Answer:
There are 6 total possibilities, 3 red faces and 3 prime numbers however, 2 and 3 are prime numbers and they are red as well so total successful outcomes = 3 + 3 - 2 = 4. This means that the answer is 4 / 6 or 2 / 3.
Step by step explanation:
You first subtract z on both sides of the equal sign
ax-bx=(z-y)
Since a and b both have a "x" you can subtract them
(a-b)x=(z-y)
then you divide "x" on both sides of the equal sign
1) Answer: 
<u>Explanation:</u>
The equation of an ellipse is: 
; where (h, k) is the center, "a" is the x-radius, and "b" is the y-radius.
Center Radius
x-axis: (10 + -10)/2 = 0 10 - 0 = 10
y-axis: (2 + -2)/2 = 0 2 - 0 = 2
Now, input the values into the equation:
************************************************************************
2) Answer: 
<u>Explanation:</u>
Vertices are: (0, 1) and (0, -5) ------> x-values are the same, y = 1, -5
Covertices are: (-1, -2) and (1, -2) ----> y-values are the same, x = -1, 1
Center Radius
x-axis: (-1 + 1)/2 = 0 1 - 0 = 1
y-axis: (1 + -5)/2 = -2 1 - (-2) = 3
Now, input the values into the equation:
Sec(90deg)
sec(x) is defined as 1/cos(x). If we measure x in degrees, then cos(90) = 0, and so sec(90) = 1/0, which is undefined
I will use the letter x instead of theta.
Then the problem is, given sec(x) + tan(x) = P, show that
sin(x) = [P^2 - 1] / [P^2 + 1]
I am going to take a non regular path.
First, develop a little the left side of the first equation:
sec(x) + tan(x) = 1 / cos(x) + sin(x) / cos(x) = [1 + sin(x)] / cos(x)
and that is equal to P.
Second, develop the rigth side of the second equation:
[p^2 - 1] / [p^2 + 1] =
= [ { [1 + sin(x)] / cos(x) }^2 - 1] / [ { [1 + sin(x)] / cos(x)}^2 +1 ] =
= { [1 + sin(x)]^2 - [cos(x)]^2 } / { [1 + sin(x)]^2 + [cos(x)]^2 } =
= {1 + 2sin(x) + [sin(x)^2] - [cos(x)^2] } / {1 + 2sin(x) + [sin(x)^2] + [cos(x)^2] }
= {2sin(x) + [sin(x)]^2 + [sin(x)]^2 } / { 1 + 2 sin(x) + 1} =
= {2sin(x) + 2 [sin(x)]^2 } / {2 + 2sin(x)} = {2sin(x) ( 1 + sin(x)} / {2(1+sin(x)} =
= sin(x)
Then, working with the first equation, we have proved that [p^2 - 1] / [p^2 + 1] = sin(x), the second equation.
