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Find all real numbers x and y such that 2x^2 - 2xy + y^2 = 2 and 3x^2 + 2xy - y^2 = 3.

Theoretically, there can be green hot chili peppers...

EDITING DNA BE LIKE

Okay.

What are the possible numbers that would qualify?

1,2,3,4,5....

But the smallest are 

1 and 2.

Plug it in and figure out the values that make it smaller.

Here's a hint:

You can find the interior angle by using the following formula:

180(n-2)/n.

Can you figure it out from there?

The equation of the line is y = -2x + 7.

a = -33/7.

The radius of the circle is sqrt(47).

r = 4*sqrt(5).

The slope of line AB is -11/15.

(a,b) = (-3/7,11/7).

(a) The sum of the numbers in the first row is 1/(1 - a).

The sum of the numbers in the second row is ab/(1 - a).

The sum of the numbers in the third row is (a^2 b^2)/(1 - a).

So, the sum of the numbers in the rows form a geometric sequence, which adds up to

1/(1 - a) + ab/(1 - a) + (a^2 b^2)/(1 - a) + ...  = 1/((1 - a)(1 - ab)).

(b) Since the colors of the chessboard alternate white and black, the sum of the numbers on the black squares is equal to the sum of the numbers on the white squares, except for the numbers that are on every other white square, and every other black square.

The sum of the numbers that are on every other white square is a/((1 - a)(1 - ab)), and the sum of the numbers that are on every other black square is b/((1 - a)(1 - ab)), so to find the sum of the numbers on the black squares, we take half the difference, which gives us a sum of

1/((1 - a)(1 - ab)) + 1/2*a/((1 - a)(1 - ab)) + 1/2*b((1 - a)(1 - ab)) = (a + b + 2)/(2(1 - a)(1 - ab)).

(a) The sum of the numbers in the first row is 1/(1 - a).

The sum of the numbers in the second row is ab/(1 - a).

The sum of the numbers in the third row is (a^2 b^2)/(1 - a).

So, the sum of the numbers in the rows form a geometric sequence, which adds up to

1/(1 - a) + ab/(1 - a) + (a^2 b^2)/(1 - a) + ...  = 1/((1 - a)(1 - ab)).

(b) Since the colors of the chessboard alternate white and black, the sum of the numbers on the black squares is equal to the sum of the numbers on the white squares, except for the numbers that are on every other white square, and every other black square.

The sum of the numbers that are on every other white square is a/((1 - a)(1 - ab)), and the sum of the numbers that are on every other black square is b/((1 - a)(1 - ab)), so to find the sum of the numbers on the black squares, we take half the difference, which gives us a sum of

1/((1 - a)(1 - ab)) + 1/2*a/((1 - a)(1 - ab)) + 1/2*b((1 - a)(1 - ab)) = (a + b + 2)/(2(1 - a)(1 - ab)).

We need an ace in each hand of 4, how the rest of the cards go doesn't matter !

The first ace has to be in some group, each of the other aces have to fall in a different group, so the 2nd ace has 39 permissible slots out of 51, and so on.

Thus, the probability is 39/51*26/50*13/49 = 2197/20825.

You are right about Helen - last step is wrong.

And yes the very first thing Steven did showed a total lack of understanding :)

Good work!

There are a few cases.

We draw two black balls.  The probability that this occurs is 2/9.

We draw a black ball and a white ball, and then we put back the white ball, and draw another black ball.  The probability that this occurs is 4/9*3/9 = 12/81 = 4/27.

We draw a two white balls, and then we put back the white balls, and draw another two black balls.  The probability that this occurs is 4/9*5/9 = 20/81.

So the probability that we get two black balls in the end is 2/9 + 4/27 + 20/81 = 50/81.

I have attempted it and i got it incorrect my answers were 

helen got it incorrect in the last step and stephen makes a mistake with ln2 

What steps do you think are wrong?

Have you attempted this question at all?

Why don't you interact with answerers here to get more meaningful answers. Answers that you will actually learn from.

Thanks a lot!!!

If you intereract with me, and do it politely, you are much more likely to get my help into the future

Find the number of arrangements of the letters DESCARTES if no two vowels are together.

DCRTSS       There are  6!/2! = 360 ways to arange the consonents  

AEE              There are 3 ways to arrange the vowels

|*| *| *| *| *| *|   

The stars are the 6 consonants

The 7 bars are possible positons for the 3 vowels.  

So I think that there are   7C3 = 35 place combinatons for the vowels.

So the number of arrangements  that I get is  360*3*35 =  37800.

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There are never guarantees with prob questions.

I think this is correct but I am open to polite questions or criticisms.

1. I'm getting g(0) = -441.

2. f(x) = x^2 - 7x + 6.

just draw a graph

im an algebra1 student and i dont know about all of this

m too lazy to solve it just draw something. it might actually help you...

I'm sorry about that. 

I don't know how to upload PNGs so this is the only way I can show you the grid. 

The image is really dark but you can turn up the brightness to see

      part a grid ------>     https://latex.artofproblemsolving.com/8/c/1/8c1cac67360acec4ef623054fed4ef33fd6c381a.png

       part b grid ------>       https://latex.artofproblemsolving.com/e/3/9/e3959697cb9ff1295895297696a62834311cd76f.png

Well if f(x)=-5 and g(x)=-2   then f(x)*g(x) = 10

It cant be any higher than that.     b=10   (at most)