sparky77 sparky77
  • 02-06-2020
  • Mathematics
contestada

For what values of k the relation R:{(k2 , 5), (3k, 6)} is not a function?

Respuesta :

jajumonac jajumonac
  • 07-06-2020

Answer:

[tex]k=0[/tex] and [tex]k=3[/tex].

Step-by-step explanation:

To this relation be a function, the horizontal coordinates can't be equal. So, let's find the value of [tex]k[/tex] that makes those coordinates equal.

[tex]k^{2}=3k\\ k^{2}-3k=0\\k(k-3)=0\\[/tex]

Using the null factor property, we have

[tex]k=0\\k-3=0 \implies k=3[/tex]

Therefore, the given relation is not a function when [tex]k=0[/tex] and [tex]k=3[/tex].

Answer Link

Otras preguntas

what is 1/9 of 60 equal
why a flat surface of a rectangular prism called faces?
what other information do you need to prove triangle GHK is congruent to triangle KLG by SAS
How do you pronounce the letter "X" in Spanish?
names is often given to the journey of africans who were brought to the americas as slaves?
I am a polyhedron. I am a prism. my two bases are hexagon. my two other faces are rectangles.
Describe what happens when blood in capillaries flows past cells
in what way is a gene pool representative of a population
The net force acting on a 50 kg crate is 0 N. What is the crate's acceleration
find an equation of the tangent line to the circle x^2 +y^2=24 at the point (-2 square root 5, 2).