JQ, KR, LS, and MT are parallel.
Answer:
Alternative C is the correct answer
Step-by-step explanation:
The first step is to determine the composite function;
![f[g(x)]](https://tex.z-dn.net/?f=f%5Bg%28x%29%5D)
![f[g(x)]=cos[cot(x)]](https://tex.z-dn.net/?f=f%5Bg%28x%29%5D%3Dcos%5Bcot%28x%29%5D)
We then employ a graphing utility to determine the range and the domain of the new function.
The range is the set of y-values for which the function is defined. In this case it is;
![[-1,1]](https://tex.z-dn.net/?f=%5B-1%2C1%5D)
On the other hand, the domain refers to the set of the x-values for which the function is real and defined. In this case; it is the set of real numbers x except x does not equal npi for all integers n.
Answer: Angles ABD and BAD (B)
Step-by-step explanation: Angles ABD BAD are not congruent angles, since they are the same angles.
Congruent angles are angles that both measure the same in degrees.
In this case, B is your correct answer since that answer has the same angle and not comparing any 2 angles, just the same one.
Hope This Helped, Have A Great Day!
Step-by-step explanation:
(x10,y-2), reflection over y=1
Ok so linear equations come in the form
y = mx + b
B is the y-intercept. In this equation the y intercept is 40. A y- intercept is where the graph crosses the y axis so at the point (0,40) the graph crosses the y axis.
M is the slope which is rise over run so if the slope was 10 (which it is) 10 is equivalent to 10/1 so you move up 10 units for every 1 unit you move across.
So to graph this equation, you would draw your first point at (0,40). For your next point, you would move right one unit, and up 10 units. Draw a point there which would be (1, 50). Hopefully you understand. For going left from the y intercept point, you would move left 1 and down 10.
Actual graph above.
Hope this helps C: