Otwo decimal places, find the value of k that will make the function f(x) continuous everywhere. f of x= 3(x) + k for x less than or equal to 3 and is equal to k(x) squared - 6 for x greater than 3

  k ≈ 1.88

Step-by-step explanation:

In order for the function to be continuous, the pieces of the function must have the same value at x=3. That is ...

  f(3) = 3(3) +k = k(3^2) -6

  15 = 8k . . . . . add 6-k, simplify

  k = 15/8 = 1.875 ≈ 1.88

To two decimal places, the value of k that makes f(x) continuous at x=3 is 1.88.

first option

a function is a relation in which an input is paired with exactly one output.

so, let's find the right answer.

in our first option, our inputs are january, february, and march. the outputs are 30, 28, and 25. each input has it's own output. this is correct!

our second option we know is incorrect, because ms. valverade has two outputs, dogs and cats. this is not a function.

our third option is incorrect as well. the yellow square has two ouputs, the square and the triangle.

lastly, in our fourth option, we see the number 7 two times with a different output. therefore, this is not a function.

remember: each input must be paired with only one output!

idk

step-by-step explanation:

i don't know