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3 years ago

Can a cloud be a line of symmetry?

Mathematics

1 answer:

Molodets [167]3 years ago

8 0

Answer:

No, it can not

Step-by-step explanation:

Hope this helped have an amazing day!

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Three companies sell their apple juice in different sized bottles jada is calculating the unit prices of each bottle and will co

kati45 [8]

The answer is dividing each bottle size by its cost I took the test and I got it right

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2 years ago

Read 3 more answers

A drawer contains 2 red shirts and 4 blue shirts. A second drawer contains 3 pairs of grey pants and 2 pairs of blue pants. A th

viktelen [127]

Answer:

b . may be

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2 years ago

Mrs. Williams is deciding between two field trips for her class. The Science Center charges $135 plus $3 per student. The Dino D

Marianna [84]

Set it up like a regular equation. I used:
$3x+135 \ \textless \ 6x$

As you can see, however, I inserted the "less than" symbol rather than the equal sign because values greater than a certain value can also equal "x". Solve it like a normal equation and you will arrive at:

 $x \ \textgreater \ 45$

This means that, as long as x is greater than 45, "6x" will be greater than "3x+135". Thus, if you bring 46 students, the Science Center will cost less.

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3 years ago

Use series to verify that <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.

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2 years ago

Please help meeee it’s question #7

Ugo [173]

Answer:

In my own opinion i time 104 and 72and if gives me7488

3 0

3 years ago