10:3=3.3333~ /a:b=c / why is c*b not a?
337736
Its because strictly speaking c*b is NOT defined.\nWhy? Multiplication is Not defined for a number with infinite decimals, ie, c = 3.3333...\n\nThis is an old trick I used before.\n\nI'll prove to you that \n1 = 0.999... (the 9's goes on forever ) you can try it. \nLet y = 0.999 ... (1) \nNow multiply both sides by 10 to get,\n  10y = 9.999 ... (2) \nNow (2) - (1), gives us, \n   9y = 9 or \n    y = 1 \n\nbut in the begining we assumed that y = 0.999... \nand thus we have proved that 1 = 0.999... \n\nThe answer is we cannot multiply or divide numbers with infinity of digits by our usual way of definition. \n\nNevertheless, we use the "techniques" above a lot in approximation for example, the sum to infinity of a geometric series with common ratio less than 1.\n\nCertain rational numbers allow their decimal representation repeat in a pattern. Like above, after the decimal point, the pattern is "3". Rational numbers are numbers of the form p/q, where both p and q are integers.\n\nOther example is 56/11 = 5.090909... where the repeating pattern after the decimal point is "09".\n\nWorking backwards, one can show that the number \nm = 5.090909...(1) is actually a rational number. \nConsider, 100m = 100* 5.090909... \n               =  509.090909...(2)\n\n(2) - (1), gives us,\n \n   100m - m  = 509.090909... - 5.090909...\n        99 m = 509 - 5 = 504\n\nand thus, m = 504/99 = 56/11 done.\n\nA careful observer may ask, the two "methods" look the same! How come in proving the former ( 1=0.999... ) this "method" is considered wrong while in proving the latter ( 56/11 = 5.090909... ) the method is viewed acceptable? I cant answer that with absolute certainity. I dont think anyone can. Thats where the mystery is. And I am a mathematics student.\n\nAfter some thought, I conclude that one possible reason is because of the idea of writing 56/11 as 5.090909... is "created" by us. The idea of "rewriting a fraction in decimal form". And the weirdness is due to the varying effects of the concept of "infinity". In some cases, "inifinity" can mean this when a mathematical operation is done on it, while on others it may mean that...\n\nWell, all this a product of human thought.\nIt shows the frailty of the human defined concept.\n\nRegards.
1060440
I'm assuming that C' means not-C, rather than another variable.\n\nABC+ABD+C'D\n= ABC + ABD(C+C') + C'D\n= ABC + ABCD + ABC'D + C'D\n= ABC(1+D) + C'D(AB+1)\n= 1ABC + 1C'D\n= ABC + C'D\n\nWrite out the truth table and you can easily see that this is the case.
841108
a. X x 3\n\nb. X / 3\n\nAnswers:\na. 19683\nb. .0018289895 or 4/2187
50939
Use a division sign in parentheses. That way three fourths will look like (3/4).
1154468
make a common denominator\n\na/b - c/d = a*d/b*d - c*b/d*b = (a*d-c*b)/b*d
955822
a+b-c=0\nDivision by a+b-c is division by zero, an "illegal" operation.\n\nI think that the "proof" is saying that the only way to make the process true is if a+b=0.  The "proof" stands apart from the assignment of values (a=4000, b=6000).
955805
That simplifies to 3[2x-1]>12\n\n[2x-1]>4\n\n2x-1>4    2x-1<-4\n\nx>5/2  or x<-3/2\n\nWhen you have absolute values, you have to look at both ends of the spectrum
914486
just plug -1.5 in for x and see if it equals 1.5.       \n2 X -1.5 = -3                  -3 does not equal 1.5\nso the answer is false
338816
As in many other pseudo-demonstrations of this kind, the error is to divide by zero (2-2) both sides of the identity. You can't do this because the division by zero is an undefined operation.\n\nFrom a well-known logic theorem, starting from a false statement, you can consistently derive any other false statement. This is the common source of many apparent paradoxes.
1032503
3/5*10/9 = 2/3;  9/3 = 3 & 10/5 = 2; so b = 3/2 or 1-1/2.\n\nI don't think you wrote the first problem correctly...I think it is supposed to ber:  8/3b = 4/9.  The negatives are going to cancel each other out, so we can ignore them.\n\nso, then you have b = 4/9*3/8 = 4/8*3/9 = 1/2 * 1/3 = 1/6.
1109323
The others are correct:  3x is one such example.\n\nIf you want to be pedantic about it, there are an infinite number of functions that fulfill this.  Just add any constant:\ny=3x+c.
535279
turn it around and divide 5 into 3. Three is the same as 3.0 so 5 into 3.0= .6 (Don't forget to bring the decimal up)
524337
OK - let's assume that Large Chicken is A, Medium Chicken is B, and Small Chicken is C. \n\nA+B=10.6\nA+C=8.5\nB+C=6.1\n\nNow, logically: 10.6+8.5 = (A+B)+(A+C), or 2A+(B+C)\n\nSo - 10.6+8.5=2A+6.1 ~ 19.1=2A+6.1 ~ 19.1-6.1=2A ~ 13=2A ~ A=6.5\n\nSo, if A=6.5, then (10.6-6.5)=B ~ B=4.1\nand (8.5-6.5)=C ~ C=2.0\n\nTHEREFORE:\nA=6.5 - Large Chicken\nB=4.1 - Medium Chicken\nC=2.0 - Small Chicken
991923
For the first one, 4<-2<6 does become -2>x>-3 (you do have to flip the signs) and yes, it is still an "And" statement, just in reverse (or in other words, -2>x>-3 is the exact same thing as -3<x<-2). For the second one, you don't actually have to multiply by 10, but you should get the same answer either way. I got 0.84<x<1.76.
916165
Sorry, there isn't one.  If you meant "3x + 9" instead of 3x, then x + 3 is a common factor.
882621
My old math teacher showed me this.\n\nSince A=B, then A-B=0, and you can't divide by 0
279415
it goes like\n3/8= 0.37....\n3/9= 0.33....\n3/10= 0.3\nso, 3/10,3/9,3/8, is the order\nu can also think that since numerator is the same, so larger the denominator, smaller is the no. as a whole