Mathematics desk | ||
---|---|---|
< June 2 | << May | June | Jul >> | June 4 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
Problems like this are relatively easy to solve with a little planning and some brute force. I've been working on a similar problem, but with the medians drawn in for each of the sixteen smallest triangles. (Sorry, no image.) With so many triangles, it's much harder to develop a strategy that avoids double counting. Any thoughts about how I should approach this? Is there a general solution or strategy for problems like this? For what it's worth, I've found 482 so far. 24.1.231.240 ( talk) 06:08, 3 June 2009 (UTC)
I'm in the market for a used car, and I've been trying to figure out a simple score that (in a handwavey sense) captures something approximating "value". It's just a first approximation (naturally condition and history will affect the real value), but as the dealers are all rather far away (in vexingly diverse directions) this should hopefully allow me at least to reject the overly ambitious. Given the same model (and ignoring age) leaves mileage and asking price.
I've been calculating the difference between the new price of the model (it's still made) minus the asking price; call that the saving. A large saving is good, a small one bad. Then I'm simply dividing the saving by the mileage (as a big mileage is bad, a small one good). Assuming that car prices vary linearly with mileage (they don't, it's of a gentle exponential) does the logic of my calculation hold? That is, if V is (original-asking)/mileage, is it true to say that a higher V is (practical things aside) a better "value"? 87.115.17.103 ( talk) 13:14, 3 June 2009 (UTC)
That all said, this is probably about as good a time as it gets to buy a new car (dealers are desperate to sell). After a series of used cars I bought my first new one several years ago (I'm still driving it) and although I knew it that decision wasn't optimal in pure financial terms, I found it to be a much more satisfying experience than buying a used car, due to freedom from uncertainty about undisclosed problems with the car, etc. 207.241.239.70 ( talk) 03:12, 4 June 2009 (UTC)
Okay, I haven't learned how to do this. I keep getting "parse error" or something similar when I try to make it appear.
In my calculus book, the first integral on the list of integrals is (integral sign) u dv = uv - (integral sign) v du. List of integrals on Wikipedia doesn't have it, so it must be somewhere else.
I was looking to see how Wikipedia explained this, just out of curiosity. I know how my calculus book does. Vchimpanzee · talk · contributions · 21:44, 3 June 2009 (UTC)
Mathematics desk | ||
---|---|---|
< June 2 | << May | June | Jul >> | June 4 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
Problems like this are relatively easy to solve with a little planning and some brute force. I've been working on a similar problem, but with the medians drawn in for each of the sixteen smallest triangles. (Sorry, no image.) With so many triangles, it's much harder to develop a strategy that avoids double counting. Any thoughts about how I should approach this? Is there a general solution or strategy for problems like this? For what it's worth, I've found 482 so far. 24.1.231.240 ( talk) 06:08, 3 June 2009 (UTC)
I'm in the market for a used car, and I've been trying to figure out a simple score that (in a handwavey sense) captures something approximating "value". It's just a first approximation (naturally condition and history will affect the real value), but as the dealers are all rather far away (in vexingly diverse directions) this should hopefully allow me at least to reject the overly ambitious. Given the same model (and ignoring age) leaves mileage and asking price.
I've been calculating the difference between the new price of the model (it's still made) minus the asking price; call that the saving. A large saving is good, a small one bad. Then I'm simply dividing the saving by the mileage (as a big mileage is bad, a small one good). Assuming that car prices vary linearly with mileage (they don't, it's of a gentle exponential) does the logic of my calculation hold? That is, if V is (original-asking)/mileage, is it true to say that a higher V is (practical things aside) a better "value"? 87.115.17.103 ( talk) 13:14, 3 June 2009 (UTC)
That all said, this is probably about as good a time as it gets to buy a new car (dealers are desperate to sell). After a series of used cars I bought my first new one several years ago (I'm still driving it) and although I knew it that decision wasn't optimal in pure financial terms, I found it to be a much more satisfying experience than buying a used car, due to freedom from uncertainty about undisclosed problems with the car, etc. 207.241.239.70 ( talk) 03:12, 4 June 2009 (UTC)
Okay, I haven't learned how to do this. I keep getting "parse error" or something similar when I try to make it appear.
In my calculus book, the first integral on the list of integrals is (integral sign) u dv = uv - (integral sign) v du. List of integrals on Wikipedia doesn't have it, so it must be somewhere else.
I was looking to see how Wikipedia explained this, just out of curiosity. I know how my calculus book does. Vchimpanzee · talk · contributions · 21:44, 3 June 2009 (UTC)