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stellarik [79]
3 years ago
6

Circle X with a radius of 6 units and circle Y with a radius

Mathematics
1 answer:
asambeis [7]3 years ago
8 0

Answer:

Option C.

Step-by-step explanation:

It is given that radius of circle X is 6 units and radius of circle Y is 2 units.

Here, circle Y is smaller and circle X is larger.

\text{Scale factor}=\dfrac{\text{Radius of circle X}}{\text{Radius of circle Y}}

\text{Scale factor}=\dfrac{6}{2}

\text{Scale factor}=3

If we translate the circles so they share a common center point, and after that we dilate circle Y by a scale factor of 3, then we get circle which is equal to circle X.

We know that, after dilation the figure and its image are similar.

Hence, we can say that both circles are similar.

Therefore, the correct option is C.

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Find the expansion of tan x about the point X = 0.
yan [13]

Answer:

f(x) = x +\frac{1}{3}x^{3}+....

Step-by-step explanation:

As per the question,

let us consider f(x) = tan(x).

We know that <u>The Maclaurin series is given by:</u>

f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......

So, differentiate the given function 3 times in order to find f'(x), f''(x) and f'''(x).

Therefore,

f'(x) = sec²x

f''(x) = 2 × sec(x) × sec(x)tan(x)

      = 2 × sec²(x) × tan(x)

f'''(x) = 2 × 2 sec²(x) tan(x) tan(x) + 2 sec²(x) × sec²(x)

       = 4sec²(x) tan²(x) + 2sec⁴(x)

       = 6 sec⁴x - 4 sec² x

We then substitute x with 0, and find the values

f(0) = tan 0 = 0

f'(0) =  sec²0 = 1

f''(0) = 2 × sec²(0) × tan(0) = 0

f'''(0) = 6 sec⁴0- 4 sec² 0 = 2

By putting all the values in the Maclaurin series, we get

f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......

f(x) = 0 + \frac{1}{1}\cdot x+ \frac{0}{2}\cdot x^{2}+\frac{2}{6}\cdot x^{3}+......

f(x) = x +\frac{1}{3}x^{3}+....

Therefore, the expansion of tan x at x = 0 is

f(x) = x +\frac{1}{3}x^{3}+.....

8 0
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