Let f(x) = p(x)/q(x), where p and q are polynomials and reduced to lowest terms. (If p and q have a common factor, then they contribute removable discontinuities ('holes').)
Write this in cases:
(i) If deg p(x) ≤ deg q(x), then f(x) is a proper rational function, and lim(x→ ±∞) f(x) = constant.
If deg p(x) < deg q(x), then these limits equal 0, thus yielding the horizontal asymptote y = 0.
If deg p(x) = deg q(x), then these limits equal a/b, where a and b are the leading coefficients of p(x) and q(x), respectively. Hence, we have the horizontal asymptote y = a/b.
Note that there are no obliques asymptotes in this case. ------------- (ii) If deg p(x) > deg q(x), then f(x) is an improper rational function.
By long division, we can write f(x) = g(x) + r(x)/q(x), where g(x) and r(x) are polynomials and deg r(x) < deg q(x).
As in (i), note that lim(x→ ±∞) [f(x) - g(x)] = lim(x→ ±∞) r(x)/q(x) = 0. Hence, y = g(x) is an asymptote. (In particular, if deg g(x) = 1, then this is an oblique asymptote.)
This time, note that there are no horizontal asymptotes. ------------------ In summary, the degrees of p(x) and q(x) control which kind of asymptote we have.
I hope this helps!
Answer:
Step-by-step explanation:
<u><em>the mean in period</em></u> 1 :
(2.3+2.1+2.2+2.2+2.2+2.1+2.4+2.5+2.2+2.0+1.9+1.9+2.1+2.2+2.3)÷15=21.733...
<u><em>the mean in period</em></u> 2 :
(2.3+2.1+3.3+1.5+3.6+1.6+3.0+1.1+4.7+2.1+2.4+1.9+2.8+0.5+2.3)÷15=23.466...
Since 23.466 > 21.733 then “The mean in period 2 is higher than the mean in period 1”.
Answer:
451, 551, 651, 751, 851, 951, 1051, 1151,
Step-by-step explanation:
you're adding 1 each time you progress through the pattern. Btw pls rate me BRAINLYIEST im in 8th grade.
Answer:
neecto por que los originales ves averlos
es duhStep-by-step explanation:
