Let f(x)=20/1+9e^3x .

what are the asymptotes of the graph of f(x) ?

select each correct answer.

y = 3

y = 0

y = 9

y = 20

y = 0 and y = 20

Step-by-step explanation:

The given function is

There is no vertical asymptotes because the denominator will never be zero.

Horizontal asymptotes are given by

Take the constant 20 out of the limit

Now, we can further simplify as follows

Similarly, the second horizontal asymptote is

We can simplify this limit as we did the previous one

Hence, asymptotes of the graph of f(x) are

y = 0 and y = 20

answer: i feel you honey.

step-by-step explanation:

the polynomial is a perfect square trinomial of the form a^2-2ab+b^2:

option a. true

step-by-step explanation:

9x^2-12x+4

comparing with the form a^2-2ab+b^2

1.) a^2=9x^2

solving for a: square root both sides of the equation:

sqrt(a^2)=sqrt(9x^2)

a=sqrt(9) sqrt(x^2)

a=3x

2.) b^2=4

solving for b: square root both sides of the equation:

sqrt(b^2)=sqrt(4)

b=2

to be a perfet square trinomial 2ab must be equal to 12x. let's check:

2ab = 2 (3x) (2)

2ab = 12x

correct