jxc516

Nevermind, I solved it. Thank you!

First of all, I know that is incorrect because of the fact that n choose n is indistiguisable items, but the cars are distinguishable

So the answer given there is incorrect

2. We know that the median is the top-bottom divided by 2, so 17-5/2=6, and 6/2 is 3, and 3+5=8

\(\binom{5}{3}\)=10 ways to arrange

There is 1 way to make XOXOX

the answer is 1/10

Thank you!

Oh wait I think I understand

When you use foil to expand, you multiply the variables together, then the sum of the products of the inside and outside terms, and the constants. Example: (x+2)(x-3)=x^2-x-6. 

Because of this, we need to factorize 2008, so for the middle term we get  x+2008x=2009x, 2x+1004x=1006x, 4x+502x=506x, 8x+251x=259x, and we take the smallest coefficient and that's our answer

I believe ive done this problem before, but I might have forgotten the approach :(. All I need is a push on how to start. Thanks!

I think you factor 2008, and using FOIL you find that (8,251) is the smallest

I might be wrong though

ok lemme get my paper...

\(\frac{-b+\sqrt{b^2-4ac}}{2a}-\frac{-b-\sqrt{b^2-4ac}}{2a}\Rightarrow \frac{-b+b+\sqrt{b^2-4ac}+\sqrt{b^2-4ac}}{2a}\Rightarrow\frac{2\sqrt{b^2-4ac}}{2a}\Rightarrow \frac{\sqrt{b^2-4ac}}{a}\Rightarrow \frac{\sqrt{61}}{3} \implies \boxed{a=3}\)

I might have explained it a little thinly on accident