How do you know whether a relation is a function?(1 point) 1. By determining if each value from one set maps to another set such that each element of the domain pairs with exactly two elements of the range.
2. By determining if each value from one set maps to another set such that each element of the domain pairs with exactly one element of the range.
3. By determining if each value from one set maps to another set such that exactly one element of the domain pairs with exactly two elements of the range.
4. By determining if each value from one set maps to another set such that exactly one element of the domain pairs with exactly one element of the range.

Step-by-step explanation:

Answer options, incorrect statements are underlined below:

1. By determining if each value from one set maps to another set such that each element of the domain pairs with exactly two elements of the range.

2. By determining if each value from one set maps to another set such that each element of the domain pairs with exactly one element of the range.

3. By determining if each value from one set maps to another set such that exactly one element of the domain pairs with exactly two elements of the range.

4. By determining if each value from one set maps to another set such that exactly one element of the domain pairs with exactly one element of the range.

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

for this case we have by definition, if two lines are perpendicular, then the product of their slopes is -1.

then, given the following line:

so,

we are looking for m_ {2}:

then, the line is given by:

we find "b" replacing the given point:

finally, the equation is:

option c