Answer:
if you noticed, it is reduces by 2
so the sequence would be
93, 91, 89, 87, 85, 83
-------------------------------------------------
Hope this helps!
The Reactance of a Capacitor is t<u>he resistance it has when alternating current (A/C) with a specific frequency (f) passes through it. </u>This is measured in ohms (Ω) and its formula is:
(1)
Where:
is the Reactance in ohms (Ω)
is the frequency of the alternating current in Hertz (
)
is the Capacitance of the Capacitor in Farad (
)
In this case we have a 25 mF capacitor with an applied frequency of 400 Hz.
Note 
Well, with the given data we have to solve (1):
Then:
XC=0.0159 Ω This is the Reactance of the Capacitor
Answer:
1
Step-by-step explanation:
note that 
= 1 for any value of n, thus
= 1
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!
