Recent content by MLeszega

Ahh, I see. That really cleared things up for me, thanks. I do wish they made some sort of side note for the problem though, saying something like 'a and b cannot both equal zero,' because then you actually can have all real solutions...

Problem proving x^4 + ax + b... So I started working on some of these http://www.math.vt.edu/people/plinnell/Vtregional/exams.pdf problems for fun. The relevant problem is #7 from year 1983. The problem: If a and b are real, prove that x^4 +ax + b = 0 cannot have only real roots. I...

Ok, so i can take my equation 2 = p3 + 3p2q\sqrt{r} +3pq2r + q3r3/2 and rewrite it as: 2 = p3 + (3p2q + 3pq2\sqrt{r} + q3r3/4)\sqrt{r} Set P = p3 and Q = 3p2q +3pq2\sqrt{r} + q3r3/4 I now have the form 2 = P + Q\sqrt{r} This is back to the original form of p + q\sqrt{r} where we...

Ok, I think I understand what you guys are talking about now. So I will set \sqrt[3]{2} = p + q\sqrt{r}, assume that \sqrt{r} is irrational and show that this leads to a contradiction? Would it be something like this: 2 = (p +q\sqrt{r})3. Then you expand the RHS to get: 2 = p3...

I really don't think that assuming \sqrt{r} is irrational is a reasonable assumption. There are plenty of rational numbers whose square roots are also rational numbers, i.e 4/9, 1/16 etc

It isn't a homework question, I don't go to school anymore. I did find it in a notebook that I used in school though. I just like working problems out for fun now. Anyways, I actually did try proving it by contradiction. You get 2=(p+q\sqrt{r})3, which I then expanded. It didn't seem to help...

Hey guys. I forget where I found this problem but it goes as follows: Prove that \sqrt[3]{2} cannot be represented in the form p+q\sqrt{r} where p,q, and r are rational numbers. It is easy to show that \sqrt[3]{2} is irrational, so it cannot be put in the form m/n, where m and n are...

Ok, I figured out the solution today while at work. Ty for the hints.

Ok. I guess my question would then be whether or not you can assume that the gcd(n,a-b)=1. The problem is simple if you can assume that, but doesn't seem to be if you can't.

Ahh, ty very much!

So v0 = 600sin(30) for the vertical component and 600cos(30) for the horizontal component?

I am pretty bad at physics so I am trying to follow the problem but I get stuck here. I thought that v0=600m/s. How come you wrote 300? Is it because v0 is m/s while gravity is m/s2?

There is a process you can do to solve this (although some of these problems are easy enough to be solved through inspection). You want to write a as a linear combination of b and c, so write it like this: x\vec{b} + y\vec{c} = \vec{a} Then you solve for x and y. This can be done easily by...

Any update on this problem? I am interested in its solution.

Ok so for #24 you need to build an equation to help find the maximum area. It says that perimeter is 1200ft. What kind of equations can you think of that would be helpful? How about the equation for perimeter and area? If you find those two equations you are finished.