Which statement is true about the discontinuities of the function f(x) x+1/6x^2-7x-3
there are asympotes at x=3/2 and x=-1/3
there are holes at x= 3/2 and x=-1/3
there are asymptotes at x=-3/2 and x= 1/3
there are holes at x=-3/2 and x= 1/3

Option A.

Step-by-step explanation:

The given function is f(x) =

Now we will factorize the denominator first.

6x² - 7x - 3 = 6x² - 9x + 2x - 3

                  = 3x(2x - 3) + 1(2x - 3)

                  = (3x + 1)(2x - 3)

If (3x + 1) = 0

x =

If (2x - 3) = 0

x =

Since numerator and denominator has no common factor

So, the given function has the vertical asymptotes.

Option A. is the correct option.

use formula for the circle a=3.14

step-by-step explanation:

answer: i guess 5 units?

step-by-step explanation: