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valentinak56 [21]

3 years ago

6

Circle O is shown. Line segments E O and E O are radii with length 24. Angle D O E is 45 degrees. The area within sector D O E i

s shaded.

2 answers:

AfilCa [17]3 years ago

7 0

Answer:

Its 72Pi. (Option C)

Step-by-step explanation:

Took it on Edge and got it right

fiasKO [112]3 years ago

6 0

Answer:

C 72pi

Step-by-step explanation:

just took the test

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3 years ago

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2 years ago

Use the definition of Taylor series to find the Taylor series, centered at c, for the function. f(x) = sin x, c = 3π/4

anyanavicka [17]

Answer:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Step-by-step explanation:

Given

$f(x) = \sin x\\$

$c = \frac{3\pi}{4}$

Required

Find the Taylor series

The Taylor series of a function is defines as:

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

We have:

$c = \frac{3\pi}{4}$

$f(x) = \sin x\\$

$f(c) = \sin(c)$

$f(c) = \sin(\frac{3\pi}{4})$

This gives:

$f(c) = \frac{1}{\sqrt 2}$

We have:

$f(c) = \sin(\frac{3\pi}{4})$

Differentiate

$f'(c) = \cos(\frac{3\pi}{4})$

This gives:

$f'(c) = -\frac{1}{\sqrt 2}$

We have:

$f'(c) = \cos(\frac{3\pi}{4})$

Differentiate

$f"(c) = -\sin(\frac{3\pi}{4})$

This gives:

$f"(c) = -\frac{1}{\sqrt 2}$

We have:

$f"(c) = -\sin(\frac{3\pi}{4})$

Differentiate

$f"'(c) = -\cos(\frac{3\pi}{4})$

This gives:

$f"'(c) = - * -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

So, we have:

$f(c) = \frac{1}{\sqrt 2}$

$f'(c) = -\frac{1}{\sqrt 2}$

$f"(c) = -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

becomes

$f(x) = \frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}(x - \frac{3\pi}{4}) -\frac{1/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{1/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Rewrite as:

$f(x) = \frac{1}{\sqrt 2} + \frac{(-1)}{\sqrt 2}(x - \frac{3\pi}{4}) +\frac{(-1)/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{(-1)^2/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Generally, the expression becomes

$f(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Hence:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

3 0

2 years ago

I need help with this work

STatiana [176]

Answer:

G. 245.04cm²

Step-by-step explanation:

The shortest way to solve this problem would be:

total surface area = 2 * π * r² + (2 * π * r) * h

or

total surface area = 2 * π * r * (r + h).

Diameter = 6

Radius = 3

Height = 10

The radius is half the diameter

TSA = Total surface area

TSA = 2 π 3(3+10)

TSA = (2)(3.141593)(3)(3+10)

TSA = (6.283185)(3)(3+10)

TSA = 18.849556(3+10)

TSA = (18.849556)(13)

TSA = 245.044227

TSA = 245.04

Or you could do the longer way, which would be:

Step 1: Find the surface area of the curved part of the cylinder

Surface Area (Curved) = 2 x π x Radius x Height

Step 2: Find the surface area of the two ends of the cylinder

Surface Area (Ends) = 2 x π x Radius²

Step 3: Add the surface are of the curved part to the surface area of the ends

Surface Area (Total) = Surface Area (Curved) + Surface Area (Ends)

For the cylinder in the picture:

A cylinder has a height of 10 and a diameter of 6

Hence, Radius = d/2 = 6/2 = 3 cm

Surface Area (Curved) = 2 x pi x 3 x 10

Surface Area (Curved) = 188.495559

Surface Area (Ends) = 2 x pi x 3²

Surface Area (Ends) = 2 x pi x 25

Surface Area (Ends) = 56.548668

Surface Area (Total) = 245.044227

245.04 is the result of rounding 245.044227 to the nearest hundredth

7 0

1 year ago

Use the formula to find the value of $400 invested in 4%

Anna [14]

Answer:

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Step-by-step explanation:

3 0

2 years ago