In the complex field, the domain and range are both the whole of the complex field.If restricted to real numbers, the domain is x >= 4 and y can be all real numbers >= 0 or all real numbers <= 0 [or some zigzagging pattern of that set].
The domain and the range depends on the context. For example, the domain and the range can be the whole of the complex field. Or I could define the domain as {-2, 1, 5} and then the range would be {0, 3, -21}. When either one of the range and domain is defined, the other is implied.
The domain is what you choose it to be. You could, for example, choose the domain to be [3, 6.5] If the domain is the real numbers, the range is [-12.25, ∞).
The domain could be the real numbers, in which case, the range would be the non-negative real numbers.
3x - 12 = 10x - 15 - 6x -10 3x - 12 = 4x -25 3x + 13 = 4x 13 = 4x - 3x 13 = x None?
so, if 2 minus Ln times 3 minus x equals 0, then 2 minus Ln times 3 equals x, therefore 2 minus Ln equals x divided by three, so Ln + X/3 = 2 therefore, (Ln + [X/3]) = 1
The domain and the range depends on the context. For example, the domain and the range can be the whole of the complex field. Or I could define the domain as {-2, 1, 5} and then the range would be {0, 3, -21}. When either one of the range and domain is defined, the other is implied.
The domain is what you choose it to be. You could, for example, choose the domain to be [3, 6.5] If the domain is the real numbers, the range is [-12.25, ∞).
x
The domain is (-infinity, infinity) The range is (-3, infinity) and the asymptote is y = -3
Assuming you mean sqrt(x-3) rather than sqrt(x) - 3, the domain can be any subset of of x ≥ 3. The range will depend on the domain but needs to be divided in two so that it contains only one of the two roots.
The domain could be the real numbers, in which case, the range would be the non-negative real numbers.
y=(√1)-x2 The domain is the set of numbers that "x" can be. In this equation "x" can be any real number. The domain for this problem would be (-inf,inf) *Inf= Infinity*
D = {x [element of reals]}R = {y [element of reals]|y >= 4}
Assuming the standard x and y axes, the range is the maximum value of y minus minimum value of y; and the domain is the maximum value of x minus minimum value of x.
it is greater than and equal to 3
Plus or minus the base. If the base is X and you square it, you get X2. If you take the square root of that, you get Plus or Minus X. This is because X*X equals X2 and -X*-X also equals X2.
No. But sin2a equals 1 minus cos2a ... and ... cos2a equals 1 minus sin2a