What is the basis to distinguish X from Y?
359986
Student's t-test can be used to determines the probability that 2 random variables have the same mean. This can then be used to test whether two sets of data come from the same source.
935707
x<0\nso for all values of x, x will be a negative. when you absolute x, the positive absolute will always be bigger than the negative x
979117
From a table -- see which y-value corresponds to x=0\nFrom a graph -- see where the graph crosses the y-axis\nFrom an equation -- get the equation in y=mx+b form; the y-interecept is b
793357
Another way to look at is follows.\n\nThink of d/dx () to mean "differentiate whatever is in the parens with respect to x.  So, for example d/dx (x)=dx/dx=1.   If y=y(x). y'(x)=d/dx (y(x))=dy/dx where the argument of y is omitted for convenience.\n\nSo, d/dx (y')=d/dx (dy/ dx)=(d d y)/(dx dx)=d^2y/dx^2. \n\nThe notation is ambiguous.  One would usually think that the x is being square in the denominator, but "dx" is a single symbol, so in fact, in this context, dx^2=(dx)^2.\n\nI hope this helps!\n\nGood luck. :)
994352
x=144 and y =1 or vice versa
914223
difference of squares rule:\n\nx^2 - y^2 \n\n(x-y)(x+y)
518633
Call  Y of x like Y(x) \n\nY(x) = Y(x-1)+ (3*4^x)-1(1/9)x\nY(x) = Y(x-2)+ (3*4^(x-1))-1(1/9)(x-1) + (3*4^x)-1(1/9)x\nY(x) = Y(x-3)+ (3*4^(x-2))-1(1/9)(x-2) (3*4^(x-1))-1(1/9)(x-1) + (3*4^x)-1(1/9)x\n...\nY(x) = Y(1) + 3(4^x+...+4^2) - (1/9)(x+...+2)\nY(x) = Y(1) + 3*4[4^(x+1)-1]/[4-1] - [x(x+1)/2-1]/9\nY(x) = (sqrt3)/4 + 4^(x+2) - 4 - [x(x+1)/2-1]/9\nY(x) = 16 * 4^x - x(x+1)/18 - (37/9 - (sqrt3)/4)
1071898
it's x, assuming that x is zero or postive. \n\nif x is negative, it is either undefined, \nor if you use complex numbers, it is |x|=-x.
44938
Let \nBlue = x\nYellow = y\n\nThen:\n2x + y = Green   [Given]
863581
You have a couple of typos, I think.  As for the "proof", dividing by x - y is dividing by zero.  That is a no-no.  (In other words, a * 0 = b * 0 does not imply a = b.)
596471
X & Y is used to find a point on a graph.  X is the number across, and Y is the number up or down.\n\nThat's all I have.\n\nFYI, ur question is a little vague...
823256
A. RY,rY\nB.Ry\nc.Ry,RY,ry,rY\nd. r,Y,ry\njust the differerent possiblities of rs and ys
922643
It depends on the graph. If the values of each notch differ by x, then the interval is x.\n\nJust look at two numbers on adjacent notches, and subtract.
739421
x<1/2(x+y)<y can be proved by splitting it in two parts and proving those:\nx<1/2(x+y) is true because 1/2(x+y) > 1/2(x+x) = x\n1/2(x+y)<y is true because 1/2(x+y) < 1/2(y+y) = y
819996
T1=m1*a, no matter are there m2 & m3 or not.\nT2=(m1+m2)*a, no matter whether m1 & m2 are connected with a string or are in one lump, the presence of m3 being unimportant too.
938489
I'd cheat and take the cross product to get a vector normal to the plane of those two vectors.  Then to prove that it's a basis, you must prove that they are linearly independent.\n\nOr\n\nnotice the zero component in v1?  It lies in the x-z plane.  So, pick a vector in the x-y or y-z plane.  Since v2 is not in any of those planes you should be OK.  Also, vectors with zero components are easy to visualize and use.
148466
You need to start from the definintions of x < y and x > y.\n\nThey probably go like this\nif x < y, then there exists 0 < z such that x+z = y.\nif x > y, then there exists 0 < w such that x = y+w.\n\nPartial proof:\nif x < y, and x = y, then for some z, 0 < z, x+z = y.\nsince you have x = y, too, y + z = y, z = 0. That contradicts 0 < z. Proving the other two parts is similar.\nx+z = y,