Which algebraic expression can be used to find the nth term in the following sequence?

Mathematics

1 answer:

Agata [3.3K]3 years ago

8 0

Answer:

n+4 is the final answer.

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Add the two expressions -4x + 3 and 2x - 6 enter your answer in the Box PLZ HELP FASt

Fed [463]

-4x+3+2x-6
The answer is -2x-3

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3 years ago

Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based

Marina CMI [18]

In this question (brainly.com/question/12792658) I derived the Taylor series for $\mathrm{sinc}\,x$ about $x=0$ :

$\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$

Then the Taylor series for

$f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt$

is obtained by integrating the series above:

$f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

We have $f(0)=0$ , so $C=0$ and so

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

which converges by the ratio test if the following limit is less than 1:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}$

Like in the linked problem, the limit is 0 so the series for $f(x)$ converges everywhere.

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3 years ago

Does it help to remember and/or list out your square roots

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It is a good thing to remember. Of course, not all, but some basic numbers that appear all the time such as the numbers from 2 to 13 should be remembered as they appear in numerous assignments and tests.

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3 years ago

A 45- 45- 90 triangle has a hypotenuse of length 14 units. What is the length of one of the legs? If necessary, round your answe

xeze [42]

Leg A squared + leg B squared = Hypotenuse Squared

You are given the hypotenuse, which is 14.

Leg a = <span>7√2
</span>
Remember to rate brainliest please, if its correct!! :D

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Write the expression as the sine or cosine of an angle sin(pi/7) cos(x) + cos(pi/7) sin(x)

Natasha_Volkova [10]

Answer:

$\sin(\frac{\pi}{7}+x)$