104
(x^2)^(1/2) equals x, therefore, y = x+4, which has a range and domain of all real numbers. The graph is a straight line, slope of 1, y-intercept of 4. Are you actually saying y = (x^2+4)^(1/2). If so, the range and domain will also be all real numbers because x^2+4 will never result in a negative number.
2 squared plus 2 x 3 = 10, 7 squared plus 7 x 2 = 63, 6 squared plus 6 x 5 = 66,8 squared plus 8 x 4 = 96 so 9 squared plus 9 x 7 = 81 + 63 = 144.
The function is a simple linear function and so its nature does not limit the domain or range in any way. So the domain and range can be the whole of the real numbers. If the domain is a proper subset of that then the range must be defined accordingly. Similarly, if the range is known then the appropriate domain needs to be defined.
you use A squared plus B squared equals C squared
The answer depends on the domain for x. For example, if the domain is x = 7, then the range is 55. If the domain is all Real numbers, then the range is y >= 6.
all real numbers
The range depends on the domain.
It depends on the domain but, if the domain is the real numbers, so is the range.
The domain and range are both [-6, +6].
The graph is a circle with a radius of 6, centered at the origin
Domain : set of all reals Range: set of all reals
The domain and range of Y = 1.3x + 8 are both [-infinity, +infinity]
The range depends on the domain and conversely.
The domain and range of the equation y = 2x+8 are both [-infinity,+infinity].
If the domain is the real numbers then so is the range.
If the domain is the set of real numbers, so is the range.