Answer:
A
Step-by-step explanation:
2(1)+ 1(2)+ 2(3)+2(4)+ 3(5)/ 10
= 2+2+6+8+15/10
= 33/10
= 3.3
or A
if my answer helps please mark as brainliest.
Answer:
4. < ACB = 90° and < CBA = 52°
5. <SPQ+ <SRQ = 180°
so, <SPQ= < QRT = 100°
Answer:
1/ x = 22.5°
2/ x = 60°
Step-by-step explanation:
1/ We have: 3x + 2x + 3x = 180°
=> 8x = 180° => x = 22.5°
2/ We have: x + x + x = 3x = 180°
=> x = 60°
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!
