Yes.
If th equestion meant: (x+y+z)^2The expansion is:(x+y+z)^2= x^2+2xy+y^2+2yz+z^2+2zx
xy + y = z xy = z - y (xy)/y = (z - y)/y x = (z - y)/y
x + y + z = 12 y = 1 x - y - z = 0 Substitute y = 1 in the other two equations: x + 1 + z = 12 so that x + z = 11 and x - 1 - z = 0 so that x - z = 1 Add these two equations: 2x + 0z = 12 which implies that x = 6 and then x + z = 11 gives z = 5 So the solution is (x, y, z) = (6, 1, 5)
(x+y)^2+z^2=x^2+y^2+z^2+2xy or ((x+y)^2+z)^2= (x^2+y^2+2xy+z)^2= x^4+y^4+z^2+6x^2y^2+4x^3y+2x^2z^2+4xy^3+4xyz^2+2z^2y^2
What are we solving here for, Z, X, or Y?If solve for Z then the answer would be z = x + yIf solve for X then the answer would be z = z - yIf solve for Y than the answer would y = z - xHope this help.
x + y + z = x + z + y is the commutative property of addition.
If th equestion meant: (x+y+z)^2The expansion is:(x+y+z)^2= x^2+2xy+y^2+2yz+z^2+2zx
(y - z)(y - 5z)
To find the cube of a trinomial ((a + b + c)^3), you can use the formula ((x+y+z)^3 = x^3 + y^3 + z^3 + 3(x+y)(y+z)(z+x)). First, calculate the cubes of each term: (a^3), (b^3), and (c^3). Then, find the products of the pairs of terms and multiply by 3. Finally, combine all these results to get the expanded form of the cube of the trinomial.
That depends on the values of 'z', 'y', and 'x'.
True
y=84-72=12 z=84-52=32 x=84-12-32=40
(1/x) + (1/y) + (1/z) is a minimum value when x=y=z=10. Symmetry gives either maximum or minimum value.
No, this is the commutative property. For addition, the associative property is: x + (y + z) = (x + y ) + z
xy + y = z xy = z - y (xy)/y = (z - y)/y x = (z - y)/y
x + y + z = 0 x = a - b, y = b - c, z = c - a, therefore a - b + b - c + c - a = ? a - a + b - b + c - c = 0
(2+x+y+6+z) += 9x,y and z are variables, there are 3.