+0  
 
0
78
1
avatar+230 

Choose the correct answer

 Nov 13, 2018
 #1
avatar+4030 
0

\(\text{The line has slope }m=\dfrac 1 3 \text{ and lies in the I and III quadrants}\\ \text{As }\sin(\theta)<0,~\theta \text{ must lie in the III quadrant and thus }\cos(\theta) < 0\\ \text{the III quadrant point }(-3,-1) \text{ lies on the line and from this we know that }\\ \cos(\theta) = \dfrac{-3}{\sqrt{(-3)^2+(-1)^2}} = -\dfrac{3}{\sqrt{10}}=-\dfrac{3\sqrt{10}}{10}\)

.
 Nov 13, 2018

8 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.