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
Answer:
4x+12/x^2+9
Step-by-step explanation:
Perimeter
x+3+x+3+x+3+x+3
Area
(x+3)(x+3)
Answer:
36
Step-by-step explanation:
2 *3 sqare root 3 * 12
Solving gives:
6 sqare root 36.
sqare root 36 = 6
2 * 6 * 3
So our answer is 36.
Answer:
Step-by-step explanation:
srry if its wrong but I think it's right :)
btw for the first one
p=19.53
