Answer:
We can use slope intercept form to get the points needed. Y= -7+1/3x The points are (0,-7) and (3,-6)
Step-by-step explanation:
Subtract 2x from the left side and place it over to the right side with the 42. Now we have -6y= 42-2x. From here we divide by -6 and we get y= -7+1/3x. We know that are slope is 1/3 which the one is the rise and the 3 is the run. We also know that our y intercept is -7. We plot the points at (0,-7) and (3,-6)
Answer: A
Step-by-step explanation:
First, the problem is g(f(x)). You would plug in f(x) wherever you see an x in g(x). To find the domain, you take the bottom function, and set it equal to 0.
When you solve that, you get x=2. You know your domain is x≥2, but there is as asymptote at x=11. That means the graph never reaches x=11, but gets very close. You find that by setting the entire equation equal to 0 and solve from there.
I think the answer is c, 88
Hope this helps :)
By definition of tangent,
tan(2<em>θ</em>) = sin(2<em>θ</em>) / cos(2<em>θ</em>)
Recall the double angle identities:
sin(2<em>θ</em>) = 2 sin(<em>θ</em>) cos(<em>θ</em>)
cos(2<em>θ</em>) = cos²(<em>θ</em>) - sin²(<em>θ</em>) = 2 cos²(<em>θ</em>) - 1
where the latter equality follows from the Pythagorean identity, cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1. From this identity we can solve for the unknown value of sin(<em>θ</em>):
sin(<em>θ</em>) = ± √(1 - cos²(<em>θ</em>))
and the sign of sin(<em>θ</em>) is determined by the quadrant in which the angle terminates.
<em />
We're given that <em>θ</em> belongs to the third quadrant, for which both sin(<em>θ</em>) and cos(<em>θ</em>) are negative. So if cos(<em>θ</em>) = -4/5, we get
sin(<em>θ</em>) = - √(1 - (-4/5)²) = -3/5
Then
tan(2<em>θ</em>) = sin(2<em>θ</em>) / cos(2<em>θ</em>)
tan(2<em>θ</em>) = (2 sin(<em>θ</em>) cos(<em>θ</em>)) / (2 cos²(<em>θ</em>) - 1)
tan(2<em>θ</em>) = (2 (-3/5) (-4/5)) / (2 (-4/5)² - 1)
tan(2<em>θ</em>) = 24/7
