show that every symetric positive definit martix is convertable?
102340
I assume you mean invertible, not convertible.\n\nIt depends on how detailed you want to get. If you do not want to dig in extremely deeply, you can just do it by construction. \n\n1. A symmetric matrix M has an orthogonal decomposition\n\nM=O'DO\n\nwhere O is an orthogonal matrix, O'O=I, the identity, and D is a diagonal matrix with eigenvalues of M down the diagonal.\n\n2. If M is positive definite, then the diagonal elements of D are >0\n\n3. Produce the matrix \n\nA=O'(D^(-1))O\n\n\n4. Verify that A is the inverse of M\n\ni.e., \n\nA*M=M*A=I
1070589
Assuming you mean as 4 shapes equivalent to the original L shape, the answer is given in the attached image.
856009
false,\nuse the identity and -identity,\nboth are invertible, but the sum is the zero matrix, that is NOT invertible.....
81607
z = kxy/w where k is the constant of proportionality
130202
regular, reverse and atomic
829736
what is "hotelling"? and is "T" a constant?\n\ni only know that any squared value is ALWAYS positive...
163366
Yes with a gaussmeter.
980349
Domain:  all real numbers\nRange:  [2, positive infinity]   (all real numbers from 2 to positive infinity)
780025
x ≤ 5\n\n-- - - - - - -s-
44938
Let \nBlue = x\nYellow = y\n\nThen:\n2x + y = Green   [Given]
1069671
yep...scalars multiply vectors component by component just like you have shown.  you can also show it by plotting out the original vectors, scaling them appropriately, adding them, and finding the resukltant components.
972809
Consider the left-hand side(LHS) [cot(x) + 1] / [cot(x) – 1]. We want to prove that this is equal to the right-hand side (RHS), that is, [1 + tan(x)] / [1 – tan(x)]. Thus, we have the following:\n\nLHS = [cot(x) + 1] / [cot(x) + 1]\n        =  { [cot(x) + 1] / [cot(x) – 1] }* { tan(x) / tan(x) }.\n\nNOTE: We multiply tan(x) to cot(x) + 1 and the other tan(x) is multiplied to cot(x) – 1.\n\n        =  { tan(x) [cot(x) + 1] } / { tan(x) [cot(x) – 1] }\n\n        =  { tan(x) cot(x) + tan(x)] } / { tan(x) cot(x) – tan(x)] }, by distributive property of multiplication over addition (subtraction)\n\n        =  { 1 + tan(x)] } / { 1 – tan(x)] }, since tan(x) cot(x) = 1 by reciprocal identity.\n\n        =  RHS.\n\nIndeed, [cot(x) + 1] / [cot(x) – 1] = [1 + tan(x)] / [1 – tan(x)]
433823
Revolving about the x-axis yields a cone whose tip is at the origin and opens in the positive x direction.  Taking discrete x slices (dx) yields a disk whose volume is given by pi*(kx)^2 dx.  Integrating from 0 to h gives us (where "S" is used as the integration symbol):\n\nV = S pi*(kx)^2 dx\n= pi * k^2 S x^2 dx\n= (pi * k^2) * (x^3)/3\n\nEvaluating from 0 to h gives us\n= (pi * k^2) * (h^3)/3 - (pi * k^2) * (0^3)/3\n= (pi * k^2 * h^3)/3
766469
assigning the values will be fool proof
46351
Sy   = |x| - x^2 + 2   \n     = (x^2/2) - (X^3/3) + 2X
813100
ΔH =ΣΔHf°prod-ΣΔHf°react
576001
By the transitive property, a=c