Of course the easiest way is with technology: Maple tells me that ... 9.000042868 which works for me. A more difficult option would be to get enough terms in...
i have a problem if u solve this immedately i will be happy,thanx show that every positive integer n bigger or equal than 24 can be written as a sum of 5's...
I think this works: The cases from 24,25, ... , 28 all work (check them, they are 7+7+5+5, 5+5+5+5+5, 7+7+7+5, 7+5+5+5+5, 7+7+7+7). After that, all numbers can...
... Let (*) be the center expression in your problem. Let f(x) = (x^4 - 1)^{1/4}. Since f is strictly increasing with f(1) = 0, we have that \int_1^3 {f(x) dx}...
... how ... worked ... (1) ... Plugging in a few more digits, (*) < 9.00004286888 Using convexity, we can also get a lower bound. The function (x^4 + 1)^{1/4}...
kingaragorn2001 <no_reply@yahoogroups.com> wrote: i=sqrt(-1) Find the number of real number pair (a,b) such that (a+bi)^2002=a-bi. i think its proper solution...
if there's a class contains some students and all of them have a degree of math exam. u want to compute the average degree of them but no one of them want to...
Have each person reveal his degree to an outsider who isn't in the class. The outsider will compute the average. ... degree of math exam. ... want to say his...
If a+bi=r*e^(i*t) in exponential form, then (a+bi)^2002=r^2002*e^ (2002it) and a-bi=re(-it) so (a+bi)^2002=a-bi iff r^2002=r and 2002it=-it+2k*pi*i for some k ...
haytham ragab <haytham_ragab@...> wrote: if there's a class contains some students and all of them have a degree of math exam. u want to compute the...
Add up all the grades, and then divide that total by the number of students. This will give you the average grade. ... degree of math exam. ... want to say...
haytham ragab <haytham_ragab@...> wrote: if there's a class contains some students and all of them have a degree of math exam. u want to compute the...
So if each student is unwilling to reveal his/her grade, then surely they will reveal their grades to someone else? And it is from these someone elses that...
... As others have pointed out, one set of solutions is the 2003 roots of (a+bi)^2003 = 1; however, there is one additional solution: a+bi = 0. So there are...
... i ... Each student writes his grade on the back of a $20 bill and mails it to me. I then compute the average and post it to mathforfun. I promise not to...
Hello, I am new here, I am a student in grade 9, I like science very much, especially math, I am in China, I want to study with everyone and change our...
Hello, everyone, I have a very hard problem, it comes from my book. I can't solve it including my teachers, I hope you can help me, I have created it to the...
Hello, This email message is a notification to let you know that a file has been uploaded to the Files area of the mathforfun group. File :...
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Jan 6, 2003 1:32 pm
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... wrote: Show that for 0 <= a < b (b-a)/(b+1) <= ln((b+1)/(a+1)) <= (b-a)/(a+1) ... First we prove that for all real x such that x >= 0 x/(1+x) <= ln(1+x) <=...
... The problem: a,b,c are all positive and integer number, and parabola "y=ax^2+bx+c" has two different points of intersection "A" and "B" with x-axis. If the...
I need help thinking something through (this should be embarrassingly easy for you guys, forgive my ignorance). Here is the question: If over the last year the...
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