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 cant see the numerators on the numbers on the top one but the last one is 53/60.
Standard form: -3x +5y= -1
3.(6^(12-10)) = 3× 6^2 =3× 36 =108
An imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit i, which is defined by it's property i² = −1. The square of an imaginary number bi is −b². For example, 5i is an imaginary number, and it's square is −25.
