Answer:
615.75216....
Step-by-step explanation:
The formula is A = π * (ø/ 2)^2
A = Circle area
π = Pi = 3.14
ø = Circle diameter
A= 3.14 (28/2)^2
Answer:
(11, 12)
Step-by-step explanation:
<u>Mid point coordinates:</u>
- x = (x₁ + x₂)/2
- y = (y₁ + y₂)/2
<u>Since midpoint and one endpoint given (10, 4) and (9, -4), the other endpoint:</u>
- x₂ = 2x - x₁ = 2*10 - 9 = 11
- y₂ = 2y - y₁ = 2*4 - (-4) = 12
<u>So the other endpoint is: </u>(11, 12)
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).