Melody

Hi guest,

You have chosen them in the order red blue white green. but they can be chosen in any order 

So I think you answer should be multiplied by  4!

Now that takes me back :))

Actually the answer is not bad Ginger.

The answerer mistook the symbol for a minus sign. That is not unreasonable at all.

Am I missing something ?

Once or twice.  

To be honest Ginger I did not, and still have not, looked at the answer.

I don't have the time. So mostly these days I give points for 'perceived' effort.

As for the check mark, well the poster can jump up and down and ask for another answer. 

Or if they are polite they can just question the answer.

99% of them do not care enough to do so. Until someone complains I suppose I figure that an answer is an answer.

If they are wrong sometimes it might discourage the student from having blind faith in this forum as a means of cheating.

I know there are exceptions but it seems to me that if a asker is really interested in learning they will be a member and they will become known to us in order to increase our desire to help them.

Nice work EP

Hi,

Have you tried doing a rough graph?

I'd find the 3 points of intection, they are your vertices.

Y=1 is a horizontal line, I'd would use that as the base.  The length of this side will be easy to find.

The height from this line is also very easy to get.

Then just use the area of a triangle formula  half*base*height 

\(16x^4 -96x^3y +216x^2y^2 -216xy^3 +81y^4=(ax+by)^n\\~\\ \text{If we just assume that this is actually possible then the answer is obvious}\\ n=4\\ a^4=16\;\; so\;\; a=\pm2\\ b^4=81\;\; so\;\; b=\pm3\\ \text{If a is positive b is negative and vise versa.}\\ (2x-3y)^4\qquad or \qquad (-2x+3y)^4 \)

But if you want to prove it is an identity then you have to go further.

\(16x^4 -96x^3y +216x^2y^2 -216xy^3 +81y^4=(ax+by)^n\\~\\ \text{Since the largest power of x and y is 4, }\;n=4\\ (ax+by)^4\\ =(ax)^4+4C1*(ax)^3(by)+4C2*(ax)^2(by)^2+4C3*(ax)^1(by)^3+(by)^4\\ =(ax)^4+4*(ax)^3(by)+6*(ax)^2(by)^2+4*(ax)^1(by)^3+(by)^4\\ a^4=16\quad so\;\;a=\pm2\\ b^4=81\quad so\;\;b=\pm3\\ =16x^4+4*(ax)^3(by)+6*4x^2*9y^2+4*(ax)^1(by)^3+81y^4\\ =16x^4+4a^3bx^3y+216x^2y^2+4ab^3xy^3+81y^4\\ \)

In order to get the negatives, when a is negative then b is positive and vise versa so\(4a^3b=-4*8*3=-96\\ 4ab^3=-4*2*27=-216\\~\\ (2x-3y)^4=16x^4-96x^3y+216x^2y^2-216xy^3+81y^4\\ and\\ (-2x+3y)^4=16x^4-96x^3y+216x^2y^2-216xy^3+81y^4\\\)

Here's the question: What is the sum of the first 893 EVEN numbers?

This is an Arithemetic progression.

Even numbers can be negative, I assume you do not mean thoses.

So do you want to include 0 or not?

Ok, lets not include 0, the first one can be 2

a=2

n=893

d=2

\(S_n=\frac{n}{2}[2a+(n-1)d]\\ S_{893}=\frac{893}{2}[2*2+(893-1)*2]\\ S_{893}=\frac{893}{2}*2[2+(893-1)]\\ S_{893}=893*[2+(892)]\\ S_{893}=893*[894]\\ S_{893}=798 \;342\\ \)

ahhh 

1/2    not   12

We have another forensic mathematician here. :)

Thanks Gavin :)