Answer:
C. 
Step-by-step explanation:
Given:
Add the two equations, we get
Therefore, the correct option is option C.
by the double angle identity for sine. Move everything to one side and factor out the cosine term.
Now the zero product property tells us that there are two cases where this is true,
In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of
, so
where
![n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}](https://tex.z-dn.net/?f=n%5B%2Ftex%20%5Dis%20any%20integer.%5C%5CMeanwhile%2C%5C%5C%5Btex%5D10%5Csin%20x-3%3D0%5Cimplies%5Csin%20x%3D%5Cdfrac3%7B10%7D)
which occurs twice in the interval
for
and
. More generally, if you think of
as a point on the unit circle, this occurs whenever
also completes a full revolution about the origin. This means for any integer
, the general solution in this case would be
and
.
D) Definition of Equilateral Triangle
e) Vertical Angles
f1) ΔTVS <span>≅ </span>ΔUVR
f2) Side Angle Side Postulate
g) Definition of Congruent Triangles
Hope that helps!
<span> Move to the right side of the </span>equation<span> by adding to both sides.</span>
Answer:
Range is 3 <= g(x) <= 27.
Step-by-step explanation:
The range is the values of g(x) for the given domain.
When x = 0 g(x) = 3(0)^2 - 6(0) + 3 = 3.
When x = 4 g(x) = 3(4)^2 - 6(4) + 3 = 27.
