divide by I and r so t=p/(ir)
parallel lines have the same slope
The slope-intercept form of a linear equatio is y=mx+b, where m stands for the "slope of the line" and b stands for the "y-intercept of the line"
They give you the equation y= -5/6x+3 Notice this is already on the slope-intercept form, so in this case the slope is -5/6 and the y-intercept is 3
You want an equation of the line that is parallel to the given line. The slopes must be the same, so m=-5/6
So far we have y=-5/6x + b
We don't have b yet but that can be found using the given point (6,-1) which tells you that "x is 6 when y is -1"
Replace that on the equation y=-5/6x + b and you get
-1 = (-5/6)(6) + b
-1 = -5 +b
4 = b
b = 4
We found b, or the y-intercept
Go back to the equation y = -5/6 x + b and replace this b with the b we just found
y = -5/6x + 4
Answer:
and 
Step-by-step explanation:
Assume that the terminal side of thetaθ passes through the point (−12,5).
In ordered pair (-12,5), x-intercept is negative and y-intercept is positive. It means the point lies in 2nd quadrant.
Using Pythagoras theorem:
Taking square root on both sides.
In a right angled triangle
In second quadrant only sine and cosecant are positive.
and
Answer: 1
Order of operations:
add first (2 + 2)
Multiply ( 2 x (4))
Divide ( 8/(8))
= 1
