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3 years ago

What is 562949953421312 multiplied by 2

Mathematics

1 answer:

Nesterboy [21]3 years ago

6 0

I used and got 1.1258999e+15 lol Im sorry Im only in middle school

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At a doctor's appointment, a baby weighed 10 pounds. How many ounces did the baby weigh?

Sergio [31]

Answer:

The baby would be 160 ounces.

Step-by-step explanation:

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3 years ago

Find the volume of the following cone. Use 3.14 for n and

fenix001 [56]

28 m because it is the smallest number

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2 years ago

PLEASE ANSWER FAST AS YOU CAN and answer like

OLga [1]

Answer:

number 1. answer is 5

number 2. answer is 2 1/2

number 3. answer is 9/16

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2 years ago

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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

3 years ago

A. 180 B.82 C. 46 D. 98 Please answer help me

Doss [256]

Answer:

Step-by-step explanation:

Because the angle with measure 82 and angle 1 are corresponding angles, their angles measure must be equal, and therefore angle 1 has measure of 82 as well. Since angles 1 and 2 are a linear pair, they must be supplementary, and angle 2 is therefore 180-82=98 degrees. Hope this helps!

8 0

3 years ago

Read 2 more answers