10g^3 - 8g^2 + 5g - 14 + 10g^2 + 12g
its just a matter of combining like terms
10g^3 + 2g^2 + 17g - 14
Answer:
I'm not ok. that is too expensive
Step-by-step explanation:
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).
1) combine like terms (k)
0 = 7k
k = 0/7 = 0
zero divided by any numbers will be zero
2) combine the like terms (the constant of -4 + 1)
9 = 6x - 3
add 3 to both sides
12 = 6x
x=2
3) -3+3=0
-4=v
4) 4+3=7
8=k+7
k=1
5)x-5x = -4x
16= -4x
x = -4
Answer:
200 minutes
Step-by-step explanation:
$20/$0.10=200