Help with limits will give brainliest

How can an expression using the distributive property can be simplified to 12a + 18b - 6c

elena-s [515]

Look for a common number in all the terms

6(2a+3b-c)

3 0

3 years ago

A documentary is a film that

DanielleElmas [232]

Answer:

1. the answer is nonfictional story about a person, topic, or event

Step-by-step explanation:

hope this helps

3 0

2 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.

7 0

3 years ago

peter has 6/7 of a stack of baseball cards left. If he is planning on splitting what he has left into stacks that are each 3/28

DIA [1.3K]

<h3>Answer: 8</h3>

=======================================================

Work Shown:

Let's say there are 28 cards in a full stack.

6/7 = 24/28 after multiplying top and bottom by 4

Since he has 24/28 of a stack left, this means he has 24 cards left.

He wants to arrange the remaining cards into piles so that each pile consists of 3/28 of a full stack. In other words, he wants each pile to have 3 cards.

So this must mean he will get 24/3 = 8 piles

(8 piles)*(3 cards per pile) = 8*3 = 24 cards total

3 0

2 years ago

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In a two-player game, five cards, numbered 1 through 5, are placed in a bag. A card is drawn at random, and the players look at

IceJOKER [234]

Answer:

If it is less than 3, Player 1 earns 3 points.

If not, Player 2 earns 2 points.

Step-by-step explanation:

<u>Player 1</u> :

p(N < 3) = p(N = 1 or N = 2) = 2/5

<u>Player 2</u> :

p(N ≥ 3) = p(N = 3 or N = 4 or N = 5) = 3/5

<u>We notice that</u> :

p(N < 3) × 3 = (2/5) × 3 = 6/5

On the other hand,

p(N ≥ 3) × 2 = (3/5) × 2 = 6/5

since ,the probability player 1 win multiplied by the associated number of points (3)

is equal to

the probability player 2 win multiplied by the associated number of points (2).

Then the game is fair.

8 0

1 year ago

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