Which function has an inverse that is also a function? {(–4, 3), (–2, 7), (–1, 0), (4, –3), (11, –7)} {(–4, 6), (–2, 2), (–1, 6), (4, 2), (11, 2)} {(–4, 5), (–2, 9), (–1, 8), (4, 8), (11, 4)} {(–4, 4), (–2, –1), (–1, 0), (4, 1), (11, 1)}

First Option  

Step-by-step explanation:

A relation is a function if and only if each value of the set "domain" has assigned only one value in the set "range".

That is, there is an output value and for each input value x.

Then the inverse of a function is denoted and the domain of this function is the range of the function . Similarly, the range of is the domain of .

That is, if contains the points Then contains the points .

So if the inverse of a function is also a function then must be a "one-to-one" function. This means that there can not be an equal value of y assigned to two different values of x.

For example

is a function, but is not a function.

To solve this problem, identify only the set of ordered pairs that is "one-to-one"

{(–4, 3), (–2, 7), (–1, 0), (4, –3), (11, –7)}   It is a "one-to-one" function

{(–4, 6), (–2, 2), (–1, 6), (4, 2), (11, 2)}   It is not a "one-to-one" function

{(–4, 5), (–2, 9), (–1, 8), (4, 8), (11, 4)}      It is not a "one-to-one" function

{(–4, 4), (–2, –1), (–1, 0), (4, 1), (11, 1)}       It is not a "one-to-one" function

Only the inverse of is a function

i dont know

step-by-step explanation:

step-by-step explanation:

i believe its "c" because all of the other options include a but it could be d because there is a way of financing with cash at certain places but its uncommon.