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Evgesh-ka [11]
2 years ago
11

How many millilitres are there in two point four litres?

Mathematics
2 answers:
Naily [24]2 years ago
6 0

Answer:

1 litre-1000 ml

so it is24000ml

vlada-n [284]2 years ago
4 0

Answer:

The answer is 2400

Step-by-step explanation:

1l = 1000 mal

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Consider the function ​f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0
dybincka [34]

I suppose you mean

f(x)=\cos(x^2)

Recall that

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

which converges everywhere. Then by substitution,

\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}

which also converges everywhere (and we can confirm this via the ratio test, for instance).

a. Differentiating the Taylor series gives

f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}

(starting at n=1 because the summand is 0 when n=0)

b. Naturally, the differentiated series represents

f'(x)=-2x\sin(x^2)

To see this, recalling the series for \sin x, we know

\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}

Multiplying by -2x gives

-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}

and from here,

-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}

-2x\sin(x^2)=\displaystyle 4x\sum_{n=0}^\infty(-1)^n\frac{nx^{4n}}{(2n)!}=f'(x)

c. This series also converges everywhere. By the ratio test, the series converges if

\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}

The limit is 0, so any choice of x satisfies the convergence condition.

3 0
4 years ago
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makvit [3.9K]
The shipping fee is $30
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We can use the following formula to solve this exercise:

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