Given:
To find:
The quadrant of the terminal side of 
and find the value of 
.
Solution:
We know that,
In Quadrant I, all trigonometric ratios are positive.
In Quadrant II: Only sin and cosec are positive.
In Quadrant III: Only tan and cot are positive.
In Quadrant IV: Only cos and sec are positive.
It is given that,
Here cos is positive and sine is negative. So, 
must be lies in Quadrant IV.
We know that,
It is only negative because lies in Quadrant IV. So,
After substituting , we get
Therefore, the correct option is B.
The answer should be 26 hours
It is 22 my guy because i think it is 22
2-3x=-x-8
-3x=-x-10
-2x=-10
X=5
- Slope-Intercept form: y = mx + b, with m = slope and b = y-intercept
The easiest method to find the slope and y-intercept is to convert this standard form equation to slope-intercept form. Firstly, subtract 8x on both sides of the equation: 
Next, divide both sides by 12, and your slope-intercept form is: 
Now looking at this slope-intercept form, we can see that the <u>slope is -2/3 and the y-intercept is (0, 3/4).</u>
