When you are looking at a graph, a minimum point would be where the curve is decreasing, then begins to increase. Right at the point where it switches, the slope is a horizontal line, or 0. We can take the derivative is f(x), then look for all the x values where the slope (which is equal to the first derivative) is equal to zero.
f'(x) = 2 * -4sin(2x - pi)
The 2 comes from the derivative of the inside, 2x-pi.
So now set the derivative equal to 0.
-8sin(2x-pi) = 0
We can drop the -8 by dividing both sides by -8.
sin(2x-pi) = 0
This can be rewritten as arcsin(0) = 2x-pi
So when theta equals 0, what is the value of sin(theta)? At an angle of 0, there is just a horizontal line pointing to the right on the unit circle with length of 1. Sine is y/h, but there is no y value so it is just 0. If arcsin(0) = 0, we can now set 2x-pi = 0
2x = pi
x = pi/2
This is a critical number. To find the minimum value between 0 and pi, we need to find the y values for the endpoints and the critical number.
f(0) = -4
f(pi/2) = 4
f(pi) = -4
So the minimum points are at x=0 and x=pi
Answer:
m/1 is parallel.
Step-by-step explanation:
Answer:
the area is 68 feet
Step-by-step explanation:
-first pretend that one spot is filled and if you do that the whole rectangle is 8 by 10.
- the area of something is length times width and so 8 times 10 is 80
- however, we can't pretend that we don't have a hole anymore so to figure out that tiny rectangles area we have to find its width and length which is 4 and 3.
- 4 times 3 is 12 and since we have that we subtract 12 from 80 and we get 68.
Answer:
I think it's B.
Step-by-step explanation:
The line equation in slope-intercept form is y=mx+b
You have one line with a y-intercept at 1 and a slope of -3 -> y=-3x+1
and another with a y-intercept at -1 and a slope of 
-> 
The area marked is below the first line and above the second one -> replace = with < and >:
y<-3x+1
Which is the system of linear inequalities shown in the graph
