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.
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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
<span>I think the answer is The function f(x) = 9,000(0.95)x </span>
3(2x^2 - 3x-2)
3(x-2)(2x+1)
A
Answer:
27 pages.
Step-by-step explanation:
Let l be the number of pages in the long paper and s be the number of pages in the short paper.
We have been given that the total number of pages for both papers is 40. We can represent this information in an equation as:
We are also told that the number of pages in the long paper is one more than two times the number of pages in the short paper. We can represent this information in an equation as:
We will use substitution method to solve system of linear equations.
From equation (1) we will get,
Upon substituting this value in equation (2) we will get,
Upon adding 2l to both sides of our equation we will get,
Let us divide both sides of our equation by 3 we will get,
Therefore, there must be 27 pages in the long paper.
