X and A are Counting Numbers except 1 (2,3,4,5...ext)\nIf X=A*A then\nWhich number should we add to X to find the result of (A+1)*(A+1)?\nLet me simplify it....\nX + ? = (A+1)*(A+1) if X=A*A what is ?
125669
x=A*A\n\nSO\n\nA*A + ? = (A+1)*(A+1) = A*A+2A+1\n\nSO ? = 2A+1\n\n\nand sice you defined X=A*A , A=√(X)\n\nSO ? = 2A + 1 = 2√(X) + 1
329228
start: (a^2+2a+1)(a+1)\n\nyou basically multiply all the terms\n\na^2(a)+a^2(1)+2a(a)+2a(1)+1(a)+1(1)\n\nwhich is a^3+a^2+2a^2+2a+a+1\n\nwhich is a^3+3a^2+3a+1\n\ntoo bad i've completely forgotten how to go the other way...
184040
In your second step, you create a -1 for each and every +1.  Then, in the final step, you miscount, leaving off a -1.  That's the nature of working with infinity.  You could just as easily rearrange the numbers as \n\n= 2 - 2 + 2 - 2...\nor\n= 157 - 157 + 157 - 157...\n\nThus "proving" that every number equals every other number.  Infinity resists such manipulations.
333425
find whole (integer) number a,b, and c such that a^3+b^3=c^3.\nDon't remember the answer anymore,but it should be easy - all numbers were less that 10.
460869
1) 4-3x<_8+x\n4+8<_x+3x\n12<_4x\n3<_x or x in the interval [3,plus infinity)\n\n2) 3x-3>-x+1\n3x+x>1+3\n4x>4\nx>1 or x in the interval (1, plus infinity)\n\n3) 3(x-1)_<4x\n3x-3<_4x\n-3<_x or x in the interval [-3, plus infinity)\n\n4)1/2x<x+1\n-1<x/2\n-2<x or x in the interval (-2, plus infinity)\n\n5)-1/2x>1-x\nx/2>1\nx>2 or x in the interval (2, plus infinity)
205680
The answer is X = -2\n\nYou were right the first time.  You just made a calculation mistake when you tried to check it.\n\nHere's the check I got (I entered periods as spacers):\n\n_1_...+..._2_....=....0\n2+3x......2-3x  \n\n___1__...+...___2__....=....0\n2+3(-2).........2-3(-2)\n\n_1_...+..._2_....=....0\n2-6........2+6\n\n_1_...+..._2_...=...0\n-4...........+8\n\n_1_...+.._1_...=...0\n-4..........+4\n\n0...=...0\n\nTo get to the x = -2, we do what you did with (2 + 3x)(2 - 3x) as the common denominator.  When we multiply it out (multipying the common denominator to both sides of the equation), we get:\n\n2 -3x + 4 + 6x = 0\n6 + 3x = 0\n3x = -6\nx = -2\n\nYou asked a question above about the result of multiplying the common denominator to the right half of the equation.  Of course, multiplying anything times "0" always equals "0".  Right?\n\nGood luck!
3592
These numbers seem to be defined as : x,x+2,x+3 for every triplet. \nTherefore, 1,3,4,5,7,8,9,11,12,13,15,16,...
156921
Let the number be x.\n\nThen 7 less than a number will be: (x-7)\nTwice the number is: 2x\nThe difference is:12\nSo :\n\n2x - (x-7) = 12\n\n2x - x + 7 = 12\n\nx= 12 - 7 = 5\n\n(2*5) - (5-7) = 12\n10 - (-2) = 12
315513
6x - 4x -3x = 5x becomes 6x - (4x -3x) = 5x\n6x - 4x -3x=`x I'm not sure about this one.  Is it supposed to be 6x - 4x -3x = -x? then it would be (6x-4x) - 3x = -x\n7 + 5p - p = 11p becomes (7 + 5)p - p = 11p\n7 + 5p - p = 7 becomes 7 + 5(p - p) = 7\n7 + 5p - p = 7 + 4p becomes 7 + (5p - p) = 7 + 4p\nYou do not actually need parentheses for the previous one.\n7 + 5p - p = 0 I cannot figure this one out.  Is something typed wrong?
226466
a+a^20 DOES NOT =a^21. Don't confuse how to add and subtract exponents with real numbers.  a+a^20= a+(a*a*a*...*a)--where there are 20 a's inside the parentheses. -a+a^20 and a+a^20 are equal to a(-1+a^19) and a(1+a^19) respectively, using the distributive law.  Maybe this helps you see the difference, you can't turn that + sign into a * sign.  Lastly.  Working with exponents.  a^3*a^4= (a*a*a)*(a*a*a*a)=a^(3+4)=a^7.  That's the way it works, multiply "a's", and add the exponents of "a".  Now to multiply exponents, we take for example a^3 and raise it to the 4th power like this: (a^3)^4=(a^3)*(a^3)*(a^3)*(a^3)=a^(3*4)=a^12. I hope this helps.
460710
(x-7)(x-5)\nso x= +7, or +5
23847
This answer can be easily explained using algebra. Let x=your number. Then step by step here is your expression:\n\nx\n2x\n2x+5\n2x+5+12=2x+17\n2x+17-3=2x+14\n(2x+14)/2=x+7\nx+7-x=7\nSo the answer is always 7, no matter what x is.
179507
X  = Nothing \nor X  = 24th letter of the English Alphabet.\n\nAm I too clever or do I get the 10 points for being a smart ass?
458284
1. 2(n-1) -1= x, where n starts as 10 then n-1 will be equal to 9 and then becomes your answer, or x. 2(9)-1=17, then n is 17 the second time you use this equation....the first 5 numbers are 17, 31, 61, 121, 241. each answer is one less than one less twice the number before it, confirming our equation.\n\n2. 3n+1=x, where n starts as 2. The answers are 7, 22, 67, 202, 607. each number is triple the number before it, plus one.......\n\nsorry if my answers are wrong.......\n\nbut hope this is correct because it is very familiar to me.......it is like an IQ test...
511991
Start out by letting (x,y) be a point in AnB.  As it is in AnB, it must be in A.  So, x=>0 and y=>x. [We've gotten the x=>0 part, now we just need the x=y bit.]  (x,y) is also in B, so x=>y.  Now we have that y=>x and x=>y.  The only way that is possible is for x=y.\n\nQED
638478
x*0=x*(0+0)=\nx*0+x*0.\nThe first equality is because 0 is an additive identity;\nthe second from the distributive law.\nNow add the negative of x*0 to both sides:\n0=x*0+(-)x*0=\n(x*0+x*0)+(-)x*0=\nx*0+(x*0+(-)x*0)\n=x*0 +0=\nx*0.\nThe first equality is from the definition of a negative;\nthe second from the previous result; the third from\nassociativity; the fourth from the definition of a negative;\nthe fifth from the fact that 0 is an additive identity.
234640
1: x<12\n2: x>257\n3: x>42\n4: x>84\n5: x>25\n6: x>179\n7: x<8\n8: x>577\n9: x<113