Need help step by step plz
1 answer:
I'm assuming D is the center of the circle. See the new figure.
Angle BDC = 106 degrees, that's the definition of arc measure.
Triangle BDC is isosceles (two radii make congruent sides) so the two remaining angles in BDC are equal, DBC=DCB=(180-106)/2=37 degrees.
Angle BAC = 106/2 = 53 degrees, subtending the arc of 106 degrees.
Angle CAD = 53/2 = 26.5 degrees because AD is a bisector, the shared hypotenuse of congruent right triangles AED and AFD.
AFD is also congruent to CFD, so angle DCA is also 26.5 degrees
Angle ACB is the sum of DCA and DCB, ACB = 26.5 + 37 = 63.5 degrees
The angle subtends the arc we seek, whose measure must be double:
10x - 23 = 2(63.5) = 127 degrees
10x = 150
x = 15
Answer: 15
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Answer:
The Taylor series of f(x) around the point a, can be written as:
Here we have:
f(x) = 4*cos(x)
a = 7*pi
then, let's calculate each part:
f(a) = 4*cos(7*pi) = -4
df/dx = -4*sin(x)
(df/dx)(a) = -4*sin(7*pi) = 0
(d^2f)/(dx^2) = -4*cos(x)
(d^2f)/(dx^2)(a) = -4*cos(7*pi) = 4
Here we already can see two things:
the odd derivatives will have a sin(x) function that is zero when evaluated in x = 7*pi, and we also can see that the sign will alternate between consecutive terms.
so we only will work with the even powers of the series:
f(x) = -4 + (1/2!)*4*(x - 7*pi)^2 - (1/4!)*4*(x - 7*pi)^4 + ....
So we can write it as:
f(x) = ∑fₙ
Such that the n-th term can written as: