What do they mean when they say to find the prime factorization of the #s 352, 89, and 156? Put in exponential form?
135527
The question relates to two things:  the first is to determine basically what numbers it takes multiplied together to create the final number in question.\n\nTo do this, you start with the prime numbers, i.e., numbers which are only divisible by 1 and themselves.  Several examples of these would be 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, and so on.\n\nTake 352, for example:\n352/2 = 176\n176/2 = 88\n88/2 = 44\n44/2 = 22\n22/2 = 11\nYou have exhausted the number 2.  Now move on to the next prime number by which 11 is divisible.  As 11 itself IS a prime number, the answer is 11.\n\nThe second part of the question relates to exponential form.  How many times did you divide 352 by 2 until you hit an odd number?  Five times.  So, the exponential expression of the prime factorization of 352 is 2 to the 5 (which is not posible to illustrate here.  Write the number 2 and then write a small 5 to the upper right,) * 11\n\n2(5) * 11 = 352\n\nNow take the other two numbers.  If you look at the list I provided above, you will note that 89 is a prime number.\n\n1 * 89 = 89\n\nFinally, take 156.\n\n156/2 = 78\n78/2 = 39\n39/3 = 13\n\n2(2) * 3 * 13 = 156\n\nHope this helps.\n\nPlease take time to pick the best answer.\n\nI tip the hat to the people who chose to use the "^" key.  I could not think of a way to describe it otherwise.  Their representation is probably more valid than my use of parentheses, which would actually represent something else altogether.  Note well though that I indicated how to write "two to the fifth power" in my paragraph above.
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Seems like I have been doing most of your homework, SMILE4LISA *wink.  But I do not mind.  I noticed a couple of errors in your question, so I have taken the liberty of fixing them myself.\na) Should read: factor a^6 + 1 = (a^2)^3 + 1 as a sum of two cubes.\nANSWER: Using the formula, a^3 + b^3 = (a + b)(a^2 - ab + b^2), I can write a^6 + 1 = (a^2)^3 + 1^3.  Hence, a^6 + 1 = (a^2)^3 + 1^3 = (a^2 + 1)((a^2)^2 - a^2 + 1)\nb) Simple plug n chug: \n2^6 + 1 = (2^2)^2 + 1^3 = ((2^2) + 1)((2^2)^2 - 2^2 + 1), both check out to be 65 when simplified.\n2^12 + 1 = (4^2)^3 + 1^3 = ((4^2) + 1)((4^2)^2 - 4^2 + 1), both check out to be 4097.\nc)It should read: show that a number of the form 2^n + 1 (n > 0) cannot be prime if n is a multiple of 3 (ie if n = 3k for some integer k)\nANSWER: 2^n + 1 = 2^3k + 1 = (2^k)^3 + 1^3 = (2^k + 1)(2^2k - 2^k + 1) and this implies that it is divisible by the factors of (2^k + 1) and (2^2k - 2^k + 1); primes can only be divisible by 1 and itself.
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120(14-6)\n\nStep 1: 120(8) <---- because in order of operations, the "p" which is parentheses in "pemdas" is first\nSteo 2: 120 x 8= 960 <------------- because when a number is outside the parentheses yet doesn't have any operation sign between them, u do multiplication
520663
If P is an odd prime, then half of the time we have P = 1 mod 4, and the other half of the time we have P = 3 mod 4.\n\nThis is the same as saying that half of the odd primes are of the form 4n + 1, and the other half are of the form 4n + 3.
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The answer is 55, but your question says that you want to know HOW to do this.\n\nDave's method works, but it could take a long time if you're working with large numbers.\n\nIf you're able to be a bit more sophisticated, you can save some time by applying this method:\n\n1. Find the prime factors of each of the numbers. Remember that you need to consider not only which prime numbers are factors of the numbers, but also HOW MANY TIMES each prime number is used. For example, 2 is the only prime factor of 8, but it occurs 3 times, so the correct prime factors of 8 are 2, 2, 2.\n\n2. Make a list of the prime factors you will need in the least common multiple (LCM). This list includes every prime factor for one of the two numbers you started with, and also every prime factor that occurs in the second number but not the first. Importantly, if a factor occurs in the second number MORE TIMES than it occurs in the first, you have to include it in your LCM that many times.\n\nExample:\nFind the LCM of 168 and 180.\nPrime factors of 168:  2, 2, 2, 3, 7\nPrime factors of 180:  2, 2, 3, 3, 5\n\nPrime factors of LCM: 2, 2, 2, 3, 3, 5, 7 \n\nExplanation: 2, 2, 2, 3, 7 are all the prime factors of 168.\nWe then need another 3 (the factors of 180 include a second 3) and a 5 (since 180 had a 5, which 168 did not).
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36x^2 yz^3 is a single expression.  You cannot factor it.\n\nThe gcf of 21ed^2 and 35de^3 is 7ed\nBoth 21 and 35 are divisible by 7.\nSInce d and e appear in both terms, you can take out the lowest power of each one.  The lowest power of d is one and the lowest power of e is one.
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One way is to write down what prime factors are required. For each prime find the most number of times that prime divides one of the number and include that number as a factor of the least number.\n\n2 divides 2, 4, and 8, and it divides 8 three times: 8 = 2*2*2. So the least number has a factor of 8.\n3 divides 3 and 9. 9 is 3*3, so the least number has a factor of 9.\n5 divides 5 and 10, but only once in each. So the least number has a 5.\n7, 11, and higher primes don't divide these numbers.\n\nSo the number you want is 8*9*5 = 360. \n\nIf you include 7, then we must account for that prime number. It occurs once in itself. So the least number must have 7 as a factor. So the number you want is 8*9*5*7 = 2520.
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.
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Everyone so far is wrong.  \n\nThe form of the exponential equation is y=Ae^(Bx)+C, where C=3.\n\nStart by substituting in the values of (x,y) given, (2,200) and (3,103)...\n\n1) 200=Ae^(2B)+3 or 197=Ae^(2B)\n2) 103=Ae^(3B)+3 or 100=Ae^(3B)\n\n...this substitution generates 2 equations with 2 unknowns (A, B), which you can now solve in various ways: you can use a matrices, or substitution, or algebraic simplification of the equations by taking the ln of both sides of each equation...however you do it, the answer becomes:\n\nA=764.53730   B=-0.6780335\n\nwhich gives you:  y=764.53730e^(-0.6780335x) as your final answer.\n\nYou can check the answer by plugging in the x value into the equation and solving for the y value:  i.e, by observing that the following are both true:\n\ny(2)=200 and y(3)=103  (look familiar?)\n\nThis check illustrates that the equation indeed goes through the 2 given points, and is an exponential equation, so your done.  Furthermore, if you wanna know the y-intercept, just take y(0), which turns out to be \n\ny(0)=764.53730 - which is the value of the unknown A in the original equation you set up.
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5/6 = 5/2*3\n10/25 = 2*5/5*5\n18/36 = 2*3*3/2*2*3*3\n15/50 = 3*5/2*5*5\n\n18/2 = 9/3 = 3 so the prime factorization is 2*3*3\n105/3 = 35/5 = 7 so the prime factorization is 3*5*7\n711/3 = 237/3 = 79 so the prime factorization is 3*3*79
51367
http://en.wikipedia.org/wiki/E_%28mathematical_constant%29\nfor the significance of e and how it is used, use this link
35727
I'll take it as far as I can.\n\n3 - Add all the numbers and if the result is divisible by 3 then the entire number is 9+1+2+0+3+1+2=18 which is divisible by 3.\n4 - The last 2 numbers are divisible by 4. 12 is divisible by 4 so the entire number is.\n6 - It's divisible by 2 and 3.\n8 - If the last 3 numbers are divisible by 8, then the entire number is. 312/8=39. \n9 - Like for 3, except it has to be divisible by 9. The result, from above, is 18 which is divisible by 9.\n12 - If divisible by 3 and 4.\n18 - If divisible by 2 and 9.\n24 - If divisible by 8 and 3.\n36 - If divisible by 4 and 9.\n72 - If divisible by 8 and 9.
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i don't remember the notation, but it is\n\nthe set of Integers not found on pages p sub 1 through p sub n
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p^5 +2p^4 -2p^3 + 3\n\nNote: P times P^4 is P^5, b/c when you multiply exponents, you add the exponential numbers.  Since P is really P^1, 1+4 = 5, giving you P^5.
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The answer is i.\n\nFirst, look at i^0(i to the 0th power). It is 1, because anything to the 0th power is 1.\nNow, look at i^1(i to the 1st power). It is simply i.\nNext, look at i^2, which is -1, by the definition of i.\ni^3 = i^2 * i, right, which is -1 * i = -i.\ni^4 = 1 etc..\n\nObserve i^n, where n is any value over 3. We can then write i^n as i^4 * i^(n-4) = 1*i^(n-4) = i^(n-4). So i^n = i^(n-4). So we can just keep subtracting 4 from n until we are at 0,1,2 or 3, which we now know! We see that this number (0,1,2 or 3) is simply the remainder when we divide n by 4.\n\nSo to find powers of i, we simply find the remainder when it is divided by 4, and pit it against this chart we have made:\n\n0:1\n1: i\n2: -1\n3: -i\n\nSo for this problem, if we do the long division, we see that 21/4 = 5 + 1/4, so the remainder is 1. Therefore, the answer to this particular problem is i.\n\nSo the answer is i.
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Proving irreducibility over Q is rather straightforward based on what the form of a nontrivial factorization would have to look like.\n\nReduction modulo every prime follows from X^4 - 10x^2 + 1 being equal to all three of the following:\n\nx^4 - 2x^2 + 1 - 8x^2 = (x^2 - 1)^2 - 2(2x)^2\nx^4 + 2x^2 + 1 - 12x^2 = (x^2 + 1)^2 - 3(2x)^2\nx^4 - 10x^2 + 25 - 24 = (x^2 - 5)^2 - 6(2^2)\n\nIf 2 is a quadratic residue modulo p, then the first is a difference of two squares.\n\nIf 3 is a quadratic residue modulo p, then the second is a difference of two squares.\n\nIf neither 2 nor 3 is a quadratic residue modulo p, then their product 6 is a quadratic residue modulo p, and so the third is a difference of two squares.
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What you want to do is factor out the same values from the numerator and denominators.  \n\nThe order in which you factor these out does not matter, the key is to continue to factor them out until there is nothing else to factor out.\n\nFor example you can factor out a^8 by turning it into:\n\n36 * (a^8) * (b^4) * 1/2\nover\n27 * (a^8) * a * (b^6) * 1/3\n\nsince (a^8) over (a^8) = 1 you can cancel them out leaving you with:\n\n36 * (b^4) * 1/2\nover \n27 * a * (b^6) * 1/3\n\nwe do the same factoring for b where we can factor out a (b^4):\n\n36 * (b^4) * 1/2\nover \n27 * a * (b^4) * (b^2) * 1/3\n\ncanceling the (b^4)s we get:\n\n36 * 1/2\nover\n27 * a * (b^2) * 1/3\n\nYou can then further reduce the fraction by multiplying out the 36 * 1/2 and the 27 * 1/3:  leaving you with:\n\n18\nover \n9 * a * (b^2)\n\nAs you can see now you can factor out a multiple of 9 leaving you with:\n\n2\nover\na*b^2