if 1,5,and 9 is in the input section,and 5,1,and-3 are in the output section then what is the rule?
139634
Since the sum of each pair is 6, the function is f(x) = 6 - x
330638
5/X+1<-2\n-1\n5/X<-3\n*X\n5<-3X\n/-3\n-5/3<X
562792
A=[1 2 3;3 4 5;7,8 9]\n\nA =\n\n     1     2     3\n     3     4     5\n     7     8     9\n\n\n>> B=(A([1 3],:))\n\nB =\n\n     1     2     3\n     7     8     9\n\n\n>> B=(A(:,[1 3]))\n\nB =\n\n     1     3\n     3     5\n     7     9\n\n\ngood luck i wish i could helped u
302002
Basic subtraction. Ok, you are taking stuff away from what you have, so...\n\n(-3)-3 you have -3 take away three, counting down that is -6. Or you can think of it as the two -s forming a +. \n\n Ok if you have the problem...(-5)-(+3)= change the subtraction sign to + and the 3 to the oppisite in this case -3\n\nExaples\n1. (-8)-+3=   -8+-3 now it is a adding problem = -11
926416
your getting a little anxious in your coding:\ntry changing the section below:\n\ncout << "Enter a number Negative, Positve, or Zero.";\ncout << '&#92;n';\ncin >> num;\n\nif (num > 0)\n{\n    pos(num);\n}\nelse if (num < 0)\n{\n  neg(num);\n}\nelse\n{\n  zero(num);\n}
310681
Basic Input Output System.
1119698
5 > x+1/-3 > 0\n5 > x-1/3 > 0\n5 1/3 > x > 1/3
160639
good 0, 3, 7, 4\n\n    bad 2, 9, 5\n\n    neutral 1, 8, 6
877986
it could be + 5 or -5 . the absolute value makes a number positive so 4+5=9 and 4 + absolute value of -5 =9
306730
well we know the 3 space must be even, in order to avoid the last space resulting in another 5\n\nhmmm... i'l work on it, provided you're working as well!\n\n..and #3 mustn't be 0, to avoid another 0\n..lol i'd have had it if i had been home...i made a program to do it, and then left before i finished/ran it lol\n\nbut that, is of course, the correct asnwer, 396*45=17820\n\nhaha thanx for the challenge!
136546
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...
8884
It would be the third set - (0,-9) (2,0) (1,-4.5)
400503
since each one gives you a *-----* i would say all of them are line segments.\n\nHowever since D. is the only one with an indefinite beginning and end and not a possible end or beginning, i would definately have to go with \n\nD.) -5 <= x <= 5\n\nThat is the only one that has a defined beginning and end and not a undefined end and/or beginning.
869678
1 and 1\n9 and 3\n25 and 5\n49 and 7\n81 and 9\n121 and 11\n...
868097
-\naside from 1,\n2,3,4,6,12
211910
(-3)-[(-2)-(4)(-5)] \n(-3)-[(-2)-(-20)]\n(-3)-[-2+20]\n(-3)-18\n-21
803954
check the discriminant:\n81-4(5) = 81 -20 =61 >0,\nso there are 2 real solutions to this problem\n`