What are you trying to do here?
Solve the graph, or make it appear as something else?
First, we're going to take one sec (x) out so that we get:
sec (x) (2sec (x) -1 -1) = 0
sec (x) (2sec (x) -2) = 0
Then we're going to separate the two to find the zeros of each because anything time 0 is zero.
sec(x) = 0
2sec (x) - 2 = 0
Now, let's simplify the second one as the first one is already.
Add 2 to both sides:
2sec (x) = 2
Divide by 3 on both sides:
sec (x) = 1
I forgot my unit circle, so you'd have to do that by yourself. Hopefully, I helped a bit though!
Answer:
it's probably ours it's just a lucky guess
Step-by-step explanation:
Answer:
C, 5/2
Step-by-step explanation:
3 I2x -1I + 4 = 16
Isolate absolute value:
3 I2x -1I + 4 - 4 = 16 - 4
3 I2x -1I= 12
(3 I2x -1I) / 3 = 12 / 3
I2x -1I= 4
Since the question is only asking for the positive, all you need to solve for is the +4. Solve for x
2x -1 = 4
2x -1 + 1 = 4 +1
2x = 5
2x / 2 = 5 / 2
x = 5/2 or C
