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.

i think it's ф or x=6,-6

step-by-step explanation:

-x^2=-36

x^2=-36

ф

or

if you mean by : -(x^2)=-36

it'll be: x^2=36

x=6,x=-6

hope it'll ! have a good day!

you for asking!