ITS C - Factor 6 out of the variable terms
Answer:
B. △RTS
Step-by-step explanation:
△MON ≅ △RTS
Check the picture below.
now, we know the directrix is at y = 1, and the focus point is at 1,3, well, notice the picture, the distance between those fellows is just 2 units.
the vertex is half-way between those fellows, therefore, the vertex will be at 1,2.
the distance "p", from the vertex to either the directrix or focus, is really just 1 unit. Since the focus point is above the directrix, is a vertical parabola, and it opens upwards, like in the picture, and since it opens up, the "p" value is positive, or +1.
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,
So distance = rate × time. You would do 3/4 × 5.5, which would get you 4.125.