F(x)=x^3-7x-6 Since I don't have the graph and this is not a perfect cube, I will have to rely on Newton :P
x-(f(x)/(dy/dx))
x-(x^3-7x-6)/(3x^2-7)
(2x^3+6)/(3x^2-7), letting x1=0
0, -6/7, -.988, -.9999, -.99999999999, -1
(x^3-7x-6)/(x+1)
x^2 r -x^2-7x-6
-x r -6x-6
-6 r 0
(x+1)(x^2-x-6)=0
(x+1)(x-3)(x+2)=0
x= -2, -1, 3
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:
B) 1
Step-by-step explanation:
x+1/2(x-4)=1/2x+6(x-2)
x+1/2x-4/2=1/2x+6x-12
x+1/2x-1/2x-2=6x-12
x-2=6x-12
x-6x-2=-12
-5x-2=-12
-5x=-12+2
-5x=-10
5x=10
x=10/5
x=2