can u divide a number by prime number multiplied by the integral of pi squared with a power of the exponential
127246
doesn't pi have an e in it?
514571
3e^x. Pull the constant outside the integral sign\nand just let it sit there!
151170
no this is how to work it out. I am assuming you know how to take the logarithm of a coplx number and find the modulus and argand of it.\n\ni^i = e^(logi^i) = e(ilogi)\n\nnow log i = log|i| + i(argi)\n= 0 + iπ/2 + i2kπ\n\nwhere k is an integer\n\nso substituting this is\ni^i = e^(i(iπ/2 + 2ikπ)\n= e^(-π/2-2kπ)]\n= e^(π[-1/2-2k])
75511
Id & prime are both correct on the formulas.  (mv^2)/2 is the same as (0.5)m(v^2), which you could just write without the parentheses.  (Note (mv^2) does not square m.) \nId's math had an error.  350474 is correct.
98692
i think this will work...\n\nint prime=43;\nint isprime=0;\nfor(int i=2 ;i<prime;i++)\n{\n   if((prime%i)==0))\n   {\n       isprime=1;\n       break;\n   }\n}\nif(isprime==1)\n   printf("Not prime");\nelse\n   printf("Prime !");\n\n\n=====\nGot it na??
58498
Try http://integrals.wolfram.com/
611684
you can simply the first one to (x-a)(x-a) then you can divide that by (x-a) and it equals (x-a)\n\nDo you know synthetic division; cuz you can divide the second problem by (x+2)^2 if you mulitply (x+2)^2 out.
370003
A bit tricky to integrate, but is like this (i could mistake!):\n\na)\n9^(1/5)\nS pi ( (3 sqrt(x))^2 - x^3^2) dx\n0\n\nb)\n9^(1/5)\nS pi ( (3 sqrt(x)-6)^2 - (x^3-6)^2) dx\n0\n\nc)\n9^(3/5)\nS pi ( (y^2/9)^2 - (y^(1/3))^2) dy\n0\n\nd)\n9^(3/5)\nS pi ( (y^2/9-4)^2 - (y^(1/3)-4)^2) dy\n0\n\nRegards
176771
split the integral up\n\nyou have:\n(1-2x^2)/x = 1/x - 2x^2/x = 1/x - 2x\n\nintegrating this gives:\n\nIn(x) - 2x^2/2 = In(x)-x^2
10589
I just had a math packet dealing with this topic.  There are many equations out there that deal with finding primes, but all so far fail at a certain point in the sequence.  The method you refer to, called the Sieve of Eratosthenes, was developed by its namesake in Greece sometime between the 500s and 350 B.C. and is so far the only proven method to find primes.  Sorry, but unfortunately, there is no other accurate choice.
673193
It looks like that first 1 should be an i. In this case,\ni^z=exp(i*pi/2*z),\nwhere I am using the principal branch of the logarithm to define the exponential.\nThen the derivative will be \n(i*pi/2)*exp(i*pi/2*z)=\npi/2*i^(z+1)\n\nEasy.
303027
You can noy factor it out... it makes it easier to figure out when you put it into lowest terms... so it would be \n\n   x^2+7x-25..... so you cant factor it.
121348
Yes it is, because you did not multiply by an approximation of pi, you left the pi whole, so to speak :).  It's more accurate that way.
658462
The problem is not that you can't take the log of a complex number. The problem is that log is multiple valued in the complex plane. In other words, the exponential function is not one-to-one. You can take the 'principle branch' of the logarithm, but then you can lose certain of the rules of logarithms.\n\nThe upshot is that when you take the log of -1, you either get an infinite set of values (.., -pi*i,  pi*i, ...) and another infinite set of possibilities for the log of 1 OR you lose the property that log(a^b)=b*log(a) that you used in your first paradox. This is just a fact of life when working with complex numbers. In the second, BOTH -pi*i and pi*i are logarithms of -1, so the argument fails. It's somewhat like saying that the squares of -1 and 1 are the same, so -1=1.
69527
Unfortunately I don't know of a quick method but I can help you a little here. \n\nFirst, as you probably already know, prime numbers can only be evenly divided by 1 and the number itself.\n\nSecond, since all numbers are a product of prime numbers you can save time in by not testing division by non-prime numbers like 4 and 6 and 9 and so on.\n\nThird, you only have to test those prime number that are smaller than the square root of the number.\n\nIn your example, 117, you need only test the prime numbers less than 10.81.... since that is the square root of 117. So do a trial divide of 117 by 2, 3, 5, and 7.\n\nOf course there are a few other rules that can help you eliminate the first three choices. If the number is odd, it can't be evenly divided by 2. If the sum of digits isn't divisible by 3 then the number isn't either. And all numbers that are divisible by 5 end in 5 or zero.\n\nFor 117, the number is odd so that eliminates 2. The sum of 1 and 1 and 7 is 9 which is divisible by 3 so 117 is divisible by three and therefore not prime.\n\nSo now you may notice that 117 is divisible by 3 and 39 but I said you only need to test numbers smaller than the square root. What about 39? Well, since the number is divisible by three that alone proves the number is not prime. If you want the full prime factorization, you need to keep dividing.\n\nGood luck.
169504
this is exactly the same as a previous question asked about an hour ago. Look back on the open questios, your answer is there\n\nh = 100/pi*r^2\n\nh=7.96
181164
1/(1+e^x)=(1+e^x -e^x)/(1+e^x)=1-  e^x/(1+e^x)=\n1- (1+e^x)'/(1+e^x)\n& u'll integrate