I have an assignment due next Tuesday, and I have some difficulties writing the dichotomous keys. My messeger ID is goldman_equation. If you could help me through it, it will be highly appreciated.
215425
Dichotomous keys are ways of identifying something (plant or animal usually) by asking 50-50 questions.  For example, lets think of eight animals and do this NOT biologically for simplicity:\n\npig, cow, horse, sheep, chicken, dog, cat, human.\n\nIf YOU have to write a key, then first thing is to think of a question which will split these animals into two groups.\n\nFor example:\n1.We usually eat it (2)\n1.We don’t usually eat it (5)\nso this will tell you to go to part 2 if yes (pig, cow, sheep, chicken) or part 5 if no (horse,dog,cat,human)\n\nThen you do the next question that will cut the options in half again, eg:\n2.We use its milk (3)\n2.We don’t use its milk (4)\nso this will cut the options down to two (cow and sheep)\n\nThen you do one question that will cut it down to one, eg:\n\n3.We use its wool – SHEEP\n3.We don’t use its wool – COW\n\nWhen we get down to the animal we just write it.  When you have finished that line of questions, you do the others, eg:\n\n4.It is a bird – CHICKEN\n4.It is not a bird – PIG\n5. It is a pet in the home (6)\n5. It is not a home pet (7)\n6. It hunts alone – CAT\n6. It hunts in PACKS – DOG\n7. We make glue from its bones – HORSE\n7. We don’t make glue from its bones – HUMAN\n\nO.K. this is silly and simple but it is just a start – you would have to write serious questions like; is it a mammal? is it a meat eater? is it a bovine? but this is just to get the idea of writing them – write questions that cut the options down by half each time.
279287
yes here you can consider in terms of set like :\nHumanity={course1,course2,......}\nSociology=(course1,...........}\nScience=(course1,...........}\nMaths=(course1,...........}\nMusic=(course1,...........}\n\nhere n(Humanity)=3\nn(Sociology)=4\nn(Science)=5\nn(Maths)=6\nn(Music)=3\n\ntherefore no of ordered pairs will be equal to=3 x 4 x 5 x 6 x 3= 1080
220606
The previous poster didn't see the whole problem. This is a question about modular arithmetic, not standard algebra. The equations are to be done modulo n and the solutions are to be integers.\n\nBy adding 3 times the first to the second, we need 5x=8. By subtracting twice the first from the second, we get -5y=-2, so 5y=2. So all that is required is that 5 be invertible modulo n, i.e., that 5 does not divide n. For example, if n=4, x=0 and y=2 works.
382050
Looks as though this is some kind of assignment. No? Maybe you ought to think it all through and provide your own answers. You can do it!!\n\nChow!!
622611
m0= A'B'C'D'\nm1= ABSENT\nm2=A'B'CD'\nm3=ABSENT\nm4=A'BC'D'\nm5= ABSENT\nm6= ABSENT\nm7= A'BCD\nm8=AB'C'D'\nm9= ABSENT\nm10= ABSENT\nm11= AB'CD\nm12=ABC'D'\nm13=ABSENT\nm14=ABCD'\nm15= ABSENT
331911
go online, click on search and type in MATH 012 : Intermediate algerbra course. something should pop up. and then write down the evidence
373799
Problem 1:\nWhich one is a RELATION but not a function\n\nChoices:\n1) {(-10,1),(1,8),(-4,-2)}\nfunction\n2) {-4,1),(1,1),(-2,-2)}\nfunction\n3) {(-5,1),(1,-7),(21,-2)}\nfunction\n4) None\n\n\nProblem 2:\nWhich of the following is a function\n\nChoices:\n1) {(-4,1),(1,8),(-4,-2)}\nrelation\n2) {(-4,1),(1,-8),(-2,-2)}\nfunction\n3) {(-4,1),(1,-8),(1,-2)}\nrelation\n4) None\n`
234993
Hi Lauren!  Lauren here, lol!!!!  (yes thats my real name too).  the answer is........cdefgabc  - not to mention the sharps, flats, etc. which are the black keys.  Need more info?  e-mail me.  I too have a keyboard with 36 keys.  and I can go more into detail.   Good Luck Lauren,  sincerely, Lauren.....p.s. luv yer name!!!!!  LMAO !!
58482
- -----------O\n------------ . .-----------  _\n  K+ ( O: Mn  :O )\n-------------. .\n------------ O\n\nomit the long dotted lines
169806
Check this site out, it's neat\nhttp://www.utdallas.edu/~mbaron/3341/Notes/mcarlo.html\n\nhttp://ic.net/~jnbohr/java/CdfDemoArgs.html
54048
here's an output step by step:\n\nthis is the initial state of deque D\nD :: first -> <- last\n\naddFirst(3)\nD :: first -> 3 <- last\n\naddLast(8)\nD :: first -> 3 8 <- last\n\naddLast(9)\nD :: first -> 3 8 9 <- last\n\naddFirst(5)\nD :: first -> 5 3 8 9 <- last\n\nremoveFirst()\nD :: first -> 3 8 9 <- last\n\nremoveLast()\nD :: first -> 3 8 <- last\n\nfirst()\nfirst = 3\n\naddLast()\nD :: first -> 3 8 <- last\n\nremoveFirst()\nD :: first -> 8 <- last\n\nlast()\nlast = 8\n\nremoveLast()\nD :: first -> <- last\n\nhope this helps...
406747
ABCDEFGHJ = 256973148\n\n\nFirstly...\nC - B = 1.......C = B + 1\nH - G = 3.......H = G + 3\n\nThe equation can be simplified as:\n\nA * [ (2B + 1)^(D - E) - F^(G(G+3)) ] / J = 10\n\n1. The difference between the numbers within the square brackets cannot be more than 100 since dividing by J, which is a single digit will not give 10 (assuming that A was 1). Therefore both numbers are supposed to be weighted similarly.\n\n2. Consider the value of 'A'. If the difference were even, then it must be multiplied by 5 before dividing in order to get 10. If it were 5, then 'A/J' has to scale it down accordingly. If it were odd, there are no possible values 'A' can take. Thus the difference can be either be even or a multiple of 5.\n\n3. I assume F is an odd number! The reasoning behind this is that the difference between the 2 numbers can be even [Exception is if the difference were an odd multiple of 5. If it doesn't work then consider the other alternative]. Since (2B + 1) raised to any power is odd, then F must be odd to obtain an even difference.\n\n4. Look at all powers that have values in the range of 1 - 256, since I assume that the values are small and close in value. I consider 1^2 to 15^2, 1^3 to 6^3, 1^4 to 4^4 and 1^5 to 3^5. Look at those with odd bases [also easier to start off with since trial and error must take place now].\n\n5. I obtain and test 11^2 - 3^4 = 121 - 81 = 40. From this, (2B+1) = 11-->B = 5, D-E=2, F = 3 and G^2 + 3G = 4 --> G=1\n\nusing 40...A = 2 and J = 8 in order to get a solution of 10\n\nIf G = 1 then H = 4 from original equation\nIf B = 5 then C = 6 from original equation\n\nThe remaining numbers are 7 and 9, which allows D = 9 and E = 7 to satisfy the final equation of unknowns [D-E=2]\n\nI hope you understand and it helps!
490523
β=10 log I / Iο\nβ2=10 log I2 / Iо\nβ2-β=10 log I2 / Iо - 10 log I / Iо\n         =10 log I2 / Iо / I / Iо\n         = 10 log I2 / I\n         = 10 log 2 I / I\n         = 10 log 2 = 10 x o.3 = 3\n\nI hope i helped you, sorry if i have done any mistake.
238896
It's hard to show it clearly here but the solution is actually:\n\n[-b±√(b²-4ac)]/2a\n\nthere will usually be two solutions!\nThe site below has it laid out more clearly...\n\n("invited" to leave school early.. 1974)
534479
#1.)-----------------------------------\n  \n25b^ -20by + 4y^\n\nMultiply 1st and 3rd term together to get 100b^y^.\n\nFactor 100b^y^ into two factors whose sum = the middle term of -20by.\n\n1by   X 100by\n2by   X 50by\n4by   X 25by\n5by   X 20by\n10by X 10 by   ....................BINGO!\n\n-10by + (-10by) = - 20by   &   (-10by)(-10by) = 100by\n\nThis is a really kewl factor by grouping trick, I LOVE it!\n\nBring down the first term of: (25b^)\n\nNext, write the SMALLER of the two factors 1st.  Followed by the LARGER of the two factors.  In this case, they're both the same, so it doesn't matter which one goes first: (-10by)\n\nFinally, bring down the third term of: (+4y^)\n\n25b^ -10by -10by + 4y^\n\nNow, lets factor by grouping! \nFactor out the GCFs.\nGCF#1 = 5b\nGCF#2 = -2y\n\n(25b^ -10by) + (-10by +4y^)\n5b(5b - 2y) -2y(5b - 2y)\n\nGCF#3 = (5b - 2y)\n\n(5b - 2y)(5b - 2y)\n\nSimplifies further as: (5b - 2y)^   ....Whatdayaknow?! \n.......................................................................................................It's a Perfect Square Trinomial.\n\n#2.)------------------------------------------------------------------\n\n100 - 9b^\n\nAh, the difference of two perfect squares.\n\n---------------------------------------------\n\nFORMULA:\na^ - b^ = (a-b)(a+b)\n\n---------------------------------\n\nSOLUTION:\n100 - 9b^\n(10)^ - (3b)^\n\n-----------------------\n\nANSWER:\n(10-3b)(10+3b)
309891
go to mypyramid.gov it's very very helpful
667822
Doesn't sound like you know what ID means let alone irreducible complexity or paley's watchmaker argument to back it up at all and what the hell is your question?  Seems to just be a ignorant claim without any good supporting evidence or logic to back it up.  Keep reading and maybe you'll learn something.