The area of the circle by definition is given by:

For f (r) = 154:

Clearing r we have:

For f (r) = 616:

Clearing r we have:
Answer:
a reasonable domain for f (r) is:
7 <r <14
Answer:
a) x = 2
b) x = 4
Step-by-step explanation:
a = (b₁ + b₂)/2 * h
----------------------------
a)
48 = (10 + x)/2 * 8
Divide both sides by 8
6 = (10 + x)/2
multiply both sides by 2
12 = 10 + x
Subtract 10 from both sides
2 = x
x = 2
---------------------------------
b)
18 = (7 + 2)/2 * x
18 = 4.5 * x
divide both sides by 4.5
4 = x
x = 4
The domain is three and the range is 68
Now, the cosecant of θ is -6, or namely -6/1.
however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.
we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

recall that

therefore, let's just plug that on the remaining ones,

now, let's rationalize the denominator on tangent and secant,