Explain why its important to use the order of operations when evaluating an algebraic expressionUse an example
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(1+1)-2*3/2^2+2\n\nWithout useing Order of Operations: 2\n\nWith Order of Operations: 2.5
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An associative property of an operation says that if the operation is performed multiple times successively, the order in which it is done is not important, that is, the operands can "associate" with each other without affecting the result; eg. 2 + (3 + 4) = (2 + 3) + 4.
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Philo's answers don't use each operator only once.  Simple fixes to his answers:\n4 = sqrt(4*4)^(4/4)\n7 = sqrt(4)*4-(4/4)\n\nHere's another way to do 2:\n2 = (4*4/4)-sqrt(4)\n\nTo make it even harder, I decided to try requiring them to all use a factorial (and also a power and a square root when possible):\n0 = (4-4)/[4!^sqrt(4)]\n1 = [4!*sqrt(4)]^(4-4)\n2 = sqrt(4)+(4-4)^4!\n3 = sqrt(4)+4!^(4-4)\n4 = 4+[(4-4)^sqrt(4)!]\n5 = 4+[sqrt(4)!^(4-4)]\n6 = (4!*4)/sqrt(4^4)\n6 (also) = 4!*sqrt(4)/(4+4)\n7 = (4!+4)/sqrt(4*4)\n7 (also) = 4*[sqrt(4)!]-4/4\n8 = [4!+4*sqrt(4)]/4\n8 (also) = 4^[4!/(4*4)]\n9 = 4*[sqrt(4)!]+4/4\n\nNote that a lot of the parentheses are implied to not exist if you were to write these out on paper, and therefore shouldn't count as precedence operators.
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The order of operations gets messed up.  You have to add x+y before you square it.\n\n PEMDAS \n  (You might remember this as "Please excuse my dear Aunt Sally.")\n\nOrder of Operations\n\nParentheses \nExponents \nMultiplication and Division \nAddition and Subtraction
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You use the Orders of Operations\n\nParenthesis\nExponets\nMultiplication\nDivision\nAddition\nSubtraction
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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
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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...
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remember the order of operations. for this particular problem, division and multiplication come first.  since they are both present in the problem, and neither one is supposed to be performed before or after the other in the order of operations, you should start from left to right, the same way you read.  \n\ntimes the 80 and the 2 together, and then divide it by 5. \n\n now do the addition and subtraction. again, neither subtraction nor addidtion is supposed to be performed before or after each other in the order of operations, so you start from the left to the right. \n\nsimply complete the steps.
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basically, think of a function as a machine:\n\ntake this an an example: f(x) = 2x +3\nsee the x in brackets?\n\nif x = 1, then f(1) = 2(1) + 3 = 5\nif x = 2, then f(2) = 2(2) + 3 = 7\nif x = 3, then f(3) = 2(3) + 3 = 9\n\nsee whats happening? the answer DEPENDS on number inputted in the brackets.\n\nso therefore, the definition you quoted "a consistent, predictable, relationship between the independant and dependant variable" is true.
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There are several reasons. One of the most basic ones is that this representation provides a "polar" representation of complex numbers. Any complex number z=a+ib can be written as\n\nz= r exp(i theta)\nr= sqrt(a^2+b^2)\ntheta = atan(b/a)\n\nThis helps because, for example, a product of z1 and z2 would give\n\nz1 z2 = R exp(i phi)= r1*r2 exp (theta1 + theta2)\nR=r1*r2\nPhi = theta1 + theta2\n\nSo multiplying by z2 "stretches" z1 by amount r2 and "rotates" z1 by theta2. Such interpretations are convenient in many instances. Also, exponential functions are easier to deal with than regular a+ib type expressions in other cases.
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You put a round bracket "()" on the number line when the sequence doesn't include the number where that bracket goes. For example, if the final answer you get is "all the numbers between 2 and 8, excluding 2 and 8" then you use the round brackets. You'll know your answer excludes a number or numbers because it will be "bigger than" or "smaller than" only, not "bigger than or equal to" or "smaller than or equal to", in other words, ther will be no line underneath the inequality sign. \nNow, if you do get something like "x is bigger than or equal to  8 and smaller than or equal to 2" than you know your answer includes 2 and 8, so you use the brick bracket, "[ ]". Good luck.
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A1 k is 0 because the expression got short circuited.  Since i and j are non zero values, the value of the || will always be true, so the right side does not need to be evaluated and k is never incremented.\n\nA2 when you switched to put the || first, no short circuiting can occur (since i and j are non zero, k still needs to be evaluated because of the &&).  The value of m is non-determinate in both cases.
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Choose the most repeated operation and determine of what its repeat is dependent on (amount of input data).\n\nFor example: for (int i=0; i<N; i++) { /* do something */ )\nHas complexety N because /* do something */ is repeated N times. Here N is supplied to the algorightm.\n\nFor example: Two nested loops on of N and another of M operations have complexity of N*M.\n\nYou can ignore constant additions and substractions if they dont depend on the amount of the input data (as the are constant).\n\nFor example if you initialize an array of 100 elements in the begining of the algorithem, you can ignore it instead of include it as 100+N for example :)
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Gopal is correct, except (a) he meant x =  -1should be excluded from the domain, and (b) x = 0 should also be excluded from the domain (since g(0) is undefined).
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There's a very explicit algorithm which explains\nhow to handle number, operators (including left\nor right precedence), functions, etc. on the \nWikipedia site for RPN (see reference #1).\n\nMostly, it's very straight-forward.  If it's an operand\n(ie. a number), then push it on the stack.  If it's an\noperator, then pop off as many operands are needed and\nevaluate.\n\nBut do look at the full algorithm for the details.\n\nOh yeah, for a full-blown implementation, see the source\ncode to the Unix "dc" (desk calculator) program.  The\nSolaris source is at reference #2.
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Concentrate on the ^2, because that will have the most significant impact on the value of the expression being raised to the second power:\n\nFirst I tried just raising the 4 to the second power.  4^2 = 16, so "n" / 16 = 6, therefore n = 96.  If you can group the other numbers to reach 96, you have an answer.  But working that out for a while got me nowhere, so I started working leftwards to bring the next number and operator into the square:\n\n(6/4)^2 = 2.25\n\nThat is not an integer so it probably won't work.\n\nEven bringing the next number and operator in, which is 2 -, I still end up with a non-integer:\n2 - (6/4)^2 = -0.25\n\nBut look -- the next number and operator are "8 x " so lets work that out:\n\n8 x (2 - (6/4)^2) = 8 x (2 - 2.25) = 8 x (-0.25) = -2\n\nso that might work if we can come up with a number which works out for ___ + (- 2) = 6\n\nWhat plus -2 is =6?  It is 8 of course.  Now we just need to group 3 x 5 - 3 + 2 to equal 8:\n\n(3 x (5-3)) + 2\n(3 x (2)) + 2\n(3 x 2) + 2\n6 + 2\n8\n\nAha!  We have it.\n\nSo then, or grouping is:\n(3 x (5-3)) + 2 + (8 x (2 - ((6 / 4) ^2))) = 6\n\nTechnically some of those parenthesis are extras because the order of operations says multiplications and divisions are always done first, you don't need to explicitly show that they are (but it helps keep things clear).  So this would be a simpler form:\n3 x (5 - 3) + 2 + 8 x (2 - (6 / 4) ^2) = 6\n\nBut anyway, putting it back in the more complex form and proving it true works like this:\n(3 x (5-3)) + 2 + (8 x (2 - ((6 / 4) ^2))) = 6\n(3 x (5-3)) + 2 + (8 x (2 - ((6 / 4) ^2))) = 6\n(3 x ( 2  )) + 2 + (8 x (2 - (( 1.5 ) ^2))) = 6\n(3 x   2   ) + 2 + (8 x (2 - (    2.25   ))) = 6\n(3 x   2   ) + 2 + (8 x (      -0.25        )) = 6\n(6) + 2 + ( -2) = 6\n6 + 2 - 2 = 6\n6 = 6 <<<<< TRUE!\n\n\nHere is another grouping that also works out:\n3 x (5-3) + 2 + 8 x ((2 - 6) / 4^2) = 6\n3 x    2    + 2 + 8 x     (-4 / 16) = 6\n3 x    2    + 2 + 8 x (-0.25) = 6\n3 x    2    + 2 - 2 = 6\n6 + 2 - 2 = 6\n\nand another, this one involves only integers in the proof so might be more what you are looking for:\n3 x (5-3) + 2 + (8 x (2 - 6)) / 4^2 = 6\n3 x 2 + 2 + (8 x (-4)) / 16 = 6\n3 x 2 + 2 + (-32) / 16 = 6\n3 x 2 + 2 + (-2) = 6\n3 x 2 + 2 - 2 = 6\n6 + 2 - 2 = 6\n6 = 6
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When an iterative calculations with predefined conditions are to be done, the need of loop arises.\n\nThe following are the loop constructs...\n\n1. While Loop:\n\nSyntax: \n\nwhile( x < 100 ) {\n     if( x < 0 )\n       break;\n     cout << x << endl;\n     x++;\n   }   \n\n2. for loop\n\nSyntax: \n\nfor( initialization; test-condition; increment ) {\n  statement-list;\n  }\n\n   for( int i = 0; i < 10; i++ ) {\n     cout << "i is " << i << endl;\n   }\n   int j, k;\n   for( j = 0, k = 10;\n        j < k;\n        j++, k-- ) {\n     cout << "j is " << j << " and k is " << k << endl;\n   }\n   for( ; ; ) {\n     // loop forever!\n   }            \n\n3. do \n\nSyntax: \n  do {\n  statement-list;\n  } while( condition );