1.)  How to solve <<< and >>> bitwise operator.       2.) solve -14<<2 bitwise operator.....
138377
AllI can see of your question is:\nHow to solve <<< and >>> bitwise operator. 2.) solve -14<<2 bitwise operator.....\n\nThat's not enough to go on!  There is a way to extend your question info so it's not all trying to fit into the title space.  Use it.\n\nNote that results of bitwise operations on numbers (e.g., integers) depend on the machine's representation of numbers (2s complement or whatever).
856711
because this has 2 unknowns it cannot be solved.\n\nbut I can write  x in terms of b\n\n2x = 13-b\nor x =(13-b)/2
40850
-3 <4x\n x < -3/4\n solution set : {- infinity ,.............,-3/4}
447443
Ok...you have to solve two equations\n\nFIRST ONE\n\nif (8-4x) < 5\n\nthen (8-4x-5) < 0 (by subtracting 5 from both sides of the equation)\n\nthus (3-4x) < 0 (obviously)\n\nso 3 < 4x (by adding 4x to both sides of the equation)\n\nso 4x > 3\n\nso......... x > 3/4\n\nSECOND ONE\n\nif -(8-4x) < 5\n\nthen -8+4x < 5 (obviously...see that -*- = +)\n\nthen -8+4x-5 < 0 (by subtracting 5 from both sides of the equation)\n\nso -13+4x < 0 (obviously)\n\nand 4x < 13 (by adding 13 to both sides of the equation)\n\nso......x < 13/4\n\nFINAL ANSWER IS.......(3/4 < x < 13/4)
978426
for example:\n\n-1 < 2\nthen\n1 > -2\n\nanother:\n x < -5\nthen\n-x > 5\n\n-2 x > 9\nx< -9/2\n\ns
110586
Multiplication/Division\n+ and + give you a positive answer.\n- and - give you a positive answer.\n+ and - give you a negative answer.\nIf you have an odd number of negative signs your answer will be negative.\nAddition/Subtraction (subtraction is just adding a negative)\n+ and -, the sign will be the sign of the integer with the larger absolute value.  (ex. -4 + 6 = 2, since 6 is bigger than 4 we give the answer a positive sign).\n+ and +, answer will be positive.\n- and -, answer will be negative.\nIt also helps to think about a number line.  Mentally place your first number on the number line; adding a + number moves right and adding a - number moves left on the line.\nHope this helps :-)
665576
x^2<4x\n\ncase 1: x<=0\nno solution\n\ncase 2: x>0\ndivide by x\nx<4\n\ntherfore the solutions are 0<x<4
165476
How about (a+c)/(b+d)  ? more interesting!!\n\nbtw this only work when a/b != c/d
1062229
Technically there are only 4 answers that will work, and they all rely on parentheses to modify the natural precedence rules:\n\n(5+5/5)*5\n(5/5+5)*5\n5*(5+5/5)\n5*(5/5+5)\n\nAs you can see they are all variations on (5 + 1) * 5, where you use 5/5 = 1.  So there is really just one answer.\n\nYou can easily solve problems like this yourself using the link below...\n\nFor example, can you make 24 using 3, 3, 8, 8?  Try it before you cheat... also, note that the solver allows % (mod) as an operator, so you may have to filter out those results.
762440
Just split it into parts - when is x-x^2<0, and when is it >0?\nThat bit has roots 0 and 1, so it is positive between 0 and 1, negative everywhere else.\n\nThus, ∫(x=-1 to x=0) (x^2-x) dx + ∫(x=0 to x=1) (x-x^2) dx + ∫(x=1 to x=3) (x^2 - x) dx\n\n= [(1/3 x^3 - 1/2 x^2) from -1 to 0] + [(1/2 x^2 - 1/3 x^3) from 0 to 1] + [(1/3 x^3 - 1/2 x^2) from 1 to 3]\n= (0- -5/6) + (1/6 - 0) + (9/2 - -1/6)\n= 17/3.
743387
Working above is good until this step:\n\n2 + 0.8x >0.9(x + 6)\n2 + 0.8x >0.9x + 5.4)\n0.1x > -3.4 ← Symbol error in this line\n\nShould read 0.1x < -3.4\nie x < -34
855482
You set up two inequalities and solve each inequality separately.\n\nInequality one:  5x-1 <= 14  (drop the absolute value symbols)\n\nInequality two:  5x-1 >= -14  (drop the absolute value symbols, reverse the inequality and make the 14 negative)\n\nBy the way, you have a contradiction in your question.  < is the less than sign, but you refer to it as greater than.  Just a little confusing.\n\nHope this helps.  :)
759105
False... y has to be 2...\n(3,2) is a solution, but not (3,1).
730496
iteration1: a = 3, b = 2\niteration2: a = 13, b = 8\niteration3: a = 51, b = 30\nloop exit:  a = 81, b = 30 // condition a = a + b executed before exit
228750
x xxxxx x\n    x----->\n\n<--x    x x--->\n    x
920569
It looks like you are trying to solve a literal equation, which is an equation with more than one variable, and you have to solve for a specified variable.  You can use the SAME rules to solve this kind of equation as you do with regular equations (those with numbers).  For example, if you had 8=2(b)(9), you would divide by 2 and then again by 9 (which is like dividing by 18).  \n\nIf your problem is a=(1/2)bc^2, then you would divide by 1/2 and then again by c^2.  Remember that to divide by a fraction, you multiply by the reciprocal.
846187
if you are using a "while" or "do while" loop\n\nwhile ((x1 - x0) > 0.001)\n{\n//your code here\n}\n\nor you can do this\n\ndo\n{\n//your code here\n}\nwhile ((x1 - x0) > 0.001);