here is a question for a 200 thousand dollars. if dont believe me take my cridt card no.74683264.prove x-y=y-x
489863
HEY HERE IS THE ANSWER MAN  200 THOUSAND dollars are mine.\n\nX-Y=Y-X\nit is given that we can add any symbol to Y\nthere fore\nX-XY=XY-X\nX(1-Y)=X(Y-1)\nx(1-Y)=-X(1-Y)\nTHIS IMPLIES\nx=-x\nSQUARING BOTH SIDES\nX^2=X^2\nL.H.S.=R.H.S.\n\nHENCE PROVED
511948
There are an infinite number of ordered pairs that would satisfy this equation. If you want to find a few of them, assign a value to x and then solve for y.
781527
y=mx+b\n\ndid you get any actual integers in the question>
699429
c'mon dude, these are easy.\n\nAdd second to first.\n\n2x = 2, x = 1, therefore y = -2.
953599
y = mx + b\n\nCharge = $2(x) + $10\nx = number of rentals\n\ny = 2x + 10
112971
There's a problem with equation. Go back and check it.\nAs it stands, no x would satisfy this.\nInfity is an abstraction, not a number.
994352
x=144 and y =1 or vice versa
871673
So call the number\nB = {b_0,b_1,b_2,b_3,b_4,b_5,b_6,b_7,b_8,b_9}\n(this reverse indexing makes things easier).\n\nYou must have b_0 != 0, since you were told it was a ten-digit number. So b_0 is between 1 and 9.\nSo b_0 = 0 - (number of the b_i which are non-zero)\n\nAlso note that Sum_i=0..9 (b_i) = 10 since B has ten digits, and each b_i has to have a single value between 0..9\nHence Sum_i=1..9 (b_i) = (10 - b_0) for any particular choice of b_0\n\nTo prove the number is unique, see below:\n\nIt's easy to show 6210001000 is one possible answer.\nb_0 = 6 since there are six 0's in B\nb_1 = 2 since b_2, b_6 =1\nb_6 = 1 since b_0 = 6\nAll other b_i are 0 since there are no other non-zero b_i.\n\nTo prove the number is unique, the easiest way seems to be to show that the other b_0 between 1..5 and 7..9 are impossible by (proof by contradiction).\n\nb_0 = 9 is impossible since then b_9 must be 1, which gives a contradiction giving '9000000001', which only has 8 zeros, also b_1 must be 1, et.c\n\nb_0 = 8 gives '8000000010', but then b_1 must be 1, and it has one less zero, so that becomes '7100000100'...\nbut that has two ones, so b_1++, and b_2 must be 1, and you have again one less zero so b_0=6, giving '6210001000', which finally works.\n\nHarder to show that b_0 = 1..5 are impossible.\n\nLet's start from the opposite extreme, b_0=1.\nThen Sum_i=1..9 (b_i) = (10 - b_0) = 9\nThis constraint means each b_i can only be 0, 1 or 2.\nSince we set b_0=1, this means we have one zero and eight ones e.g. B='11111111110', doesn't matter where you put the 0. But that doesn't work since clearly b_1 must >=8, thus Sum_i=1..9 (b_i) = 8+7>9 which is a contradiction.\n\nIt should be trivial to crank out the cases b_0=2..4, maybe there is a more compact proof.
885282
Let x and y be the other two sides \n\n18 is hypotenuse \n=>x<18,y<18\n\nx+y+18=40\nx+y=22------------------1\n\nx^2+y^2=(18)^2=324----------2------Pythagoras theorem\n\nx^2+(22-x)^2=324---------------------Put 1 in 2\n\nSimplify,\nx^2 + 484 -44x +x^2=324\n\n2 x^2  - 44x +160 =0\n\nx^2 -22x +80=0\n\nSolve to get x,\nonly 1 of two value is possible\n(Other is discarded if<0 or >18\nor  will be y value if 2 values add up to 22).
536743
The sum of two numbers is 30\nx+y = 30\n\nThier difference is 6\nx-y = 6\n\nEliminate Y through addition (it's practically eliminated itself)\n\nx+y = 30   (+)\nx-y = 6\n_________\n2x = 36\n\nx = 18\n\nnow \nx + y = 30 right?\n18 +y = 30 (x = 18 so 18 replaces x)\n-18    -18\n___________\ny = 12
160067
When you substitute with y=x^(1/6), you're assuming :\nI) x>0 (which should be put as a starting condition)\nII) y>0 . This is to be considered after solving the 2nd degree equation, as you get a fake solution y= -3 that you must reject as non-acceptable.\n\nThe only acceptable solution is x= 1/64
866301
because y- x > 0\n\n|y-x | = y- x (positive value)\n\nx-y < 0\n\nso | x-y| = y-x (as absolute value positive)
879699
X+Y=100               X   is good land            Y   is bad land\n300 X + (500/7) Y = 10,000\nNOW TO THE ELIMINATION\nY=100 - X\n300 X + (500/7) (100-X) = 10,000\n300 X + (50,000/7) - (500/7) X = 10,000\n300 X - (500/7) X = 10,000 - (50,000/7)\n(1600/7) X = 20,000 /7\n1600 X = 20,000\nX = 12.5\nY = 100 -12.5 = 87.5
965748
|x - x²| \n= x - x² for x - x² ≥ 0\n= x² - x for x - x² < 0\n\nx - x² = x(1 - x) = 0 when x = 0 and x = 1\n\nSo x - x² ≥ 0 when 0 ≤ x ≤ 1\nand x - x² < 0 when x < 0 and x > 1\n\nSo  dx under the interval [-1,3]\n\n= ∫(x² - x) dx under the interval [-1, 0] +\n∫(x - x²) dx under the interval [0, 1] +\n∫(x² - x) dx under the interval [1, 3]\n\n=  [⅓x³ - ½x² ] [-1, 0] + [½x² - ⅓x³] [0, 1] + [⅓x³ - ½x² ] [1, 3]\n\n= [(⅓0³ - ½0²) - (⅓ (-1)³ - ½ (-1)²) ] + [(½(1)² - ⅓(1)³) - (½0² - ⅓0³)] + [(⅓ 3³ - ½ 3²) - (⅓1³ - ½1²) ]\n\n= [5/6] + [1/6] + [9/2 + 1/6]\n\n= 5 2/3
467967
this is probably not a perfect answer but this site may give you some clues.\n\nhttp://www.gomath.com/Questions/question.php?question=4751\n\naltrnatively type 'prove v*v = 2gr*r/s+v0*v0-2gr' into a search engine and see what comes up.
799272
is it asking you to graph that. otherwise it is not solvable. if it's asking you to graph it then you have to plug into your graphing calculator y=x-9
1028088
x=y+150\n\nYou will need another statement with x & y or just insert a value for x or y and resolve the question above.