Please help me with this math problem

Mathematics

1 answer:

balandron [24]3 years ago

6 0

The answer is 21
2.7x8=21

You might be interested in

Charles is planning his son’s birthday party and has a budget of $100. The Go Cart Corral charges $8 per child. How many childre

Montano1993 [528]

<h3>12 children</h3>

because

12 × 8 = 96

hope that helps :)

5 0

3 years ago

lin drew a triangle and a dilation of a triangle with scale factor 1/2. what is the center of the dilation?

mylen [45]

Bbjjjkkkmnnnjhhhhhgvvvggg

4 0

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 $y = e^x$

-----------------------

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.