Is d(x,y)=|x^2+y^2| is metric  over  R^2?
467207
I don't believe so.  Let's say d(x,y) : R x R to R  (it takes two points on the number line and gives you a number), so really you are asking a question about a metric over R^1.\n\nOne of the conditions of a metric is that d(x,y) = 0 if and only if x=y.\n\nConsider x=1 and y=1.  Then d(x,y) = 1^2+1^2 = 2.  But it should be 0.  This means d(x,y)=|x^2+y^2| cannot be a metric.\n\nIt does fulfill the various other properties of being a metric; e.g., it's symmetric, fulfills the triangle inequality, and so on, but the notion above is important.  A metric should represent a "distance" between two points.  If you are trying to compare the same point, that "distance" should be 0.\n\nIf you are trying to compare two points on the plane, w=(w1,w2) and z=(z1,z2), the following is a metric between these two points which looks similar to your question (this is called the L1 distance, or the city block distance):\n   d(w,z) = |w1-z1| + |w2-z2|\n\nAnother metric is the standard Euclidan metric on R^2:\n   d(w,z) = sqrt( (w1-z1)^2 + (w2-z2)^2) )\n\nwhere sqrt() is the square root function.
782051
Given 2 functions f and g with the domains Df and Dg respectively.\n\nAny composite function of f and g will as domain have Dfg=Df ∩ Dg &#92;{ those interval or point where the composite functions has poles coming from the way the functions is composited}\n\nEX 1)\nf=x^2 ; Df={x in R l 0<x<20} by own choice\ng=x ..; Dg={x in Rl x belongs to [-5,5[} by own choice\n\na composite function fg=f/g+1/f-g\nfinding the domain Dfg:\nDfg={ x in R l x belongs to Df ∩ Dg &#92; {those interval or point where the composite functions has poles coming from the way the functions are composited} }\n\nDf∩Dg=\n{x in R l 0<x<20}∩{x in Rl x belongs to [-5,5[}=\n{x in R l 0<x<5}\n--------------------\n\n\nfinding: \nPole={those interval or point where the composite functions has poles coming from the way the functions are composited}\n\nlook at the composite function: fg=f/g+1/f-g.\nWe easily see the composite function has pole for g=0=x => x=0 and for f-g=0=x^2-x=x(x-1)=> x=0 or x=1=>\nPole={ x in R l x=0 or x=1}\n----------------------------------\nnow we can find the domain for the composit function fg\n\nDfg=Df ∩ Dg &#92; Pole ={x in R l 0<x<5} &#92; { x in R l x=0 or x=1} }= {x in R l 0<x<5 & x≠1}\n=======================
937424
Area = pi r^2\nd Area/ dr = 2 pi r\n\nd Area = 2 pi r dr\n\nGiven d Area = 2 dr,  so substitute to get 2 dr = 2 pi r dr\n\n1 = pi r\n\n1/pi = r
201432
x=2 and y =2,\n\n2^2=2^2 and 2*2 = 2 + 2 =4
1099133
Let Y =  0 + x/(0 + x/( ...\nthen we can do a sneaky substitution and say that:\nY = 0 + x/Y\nSo, Y = x/Y\nx = Y^2\nY = +/- sqrt(x)\ng(x) = +/- sqrt(x)\nSince all the terms are positive, this is positive, so g(x) = sqrt(x).\nExcept for g(0), because then you would be dividing by 0, so g(0) is undefined.
958486
Rs=Im*Rm/Is\n\nRs*Is = Im*Rm\nRm= (Rs*Is)/Im\n\nc=(8.85*10^-12)(K)(A/d)\nc*d = (8.85*10^-12)(K)(A)\nd= [8.85*10^-12)(K)(A)]/c\n\ny=mx+b\nb=y/(mx)\n\nx= (y-b)/m\n\nLm=(Lt-L1-L2)/2\n\nE=Q/C\n\n1/Rt - 1/R2 = 1/R1
546400
D=square root ((x1-x2)^2 + (y1-y2)^2)
565556
This is counterexample is posted on Wikipedia; the link is below.  They don't explain all of the details, so I'll do that here.\n\nConsider the function f: R^2 -> R defined by\n\nf(x,y) = x y (x^2 - y^2) / (x^2 + y^2)\n\nI'm going to show that D_x (D_y f) is different from D_y (D_x f) at (x,y) = (0,0).\n\nThis function is continuous, because it is easy to check that lim_{(x,y) -> (0,0)} f(x,y) = 0.  Its derivatives are\n\nu(x,y) = D_x f(x,y) = y (x^4 + 4 x^2 y^2 - y^4) / (x^2 + y^2)^2\nv(x,y) = D_y f(x,y) = x (x^4 - 4 x^2 y^2 - y^4) / (x^2 + y^2)^2\n\nAgain, these functions are continuous, because lim_{(x,y) -> (0,0)} u(x,y) = lim_{(x,y) -> (0,0)} v(x,y) = 0.  Now to compute the mixed derivatives at (x,y) = (0,0), we need to compute the following limits.\n\nD_x (D_y f) @ (0,0) = lim_{h -> 0} ( v(h,0) - v(0,0) ) / h = lim_{h -> 0} h/h = 1\nD_y (D_x f) @ (0,0) = lim_{k -> 0} ( u(0,k) - u(0,0) ) / k = lim_{k -> 0} -k/k = -1\n\nHence these partial derivatives are *not* equal at (x,y) = (0,0) as claimed.\n\n-------------\n\nBy the way, I should point out that this gives a counter example to the famous statement\n\ndiv . curl (F) = 0\n\nIndeed, take F(x,y,z) to be that vector valued-function which is f(x,y) in the z-direction.  Then\n\ndiv . curl(F) = D_x (D_y f) - D_y (D_x f)\n\nIf you choose f(x,y) as above, then this expression is not zero at (x,y) = (0,0).
726082
Yes since ∫f(x) dx = ∫x dx\n= ½x² + c where c is ANY constant.
192169
Metric spaces are abstractions that allow you to extend the ideas that you have proven for 1,2, and 3 dimensional Euclian space to other, more complex objects.\n\nGiven a set and a distance fuction (metric) defined on that set, if you can prove that they satisfy the requirements of a metric space, then you can apply all the theroems that have been proven for abstract metric spaces. That is, you can prove the same theorem over and over, or you can prove it once for metric spaces and carry it over to every metric space.\n\nWhen you have a theorem stated for an arbitrary metric space, think about what it says for the real line or the plane.
567245
simple x=r*cos(theta), y=r*sin(theta), r^2=x^2+y^2\nsubstitude, not hard
747583
y= √(1/(x² + 1) – 2) \n\nWell 1/(x² + 1) – 2 ≥ 0\nie 1/(x² + 1) ≥ 2\nSince x² + 1 ≥ 0 for all x\nOn taking reciprocals\nx² + 1 ≤ ½\nx² ≤ -½ This is not possible as x² ≥ 0 for all x \nSo there is no domain\n----------------------- -----------------------\nWell that makes sense as  x² + 1 ≥ 1\nSo 1/(x² + 1) ≤ 1\nSo 1/(x² + 1) - 2 ≤ -1\n\nSo taking square roots gives no real solutions.
411840
x^2 - y^2 = (x+y)(x-y)\n\nso 25u^2-81= (5u+9)(5u-9)
699742
-(3x+y)(3x-y)\n-(9x^2-y^2)\ny^2-9x^2
914223
difference of squares rule:\n\nx^2 - y^2 \n\n(x-y)(x+y)
554946
Use the equation d = sqrt((y2-y1)^2 + (x2-x1)^2)\nyou get d = sqrt((5-2)^2 + (3-(-1))^2)\n=> d = sqrt(9 + 16)\n=> d = 5 units
919933
I assume this is equal to y?\n\nIf it is, it's definitely a maximum because it's a parabola that opens downward\n\nNow I'll work on the vertex