Recent content by Suprin

Sorry if this is not in the correct section. I have an engineering economics exam this week. I'd like to know if anybody (particularly professors and/or tutors) know of a website or have a document they could share with me with a few practice problems. I have attached a PDF document with the...

Sorry. It wasn't obvious to me since I didn't know it was fine to have one equation ending in dollars and the other in another unit (in this case screws). I had in fact written it on a piece of paper (somewhere), but the previously stated reasoning was what stalled me. What Dick and Mark said...

100 screws total for a total of $100. My best guess is that the second equation would be: L + M + S = 100 screws My guess is that the Small screws purchased are a multiple of 20. Problem is that the professor does want us to show the procedure. I wasn't able to get a hold of her via e-mail...

I don't have 2 equations. I only wrote one.

To mafagafo: That's what a friend just helped me realize on the phone. Still, I *have* to express this as a system of equations. It's the very first instruction on the paper. Solving it should be simple enough.To Ray: It's all even more detailed on the papers. I'm bogged down because I can't...

I did in the very first post. Or are you referring that I should've added the words "number of" or "amount of" as well?

Large, Medium, and Small. The respective amount of each size of screw.

That's the part I'm seriously stuck at; the whole "buys 100 screws" ordeal. I'm pretty sure I know how the solution for the system will come out; more than likely a free variable since there will be a good number of possible solutions.

Homework Statement It's a verbal problem. It goes like this: "Peter needs to buy some screws: large, medium, and small. He goes to the hardware store and orders 100 screws for a total of $100. For each large screw he was charged $5.00, $1.00 for each medium one, and $0.05 for each small one...

Sorry, but I really just don't get it. Gave up on it because I have quite a few more problems to solve in my papers. I don't know if I wrote down something wrong or if the professor made a mistake at some point. The table I have says to let u = a * sin \theta while du = a * cos \theta...

I give up on this one then. Thanks for your help though. Professor always uses the triangle right at the end though.

Okay, so... I break it apart. It being: \int \frac {cos\theta}{sin^2 \theta} d\theta Which pretty much becomes this: \int (\frac {1}{sin \theta}) (\frac {cos \theta}{sin \theta} d\theta Applying identities, yada yada yada... \int csc\theta cot\theta d\theta Which can be easily...

This is what I did afterwards and where I pretty much left the problem at before moving on to other problems in my paper: \int \frac {\sqrt{1 - x^2}}{x^4} dx Let u = x, du = 1dx \int \frac {\sqrt{1 - u^2}}{u^4} dx Now, let u = sin\theta, du = cos\theta d\thetaSo I end up with: \int...

That -4 was a typo. Fixed.

Reviewing my notes after remembering something about trigonometric substitutions I found: \sqrt {a^2 - u^2} , let u = a \sin\theta and du = a \cos\theta d\theta Is that what you were referring to, Mark44?