Answer:
Step-by-step explain
Find the horizontal asymptote for f(x)=(3x^2-1)/(2x-1) :
A rational function will have a horizontal asymptote of y=0 if the degree of the numerator is less than the degree of the denominator. It will have a horizontal asymptote of y=a_n/b_n if the degree of the numerator is the same as the degree of the denominator (where a_n,b_n are the leading coefficients of the numerator and denominator respectively when both are in standard form.)
If a rational function has a numerator of greater degree than the denominator, there will be no horizontal asymptote. However, if the degrees are 1 apart, there will be an oblique (slant) asymptote.
For the given function, there is no horizontal asymptote.
We can find the slant asymptote by using long division:
(3x^2-1)/(2x-1)=(2x-1)(3/2x+3/4-(1/4)/(2x-1))
The slant asymptote is y=3/2x+3/4
i put it in coments sence it says it is inapropreait
Answer:
The Answer is 3 - sqrt7
Step-by-step explanation:
2/3 + sqrt7 the starting equation
You multiply it by the conjugate which is 3 - sqrt7
= 2(3 - sqrt7) / (3 + sqrt7) (3 - sqrt7) the equation after you multiply by 3 - sqrt7
(3 + sqrt7) (3 - sqrt7) = 2
= 2(3 - sqrt7) / 2 the equation after you simplify (3 + sqrt7) (3 - sqrt7)
Divide out the 2's
= 3 - sqrt7/1 the equation after you divide out the 2's
= 3 - sqrt7 is THE ANSWER
