For what value(s) of k will the relation not be a function?   a = {(3k−4, 16), (4k, 32)}

We won't have a function if for same value of x in (x,y) we get different values y.

So first step: figure out k so that the first coordinate (x) is the same:

3k-4=4k    | solve for k

k = -4

no check the values y for the elements of the relation

x = 3k-4 = -12-4=-16

so at -16 we get (-16,16) and (-16, 32), which mean for k=-4 the relation is not a function.

Let me know if you have any questions.

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

or

step-by-step explanation:

we need to solve for 'x' the following equation:  

we have that:

⇒

⇒ ⇒

factorizing, we get that:

  ⇒

then, the solutions are: or

nothing extraneous has been found!