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3 years ago

5,256 divide 52 and show your work

Mathematics

1 answer:

Klio2033 [76]3 years ago

4 0

Answer:

Step-by-step explanation:

Okay, I hope this formats well.

___101.076

52 / 5256.000

400

364

360

I think I'll stop here, there are a lot of decimal places. Do you get it now?

Please mark this as Brainliest. Thanks!

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Carol walked 1 1/3 miles on Monday. 35/6 miles on Tuesday, and 2 miles on Wednesday. About how far did she walk altogether?

Dmitry_Shevchenko [17]

Answer:

9 1/6 miles

Step-by-step explanation:

Add the mixed number by the improper fraction.

1 + 1/3 + 35/6

Solve the fractions first.

In order to have both of the fractions have the same denominator, find the Least Common Multiple of both of the fractions.

1/3 = 2/6

2/6 + 35/6 = 37/6

Turn the improper fraction into a mixed number by dividing the the numerator by the denominator. When you get your quotient, use the remainder as the new numerator over the denominator.

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Now, add the 1.

6 1/6 + 1 = 7 1/6

Now, add the 2 miles that Carol walked on Wednesday.

7 1/6 + 2 = 9 1/6

So, Carol walked about 9 1/6 miles on Monday, Tuesday, and Wednesday all together.

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3 years ago

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.

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