What is the practical application of HCF and LCM?
138127
It helps decide the optimum size for packages when you have different things that you want to pack. It's part of optimization.
364954
practical HPLC method development by snyder\n\nI have this book. I do hplc almost everyday it has helped me alot.
816611
hi,\n  regarding lcm of a set of numbers.....\n\n  the lowest common multiple (lcm) of two whole numbers\n  is the smallest whole number which is a multiple of both.\n\n  consider the numbers 12 and 15-----\n  the multiples of 12 are : 12, 24, 36, 48, 60, 72, 84, ....\n  the multiples of 15 are : 15, 30, 45, 60, 75, 90, ....\n\n  60 is a common multiple (a multiple of both 12 and 15),  \n  and there are no lower common multiples.\n\n  therefore, the lowest common multiple of 12 and 15 is \n  60.\n\n  now try it yourself for 120 and 130.
947409
You mean LCM?  LCD is lowest common denominator.\n\nFor each number that you are trying to find the LCM for, break the number down into its prime factors (recall that the primes are 2, 3, 5, 7, 11, 13, 17, etc.), then gather the highest prime factor of each prime.\n\nAn example will work better here.  Let's try this:  Find the LCM of 9, 12 and 150.\n\nSTEP 1:  Get prime factors of each:\n9 = 3*3 = 3²\n12 = 2*2*3 = 2² * 3\n150 = 2*3*5*5 = 2 * 3 * 5²\n\nSTEP 2:  Take the highest prime factor of each prime:\n2 >> highest is 2² from 12\n3 >> highest is 3² from 9\n5 >> highest is 5² from 150\n7 >> we don't have any primes for 7 or higher, so we can stop\n\nSTEP 3:  Multiply the highest prime factors from step 2 together to get the LCM:\n\nLCM = 2² * 3² * 5² = 900\n\nHope that helped a little!\n\n~ ♥ ~
757059
Your first two are right.  Some notes:\n\n1.  If you don't already know, you can find the GCF by factoring the numbers and taking the factors they have in common.  In this case, they only have 1 as a common factor.  You can find the LCM by taking the factors and multiplying the highest powers of each multiple.  In this case, you are simply multiplying all three numbers together.  If the GCF of two or more numbers is 1, then the LCM is the product of those numbers.\n\n2.  Yes, though you can type |6-8| by holding SHIFT and pressing the slash key (by the ENTER key).\n\n3.  Multiply through and combine like terms (assuming x squared in the second part):\n2(4x^2 - 3x + 2) - 3(x^2 - 5x - 8) =\n8x^2 - 6x + 4 - 3x^2 + 15x + 24 =\n5x^2 + 9x + 28\nThis cannot be factored any more.\n\n4.  Solve for V by multiplying through and getting V by itself:\n6(V - 2) - 4(V + 3) = 10-42\n6V - 12 - 4V -12 = -32\n2V - 24 = -32 \n2V = -8 (add 24 to both sides)\nV = -4 (divide both sides by 2)\n\n5.  You never want to leave a radical in the denominator.  So, the expression, 3/√x, is not the correct form.  You can get rid of the radical by multiplying by the square root of x for both the numerator and the denominator:\n3/√x * √x/√x (since √x/√x = 1)\n= 3√x / √x√x\n= 3√x / (√x)^2 \n= 3√x / x\n\nNow, the radical is in the numerator and not the denominator.
5045
"HCF of A&B is 9& of B&C is 12"\n\nB is divisible by 36, because it's divisible by both 9 and 12.\nA is divisible by 9, but not by 2.  (If A were divisible by 2, then the HCF of A&B would be at least 18.)\nAnd C is divisible by 12, but not by 9.\n\n\n"LCM of A&C is 216"\n\n216=27*8\nA=27.\n\nOne solution is:  A=27, B=36, and C=24\nSo LCM of A & B is 108.\nBut B can be any value of 36*N, where N is not divisible by 2 or 3.\n\nSo another solution is: A=27, B=180 (36*5), and C=24.\nLCM of A & B is 540.
17656
Best technique is to look for where it is unbalanced and change it so it is balanced. Repeat until it is balanced.\n\neg. We need 2 hydrogen here on the left, so go for:\n\nMg + 2HCl -> MgCl + H2\n\nNow we need another 2 as we have 2 Cl on the left and one of the right.\n\nMg + 2HCl -> 2MgCl + H2\n\nNow we need to have 2 Mg on the left as we have 2 Mg on the right. So we get:\n\n2Mg + 2HCl -> 2MgCl + H2\n\nAll balanced so that is the answer.
995845
http://www.eng-tips.com/index.cfm
515976
http://www.hccs.edu/catalog/hom.html
63267
Try www.clm-services.com
803921
Use conservation of energy.\n\na.  1/2mA*vA^2 - mA*g*h = 0 \n\nwhere h =  L - Lcos60   assuming L is the length of whatever is suspending the two balls.\n\nb. Use:  1/2mA*vAi^2  + 1/2mA*vAf^2 = 1/2mB*vBi^2  + 1/2mB*vBf^2 \n\nc.  1/2mA*vAf^2 - mA*g*ha2 = 0 \n     1/2mB*vBf^2 -  mB*g*hb2 = 0 \n\nsolve for ha2  and hb2
259646
LIFR(Line Item Fill Rate)
865560
A. Just multiply CF and then subract the resulting matrix from C. Is there something specific about the required operations you don't understand?
513656
sl/fro=hsgf=(ufyh)^o\n\nI love it when you talk like that....
818533
For LCM, keep dividing by common prime factor, starting from 2, throughout till you get 1 in the end.\nInitially 6, 15, 20\nDivide by 2: 3, 15, 10 (You can't divide 15 by 2, hence you leave it as it is.)\nDivide by 2: 3, 15, 5 (Similarly, you can't divide 3 by 2, hence you leave it as it is.)\nDivide by 3: 1, 5, 5\nDivide by 5: 1, 1, 1\nLCM = 2 x 2 x 3 x 5 = 60\n\nHope this helps :)
798301
27=3x3x3\n189=3x3x3x7\nto calculate lcm, just cancel out (one each of the pair of) the numbers that appear twice; you're left with 3x3x3x7, or 189.\nApparently your other respondent thinks if he can't do it it can't be done...
188400
best I could find for you (and me as I am in the midst of working on own company issues):\nwww.tradedynamics.com/docs/pdf/sclm_showcase.pdf