Answer:
y=2/3+11
Step-by-step explanation:
Since triangle DEF = triangle JKL, m<D = m<J, m<E = m<K, m<F = m<L.
m<F = m<L = 90 degrees
m<K = m<E = 5(m<D)
but m<E + m<D = 90 degrees [right angled triangle]
5(m<D) + m<D = 90 degrees
6(m<D) = 90 degrees
m<D = 90 / 6 = 15 degrees.
Rewrite <span>2cos x + 1 = 0 as:
2 cos x = -1, and then cos x = -1/2
x must be in Quadrant II or Quadrant III, since the adj. side is negative.
Note that the angle 120 has adj. side -1 and hyp 2. So 120 degrees is one solution.
Now what about a possible 2nd solution, to be found in Quadrant III? That would be -120 degrees, which has the same terminal line as does 240 degrees.
No soap.
So, the solution is 120 degrees.</span>
Answer:
no...
Step-by-step explanation:
