Answer:
Anything in the form x = pi+k*pi, for any integer k
These are not removable discontinuities.
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Explanation:
Recall that tan(x) = sin(x)/cos(x).
The discontinuities occur whenever cos(x) is equal to zero.
Solving cos(x) = 0 will yield the locations when we have discontinuities.
This all applies to tan(x), but we want to work with tan(x/2) instead.
Simply replace x with x/2 and solve for x like so
cos(x/2) = 0
x/2 = arccos(0)
x/2 = (pi/2) + 2pi*k or x/2 = (-pi/2) + 2pi*k
x = pi + 4pi*k or x = -pi + 4pi*k
Where k is any integer.
If we make a table of some example k values, then we'll find that we could get the following outputs:
- x = -3pi
- x = -pi
- x = pi
- x = 3pi
- x = 5pi
and so on. These are the odd multiples of pi.
So we can effectively condense those x equations into the single equation x = pi+k*pi
That equation is the same as x = (k+1)pi
The graph is below. It shows we have jump discontinuities. These are <u>not</u> removable discontinuities (since we're not removing a single point).
Answer:
c
Step-by-step explanation:
=16
![\sqrt[3]{16}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B16%7D)
The answer is going to be 60m^2
Answer:
You're answer is right it is -2,2
Step-by-step explanation:
remember x comes before y
L = 5 and w = 3
Perimeter = 2(l + w)
P = 2(5 + 3)
P = 2(8)
P = 16 cm
Area = l*w
Area = 5*3
Area = 15 cm^2