we can factor the whole thing:
(2sin(x) -1)(sin(x)+1) = 0.
Therefore, sin(x) = -1 and sin(x) = 1/2.
For the first one x = 3π/2 and the second is π/6 and 5π/6
So 3π/2, π/6 and 5π/6 are the solutions.
I do kind of have a problem with this because it doesn't mention if you should go over 360°. Otherwise, you have to add in an 2nπ into the equations like 3π/2 + 2nπ; 
but I don't know if that is necessary for you.
Answer: -27
Step-by-step explanation:
It is fine that you did not include the measure of angle XYZ in your posting.
This question is testing your knowledge of the four types of transformations.
1) Translations - an item is "slid" to a new location.
2) Reflections - an item is "flipped" (usually over the x-axis or y-axis)
3) Rotations - an item is rotated, usually around the origin (the point (0,0) is the center of most rotations, especially in high school math).
4) Dilations - an item is enlarged or reduced by a certain ratio.
It the first three, the image after the transformation is congruent to the pre-image. It has the same size and shape. It is simply flipped, rotated, slid...
But... in the fourth, dilation, the image now has a different size. It is still, however the same shape.
In geometry terms, after the first three transformations, the image is still "congruent" to the pre-image. After dilation, the image is "similar" but not "congruent."
So... all that to say that when you rotate an angle around the origin, the measure of the angle doesn't change.
So the first choice is correct. The measure of the image of the angle is the same as the measure of the angle.
<span>m∠X’Y’Z’ = m∠XYZ
</span>
Answer:
the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)
Step-by-step explanation:
since the variance S² of the batch follows a normal distribution , then for a sample n of 20 distributions , then the random variable Z:
Z= S²*(n-1)/σ²
follows a χ² ( chi-squared) distribution with (n-1) degrees of freedom
since
S² > 3.10 , σ²= 1.75 , n= 20
thus
Z > 33.65
then from χ² distribution tables:
P(Z > 33.65) = 0.02020
therefore the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)