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3 years ago

Is the following relation a function? Yes No 100 points!

Mathematics

2 answers:

kirill [66]3 years ago

8 0

Answer:

Yes, it is a function

Step-by-step explanation:

It passes the vertical line test, therefore it is a function.

Vilka [71]3 years ago

6 0

Answer:NO

Step-by-step explanation:

you can do the vertical line test and i does not go through the x axis 2 times

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3 years ago

Is compound interest always greater than simple interest ?Give examples

g100num [7]

Yes compund interest is always greater than simple interest

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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:

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

Which inequality is true 1.5 > 2 1/2, 1/2 > 0.5, -2.5 > -1.5, -3 1/2 > - 4.5

yawa3891 [41]

Answer: D

Step-by-step explanation:

(A. 1.5 > 2 1/2

(B. 1/2 > 0.5

(C. -2.5 > -1.5

(D. -3 1/2 > - 4.5

Step 1: Convert everything to decimals or fractions so its easier to see and make it less likely to get the question wrong

Method #1: Fractions

1.5= 1 1/2

0.5= 1/2

-2.5= - 2 1/2

-1.5= - 1 1/2

-4.5= -4 1/2

Now we have:

(A. 1 1/2 > 2 1/2 --> This is wrong because 1 1/2 is supposed to be less than 2 1/2.

(B. 1/2 > 1/2 This is also wrong. 1/2 is equal to 1/2; they are obviously the same.

(C. -2 1/2 > -1 1/2 This is wrong because, in negative numbers, its sort of backwards. For example, -2 is actually less than -1 because its farther away from 0. But if they were normal positive numbers, 2 would be bigger. But here, -2 1/2 is less than -1 1/2.

(D. -3 1/2 > - 4 1/2 This is correct. -3 1/2 is closer to 0 than -4 1/2, so it has to be bigger.

Solving the problem this way gets me to: Answer= D

Method #2: Decimals (remember: same numbers, different way of writing them!!!)

(A. 1.5 > 2.5 False. 1.5 is less than 2.5

(B. 0.5 > 0.5 False. 0.5 is equal to 0.5

(C. -2.5 > -1.5 False. -2.5 is farther away from 0 so its actually smaller.

(D. -3.5 > - 4.5 True. -3.5 is closer to 0.

Doing it this way also leads me to Answer= D.

Hope this helps :)) Also can i please have brainiest? I worked pretty hard on this :D

6 0

2 years ago

Read 2 more answers

I need to paint a bin rectangular bin w no top... bin length is 25mm and width of 9 mm how much paint do I need to cover bin

Nikitich [7]

Answer:

900mm of paint for all four sides

Step-by-step explanation:

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3 years ago