how can i calculate UFL value for dairy cows?
450617
assume a spherical cow.
809519
Well, let's see.\n\n18 = b + s + c,  where b means bird(chicken)\n\ntotal number of legs:\nLt = 2b+4s+4c, right?\n\nLt = 3Ht +10 , total legs 10 more than 3 times number of heads\n\nI'd assume each animal only has one head, thus Ht = 18\n\nLt = 3*18+10 = 64\n\nLike most problems of this sort you must find integer solutions.\n\n18 = b + s + c\n74 = 2b + 4s + 4s\n\nA trial and error procedure is effective.  \nstart at \nb=s=c=6, -> Lt = 12+24+24 = 60, need more legs!\nb,s,c\n5,7,6, -> Lt = 10+28+24 = 62,\n4,7,7 -> Lt = 8+28+28 = 64 <------------- bingo\n\nNote that you can freely shift sheep and cows as long as the total is 14.  So, there are other solutions:\n\n4,14,0 -> Lt=8+56 =64....no cows.....
409921
This page has the UDSA database with nutritional values of thousands of foods: \n\nhttp://www.nal.usda.gov/fnic/etext/000020.html
1144126
http://www.wabdl.org/kilo_chart.htm\n\nJust what you need!  It will eliminate your need to multiply.  smiling
134946
here you go: http://finance.yahoo.com/q?s=FSL
408231
get a cow and find out
1077695
Check the usda website
751939
you can either do the math, look it up on a chart or go here:\n\nhttp://leebrewery.com/hydromet.htm\n\nand use the java-calculator.
1117478
v=u+at\n\n24=10+16*a\na=14/16 mtr/sec^2\n\n\nf=ma\nf=1200*14/16 N\nf=150*7 N\n\ntherefore f=1050 N
883460
Please see the answer above and find out the missing factor in your calculation. You are in the right track. All the best!
1045889
Holsteins make up the vast majority of dairy cattle for their ability to adapt to confinement methods. Calves are normally 90-100 pounds at birth; weanling weights generally aren't taken on dairy but as yearlings at about 800# they are bred, with adult cows not uncommon at the 1500# mark. Bulls from 1800# to just over a ton aren't uncommon. They can be black and white or red and white. \n\nAyreshire cows are more like 1200# at maturity - they were developed to be efficient grazers so body weight wasn't as big of a factor. \nGuernseys are similar to Ayrshires. \n\nBrown Swiss and Milking Shorthorn have more variance because of their dual purpose background. Swiss cattle fall between the Ayreshire and Holstein in size with bulls not uncommon at 1700#+ and cows 1,000-1200. Milking Shorthorns are very slightly smaller...not that many years ago, relatively speaking, they were a single breed - shorthorns. A demand for specialization divided milking and beef cattle into breeding for different traits \n\nJerseys are the smallest breed - calves 50-60#, adult cows 800-1200# bulls 1600-1700# \nThe loss of farms that use grazing is threatening all but the Holstein and Jersey.
803108
This is the result from one of the on-line conversion charts. I have put a link below the answer.\n\n$149.00  =£78.3575 \n\nhttp://www.xe.com/ucc/convert.cgi
498327
Multiply the number of cubic feet by 7.481 to get the number of gallons (U.S.). \n\nThis formula applies to "liquid gallons", which are the kind commonly used today.
1032535
You can get that info from Yahoo Finance, linked below
850193
Look here for the tables:\n\nhttp://wulfenite.fandm.edu/Data%20/Table_7b.html\n\nhttp://wulfenite.fandm.edu/Data%20/Table_7a.html\n\nThe values you need are in the second link.
1150368
This is a good statistic which shows US states ranked of farm productivity:\n\nhttp://www.statemaster.com/graph/eco_gsp_cha_qua_ind_cro_and_ani_pro_far-indexes-crop-animal-production-farms\n\n\nFor more information, surf around this site for other economic indicators related to farming, or try the USDA website at:\n\nhttp://www.usda.gov/wps/portal/usdahome
189782
It's a calculation called the r value.  The link below shows how to calculate it.  If you take a college level statistics class you'll learn how to do it with a good calculator.