30% of 60 is ______ help me please :)

Mathematics

2 answers:

Natasha_Volkova [10]3 years ago

6 0

Answer: 18

Step-by-step explanation:

vfiekz [6]3 years ago

5 0

I think it is just 30

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HELP ME PLEASE I REALLY NEED HELP!? :(

Aleksandr [31]

Just answers since long and many simplifiction
it's like doing those multiplication memory things that had like 100 questions but now with algebra fractions

5. fourth or $\frac{5x-15}{x^{2}-3x-28}$
6. third or $\frac{4x+25}{x^{2}+5x+4}$
7.the answe ris x+3 since it can be factored out of the 5x+15 tingie (second )
8. fourth or $\frac{x^{2}+4x}{4}$
9. fourth or $\frac{2y+6}{9y-7}$ times $\frac{3}{2}$
10. third or $\frac{(x+3)(x+8)}{(x-8)(x-8)}$
11. second or $\frac{x-5}{x-4}$
12. third or 11(x+7)=60

5 0

2 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.