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3 years ago

How many unique ways are there to arrange the letters in the word PRIOR?

Mathematics

1 answer:

horrorfan [7]3 years ago

7 0

PRIOR MEAN PREVIOUS SO THAT IS UR ANSWER CUH a b c d the answer is yes

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If I can read 54 pgs in 3 days, how many can I read in 1 day?

nignag [31]

Its 18 because you would divide 54 bu 3

7 0

3 years ago

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16=5a+a<br> Helppp please

Nostrana [21]

A=8/3

5a+a= 6a

16/6a= 8/3

3 0

3 years ago

Quadrilateral ABCD is similiar to quadrilateral EFGH. The lengths of the three longest sides in quadrilateral ABCD are 20 feet,

Anika [276]

Answer:

7.7 ft

Step-by-step explanation:

The third longest side is also the second shortest side.

The 14 ft side of quadrilateral ABCD corresponds to the 6 ft side of quadrilateral EFGH.

14/6 = 18/x

7/3 = 18/x

7x = 18 * 3

7x = 54

x = 54/7

x = 7.7

Answer: D. 7.7 feet

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3 years ago

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Morgan drove from los angeles, at an elevation of 330 feet, to death valley, at an elevation of -282 feet. what is the differenc

dlinn [17]

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3 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

4 years ago