To two decimal places, find the value of k that will make the function f(x) continuous everywhere. 3x+k

solution:

we are consider the following function,

f(x)=3x+k,x\leq 3

=kx^{2}-6,x>3

\lim_{x\rightarrow 3^-}(3x+k)=9+k

\lim_{x\rightarrow3^+}

(kx^{2}-6)=9k-6

so the left and right limits are equal.

therefore, the function is continuous at x=3

so,the therom of the function is continous at x=3

9k-6=9+k

8k=15

k=15/8

=1.875

therefore,the value of k=1.875

a. -0.3 > -0.33

a.

step-by-step explanation:

step 1 distribute 2 to every term in the parentheses

step 2 add 10 on both sides of the equal sign