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The Taylor Series expansion of f(x) = sin(x) about a = π i given by

where the c's are contants.
That is
f(x) = c₀ + c₁(x-π) +c (x-π)² + c₃ (x-π)³ + ...,
₂
The first few derivatives of f(x) are
f' = c₁
f'' = 2c₂ = 2! c₂
f''' = 3.2c₃ = 3! c₃
f⁽⁴⁾ = 4.3.2c₄ = 4! c₄
and so on.
The pattern indicates that

The derivatives of f(x) are
f' = cos(x)
f'' = -sin(x)
f''' = -cos(x)
f⁽⁴⁾ = sin(x(
and so on
The pattern indicates that
f⁽ⁿ⁾(x) = cos(x), n=1,5,9, ...,
= -sin(x), n=2,6,10, ...,
= -cos(x), n=3,7,11, ...,
= sin(x), n=4,8,12, ...,
The radius of convergence is |x-π|<1 by the ratio test.
21 because you turn the 5 yards in to feet which is 15, then you add 15+4+2=21
Answer:
The 4th system (h = 4.2 − p
; h = 2p)
Step-by-step explanation:
Your line 4 should have "angle DEC" not "angle EDC". The order of the letters is important. I marked the two angles ABC and DEC in the attached image below. They are alternate interior angles. Because DE || AB, this means the alternate interior angles are congruent.
Since C is the midpoint of EB, this means BC = CE by the definition of the midpoint. Basically EB is cut into two halfs BC and CE which are the same length.
After you establish all of what is mentioned previously, you would use the AAS congruence theorem. AAS stands for Angle Angle Side. One pair of angles are the vertical angles (see line 2). Another pair of congruent angles is the alternate interior angles (line 4). The sides are not between the pairs of angles mentioned. Again check out the attached image for more visual information.
Finally, if we know the triangles are congruent, then the corresponding parts are congruent (CPCTC). That leads to the conclusion that DC = AC.
Edit: it's actually ASA and not AAS because the sides are between the two angles (I was thinking of another diagram earlier)