/ 56  =  x  /  [ 50 - x ]        

1 / 2   =  x / [ 50 - x]      cross-multiply

50 - x   =  2x      add x to both sides

50  =  3x      divide both sides by 3

50/3   =  x   =  AX

YAH YEET

answer is 

I am friends with sssniperwolf #jointhewolfpack

 spooky!

the answer is oh hi mark........................................................................................................................................................................................