no. its an exponential with a vertical zero axis
no
Yes, the world "linear" and "line" have the same root.
No because when you place x in terms of y, you get y = 1/x ( by dividing both the sides by x). This is not linear because the power of x is (-1) and for a linear relationship, you need the power of x to be (1).
If 3x -5y = 16 and xy = 7 then by combining both equations into a single quadratic equation and solving it then the points of intersection are at (-5/3, -21/5) and (7, 1)
no. its an exponential with a vertical zero axis
Yes. You can tell by the fact that both variables are raised to the first power.
no
No. In the variable x, alone, it is linear. In the variable y, alone, it is linear. But taken together, in x and y, you have a term which contains xy - that is, a term in which the powers of the unknowns add to 2. So the equation is not linear.
Yes, the world "linear" and "line" have the same root.
No because when you place x in terms of y, you get y = 1/x ( by dividing both the sides by x). This is not linear because the power of x is (-1) and for a linear relationship, you need the power of x to be (1).
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If 3x -5y = 16 and xy = 7 then by combining both equations into a single quadratic equation and solving it then the points of intersection are at (-5/3, -21/5) and (7, 1)
You mean -1 = xy , right? If we think of y against x, then it's like y = -1/x .... It's not linear. But if we consider y against 1/x , it's linear....
No powers also, no x times y i.e. xy = 1 is not linear
equation 1: xy = 4 equation 2: y = 2x+2 Substitute equation 2 into equation 1: x(2x+2) = 4 2x2+2x-4 = 0 => x = -2 or x = 1 When x = -2, y = -2 and when x = 1, y = 4 Therefore the coordinates are: (-2,-2) and (1,4)
To find the xy-trace, set z = 0 in the equation -5x - 2y - 3z = 10. Simplifying, we get -5x - 2y = 10. This is the equation of the xy-trace for the given plane.