Help me please!!! Thanks!
1 answer:
You might be interested in
Based on <em>trigonometric</em> formulas, we can find that the options that applies to the given <em>trigonometric</em> relationship are tan θ = 5/12 and cos θ = 12/13.
<h3>How to find the exact form of trigonometric functions</h3>
There six <em>trigonometric</em> functions, which are related by a group of expressions, which will be used in this question:
Based on <em>trigonometric</em> formulas, we can find that the options that applies to the given <em>trigonometric</em> relationship are tan θ = 5/12 and cos θ = 12/13.
<h3>Remark</h3>
The statement is incomplete. Complete form is shown below:
<em>Check all that apply: </em>
<em />
<em>If sec θ = 13/12, then: </em>
<em />
<em>a) sin θ = 12/13</em>
<em>b) tan θ = 5/12</em>
<em>c) cos θ = 12/13</em>
<em>d) csc θ = 12/13</em>
To learn more on trigonometric equations: brainly.com/question/22624805
#SPJ1
Hello!
Vertical asymptotes are determined by setting the denominator of a rational function to zero and then by solving for x.
Horizontal asymptotes are determined by:
1. If the degree of the numerator < degree of denominator, then the line, y = 0 is the horizontal asymptote.
2. If the degree of the numerator = degree of denominator, then y = leading coefficient of numerator / leading coefficient of denominator is the horizontal asymptote.
3. If degree of numerator > degree of denominator, then there is an oblique asymptote, but no horizontal asymptote.
To find the vertical asymptote:
2x² - 10 = 0
2(x² - 5) = 0
(x - √5)(x + √5) = 0
x = √5 and x = -√5
Graphing the equation, we realize that x = -√5 is not a vertical asymptote, so therefore, the only vertical asymptote is x = √5.
To find the horizontal asymptote:
If the degree of the numerator < degree of denominator, then the line, y = 0 is the horizontal asymptote.
Therefore, the horizontal asymptote of this function is y = 0.
Short answer: Vertical asymptote: x = √5 and horizontal asymptote: y = 0
