Why is setting important to a story

Mathematics

2 answers:

klasskru [66]3 years ago

8 0

It’s helps you know where the story is taking place and helps you visualize where the characters are in the book

levacccp [35]3 years ago

3 0

Answer: so your readers can visualize and experience it..

Step-by-step explanation:

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What are the next three terms of the geometric sequence 32, -16, 8, …?

Marrrta [24]

Answer:

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Step-by-step explanation:

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3 years ago

Kim creates a model of a building. The scale of her model building is 1 foot to 8 inch. Some of the measurements are shown in th

Shalnov [3]

Answer:

Left

50 feet

Right

16 inches

10 inches

Step-by-step explanation:

Here, we want to make a conversion and fill in the empty spaces

The scale is 1 foot is 1/8 inch

To convert from foot to inch, we multiply by 1/8

To convert from inch to foot, we multiply by 8

Thus, we have ;

Let’s fill in the left spaces first;

6.25 * 8 = 50 feet

Right spaces, top to bottom;

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3 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.