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let's recall that the graph of a function passes the "vertical line test", however, that's not guarantee that its inverse will also be a function.
A function that has an inverse expression that is also a function, must be a one-to-one function, and thus it must not only pass the vertical line test, but also the horizontal line test.
Check the picture below, the left-side shows the function looping through up and down, it passes the vertical line test, in green, but it doesn't pass the horizontal line test.
now, check the picture on the right-side, if we just restrict its domain to be squeezed to only between [0 , π], it passes the horizontal line test, and thus with that constraint in place, it's a one-to-one function and thus its inverse is also a function, with that constraint in place, or namely with that constraint, cos(x) and cos⁻¹(x) are both functions.
First, plot the points. Point R would be somewhere in the second Quadrant, point M would be in the first quadrant 1, point B would be in the fourth quadrant, and point S would be on the negative y-axis. A property of rhombi is that their diagonals are perpendicular. One would need to calculate the slopes of the diagonals and determine whether or not they are perpendicular. Lines are perpendicular if and only if their slopes are opposite reciprocals. Example: 2 and -0.5
Formulas needed:
Slope formula:
The figure would look kinda like this:
R
M
S
B
Diagonals are segment RB and segment SM
So, your slope equations would look like this:
and
Slope of RB= -1
Slope of SM=7
Not a rhombus, slopes aren't perpendicular. But this figure may very well be a parallelogram
Formulas needed:
Slope formula:
The figure would look kinda like this:
R
M
S
B
Diagonals are segment RB and segment SM
So, your slope equations would look like this:
and
Slope of RB= -1
Slope of SM=7
Not a rhombus, slopes aren't perpendicular. But this figure may very well be a parallelogram