This question doesn't make sense since the LCM of 10 and 6 is 30.
Hope i helped!!! :)
Answer:
Vertical Asymptote:

Horizontal asymptote:
it does not exist
Step-by-step explanation:
we are given

Vertical asymptote:
we know that vertical asymptotes are values of x where f(x) becomes +inf or -inf
we know that any log becomes -inf when value inside log is zero
so, we can set value inside log to zero
and then we can solve for x

we get

Horizontal asymptote:
we know that
horizontal asymptote is a value of y when x is +inf or -inf
For finding horizontal asymptote , we find lim x-->inf or -inf



so, it does not exist
Start with the equation of a circle whose center is at (h,k) and whose radius is r:
(x-h)^2 + (y-k)^2 = r^2
Substituting the given coordinates of the center:
(x-5)^2 + (y-[-5])^2 = r^2, or (x-5)^2 + (y+5)^2 = r^2
Substituding the given coordinates of a point (6,-2) on the circle:
(6-5)^2 + (-2+5)^2 = r^2
Simplifying:
1^2 + 3^2 = r^2, or 1 + 9 = r^2, or 10= r^2. Then r = sqrt(10).
Answer:
x ≈ 4.6
Step-by-step explanation:
Reference angle = 75°
Opposite side to reference angle = 17
Adjacent side = x
Applying TOA:
Tan 75 = opp/adj
Tan 75 = 17/x
Multiply both sides by x
x*Tan 75 = 17
Divide both sides by Tan 75
x = 17/Tan 75
x ≈ 4.6