Proofs Homework Help?
823097
Can't prove it as n cannot = + or -1 and k be an interger.\nIt appears to work with other odd integers. Change your question then it may be possible to prove it.
942422
You can start from the following identity for set differences which holds true for any sets X, A, and B\n\nX − (A ∪ B)  =  (X − A) ∩ (X − B)\n\nNow, if B is a subset of A, then A ∪ B = A\n\nthe identity above becomes\n\nX − A =  (X − A) ∩ (X − B)\n\nFor this to hold true, X − B must be a superset of (X − A), or in other words, (X − A) must be a subset of X − B
703362
start with a thesis statement (a main argument that you are going to prove is correct throughout your essay).\nthen come up with 3 main points to prove your thesis.\nthen start writing! good luck.\ncheck out this site for more info. http://www.essaywritinghelp.com/
786886
Why didn't you just ask me?  You could have emailed and saved 5 points.\nMy solution did indeed contain a typographical error omitting the q from the value of m.\nAlso, I had amended the solution with a comment shortly after the question had been resolved.  (There was another small error as well.  It is very difficult sometimes typing and then proof-reading mathematical proofs in these Yahoo boxes, especially when they contain alot of special characters.)\n\nSee:\n\nhttp://answers.yahoo.com/question/index;_ylt=ApZ1XohHNUdL5xkBg8bXQEfsy6IX?qid=20061124162535AAURdJ5
268070
Check out this site. It gives you a few examples. It was helpful for my cousin when she was leaning how to do them.\nhttp://www.purplemath.com/modules/solvelin.htm
420735
When there is proof!!
1081259
Limt { [ sin^2 (3t) ] / [t^2 }\nt-->0\n\nMultiple by 9/9\n\nLimt {9/9 [ sin^2 (3t) ] / [t^2 }\nt-->0\n\nLimt {9 [ sin^2 (3t) ] / [9t^2 }\nt-->0\n\n9 Limt { [ sin^2 (3t) ] / [3t]^2 }\nt-->0\n\nsince\nLimt  sin u /u = 1\nu-->0\n\nThen let 3t = u and as t-->0 so u-->0 \n\n9 Limt { [ sin (u) ] / [u] }^2 \nu-->0\n\n9 Limt { [ sin (u) ] / [u] }* { [ sin (u) ] / [u] }  \nu-->0\n\n9 *1*1 = 9
168953
121    since(1+4)+(1+4*2)+(1+4*4)
403642
just remeber, if you want to know if any number is divisible by 4 just check to see if the last two digets are divisible by 4... that should greatly help in solving this problem\n\nNow if you would please, stop asking all of your homework questions
743138
join AC\nnow youhavetwo triangles ABC and ADC\nin triangle ABC <CAB+<B+<ACB=180*.............(1)<s of a triangle\nin triangle ADC <CAD+<D+<ACD=180*............(2)--------do--------\nadding (1) and (2)\n<ABC+<ADC+<B+<D+ACB+<ACD=360*\n=><DAB+<B+<BCD+<D=360*\nhence proved
104175
f=thanx\ng=for\nx=the\ny=two\nz=points\n\nTell your teacher you need some help with your algebra.
841265
its a square with 13 feet sde.\n\nIf a proof is to be sumbitted ask her her to find are at different side lenghts,\n\n1,12\n2,11\n......\n......\n.......\n13,13\n\n14,2 etc and hence conclude.\n\nI i dont know a better proof for fifth grader. otherwise i will nedd to use maxima minima or a quadratic equation atleast to prove that\n\n\nsorry i cant help much\n\n\n\nbythe way you dont adjust anything  like  13.00001 or 12.9999999\n\nsquare is a special case of rectangle with length=breadth
745882
C is correct but not complete... only the 3 is factored out... the real answer would be:\n\n3(x-1)(x+4)
832106
1) 1/-4<m/-4\nmultiply both sides by negative 4 and flip the inequality sign\n4>m.\n\n2) -1/8< or equal to n/-32\nmultiply both sides by negative 32 and flip the inequality sign\n4 greater than or equal to n\n\n3) x/-12>1/4\nmultiply both sides by negative 12 and flip the inequality sign\nx<-3\n\n4) -1/2x< or equal to 2\nmultiply both sides by negative 2 and flip the inequality sign\nx> or equal to -4
144290
This is a better link that the previous one posted here!! try it! \n\nhttp://pubs.logicalexpressions.com/Pub0009/LPMArticle.asp?ID=305
300657
let a = b  \nMultiply both sides by a \na² = ab  \n Add (a² - 2ab) to both sides \na² + a² - 2ab = ab + a² - 2ab  \n Factor the left, and collect like terms on the right \n2(a² - ab) = a² - ab  \n Divide both sides by (a² - ab)\n2 = 1\n(The place where the proof goes wrong is the very last step.  If a² - ab = 0, then you would be dividing by zero, which is undefined.)
311472
i can prove that\n' a = x [true for some a's and x's]\na+a = a+x [add a to both sides]\n2a = a+x [a+a = 2a]\n2a-2x = a+x-2x [subtract 2x from both sides]\n2(a-x) = a+x-2x [2a-2x = 2(a-x)]\n2(a-x) = a-x [x-2x = -x]\n2 = 1 [divide both sides by a-x] ' is wrong\nbecause if we assume that a=x, we can't divide by a-x which is 0!!!! there is no way anyone can prove that 2=1!!!! don't even try:)\n\nalso, the one with 2^0 and 1^0\nany number ^0 is 1, does it mean that 1=2=3=4=5=6=7=8=9=...\nalso, for example 2^2=4 but (-2)^2=4 too, does it mean that 2=-2????? NO!!!