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:
i would say no
Step-by-step explanation:
i would say no because quadrilateral FGIH is equal to 120 and quadrilateral SRTQ is equal to 134 and also their areas are indifferent FGIH's being 875 and SRTQ's being 1120
please let me know if this was wrong
hope this helps :) have a nice day !!
Answer:
<h2>P = 48</h2><h2>A = 144 </h2>
Step-by-step explanation:
The formula of regular polygon with <em>n</em> sides of length <em>b</em> and apothem <em>a</em>:
We have:
n = 6
b = 8
a = 6
Substitute:
The perimeter:
