Find the perimeter of this figure. Please show work.
1 answer:
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The Law of Cosine states
cos
so plugging in the numbers gives us
cos
We now have cosJ so plug that into your calculator and find arccos (arccos(cos(J)) = J)
arccos
rounding your answer will give you 34°
cos
so plugging in the numbers gives us
cos
We now have cosJ so plug that into your calculator and find arccos (arccos(cos(J)) = J)
arccos
rounding your answer will give you 34°
If the focus is at (6, 2) and the directrix is a horizontal line y = 1, then that tells us that is an x^2 parabola. Since the parabola hugs the focus, it will open upwards since the focus is above the directrix. The rule here is that the vertex is the same distance from the focus as it is from the directrix. If the focus is at a y-value of 2 and the directrix is at y = 1, then the vertex is right in between them as far as the y coordinate goes, which is 1.5. It will have the same x coordinate at the focus. The vertex is in the form (h, k), so our h is 6, and our k is 1.5. The vertex is (6, 1.5). The standard form of a parabola of this type is 
, where p is the distance from the vertex to the focus. Our p is 1/2. Using the h and k from the vertex, and the p of 1/2, we now have this as our equation, not yet simplified: 
. That will simplify a bit to 
. Depending upon how you are to state your answer, how it needs to "look" in the end will vary. I am going to FOIL the left and distribute the right and then put everything on one side and set it equal to y. That would be this: 
. And there you go!