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Hello. Is there such thing as a dissimilar symbol (e.g. maybe a slashed tilde) in comparing similar triangles? If so, like how does it look? Thanks in advance. -- Mayfare ( talk) 19:16, 16 December 2007 (UTC)
Does the function y=e^x-x have an inverse? How can you prove it doesn't if it doesn't? The reason for this question is in that I'm interested in that e^pi-pi is very close to 20, and I'm wondering what the reason for this is. Indeed123 ( talk) 19:23, 16 December 2007 (UTC)
Infinity can never be defined, therefore shouldn't it be equal to 1/0. 1/0 is greater than any imaginable number, as well as infinity. 1/0 being greater than any other number can be shown if you start taking any real number and plugging it in for the zero. When this happen the answer to the algebraic equation gets greater as the denominator gets lower. This brings me to conclude that 0 when plugged into 1/x would be the largest number, infinity. Even in the graph 1/x, it can been seen that when x is equal to zero, it crosses the y-axis at a point that is undefined, though logically infinity. Therefore, through this equation is possible to postulate that infinity and negative infinity are one and the same. The reason for this is since the equation is a function, meaning it crosses the y-axis at one point, it cannot have two y-axis crossings, unless infinity and negative infinity are one and the same. —Preceding unsigned comment added by ARedens ( talk • contribs) 23:35, 16 December 2007 (UTC)
If you are asking if 1/0 = in the realm of real numbers, the answer is no. 1/0 = undefined. a * b = c, c / a = b. Subsutiting #s for a = 0, b = 1, c = 0. 0 * 1 = 0, 0 / 0 = 1. Substing a = 0, b = 2, c = 0. 0 * 2 = 0, 0 / 0 = 2. So we have 0 / 0 = 1 and 2 at the same time. n / 0 = undefined. n / m as m approaches 0 = . 2/m is always 2x greater than 1/m but 2/m is not always greater than 1/o as m and o approaches 0. Whether 2/m is greater than 1/o depends on the rate m and o approaches 0. So infinity is defined but is not a number. Infinity is larger then all real numbers but since infinity is not a number, it is not possible to determine the largest infinity. NYCDA ( talk) 18:59, 17 December 2007 (UTC)
Sorry, NYCDA, but you don't know what you're talking about."But 2/n is always greater than 1/n even if n = 0." Is this true for n = -1/2? "Is 2/0 greater than 1/0? Maybe." That's a meaningless question. How exactly can you solve 1/0?
I was recently introduced to the differential equation question, "Solve the differential equation where , given that when ." To list out all my steps here would take some time, but to summarize: I first moved xy to the right side, then expanded the left side, then divided all terms by x and then finally divided all terms by dy, to end up with:
Integrating away the dxes and dy,
Knowing that (1,1) was a point on the curve I determined C to be -0.5 and ended up with the function
Problem is, if I put this function in my graphing calculator, it determines to be 2 at (1,1), however it should be 0.5 according to the intial question . I can't see what I've done wrong, but I doubt the calculator has made a mistake, so, what have I done wrong? Thanks -- Colonel Cow ( talk) 23:57, 16 December 2007 (UTC)
No, you shouldn't drop it.
Mathematics desk | ||
---|---|---|
< December 15 | << Nov | December | Jan >> | December 17 > |
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. |
Hello. Is there such thing as a dissimilar symbol (e.g. maybe a slashed tilde) in comparing similar triangles? If so, like how does it look? Thanks in advance. -- Mayfare ( talk) 19:16, 16 December 2007 (UTC)
Does the function y=e^x-x have an inverse? How can you prove it doesn't if it doesn't? The reason for this question is in that I'm interested in that e^pi-pi is very close to 20, and I'm wondering what the reason for this is. Indeed123 ( talk) 19:23, 16 December 2007 (UTC)
Infinity can never be defined, therefore shouldn't it be equal to 1/0. 1/0 is greater than any imaginable number, as well as infinity. 1/0 being greater than any other number can be shown if you start taking any real number and plugging it in for the zero. When this happen the answer to the algebraic equation gets greater as the denominator gets lower. This brings me to conclude that 0 when plugged into 1/x would be the largest number, infinity. Even in the graph 1/x, it can been seen that when x is equal to zero, it crosses the y-axis at a point that is undefined, though logically infinity. Therefore, through this equation is possible to postulate that infinity and negative infinity are one and the same. The reason for this is since the equation is a function, meaning it crosses the y-axis at one point, it cannot have two y-axis crossings, unless infinity and negative infinity are one and the same. —Preceding unsigned comment added by ARedens ( talk • contribs) 23:35, 16 December 2007 (UTC)
If you are asking if 1/0 = in the realm of real numbers, the answer is no. 1/0 = undefined. a * b = c, c / a = b. Subsutiting #s for a = 0, b = 1, c = 0. 0 * 1 = 0, 0 / 0 = 1. Substing a = 0, b = 2, c = 0. 0 * 2 = 0, 0 / 0 = 2. So we have 0 / 0 = 1 and 2 at the same time. n / 0 = undefined. n / m as m approaches 0 = . 2/m is always 2x greater than 1/m but 2/m is not always greater than 1/o as m and o approaches 0. Whether 2/m is greater than 1/o depends on the rate m and o approaches 0. So infinity is defined but is not a number. Infinity is larger then all real numbers but since infinity is not a number, it is not possible to determine the largest infinity. NYCDA ( talk) 18:59, 17 December 2007 (UTC)
Sorry, NYCDA, but you don't know what you're talking about."But 2/n is always greater than 1/n even if n = 0." Is this true for n = -1/2? "Is 2/0 greater than 1/0? Maybe." That's a meaningless question. How exactly can you solve 1/0?
I was recently introduced to the differential equation question, "Solve the differential equation where , given that when ." To list out all my steps here would take some time, but to summarize: I first moved xy to the right side, then expanded the left side, then divided all terms by x and then finally divided all terms by dy, to end up with:
Integrating away the dxes and dy,
Knowing that (1,1) was a point on the curve I determined C to be -0.5 and ended up with the function
Problem is, if I put this function in my graphing calculator, it determines to be 2 at (1,1), however it should be 0.5 according to the intial question . I can't see what I've done wrong, but I doubt the calculator has made a mistake, so, what have I done wrong? Thanks -- Colonel Cow ( talk) 23:57, 16 December 2007 (UTC)
No, you shouldn't drop it.