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3 years ago

Solve the problem.

Mathematics

1 answer:

miskamm [114]3 years ago

7 0

Answer:

0.7

Step-by-step explanation:

Find Explanation attached

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Whats the slope of (-1,4) and (-2,5)

12345 [234]

Answer:

the slope is -1

Step-by-step explanation:

rise over run

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2 years ago

Kathy swam 3 laps in the pool this week. She must swim more than 12 laps.

Marysya12 [62]

Answer:10

Step-by-step explanation:

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3 years ago

96.3 cm measured to the nearest tenth of a cm

rjkz [21]

Answer:

96.3 cm measured to the nearest tenth of a cm is 100 cm

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2 years ago

Write as a decimal. If a repeating decimal, round to the nearest hundredth.

Mamont248 [21]

Given:

$\frac{475}{80}$

Let's evaluate and find the quotient.

We have:

$\frac{475}{80}=5.9375$

SInce it is not a repeating decimal, we are asked to leave to answer as it is.

A repeating decimal is a decimal with digits or group of digits that repeats endlessly.

The decimal 5.937

ANSWER:

5.9375

6 0

1 year 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.

3 0

3 years ago