If you begin with the basic equation of a vertical parabola: y-k=a(x-h)^2, where (h,k) is the vertex, then that equation, when the vertex is (-3,-2), is
y + 2 = a (x + 3)^2. If we solve this for y, we get
y = a(x+3)^2 - 2. Thus, eliminate answers A and D. That leaves B, since B correctly shows (x+3)^2.
<h3>Answers are:
sine, tangent, cosecant, cotangent</h3>
Explanation:
On the unit circle we have some point (x,y) such that x = cos(theta) and y = sin(theta). The sine corresponds to the y coordinate of the point on the circle. Quadrant IV is below the x axis which explains why sine is negative here, since y < 0 here.
Since sine is negative, so is cosecant as this is the reciprocal of sine
csc = 1/sin
In quadrant IV, cosine is positive as x > 0 here. So the ratio tan = sin/cos is going to be negative. We have a negative over a positive when we divide.
Because tangent is negative, so is cotangent.
The only positive functions in Q4 are cosine and secant, which is because sec = 1/cos.
Answer:
Work shown below
Step-by-step explanation:
t = 15w + 40
He simplified expression C because that is the only one that has only multiplications.
0 < /6/ (absolute value)
That is your inequality since the subject is about the distance away from zero then using common sense you would make the 6 get put in between the 2 absolute value lines ( not trying to be rude, if the explanation seems rude to you im sorry)
