3m+8-x=-24
3m=-24-8+x
3m=-32+x
m=(x-32)/3
3m+8-x=-24
3m+8+24=x
x=3m+32
Ah, this is an infinite sum question, or a sum of geometric sequence
when will the sum reach 1cm of the edge or about 1.99m
so
2m=200cm
within 1cm means at least 1.99m
so
we will use m and not cm for consitancy
sum of geometric sequence is

a1=first term=initial jjump=1
r=common ratio=1/2
n=?, we ar solving for that
so
we want it to equal 1.99 so



divide both sides by 2

times -1

add 1 or 2/2 to both sides

take the ln of both sides


divide both sides by ln(1/2)

use your calculatro to find that n≈7.64386
so on 7th jump, it is not yet at 1cm to the edge but at 8th jump, it is past
so 8th jump
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!
you have the hypotenuse and the opposite side so you will use sin
sin58= (30/q)
multiply both sides by q
q(sin58)=30
divide by sin58 in your calculatior
q=(30/sin58)
^ I don't have a calculator with me so you will need to plug that in