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3 years ago

What is the radius of an arc length of 8 m that measures 45°?

Mathematics

1 answer:

sergeinik [125]3 years ago

7 0

The radius is 28.......

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10. Consider the diagram below. A. solve for x B. Solve for y

anastassius [24]

Answer:

x=42 y=38

Step-by-step explanation:

hope it helps, please thank, review, and mark as brainliest! tysm!

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3 years ago

Here are the fuel efficiencies (in mpg) of 12 new cars . 28, 14, 48, 22, 14, 18, 52, 36, 32, 20, 12, 55 What is the percentage o

SSSSS [86.1K]

25% assuming you don’t count the car with 18 mpg fuel efficiency.

if you add the amount of cars up that have less than 18 mpg and then divide it by the total number (12) you will get a decimal (0.25) to find the percent multiply the decimal by 100

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3 years ago

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10 fish are in a tank, 3 fish drown, how many fish are left?

taurus [48]

10. Fish don't drown (; nice try tho

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3 years ago

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Find minor arc in circle Include correct units

masha68 [24]

Answer:

Same thing here Arc DE is also 100 degreese

cause the lines are connected to the center point.

Step-by-step explanation:

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4 years ago

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Use this list of Basic Taylor Series and the identity sin2θ= 1 2 (1−cos(2θ)) to find the Taylor Series for f(x) = sin2(3x) based

notsponge [240]

Answer:

The Taylor series for $sin^2(3 x) = - \sum_{n=1}^{\infty} \frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$ , the first three non-zero terms are $9x^{2} -27x^{4}+\frac{162}{5}x^{6}$ and the interval of convergence is $( -\infty, \infty )$

Step-by-step explanation:

These are the steps to find the Taylor series for the function $sin^2(3 x)$

Use the trigonometric identity:

$sin^{2}(x)=\frac{1}{2}*(1-cos(2x))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(2(3x)))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$

2. The Taylor series of $cos(x)$

$cos(y) = \sum_{n=0}^{\infty}\frac{-1^{n}y^{2n}}{(2n)!}$

Substituting y=6x we have:

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

3. Find the Taylor series for $sin^2(3x)$

$sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$ (1)

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$ (2)

Substituting (2) in (1) we have:

$\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

Bring the factor $\frac{1}{2}$ inside the sum

$\frac{6^{2n}}{2}=9^{n}2^{2n-1} \\ (-1^{n})(9^{n})=(-9^{n} )$

$\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

Extract the term for n=0 from the sum:

$\frac{1}{2}-\sum_{n=0}^{0}\frac{-9^{0}2^{2*0-1}x^{2*0}}{(2*0)!}-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \frac{1}{2} -\frac{1}{2} -\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ 0-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ sin^{2}(3x)=-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

To find the first three non-zero terms you need to replace n=3 into the sum

$sin^{2}(3x)=\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \sum_{n=1}^{3}\frac{-9^{3}2^{2*3-1}x^{2*3}}{(2*3)!} = 9x^{2} -27x^{4}+\frac{162}{5}x^{6}$

To find the interval on which the series converges you need to use the Ratio Test that says

For the power series centered at x=a

$P(x)=C_{0}+C_{1}(x-a)+C_{2}(x-a)^{2}+...+ C_{n}(x-a)^{n}+...,$

suppose that $\lim_{n \to \infty} |\frac{C_{n}}{C_{n+1}}| = R.$ . Then

If $R=\infty,$ the the series converges for all x
If $0$ then the series converges for all $|x-a|$
If R=0, the the series converges only for x=a

So we need to evaluate this limit:

$\lim_{n \to \infty} |\frac{\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}}{\frac{-9^{n+1}2^{2*(n+1)-1}x^{2*(n+1)}}{(2*(2n+1))!}} |$

Simplifying we have:

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |$

Next we need to evaluate the limit

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |\\ \frac{1}{18x^{2} } \lim_{n \to \infty} |-(n+1)(2n+1)}|}$

-(n+1)(2n+1) is negative when n -> ∞. Therefore $|-(n+1)(2n+1)}|=2n^{2}+3n+1$

You can use this infinity property $\lim_{x \to \infty} (ax^{n}+...+bx+c) = \infty$ when a>0 and n is even. So

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } | \\ \frac{1}{18x^{2}} \lim_{n \to \infty} 2n^{2}+3n+1=\infty$

Because this limit is ∞ the radius of converge is ∞ and the interval of converge is $( -\infty, \infty )$ .

6 0

3 years ago