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3 years ago

PLEASE HELP ME IF YOU CAN!!!

Mathematics

1 answer:

galina1969 [7]3 years ago

6 0

Answer:

11.775 km

Step-by-step explanation:

<h3>Given</h3>

r = 3 km

Length of 75° arc

<h3>Solution</h3>

Circumference of the circle:

C = 2πr
C = 2*3.14*3² = 56.52 km

<u>The length of the 75° arc:</u>

56.52*75/360 = 11.775 km

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(1.2)Consider the diagram below. Identify a point.

Fynjy0 [20]

Option B is correct. <em>Point F</em><em> is the </em><em>only point </em><em>according to the option given</em>

A point is defined as an exact location with no size, length, or even width.

From the diagram shown, we have points, plane, and line segments.

The only plane that we have is plane R making option C incorrect.

A line is defined as a straight figure connected by two points. From the diagram, the line segments that we have are AN, BN, and AB falsifying options A and D.

The points are C, D, E, and F. Hence the correct option is B.<em> Point F</em><em> is the </em><em>only point </em><em>according to the option given</em>

Learn more here: brainly.com/question/13099718

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3 years ago

A kitchen floor has 15 1/2 tiles in an area of 2 2/5 square feet. How many tiles are in one square foot?

Marat540 [252]

6 2/5 is the answer, because 15 1/2 divided by 2 2/5 is 6 2/5

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3 years ago

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Fittoniya [83]

Each ratio is
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3 years ago

Jose's car can drive 12 miles on each gallon of gas. How many gallons will he need to drive 3,298 miles?✱

masya89 [10]

Answer:

274.84 gallons.

Step-by-step explanation:

Divide 3,298 by 12

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6 0

3 years ago

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:

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

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:

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

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

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

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

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